INTRODUCTIONTO
CONTROL
SYSTEM
DESIGN
EDITIO
HARRYKWATNY& BOR-CHINCHANG
Introduction
to Control
System
Design
FirstEdition
HarryKwatny
andBor-Chin
Chang
DrexelUniversity
SAN
DIEG
ii
Bassim Hamadeh, CEO and Publisher
John Remington, Executive Editor
Gem Rabanera, Senior Project
Christian
Berk, Production
Editor
Editor
Emely Villavicencio, Senior Graphic Designer
Trey Soto, Licensing Coordinator
Natalie Piccotti,
Director
of
Kassie Graves, Vice President
Marketing
of Editorial
Jamie Giganti, Director of Academic Publishing
Copyright
2021 by Cognella,Inc. All rights reserved. No part of this publication maybereprinted, reproduced,
transmitted, or utilized in any form or by any electronic, mechanical, or other means,now known or hereafter
invented, including photocopying, microfilming, and recording, orin any information retrieval system without the
written permission of Cognella, Inc. For inquiries regarding permissions, translations, foreign rights, audio rights,
and any other forms of reproduction, pleasecontact the Cognella Licensing Department at rights@cognella.com.
Trademark
Notice: Product
identification
Cover image:
Printed in the
or corporate
and explanation
Copyright
names
without intent
2019 iStockphoto
United States of America
may be trademarks
to infringe.
LP/akinbostanci.
or registered
trademarks
and are used only for
Preface
A
S an introduction to the design of automatic control systems, the books primary objective is to
provide students with the basic concepts and tools that are usedin present-day practice. Wenot
only want our students to know how to design a control system, but also to understand why
we dothings the way we do. In this way,the students will be prepared to innovate and expand current
methodsto address new automation technology challenges in the coming years. Part of this processis to
give students a sense of the evolution of control theory and practice.
Control system engineering in todays technology environment requires a much broader set of con-cepts
and tools than even a decade agoand the field continues to evolve. Classtime is fixed, the scope
of the material is extensive, and rigor is essential. Consequently, subjects having lesser importance to-day
haveless emphasis herein. Of course, such judgments are personal and not always easyto make.In
makingthese choices, we have been guided by two central objectives. First, the content mustbe aligned
with the needs of present-day applications and make use of modern computational tools. Second, the
narrative
must reflect
a respect and acknowledgement
of the historical
evolution
of control
analysis
and
design theory.
This book is intended for a two quarter/semester course. Correspondingly, the book is structured
with two main parts. Thelevel of complexity of Part I is somewhat mitigated by the fact that atypical
student will have had some prior exposure to some of the information, albeit in a different context. Part
I also contains preliminary hints of materialto come in PartII, withthe idea of smoothing the exposure
to complex new ideas. Part I introduces basic linear system analysis and model-assembly concepts. It
begins, in
Chapter 1, with a short history
its evolution,
along
with the contributions
of control system
of control theory
design, highlighting
the applications
driving
pioneers from late 19th century to the current
era.
Although practical systems can be complex and of high order, their building blocks arefirst-order and
second-order dynamic systems. Understanding the fundamental concepts of the first-order and second-order
systems is essential to the analysis and design of high-order systems. Chapter 2 is focused on
systems
with typical
speed dynamics.
first-order
dynamics
These systems,
although
such as massfriction
physically
different,
systems,
are
RC circuits,
mathematically
and
DC motor
equivalent
sharing
the same form of differential equation characterized by just two parameters: the time constant andthe
steady-state
variables
step response
value.
and Laplace transforms
This enables
us to systematically
with the learning
integrate
of the fundamental
the essentials
of complex
system concepts in both time
and
frequency domains including transfer functions, characteristic equations, poles, time constants, step
responses, sinusoidal steady-state responses, frequency responses, frequency filtering, Bode plot, and so
on, while
maintaining
a connection
with applications
and
mathematical
modeling
iv
Preface
Chapter 3 expands the
applications
to the typical
second-order
dynamics
such as
massspringdamper
systems, RLC circuits, and position control of a DC motor. Similar to the group of first-order
systems
mentioned in the previous
paragraph, these second-order
systems are mathematically
equivalent
sharing the same form of differential equation characterized byjust three parameters: the damping ratio,
the natural frequency, and the steady-state step response value. The fundamental system concepts we
learn in Chapter 2 are largely applicable to the second-order systems. One of the main differences is
that second-order
are interpreted
systems
enable the appearance
in the time
and partial fraction
domain
of complex
and the frequency
expansion introduced.
poles in a physical
domain.
The ability
context.
Transfer functions
to change pole location
Their dynamics
concepts are expanded
by using feedback is illus-trated.
Bothtime response and frequency response perspectives andinterpretations are developed in the
context
of easily understood
An essential requirement
physical
physical
of control
systems.
system design is that the engineer
behavior in order to be successful.
model of the system to be controlled.
Control design in todays
has a full
grasp of the systems
world begins
It is essential that the control
with a mathematical
design team understands
its formu-lation
and the assumptions on which it is based even if they do not assemble the model. Withthis in
mind, the next two chapters
Mechanical systems
Both approaches
deal, respectively,
modeling,
eventually
with the
modeling of
Chapter 4, addresses and compares
will lead to exactly the same dynamic
mechanical and electrical
Newtonian
systems.
and Lagrange techniques.
model equations.
However, they have
completely different road mapsto reach the same destination. The Newtonian approach involves the
geometry and vector relationship of each component in the mechanical system. Onthe other hand,the
Lagrange approach only needsthe information of the kinetic energy, the potential energy, the power
dissipation, and the external forces, but it requires differential calculus computations.
Both have advantages and disadvantages,
but they complement
each other perfectlythe
disadvan-tage
of one is the advantage of the other, and vice versa. The Newtonian approach is moreintuitive,
providing clear picture of the interactions among components inside the system, but the modeling pro-cess
can be tedious, especially for large multi-body systems. Conversely,the Lagrange approach is more
elegant without the needto worry aboutthe directions of vectors and detailed interconnection of compo-nents,
but it allows virtually no insight of the interactions among components within the system. Students
are urged to employ both approaches, especially for complicated systems, to minimize modeling errors.
Sinceit is rare to commit same modeling errors using the two fundamentally different approaches, the
modeling result is moretrustworthy if it is confirmed by both approaches.
Shortly after obtaining the nonlinear dynamics model of the inverted pendulum system in Chapter
4, wefelt obligated to digress a little bit to explain how the unstable nonlinear system is relevant to the
linear control theory. The discussion articulates the process of determining a local linear design model
from the nonlinear model as well as the integration of the linear controller with the actual nonlinear
system. Working with this relatively simple nonlinear system provides the opportunity to introduce an
intuitive state-space analysis and design approach that stabilizes the originally unstable system using a
basic pole placement concept the students havejust learned in Chapter 3. Computer tools also are in-troduced
as the examples progress, for instance, Simulink is usedto provide a computer simulation of
the stabilized
space analysis
inverted
pendulum.
Formal
and design approaches
definition
of stability,
controllability,
and
more advanced state
will be given in later chapters.
Forthe electrical system model building in Chapter 5, wefirst review Kirchhoffs voltage and cur-rent
laws (KVL and KCL), and the characteristics of fundamental two-terminal electrical elements,
which includes resistor, capacitor, inductor, voltage source, and current source. Then we demonstrate
how to employ the NTD (node-to-datum) voltages approach, the meshcurrents approach, and the
direct state-space
approach
to assemble
approaches
have two
versions:
electric system
one in time
domain,
models. The NTD voltages
and the
and the other is in frequency
mesh cur-rents
domain.
Th
Preface
frequency-domain
time-domain
version,
also called the impedance
version, is basically
v
a Laplace transform
of the
version; thus, all KVL and KCL equations have become algebraic andthe characteristics
of capacitors and inductors are now governed by the generalized
Z(s) is the impedance.
through
In the direct state-space
the inductors
Ohms law:
approach, the voltages
are selected as state variables,
V(s) = Z(s)I(s),
across the capacitors
and then the
KCL and the
where
and the cur-rents
KVL, respectively,
are employed to obtain a KCL equation at a node connecting to the capacitor, and a KVL equation
around a mesh containing
form
the inductor.
These KCL and
KVL equations
can be easily rewritten
in the
of state equations.
The Lagrange
approach
can also be applied to electrical
systems
modeling, in
which the
electric
charge q and its derivative ?q (the electric current) are considered as configuration variables. The con-structions
of the Lagrangian function and the Lagrange equation are similar to those for the mechanical
systems
except that for electrical
inductors
and the capacitors,
using the virtual
systems the kinetic
respectively,
while the generalized
work, which is contributed
In Section 5.6, weintroduce
energy and the potential
energy are stored in the
external force
vector can be obtained
by the voltage source and the dissipation
the operational
amplifier,
usually
called op amp,
in the resistors.
which is an almost ideal
electronic amplifier dueto its three special properties: extremely large voltage gain, extremely high input
impedance,
and almost zero output impedance.
a perfect building
block in interconnected
to the development of the virtual-short
These three
systems
properties
not only
due to its extremely
makethe op amp circuit
low loading
effect,
but also lead
concept approach for the op amp circuit analysis and design.
The virtual-short concept approach has madeit possible to greatly simplify the analysis and design of
op amp circuits, which otherwise would be extremely complicated. Opamp circuits can be easily built
to perform
filtering,
a variety
of functions
PID controller
like signal addition,
implementation,
substraction,
binarytodecimal
integration,
conversion,
detection,
amplification,
decimaltobinary
conversion,
common-mode disturbance cancelation, and so on.
In the beginning of Chapter 2, the DC motor system, together
circuit and the
impossible
massfriction
for its dynamics
consisting
system,
was considered
model to be so simple
of an electric circuit,
a gear train,
a typical
since the
with the simple RClow-pass filter
first-order
and a rotational
latter
which describes how the
mechanical torque is dictated
gives the back EMF (electromagnetic
force)
equation,
system. It seems to be
is a complicated
sys-tem,
mechanical system. In Section 5.7, we
first briefly review Amperes force law and Faradays law of induction.
equation,
dynamic
DC motor physically
Theformer provides the torque
by the armature electric
which explains
current, and the
how the back EMF voltage
is related to the motor rotor speed. Applying Kirchhoffs voltage law to the armature electric circuit
will give afirst-order KVL differential equation that relates the armature current to the applied control
input voltage. Meanwhile, the Newtonian or the Lagrange approach can be applied to the rotational me-chanical
system with gear train to obtain the mechanical rotational motion equation, which is another
first-order differential equation that describes how the armature current will affect the motorrotor speed.
The four
equationsthe
torque
equation,
the back
EMF equation,
the
KVL equation,
and the
me-chanical
motion equationcan
be combined and simplified to a second-order differential equation that
describes how the applied input voltage will control the motorrotor speed. Sincethe inductor impedance
of the armature coil is negligible compared to the resistor resistance, the order of the differential equation
is reduced to one. Hence,the DC motor system dynamics can be represented by atypical first-order dif-ferential
equation characterized by two parameters: the time constant and the steady-state step response
value. Thesetwo parameters are functions of the torque constant Km,the back EMF constant Kb, and
the resistor resistance Ra of the armature coil, which can be found from the manufacturers data sheet.
Thesetwo parameters also can be obtained from a simple step response experiment in the lab.
At the end of Chapter 5, we conduct an open-loop simulation to observe the open-loop step re-sponse
of a DC motor. Then we design a simple integral feedback controller Ki/s to achieve perfec
vi
Preface
steady-state
speed tracking,
and a desired transient
response
with small
maximum
overshoot
by select-ing
the integration constant Ki so that the desiredclosed-loopsystem dampingratio is ? = 0.9. Since
there is only one design parameter
the damping ratio
problem
or the
Ki in the controller,
natural frequency.
using a dual-loop
controller
Later in
structure
we can only choose a desired
value for either
Chapter 6, we will revisit this speed tracking
with two design parameters,
con-trol
which can be chosen to
achieve the desired damping ratio as well asthe desired natural frequency for the closed-loop system.
Forinstance, if the natural frequency is double while keeping the damping ratio unchanged at ? = 0.9,
then the step response
will rise up approximately
two times faster
while keeping the
maximum
overshoot
unchanged.
Thefinal chapter of Part I discussesthe assembly of modelsfor interconnected systems. Block di-agrams
and signal flow
assemble
graphs as well as Masons gain formula are introduced.
more complex
models composed
in either a state-space
the solution
of interconnected
model or a transfer
function
of the state equation and to introduce
elements is achieved.
Thus, the ability to
This construction
model. This provides the opportunity
the concept
of the state transition
re-sults
to discuss
matrix. In addition,
the construction of the transfer function from a state-space model and its reverse, the construction of a
state-space
model from the transfer
Part II is focused
are included,
(MIMO)
on linear
function,
control
are both discussed.
system
design.
and both single-input/single-output
systems
are discussed.
The first
Both frequency-domain
(SISO)
chapter in
and time-domain
meth-ods
systems and multiple-input/multiple-output
Part II,
Chapter 7, is focused
on the fundamentals
of feedback systems. It begins with a discussion of how feedback affects the system dynamics and de-scribes
the benefits and limitations of feedback. System representations in time and frequency domain
are again discussed,
discussed,
but at a somewhat
deeper level than in Part I. Stability
both bounded-input/bounded-output
(BIBO)
stability
of linear systems is formally
and internal
stability
are defined
the differences articulated. Again, this is to set the stage for deeper consideration later in the
Similarity transformation of state-space models are addressed, and both diagonal and companion
are highlighted. Naturally, this leads to another discussion of pole placement via state feedback.
is again devoted to the cart-inverted pendulum system as a meansto summarize the control
methodsintroduced at this point.
Chapter 8 provides a complete discussion of compensator design via the root locus
achieve stability,
regulation,
and a best possible transient
response implied
and
book.
forms
A sec-tion
design
method to
by pole locations.
It
be-gins
with a study of steady-state error andintroduces the concept of system type andits role in achieving
zero steady-state error. Theinternal
model principle is also briefly introduced. A cruise control exam-ple
is usedto provide an overview of the design process, including performance objectives, the role of
feedback and feed-forward in achieving those goals, as well asthe notion of performance robustness
with respect to model uncertainty. The root locus methodis presented in some detail, including the ba-sic
construction rules, whythey are useful in design, and how to dothe computations using MATLAB.
Simple examples illustrate key points throughout this discussion. More expansive examples are given in
the last three sections. First, a DC motorsinusoidal position tracking controller is usedto illustrate the
application of the internal model principle along with root locus design. The next section examines the
longitudinal
illustrate
flight
path control
the difficulty
level
A sophisticated
of the F/A18
of controlling
root locus
aircraft.
the extremely
design
with integral
is employed to achieve stability and flight pathtracking.
can be accomplished
One of the
using the flight
In this example,
low-damping,
regulation
manual control
long-period
is simulated
(phugoid-mode)
and state-feedback
to
os-cillations.
pole placement
Thelast section illustrates how altitude regula-tion
path angle tracking
controller.
mostimportant requirements in the design of feedback control systems is robust sta-bility,
since an unstable system is
The robust stability
issue
caused
not only
by time
useless but can be harmful
delay and plant
uncertainties
or potentially
cause a disas-ter.
is considered in
Chapter 9
Preface
Chapter 9 focuses
on robust stability
beginning
with a discussion
delay, and how it can degrade closed-loop stability.
Cauchys
emphasis
principle
precedes a discussion
on Nyquists
world that is
theorem
more easily
of the
accomplished
including
A discussion of complex contour
Nyquist plot and the
is not as a tool for
gain and phase margins. The relationship
of plant uncertainty,
determining
by direct
computationbut
systems
time
mapping and
Nyquist stability
closed-loop
vii
criterion.
The
stabilityin
to-days
to establish the concepts
of
between Nyquist and Bode plots as a meansto obtain these
margins is also discussed.
Gain and phase
no meaningful
margins do not extend to
definition
multiple-input/multiple-output
for the phase of a matrix loop transfer
considers the generalized stability
magnitude (or the
function.
allowable
maximum
of the
marginfor
singular
The frequency-dependent
variation
function.
The final
because there is
section
of this chap-ter
margin, an approach conceived to address this deficiency. This
new approach of defining a generalized stability
based on the
systems
magnitude (or the
MIMO systems wasdiscovered in the 1980s
value for
MIMO case) of the complementary
generalized
stability
maximum
singular
margin function
value for
gives the
MIMO systems)
sen-sitivity
maximum
of the plant
for every frequency so that the closed-loop system can still remain stable.
State-feedback design, introduced briefly in previous chapters, is the focus of Chapters 10 and 11.
The state-space approach became popular in the early 1960s beginning with the publications of Rudolf
Kalman. Instead
to earlier
of frequency-domain
methods (i.e.,
Laplace transform
methods of analysis and design using differential
approaches)
equations initiated
attention
by James
returned
Clerk
Maxwell.
The main reasons for this paradigm shift are listed in the following:
the state-space approach resolved basic theoretical
frequency-domain tools to MIMO systems;
the nonlinear
identify
system state-space representation
the equilibria
problems that had impeded the extension of
is elegant and versatile,
of the system, and to find
a local
(linear
allowing
or nonlinear)
systematic
ways to
model at each equilib-rium
of interest that can be employed in analysis and controller design;
the state-space framework makesit easier to formulate the control problems as constrained optimiza-tion
problems like the LQR (linear quadratic regulation), the LQG (linear quadratic Gaussian),the
H2 optimization, and the H8 optimization control problems;
the computing capability and the miniaturization of the digital computer have facilitated the appli-cations
of the state-space control approaches in almost every product usedto achieve automation,
precision, reliability, and performance enhancement.
The emergence of the state-space approach did not meanthe end of the frequency-domain approach.
Instead, the state-space modelframework has madeit possible to incorporate frequency-domain perfor-mance
requirements
into
and the time-domain
design of large-scale
properties
MIMO control systems.
both remain important
The frequency-domain
aspects of any system.
properties
They are inseparable.
In
fact, the time-domain responses, stability, and robustness are dictated by the pole locations
the frequency response of the system, as will be witnessed throughout the book.
and
Ourtreatment of state-space control system design consists of two parts: the state-feedback control
part in
Chapter 10, which assumes that all state variables
part in
Chapter 11 that is capable of providing
of the physical
system
model, the input,
other in a perfect fashion. The state-feedback
an amazingly
flawless
duality
an estimate
and the
are available
for feedback,
of all state variables
measured output.
control theory
These two
based on the information
parts complement
and the observer theory
or the linear
each
are related
by
relationship.
Chapter 10 focuses on the state-feedback control theory and implementation
placement
and the observer
quadratic regulator
10.1 of the previous state-feedback
(LQR)
design approaches.
content in earlier chapters,
either via the pole
After a short overview
Section 10.2 considers
a lightly
in Section
dampe
viii
Preface
pendulum
position
control
example,
illustrating
to state-feedback stabilization
is introduced
zeros into
in
Section
10.3.
input-decoupling,
the entire
linearizing
via pole placement and tracking.
MIMO system
and transmission
the nonlinear
dy-namics
The concept of controllability
poles and zeros are defined
output-decoupling
Linear quadratic regulator
process, from
zeros is
and the classification
discussed in
Section
of
10.4.
design is the subject of Section 10.5. It is an optimal control problem to
design a state-feedback controller that stabilizes the closed-loop system and minimizes a weighted
quadratic
performance
index.
and role of the performance
State-space
and observer
Stabilization
weighting
design fundamentals
design are introduced.
properties
that result
from this
approach
and the structure
matrices are discussed.
are completed
in
Chapter 11
It provides a thorough
where the concept
investigation
of observability
of observer theory
and imple-mentation
by either pole placement orthe linear quadratic Gaussianestimation (LQG) approaches. The
duality
of controllability
highlighting
and observability
the significance
is noted, and the notion
of controllability
of a minimal realization
and observability.
output feedback design (i.e., the LQG problem).
Section
is discussed
11.5 is devoted to observer-based
As a result of the separation principle, the state
feedback controller and the observer can be designed separately and then combined to form an output
feedback
control
solution
as an H2 problem, thereby
for the standard
generalizing
H2 control
its control
design problem.
The final section of the book contains five appendices intended
concepts
necessary to the
numbers, including
includes
of Laplace transforms
model linearization
models and linear
LQG
of the book.
primarily
Appendix
to review
A contains
prerequisite
a review
math-ematical
of complex
alternative representations, Eulers formula, and algebraic operations. Appendix B
a discussion
system
main topics
Section 11.6 reformulates
design capabilities.
controllers
and examples
process is provided in
fit in real (e.g.,
nonlinear)
of their
application.
Appendix
A summary
C. An understanding
applications
is a theme
of the dy-namic
of how linear
of the book.
Appendix
D provides a discussion of Masons gain formula andits application to the reduction of multi-loop feed-back
block diagrams and/or signal flow graphs. Finally, Appendix E provides a review of matrices and
vectors including a discussion of the geometry of vector spaces.
We want to emphasize the importance
throughout
the book computation
of
modern computational
is an essential
devices
part of the control system
and
methods.
design and analysis
As noted
process.
Not only do moderntools allow usto easily execute required control design computations, but they also
provide
essential
mechanisms for learning
assemble computer
simulations
and validation
of physical
systems
of controller
along
with their
design concepts.
controllers
The ability
provides
to
an important
virtual laboratory. Herestudents can gain insight into the physical processes and complete a preliminary
evaluation of the effectiveness of the control design. Throughout the book we use MATLAB/Simulink
for performing the required computations and simulations.
Preliminary
versions
of the book
were used and updated through
years 2018 and 2019 as well as the first two quarters of 2020.
Professor
Ajmal
Yousuff,
Dr.
Mark Ilg,
Dr.
Mishah Salman,
all four
quarters
Multiple colleagues
Dr. Christine
of the calendar
and friends,
Belcastro,
including
Professor
Sorin
Siegler, Professor Baki Farouk, and Dr. Hossein Rastgoftar taught multiple classes and provided impor-tant
feedback. Wegratefully acknowledge their contributions. Wealso acknowledge the valuable con-tributions
of our teaching assistants and graduate students, including Nilan Jayasuriya, Po-Chun Chan,
Mevlut Bayram, Fatih Catpinar, Brian Amin, Joseph Masgai, and David Hartman. Student response,
both solicited and unsolicited, was critically important to us. Special thanks go to Ms. Gem Rabanera
and Ms. Abbey Hastings, the editors of Cognella Inc., for their editorial guidance and assistance. Fi-nally,
we are profoundly appreciative of the support and patience of our wives Miriam Kwatny and Janet
Chang
Contents
Part I
1
Basic
Introduction
3
1.1
Control Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
A Little
4
Impact
1.4
History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
of the
Digital
A First Example:
Computer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Linear Systems Analysis I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Typical
2.1.1
2.2
2.5
Mathematical
Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equivalency
Among
First-Order
9
v
-1.............................. 11
2.2.1
Significance
of the Imaginary
2.2.2
Polar Form,
Rectangular
Formula . . . . . . . . . . . . . . . . . . . . . . . .
13
2.2.3
Geometrical Aspects of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
A Brief
Review of the Laplace
Number
Dynamic
9
11
11
Form, and Eulers
Transform
... .. ... . . ... .. ... . . ... .. .. .. . ... .. . ...
2.3.1 Laplace Transform Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Laplace Transform Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Time-Domain Response of Typical First-Order Dynamic Systems . . . . . . . . . . . . . . .
2.4.1 The Response of the Typical First-Order System Dueto Initial Condition . .
2.4.2 The Response of the Typical First-Order Systems Dueto Unit Step Input . .
Frequency-Domain Properties of Typical First-Order Systems . . . . . . . . . . . . . . . . . .
2.5.1 Transfer Functions and Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2
Characteristic Equation and System Poles. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.3 The Responsesof the Typical First-Order Systems Dueto Sinusoidal Inputs
17
20
23
23
25
27
27
28
29
Bode Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
Linear
Frequency
Systems
Responses and the
II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Typical Second-Order Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
3.1.1
Systems . . .
40
Systems . . . . . . . . . . . . . . . . .
41
Transfer Function, Characteristic Equation, and System Poles . . . . . . . . . . . . . . . . . . . . . .
Time-Domain Response of Typical Second-Order Dynamic Systems. . . . . . . . . . . . . . . . .
42
4
3.1.2
Analysis
.
.
.
.
.
.
.
.
.
17
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
2.5.4
3.2
3.3
Dynamic
2.1.2
Characterization of Typical First-Order Dynamic Systems . . . . . . . . . . . . . . . . . . .
A Brief Review of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6
3.1
First-Order
Systems . . . . . . . . . . . . .
2.3
2.4
Cruise
5
Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
3
Systems
. .. . ... . .. .. .. ... . .. .. .. ... . . ... .. ... . . ... .. ... . . ... .. .. .. . ... .. . ...
1.3
2
Concepts of Linear
Mathematical
Characterization
Equivalency
of Typical
Among
Typical
Second-Order
Second-Order
Dynamic
Dynamic
x
Contents
3.4
3.5
3.6
4
3.3.1
The Response of the Typical
3.3.2
The Response of the Typical Second-Order Systems Dueto Unit Step Input . . . .
Characterization
Underdamped
Systems
Second-Order
Conditions
..
Complex
Plane . . . . . . . . . . . . . . . . .
57
System . . . . . . . . . . . . . . . . . . .
57
3.4.2
Step Response of the
3.4.3
Graphical Interpretation of the Underdamped Second-Order Step Response . . . .
Second-Order
Design of a MassDamperSpring
Analysis and
Design of a Simple
Sinusoidal
System
DC Motor Position
50
57
Geometry of Conjugate
Underdamped
44
Systems . . . . . . . . . . . . . . . . . . . . . . .
3.4.1
Analysis and
System Poles on the
Due to Initial
.. ... . . ... .. .. .. . ... .. . ...
Control
Response and Bode Plot of Typical
System . . . . . . . . . . . . . . . .
64
Steady-State
Systems . . . .
66
3.8
Exercise Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
Modeling of Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . .
4.1 Translational Mechanical Systems . . . . . . . . . . . . . . . . . . .
4.1.1 dAlemberts
Principle . . . . . . . . . . . . . . . . . . . . . . .
4.1.2
A Brief Introduction of the Lagrange Approach . .
4.1.3
A Quarter-Car Suspension System . . . . . . . . . . . . .
4.2 Rotational Mechanical Systems. . . . . . . . . . . . . . . . . . . . . .
4.2.1 dAlemberts
Principle in Rotational Systems. . . .
4.2.2 The Lagrange Approach for Rotational Systems .
4.2.3
A Two-Rotor/One-Shaft Rotational System . . . . .
4.3
A Rotational System with a Gear Train . . . . . . . . . . . . . . .
A Simple Inverted
4.4.1
4.5
4.6
5.4
5.5
5.6
5.7
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
.
73
73
74
77
78
80
81
83
84
86
Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
4.4.2
Equilibriums and Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
4.4.3
State Feedback
Cart-Inverted
Controller
Pendulum
Pendulum . . . . .
93
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
Approach
Design to Stabilize
Modeling of the
a Simple Inverted
4.5.1
The Newtonian
4.5.2
Lagranges Approach Modeling of the Cart-Inverted Pendulum System . . . . . . . .
99
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
Cart-Inverted
Pendulum
System . . . .
97
Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Basic Electrical Circuit Elements and Circuit Conventions . . . . . . . . . . . . . . . . . . . . . . . . . 106
Basic Time-Domain Circuit Modeling Approaches and Kirchhoffs Laws . . . . . . . . . . . . 107
Circuit
Modeling
Using the 2k Equations
Approach . . . . . . . . . . . . . . . . . . . . . . . . 108
5.2.2 The NodeToDatum
(NTD) Voltages Approach
5.2.3 The Mesh Currents Approach . . . . . . . . . . . . . . . .
BasicImpedance Circuit Modeling Approaches . . . . . . . .
5.3.1 The Impedance NTD Voltages Approach . . . . . . .
5.3.2 The Impedance Mesh Currents Approach . . . . . .
The Lagrange Approach for Circuit Modeling. . . . . . . . . .
Circuit Modeling Usingthe State-Space Approach. . . . . .
Operational Amplifier Circuits. . . . . . . . . . . . . . . . . . . . . . .
DC Motor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.1
5.8
.
.
.
.
.
.
.
.
.
.
88
5.2.1
5.3
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Modeling of Electrical
5.1
5.2
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
Modeling the Simple Inverted
..
..
..
..
..
..
..
..
..
..
Second-Order
59
61
3.7
4.4
5
of the
Second-Order
Amperes
Force Law and Faradays
Equations for the
DC
..
..
..
..
..
..
..
..
..
Law of Induction
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
109
110
110
112
113
113
118
122
128
. . . . . . . . . . . . . . . . . . . . . . . 129
5.7.2
Assembling
5.7.3
Torque-Speed Relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
Motor System . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
5.7.4
A DC Micromotor
Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Contents
6
Systems
Representations
Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
6.1
Block Diagrams. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
6.2
Signal Flow
6.2.1
6.2.2
Graphs and
Signal Flow
Masons
Masons
Gain Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
Gain Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
6.3
State-Space Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
6.4
State Transition
Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
6.4.1
Theorems
6.4.2
Computing the State Transition
and Properties
of the State Transition
Matrix . . . . . . . . . . . . . . . . . . . . . 152
Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
6.5
Solution of the State Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
6.6
State-Space
6.6.1
6.7
Part II
7
and Interconnected
xi
Models and Transfer
Find Transfer
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
Function from
State-Space
Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
6.6.2
Construct a State-Space Modelin Companion Form Using Direct Realization . . 164
6.6.3
Construct a State-Space Modelfrom Interconnected Systems . . . . . . . . . . . . . . . . 166
Exercise Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
Linear
Control System Design
Fundamentals of Feedback Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
7.1 Features of Feedback Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
7.2
7.1.1
A Demonstrative
7.1.2
Performance
7.1.3
Advantages and Limitations of Feedback Control . . . . . . . . . . . . . . . . . . . . . . . . . . 180
System
7.2.1
7.2.2
7.3
7.4
7.5
7.6
7.7
7.8
Feedback
Verification
Representations
Transfer
Impulse
Control
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
by Time-Domain
Simulation
. . . . . . . . . . . . . . . . . . . . . . 178
and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Function
and
Differential
Response and Transfer
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
Function
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
7.2.3
State-Space Model and Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
7.2.4
Characteristic
Equation,
Poles, and Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
Stability of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.1 Zero-State Response and Zero-Input Response. . . . . . . . . . . . . . . . . . . . . . .
7.3.2
BIBO Stability and Internal Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Similarity Transformation in State Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4.1
Diagonalization of A Matrix Using Similarity Transformation . . . . . . . . . .
7.4.2
Obtaining a State-Space Modelin Companion Form Using Similarity
Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Pole Placement Control in State Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5.1 State-Feedback Pole Placement: Direct Approach . . . . . . . . . . . . . . . . . . . .
Pole Placement:
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
188
188
191
194
195
. . . . . . 197
. . . . . . 200
. . . . . . 201
7.5.2
State-Feedback
Revisit
Cart-Inverted
7.6.1
State-Space
7.6.2
Equilibriums of the Cart-Inverted Pendulum System . . . . . . . . . . . . . . . . . . . . . . . . 207
7.6.3
Linearized
7.6.4
Design of a Stabilizing
Pendulum
Transform
..
..
..
..
..
Model of the Cart-Inverted
State-Space
Approach . . . . . . . . . . . . . . . . . . . . . . 202
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
Pendulum
Models at the Equilibriums
Controller
for the
Upright
System . . . . . . . . . . . . . . . . . . . 206
. . . . . . . . . . . . . . . . . . . . . . . . . 209
Equilibrium
. . . . . . . . . . . . . . . 211
Routh-Hurwitz Stability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
Exercise Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
xii
8
Contents
Stability,
8.1
8.2
8.3
Regulation,
8.5
8.6
Root Locus
Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
Type of Feedback Systems and Internal
8.1.1
Steady-State
8.1.2
Type of Feedback
An Automobile
Model Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
Systems and Internal
Cruise Control
Model Principle
. . . . . . . . . . . . . . . . . . . . 227
Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
8.2.1
Assembling a Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
8.2.2
Design of the Controller K(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
8.2.3
Simulation
8.2.4
Robustness to
8.2.5
Disturbance Feed-Forward Compensation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
Results with Ideal
Engine
Model GE(s)
= 1. . . . . . . . . . . . . . . . . . . . . 241
Model Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
Root Locus Preliminaries
8.3.1
8.4
and
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Root Loci Construction
Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
8.3.2
Root Loci Construction Using MATLAB . . . .
Root Locus Analysis and Design. . . . . . . . . . . . . . . . . .
8.4.1 The Root Locus Method . . . . . . . . . . . . . . . . . .
8.4.2 Explaining the Root Loci Construction Rules.
A Sinusoidal Position Tracking Control System . . . . .
Controller Designfor F/A18 Flight Path Control . . .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
. ..
. ..
. ..
. ..
. ..
. ..
.
.
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
258
261
262
263
266
271
8.6.1
Open-Loop ManualLongitudinal Flight Controlof F/A18 by de. . . . . . . . . . . . . 273
8.6.2
PI Controller Designfor F/A18
8.6.3
State-Space
8.6.4
Comparison of the PI Controller and the Integral Controller with
State-Feedback Compensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
Pole Placement
Flight Path Control . . . . . . . . . . . . . . . . . . . . . . . . 277
and Root Locus
Design for
F/A18
Flight
Path
Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
9
8.7
Aircraft
8.8
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
Altitude
Regulation
via Flight
Path Angle Tracking
Control . . . . . . . . . . . . . . . . . 289
Time Delay, Plant Uncertainty, and Robust Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
9.1
9.2
9.3
9.4
9.5
Stability
Issues
Caused by Time Delay and Plant Uncertainty . . . . . . . . . . . . . . . . . . . . . . . 297
9.1.1
Time Delay and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
9.1.2
Plant
Uncertainty
and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
Contour Mapping and Cauchys Principle of the Argument. . . . . . . . . . . . . . . . . . . . . . . . . 299
9.2.1
Complex
Contour
Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
9.2.2
Cauchys
Principle
of the
Argument. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302
Nyquist Path, Nyquist Plot, and Nyquist Stability Criterion . . . . . . . . . . . . . . . . . . . . .
9.3.1
Nyquist Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.2
Nyquist Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.3
Nyquist Stability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.4 Stability Issue Arising from Feedback Control System with Time Delay . . .
Robust Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4.1
Gain and Phase Margins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4.2 Effect of the Gain of Loop Transfer Function on Gain and Phase Margins. .
Generalized
9.5.1
9.5.2
9.5.3
Small
Stability
.
.
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
306
306
308
310
315
322
322
325
Margins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
Gain Theorem and
Robust Stability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
Interpretation of the Generalized Stability Margins. . . . . . . . . . . . . . . . . . . . . . . . . 331
Relationship Between Gain/Phase Margins andthe Generalized Stability Margins333
9.6
Essential
9.7
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Closed-Loop
Transfer
Functions
and Loop Shaping
. . . . . . . . . . . . . . . . . . . . . . 337
Contents
10
State Feedback
and
Linear
Quadratic
Optimization
xiii
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
10.1 Brief Review of the State-Space Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
10.2
Control of a Lightly
Damped Positioning
10.2.1
A Simple
Pendulum
10.2.2
State-Feedback
Stabilization
10.2.3 Stabilization of the
10.2.4
Tracking
Placement
10.2.5
Tracking
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
Positioning
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
of the Pendulum
Motor/Propeller-Driven
Control of the Pendulum
Positioning
System . . . . . . . . . . . . 351
Pendulum Positioning System . . . . . 353
Positioning
System
Using State-Space
Pole
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
Control of the Pendulum
Positioning
System
Using PID
Control . . . . . 362
10.3 Controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
10.3.1
Controllability
Rank Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
10.3.2
The Controllability
Decomposition
Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
10.4 Poles and Zeros of MIMO Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4.1 Revisit Poles and Zeros of SISO Systems . . . . . . . . . . . . . . . . . . .
10.4.2 Physical Meaning of System Zeros . . . . . . . . . . . . . . . . . . . . . . . .
10.5 State-Feedback Control via Linear Quadratic Regulator Design . . . . . .
10.5.1 Performance Index and LQR State Feedback . . . . . . . . . . . . . . . .
10.5.2 Stabilizing Solution of the Algebraic Riccati Equation . . . . . . . .
10.5.3 Weighting Matrices Qand Rin the Performance Index Integral .
10.6 Exercise Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
Observer
11.1
11.2
Theory
and
Observability
Output
Feedback
Control
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
373
374
375
380
381
382
384
390
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
11.1.1
Observability Rank Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
11.1.2
The Observability
Decomposition
Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
Duality in State Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
11.2.1
Duality of Controllability
and Observability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
11.2.2 State-Space Modelsin Controller Form and Observer Form. . . . . . . . . . . . . . . . . . 401
11.3
11.4
Minimal
State-Space
11.4.1
11.5
Realization
and Controllability
Models and
Minimal
Direct Realization
and Observability
Realizations
Approach to
of
. . . . . . . . . . . . . . . . . . . . . . . . . 403
MIMO Systems . . . . . . . . . . . . . . . . . . . 406
Assemble a
MIMO State-Space
MIMO State-Space Modelsin Block Controller and Block Observer Forms . . . . 409
11.4.3
MIMO State-Space
Full-Order
Observer and
Models in
Gilbert
Output Feedback
Diagonal
Form . . . . . . . . . . . . . . . . . . . . . . . 415
Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418
11.5.1 Brief Review of State-Feedback Control . . . . . . . . . . . . .
11.5.2 Observer-Based Controller . . . . . . . . . . . . . . . . . . . . . . . .
11.5.3 Design of Observer-Based Output Feedback Controller
11.6 LQG Control Problem and the H2 Control Theory . . . . . . . . . . .
11.6.1 The LQG Control Problem . . . . . . . . . . . . . . . . . . . . . . . .
11.6.2 The Standard H2 Control Problem . . . . . . . . . . . . . . . . . .
11.6.3 Solutions to the Standard H2 Control Problem . . . . . . . .
11.6.4 Application of the Standard H2 Control Formula . . . . .
11.7
A
Model . . . . . . . . 406
11.4.2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
. ..
. ..
419
420
421
425
425
425
428
429
Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
Complex
Numbers
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
A.1
Definition and Significance of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
A.2
Complex
A.3
Eulers
A.4
Algebraic
A.5
MATLAB Commands for Complex Number Computations . . . . . . . . . . . . . . . . . . . . . . . . 43
Number Representations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438
Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438
Operations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439
xiv
B
Contents
Laplace
Definition of Laplace Transform and Laplace Transform Pairs. . . . . . . . . . . . . . . . . . . . . . 441
B.2
Laplace Transform
B.4
D
E
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
B.1
B.3
C
Transforms
Inverse
Laplace
MATLAB
Equilibriums
Properties
and
Transform in the
Commands for
and Linearized
Laplace
More Laplace Transform
DC Motor Position
Transform
Pairs . . . . . . . . . . . . . . . . . . . 442
Control
System
. . . . . . . . . . . . . . . 444
Computations . . . . . . . . . . . . . . . . . . . . . . . . 449
Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453
Masons Gain Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
Vectors and
E.1
E.2
E.3
E.4
E.5
E.6
Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
Vectors and Vector Space. . . . . . . . . . . . . . . . .
Matricesand Linear Operators. . . . . . . . . . . . .
Linear Algebraic Equations . . . . . . . . . . . . . . .
Eigenvalues and Eigenvectors . . . . . . . . . . . . .
Singular Value Decomposition. . . . . . . . . . . . .
Positive Definite and Positive Semi-Definite
.. . .. . . .. .
.. . .. . . .. .
.. . .. . . .. .
.. . .. . . .. .
.. . .. . . .. .
Matrices . . .
..
..
..
..
..
..
. ..
. ..
. ..
. ..
. ..
. ..
.
.
.
.
.
.
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
.
.
.. .
.. .
.. .
.. .
.. .
.. .
..
..
..
..
..
..
.
.
.
.
.
.
. ..
. ..
. ..
. ..
. ..
. ..
463
467
470
475
477
478
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
Index
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Part I
Basic Concepts of Linear System
1
Introduction
F
EEDBACK control is a centuries-old tool applied to quite primitive machinery in its earliest days.
The last century saw an explosion of applications as mankind conceived of new machines and
devices for manufacturing, communication, land and air travel, space exploration, and more.
These applications provoked new functionalities of control that required a deeper understanding of how
control systems worked. Hence, control systems emerged as a true engineering science. In this chapter,
the basic structure of control systems will be discussed, along with a brief history of the evolution of the
control
discipline
to
where it is today.
1.1 Control Engineering
Control engineering is a discipline dealing withthe design of devices, called controllers, that managethe
performance
of a system through
based solely
on the anticipated
manipulation
response
of control
of the system
inputs.
The control-input
commands
or they can adjust in accordance
can be
with observations
of system behavior. Theformer is called open-loop control and the latter is feedback control. Feedback
control is important
environmental
centuries,
whenthere is imprecise knowledge of how the system will respond or there are
uncertainties
but today they
that
can affect
are truly
behavior.
ubiquitous.
Mechanisms of this sort have been employed
Control systems
are an essential
part of chemical
for
and
manufacturing processes, communication systems, electric power plants and systems, ground vehicles,
ships, aircraft
and spacecraft,
robots
and
manipulators,
computers,
and so on.
In the last century engineering wastransformed from a craft into a science. Thoseinterested in profit-ing
from societys
error to develop
thirst for new technology
new products
or resolve
have found it impossible
problems in existing
ones.
to rely on time-consuming
Modern technologies
like
trial
and
automo-biles,
aircraft, telecommunications, and computers aretoo complex to thrive solely on vast compilations
of empirical data and decades of experience. Some intellectual constructs that organize and explain es-sential
facts and principles
are required.
So, engineering
in general and control engineering
in particular
have come to adopt the style and methodsof the natural sciences.
At the core of this point
of view is the distinction
between two thought
processes: the physical,
and
the mathematical. While engineers conceive of problems in the physical world and construct solutions
intended for application in the physical world,the solution is almost always developed in the mathemat-ical
world. Todays
domain
we work
engineers
must be comfortable
with abstractions
of the physical.
with translating
Abstraction
between them. In the
is essential
because
mathematical
most systems
or
devices involve so manyirrelevant attributes that their complete characterization would only obscure
practical solutions. Onthe other hand, abstraction can be dangerous becauseit is often easy to overloo
4
1 Introduction
important
features
and, consequently,
to develop
designs that fail to perform
adequately in the physical
world. Hereinlies the challenge and the art of engineering.
Because of its interdisciplinary nature and the breadth of its applications, control engineering is es-pecially
reliant on a scientific perspective. Unifying principles that bring together seemingly diverse
subjects
within
a single inclusive
as the ultimate
unifying
concept
principle
are of great significance.
in science and technology,
Mathematics,
is very
project
1.2
begins
with a problem
A Little
definition
engineering
there is
may be re-garded
much at the heart of control
engineering. In some circles control theory is considered a branch of applied
mathematics is a necessary part of control
which
much moreto it.
and ends with a solution implemented
mathematics. But while
A control
system design
in the physical
world.
History
It is almost certain that feedback controllers in primitive form existed manycenturies ago. Butthe ear-liest
to receive
incorporated
prominence
in
many systems
in the
Boulton-Watt
written
history is James
steam engine of 1788.
of the late 18th and 19th centuries
Watts fly-ball
Governors,
during
governor,
or speed regulators,
a device that
was
were important
in
which time several alternatives
were developed
to meetincreasingly stringent performance requirements. An article bythe noted physicist James Clerk
Maxwell, On Governors, published in [Maxwell, 1868], is considered the first paper dealing directly
with the theory of automatic control. In that paper Maxwell considers Watts governor and contempo-rary
governing mechanisms developed by J. Thompson, L. Foucault, H. C.Jenkin, and C. W.Siemens
[Siemens, 1866]. Heemphasizes the distinction between moderatorsand regulatorswhat
we now refer
to as proportional control and proportional plus integral control. In addition, Maxwell posedthe prob-lem
of stabilization of alinear feedback system and solved it for athird-order system. Asolution method
for higher order systems was developed independently by [Routh, 1877] and [Hurwitz, 1895]. Atthat
time period, the Russian engineer Vyshnegradskiy worked on similar control problems, publishing his
results in
Russian [Vyshnegradskiy,
1877].
A discussion in English can be found in [Pontryagin,
1962].
Subsequent advances through the mid-1930s dealt with turbine speed control and other applications
such as ship steering and stabilization published in [Minorsky, 1922], electric power systems static sta-bility
in [Clarke, 1926, Clarke and Lorraine, 1933], navigation and aircraft autopilots in [McRuer and
Graham, 2003]. Thesetechnology contributions also stimulated a formal definition of the basic feedback
control problem called the servomechanism design problem as described in [Hazen, 1934]. The 1930s
also saw the development of frequency response methodsfor dealing with stability issues in feedback
systems.
Whilethese techniques
systems
systems
per se, the
analysis.
The
were developed in the context
methods of [Nyquist,
1932] and [Bode,
WWII years produced
of feedback
amplifier
design and not con-trol
1940] have become basic tools
many new results.
Radar, artillery
fire control,
of control
navigation,
and communications problems pushed control system technology to its limits and beyond. For the first
time
optimal
control
most prominent
design problems
were posed and solved (using
example is the formulation
problem, now known asthe
and solution
WeinerHopfKolmogorov
The 1950s saw dissemination
of the
war years
locus method[Evans, 1948]. Forthe first time
Frequency (transform)
of affairs turned
more appropriate
upside down
for
domain
frequency-domain
of the single-input/single-output
efforts
and also new results, including
multiple-input/multiple-output
control.
con-trol
Evans root
control problems werefor-mulated.
methods were by now well entrenched.
and nonlinear
The
optimal
problem [Wiener, 1949].
when R. E. Kalman argued that time-domain
multivariable
methods).
The 1960s saw this state
or state-space
Kalman and Bucy solved the
methods were
multivariable
ver-sion
of the WeinerHopfKolmogorov
optimal control problem in the time domain [Kalman, 1960a,b,
Kalman and Bucy, 1961]. New,important concepts of controllability and observability wereintroduced
[Kalman,
1960a,
Kalman et al., 1963]. Indeed,
state-space
and optimal
control
methods seemed tailor
1.3 Impact
madefor the race to the
of the
Digital
Computer
5
moon in this decade.
Through the 1970s and 1980s the theory attempted to reconcile the frequency-domain and time-domain
perspectives. State-space methods merged and enriched the classical frequency-domain point
of view. The notions of controllability and observability contributed to resolving the definition of
multiple-input/multiple-output
1976].
(MIMO)
system
Concerns about robustness
to
zeros [Rosenbrock,
model uncertainty
1973,
came,
1974,
MacFarlane and
once again, into
Kar-canias,
focus.
While
Kalmans optimal control design provided many advantages in terms of stabilization, it wasrecognized
that some applications
The tracking
1972,
required improved
or regulator
Davison, 1972,
output tracking
problem
was formulated
Wonham and Pearson, 1974].
and robustness
with respect to
in early 1970s [Johnson,
Afeedback regulator
model uncer-tainty.
1968, 1970,
is a controller
that
Kwatny,
both stabi-lizes
the system and ensuresthat the error of selected performance variables goto zero whenthe system
is subjected to a specified
stability
class of disturbances.
and error zeroing
The robust regulator
is preserved even if selected
problem
plant parameters
adds the requirement
vary [Francis
and
that
Wonham,
1976].
A new control
design
method for linear
multivariable
systems
became a focus in the 1980s.
Known
as robust control or H8 control, this approach wasintended to reduce performance sensitivity to distur-bances
as well as plant uncertainty
for
many decades, but a renewed
and Krener, 1977,
Brockett,
[Zames,
making.
very little theory
sometimes
Doyle et al., 1989].
1978] and new technology
electric power systems. Another important
decision
1981,
focus emerged, spurred
While it
requirements,
has been a concern
in control theory
including
[Hermann
aerospace, robotics,
and
aspect of control systems was modeswitching based on log-ical
was a long-standing
until the late 1980s.
Nonlinearity
by developments
element
Modern technologies
based onlarge amounts of incoming
of control implementation
require increasingly
data and often involving
there
more complex
dynamics
had been
decisions,
as well aslogic.
Early
theoretical work involved discrete event decision making (e.g., without considering system dynamics)
[Ramadge and Wonham, 1989]. Morerecently, a hybrid system theory has emerged that integrates logic
and dynamics [Branicky et al., 1998, Bemporad and Morari, 1999].
The articles
starting
cited are noted because of their foundational
point because they are forward-looking
significance.
research
papers.
They
may represent
A more expansive
a chal-lenging
and accessi-ble
account of the history of control system design can befound in numerous morerecent publications
that look
backward
from
elucidates
the foundational
a broader
context.
contributions
of
Notable articles include
Maxwell.
the
paper [Kang,
The papers [Bernstein,
2016],
2002] and [Bissel,
which
2009]
articulate the emerging technologies that drovethe evolution of control theory. And,finally, the papers of
[Sussman and Willems, 1997] and [Pesch and Plail, 2009] deal with the development of optimal control.
1.3 Impact of the Digital Computer
Not only hasthe theory of control engineering evolved quite substantially over the last several decadesdriven
largely by a dramatically expanding domain of applicationsbut
the tools of the discipline have
also changed radically.
and, particularly,
In fact, one could argue that it is the availability
implementation
that underlies the pervasive inclusion
of new tools for analysis,
of feedback
control in all
design,
manner
of present-day systems and devices. Digital computers only became widely available in the 1960s and
workstations
and personal computers
have become so powerful
control
not conceived
Many systems
of in earlier
really
came of age in the 1980s.
and inexpensive
that they
During the
past decade
micro-processors
have opened the door for applications
of
years.
and products require
feedback
control in
order to function.
Examples include
com-puter
disk drives, robots, spacecraft, aircraft, and electric power systems, to name a few. It would b
6
1 Introduction
impractical
and often impossible
to operate
modern
manufacturing
systems
or power
plants efficiently
and safely without automatic control. But even consumer products, from washing machinesto CD play-ers
to automobiles,
control
require
are integrated
used in engines for improved
traction
or benefit in terms
in their
of cost and performance
design. In automobiles,
efficiency
and emission
for
and climate
control.
when actuation,
control
control, in airbags,
control and in suspension systems for improving
mention cruise control
example,
both ridability
So pervasive is the design of
systems
anti-lock
sensing,
and
are increasingly
brakes, and skid and
and handling quality, not to
mechanical systems integrated
with sensing, actuation, and control that the name mechatronics has been coined to identify this branch
of engineering.
What makes this possible is the advent of compact and powerful
microprocessors.
The early feedback
controllers werelarge mechanical devices. These, overthe years, weresuperseded by more compact hy-draulic,
pneumatic,
could perform
and electronic
only limited
has dramatically
devices.
computations.
reduced the cost and
But, the improved
The advancement
miniaturized
controllers
of computer
were still
physically
and semiconductor
large and
technology
the hardware required in control system implemen-tations.
The ability to rapidly execute complex algorithms in small, low power devices wastruly a game
changer.
1.4 A First Example: Cruise Control
Automotive cruise control systems present a convincing example of introducing moderncontrol concepts
for several reasons. To begin with, mostautomobile drivers are familiar with them. Moreimportantly,
cruise control systems have evolved from extremely simple devices with a single goal of manipulating
the throttle to maintain a specified speed to highly complex systems with multiple inputs (manipulators)
and outputs (sensors) and several distinct objectives. In addition, vehicles are becoming increasingly
equipped with additional driver assist controllers such as dynamic stability control, active lane keeping,
distance control,
active braking,
coordinated to avoid interference
and controlled
collision
avoidance.
All of these controllers
need to be
with each other.
In classical cruise control, speed is measuredandthe throttle is adjusted in response to speed error.
Design of the controller requires an understanding of how the vehicle speed responds to a change in
engine torque as well as disturbances, the key disturbance being road slope. Wegenerally organize the
design process by first constructing a block diagram consisting of the major elements to be considered,
asillustrated in Figure 1.1.
Fig. 1.1: Feedback configuration
of a simple
cruise control system.
To design the controller, weneedto know how the engine torque responds to throttle change. Thus,
we begin by assembling models of the plant (i.e., the vehicle andits engine). The modelsare built with
1.4
combination
of physical
principles
and empirical
A First Example:
data. Finally, the
model consists
Cruise Control
7
of a set of differential
and algebraic equations. The process of model building also involves a set of assumptions concerning
the vehicle
need to
and its environment.
make assumptions
the vehicle
aerodynamic
Some are simple,
about how it is loaded.
resistance.
and
we just
need to
know the vehicle
Others may be more complex,
In designing the controller,
mass, so
such as how
we are concerned
we
we model
with, of course, how
well speed is regulated. Is the closed-loop system, composed of the plant with controller, stable? How
does the system respond to a change in speed command
long
does that take? Is there too
performance
acceptable if the
much overshoot?
mass is
different
(i.e.,
does the speed error go to zero)?
We also need to test our assumptions.
from
what was assumed for controller
How
Is closed-loop
design?
Orif
aerodynamic coefficients or engine parameters differ from assumptions? The cruise control orthe speed
tracking
control
design
will be discussed in
Example 5.26 of Section 5.7.4, Example
6.1 of Section 6.1,
and a mid-size SUV cruise control example in Section 8.2.
In contemporary
versions
of cruise
control there are additional
have as many as nine gears. Gear shifting
at the wheels. How should we coordinate gears and throttle?
vehicle
model from throttle
decisions
to speed changes significantly.
with logic.
Another factor
of significance
For example,
some au-tomobiles
factor in the change of torque
Notethat eachtime a gear is changed, the
Ordinarily,
based on vehicle speed, engine speed, and possibly
dynamics
complexities.
becomes an important
gear change is based on logical
other factors.
is that
Now it is necessary to inte-grate
performance
criteria
has become
more
complex. Today, we are not only concerned about speed regulation, but fuel economy has become an
equally important factor, especially in long-haul trucks.
Another
new element in cruise
control.
switching
from
control is the ability
Here, again, there are logical
one
decisions
mode of operation to another.
to switch
from
speed control to distance fol-lowing
based on data from
multiple sensors that trigger
How does this affect performance?
dynamics and logic is a key component of control design in our current world
The integration
of
2
Linear Systems AnalysisI
T
HE goal of the following two chapters is to makeit easier for beginning students of control to
associate mathematical terms and abstract concepts with real practical systems. Webegin with
a few rather simple mechanical, electrical, and electromechanical systems that can be described
by typical first-order and second-order differential equations in this chapter andthe following chapter,
respectively. Thefirst-order systems will be characterized by a time constant t, andthe second-order
systems will bespecifiedin terms of dampingratio ? and naturalfrequency ?n. Althoughthese char-acterization
parameters
are related to physical
component
values, they
have the advantage
of directly
revealing essential properties of the system dynamics. Moreover,they underscore the notion that differ-ent
physical domains can be understood and examined within a common framework of abstraction.
Throughout this chapter, the simple yettypical first-order system will be employed to demonstrate
fundamental terminology, concepts, and analysis tools including the transfer functions, characteristic
equations, system poles, and the time-domain analysis that consists of the step responses andthe steady-state
sinusoidal
responses.
The frequency-domain
plot, will also be introduced.
tangible
analysis, including
responses and Bode
Readers will find these fundamental concepts and basic approaches quite
since they are all applied to simple
but real practical
will be very helpful in the study of
more complicated
2.1 Typical First-Order
the frequency
systems.
Fully
high-order
understanding
these funda-mentals
systems in later chapters.
Dynamic Systems
Although not all physical systems fall into the categories of the typical first-order systems or second-order
systems, the fundamental concepts and problem-solving skills welearn from these two typical
systems provide a necessary knowledge basethat can be easily extended to more complicated high-order
systems. In this chapter we will begin with the three simple physical systems shown in Figure
2.1: The mass-friction system in (a) is a mechanical system, the RC low-pass filter system in (b) is an
electrical
2.1.1
circuit,
and the
DC motor system in (c) is an electromechanical
Mathematical Equivalency
Among First-Order
system.
Dynamic Systems
The governing dynamic differential equation for the mass-friction system in Figure 2.1(a) can be easily
derivedfrom Newtonslaw of motionthe effectiveforce fef f(t) applyingto the massMwillcausethe
massto move with acceleration?v(t)
= fef f(t)
?M, where v(t) is the velocity
variable of the system. The
friction
force
ffri(t)=-Bv(t)
isproportional
tothemagnitude
ofthevelocity
while
itisalways
against
themotion
ofthemass.
Hence,
we
have
feff(t) =fa(t)-Bv(t),
which
leads
tothefollowing
10
2
Linear Systems
Analysis I
?v(t)+Bv(t)
M
= fa(t)
(2.1)
Notethat v(t), the velocity of the massM,is the output orthe variable of interest in the system, and fa(t),
the applied force, is the control input
by which the
Fig. 2.1: Mathematically equivalent systems: (a)
(c) DC motorsystem.
motion of the system can be changed or controlled.
Mass-friction system, (b) RClow-pass filter circuit, and
The RClow-pass filter system in Figure 2.1(b) consists of a voltage amplifier
source ea(t)
= Aes(t), a resistor
R, and a capacitor
with gain A, a voltage
C. According to the characteristics
of the capacitor,
the currentflowing (downward)throughthe capacitorisiC(t) =C
?e(t), wheree(t)is the voltagevariable
across the capacitor with the polarity specified on the diagram. Similarly, according to Ohms law, the
current
flowing
(rightward)
through
theresistor
isiR(t)=(1/R)
(ea(t)-e(t)).
Based
onKirchhoffs
current law (KCL), the current iR(t) flowing from the resistor into the junction of R and C should be
equal to the current iC(t) flowing from the junction into the capacitor. The current equivalence leads to
?e(t)=(1/R)(ea(t)-e(t)),
C
andwith
ea(t)=Aes(t)
we
have
thefollowing:
RC
?e(t)+e(t)
= Aes(t)
(2.2)
Notethat e(t), the voltage across the capacitor, C,is the output or the variable of interest in the system,
and es(t), the voltage signal at the input of the amplifier, is the control input by whichthe behavior of
the electrical
system can be changed or controlled.
A schematic diagram of a DC motor system is given in Figure 2.1(c). Thesystem consists of a DC
motor driven by an electric voltage source em(t), a gear reduction box with gear ratio N1/N2, and the
load with moment of inertia JL and rotational friction coefficient BL. The mathematical modeling of the
electromechanical system will be given later in Chapter 5. Although the modeling process will not be
trivial, under some fair practical assumptions the behavior of the system can be described by a rather
simple first-order
linear
ordinary
differential
equation
??L(t)+a?L(t)
as follows:
= bem(t)
(2.3)
where ?L(t) and em(t) arethe output angular velocity of JL andthe control-input voltage,respectively.
The coefficients
a and b are constants
determined
by the given
motor system
component
values.
Note
that ?L(t), the angular velocity of the load JL,is the output orthe variable of interestin the system, and
em(t), the voltage signal at the input of the DC motor,is the control input by which the behavior of the
electromechanical
system
can be changed or controlled
2.2
2.1.2
Characterization
of Typical
First-Order
Dynamic
A Brief
share some common
and the third is electromechanical.
properties.
represents
of interest
equations
constant and the steady-state
step response
of the three systems,
typical first-order
in the system,
which the behavior of the system can be changed or controlled.
time
11
However, these three systems intrinsically
In fact, all of the governing
the output or the variable
Numbers
first one is a mechanical system, the
2.1, 2.2, and 2.3, can be rewritten into the same form asthe following
equation:
t?x(t)+x(t)
= xssu(t)
where x(t)
of Complex
Systems
Thesethree systems obviously are different in appearancethe
second one is electrical,
Review
Equations
differential
(2.4)
u(t) is the control input
by
The parameters t and xssstand for the
of the system, respectively.
Their physical
meaning will
be clearly described later in this section.
After rewriting the mass-friction system governing equation, Equation 2.1, in the form of the typical
first-order
differential
equation,
Equation
2.4, we obtain
(M/B)?v(t)+v(t)
=(1/B)fa(t)
wherethe output variable is x(t) = v(t), the control input is u(t)
and the steady-state step response is xss =1/B.
(2.5)
= fa(t), the time constant is t = M/B,
The RClow-pass filter system equation in Equation 2.2is already in the form of Equation 2.4,
RC?e(t)+e(t)
wherethe output variable is x(t)
= Aes(t)
(2.6)
= e(t), the control input is u(t)
= es(t), the time constant is t
= RC,
and the steady-state step response is xss = A.
Rewriting the
DC motor system governing
equation,
Equation 2.3, into the form
of Equation
2.4 will
give
(1/a)??L(t)+?L(t)
wherethe output variable is x(t)
= (b/a)em(t)
=?L(t), the control input is u(t)
(2.7)
= em(t), the time constant is t
= 1/a,
andthe steady-state
stepresponseis xss= b/a.
From these discussions,
it is clear to see the
mathematical
equivalency
among these three systems.
The mathematical equivalency and the characterization
of the typical first-order
systems have
boiled down the dynamic behavior study of a group of typical first-order systems to the investiga-tion
of the two parameters t and xssin the first-order
differential equation, Equation 2.4.
2.2 A Brief Reviewof Complex Numbers
2.2.1 Significance of the Imaginary
Number
v
-1
It is well known that a complex number X = a+ib consists of two parts: the real part a and the imag-inary
part ib, where a and b are real numbers and i is the imaginary number i =
v
v
-1.Although
the
imaginary
number
-1has
eventually
become
one
ofthemost
important
inventions
inhuman
history,
originally it was created by mathematicians merely for a simple mathematical completeness purpose
so that an n-th order algebraic equation would always have n solutions. Theterm imaginary number
seemed to reflect the reality
that it
was created out of imagination,
but this
derogatory
term
might have
misledstudents to wrongly believe that the complex numbers are not real andthus not practical or useful
12
2
Linear Systems
It is quite the opposite:
Analysis I
The complex
physical science, computer
world
we enjoy today
or discovered;
reliable
the
control
mostlikely
Another
obstacle that
important
in the study
It is fair to say that the
not be possible if the imaginary
number
would have no radio communication,
system for aircraft flight,
Fig. 2.2:
is extremely
science, and engineering.
would probably
world
number
space exploration,
Geometry of complex
or industrial
of
mathemat-ics,
moderntechnical
had never been invented
no TV, no satellite,
and no
automation.
numbers and Eulers formula.
might prevent students from learning
the complex
number subject
well is the
discouraging word complex, which seemsto imply the subject is complex and complicated. As a matter
of fact, the fundamental concept of the complex number is rather simple and straightforward
if
the geometrical aspect of the complex number shown in Figure 2.2 is well understood.
Inaddition
tothei = v-1notation,
thealternative
notation
j = v-1isalsowell
adopted,
especially
in the engineering community. The reason for choosing i seemed to be related to the term imaginary
number, yet the rationale of the engineering communitys choice for j probably wasto avoid using the
vsame
notation
astheelectrical
current
i.Inthe
rest
ofthebook,
j will
represent
the
imaginary
number
-1unless
otherwise
specified.
In Figure 2.2, the complex number X
in
which the horizontal
=s+
j? is represented by a point X onthe complex plane (s-plane)
axis Re[s] is the real axis and the vertical axis Im[s]
Notethat the complex number X = s+ j?
is the imaginary
geometrically is a point or a vector (s,?)
axis.
onthe Cartesian
coordinate
plane.
Itcanbeseen
from
thegeometry
thatthemagnitude
|X|and
thephase
angle
?ofthe
vector can be computed in terms of s and ? asfollows:
?s2
?=tan-1 ?
(2.8)
s
Onthe other hand,the projections ofthe vector X onthe real axis andthe imaginary axis are s and ?,
|X| =
+?2,
respectively.
Hence,
sand?can
bewritten
asfunctions
of|X|and?,
s =|X|cos?,? =|X|sin?
(2.9)
Therefore, a complex number can be either represented in rectangular form,
X =s+ j?
or in polar form,
(2.10
2.2
A Brief
Review
of Complex
Numbers
X=|X|??=|X|e
j?
13
(2.11)
Although both ?? and ej? in Equation 2.11 reveal that the phase angle of the complex number
is ?, the latter is required in algebraic manipulations especially when differentiation and integra-tion
are involved.
The polar form
of a complex
number is at least as important
although the latter is better known to the general
and subtraction,
yet the polar form is
public.
more efficient
in
as its rectangular
The rectangular
multiplication
form
counterpart,
form is suitable
for addition
and division
computations.
The
conversion from rectangular form to polar form and vice versa can be carried out easily using Equation
2.8 and Equation
2.9, respectively.
2.2.2 Polar Form, Rectangular
Theorem 2.1 (Eulers
Form, and Eulers Formula
Formula)
Showthat the following Euler?sformula is valid for all ?:
ej? =cos?+ j sin?
(2.12)
Proof:
The complex number X can be represented in both polar form and rectangular form:
X=|X|ej? = s+j?
(2.13)
Since
sand?canbeexpressed
interms
of|X|and?,asgiven
in Equation
2.9:s =|X|cos?
and
?=|X|sin?,Equation
2.13
becomes
(2.14
X=|X|e
j? =s+j? =|X|cos?+
j |X|sin?
whichleads to Equation 2.12.
Corollary 2.2 (Eulers
Formula and Sinusoidal
Functions)
Showthat cos? and sin? can be expressedin terms ofthe complex variables ej? and e-j? asthe
following:
cos? =
ej? + e-j?
2
,
sin?
=
ej? - e-j?
(2.15)
2j
Proof: Left as an exercise.
Remark 2.3 (Complex
Number and Geometry)
Euler summarized the relationship
between the complex numbers and the trigonometric
functions
into an elegantsimple Eulers formula: ej? =cos? +j sin?. Thecomplex numbergeometry approach,
together with Eulers formula,
two-dimensional space.
Remark 2.4 (Computation
has madeit
much easier to study trigonometry
Rules of Complex Numbers)
and vector analysis on
14
2
Addition
Linear Systems
Analysis I
and subtraction
(a1+jb1)(a2+jb2)=(a1a2)+j (b1b2)
(cos?1
+j sin?1)M2
(cos?2
+jsin?2)
M2e
j?2=M1
M1ej?1
=(M1
cos?1
M2cos?2)+
j (M1
sin?1
M2sin?2)
(2.16a)
(2.16b)
Multiplication
M1e
j?1
M2e
j?2=M1M2e
j(?1+?2)
?
? 1ejtan-1(b1/a1)
? ?
(a1+jb1)(a2+jb2)=
a2
2 ejtan-1(b2/a2)
2 +b2
a2
1 +b2
a2
1 +b2
1
=
(2.17a)
(2.17b)
2 ej(tan-1(b1/a1)+tan-1(b2/a2))
a2
2 +b2
(a1+jb1)(a2+jb2)=(a1a2
-b1b2)+
j (a1b2
a2b1)
(2.17c)
Division
M1e
j?1
M1
=
M2ej?2
?
=?
?
1 ejtan-1(b1/a1)
a2
1 +b2
a1 +jb1
a2 + jb2
a2 +jb2
Example 2.5 (Practicing
Compute X1 = 2ejp/6
=
Compute X2 =(
v
1 ej(tan-1(b1/a1)-tan-1(b2/a2))
a2
2 +b2
2
+b1b2)+
j (a2b1
-a1b2)
(a1+jb1)(a2-jb2)= (a1a2
a2
2
+b2
(a2+jb2)(a2-jb2)
(2.18b)
(2.18c)
2
Computation
of Complex Numbers)
+2ej3p/4
X1 = 2cos(p/6)+
=
(2.18a)
a2
1 +b2
=
a2
2 ejtan-1(b2/a2)
2 +b2
a1 + jb1
ej(?1-?2)
M2
v
j2sin(p/6)+2cos(3p/4)+
3+j - 2+j
v
3+j)(-1+j)
v2
=(
v
j2sin(3p/4)
v
v
(2.19
3- 2)+j(1+ 2)=0.3178+
j2.4142
v
v3+1e
jtan-1(1/
3) v1+1ej tan-1(1/-1)
v2ej135? = 2 v2ej165?
= 2ej30?
X2 =
(2.20)
=-2.732+
j0.732
Compute X3 = (
v
3+j) ?(-1+j)
v?v
=2e
j30??v
X3 =
v3+1e j tan-1(1/
2ej135?
=
3)
1+1ejtan-1(1/-1)
v2e-j105?
=-0.366-j1.366
(2.21)
2.2
2.2.3
Geometrical
Since a complex
Aspects of Complex
by a point on the complex
axis as its imaginary
is an equivalent vector on the two-dimensional
on the complex
The complex
plane can be easily
number
addition
of Complex
Numbers
15
plane,
axis, a complex
which has the horizontal
number
on the complex
plane
real space. The position change of a complex number
accomplished
is similar
Review
Numbers
number can be represented
axis as its real axis and the vertical
A Brief
using complex
to vector addition,
number
addition
but the complex
or
number
multiplication.
multiplication
is
unique in a waythat a vector rotation and translation motion can be carried out by simply multiplying
the complex number vector by another complex number. The multiplication capability has madethe
complex number a powerful tool in geometry, trigonometry, and broad applications in engineering and
science.
Example 2.6 (Using
Eulers Formula to Derive Trigonometrical
Equations)
Show
that
(a)cos2?
=cos2
?-sin2?,
and
(b)sin2?
=2sin?
cos?
using
Eulers
formula.
Proof:
ej2? = cos2?
+j sin2?
=ej?ej? = (cos? +j sin?) (cos? +j sin?)
(2.22)
cos2
?-sin2
?? +j (2sin?
cos?)
=?
Exercise 2.7 (Make
Trigonometry
Easy)
Use Eulers formula to prove the following two fundamental trigonometric
identity equations:
(a) sin(a) =sinacossin cosa
(b) cos(a) =cosacossinasin
(2.23)
Note that these two essential trigonometric identities wereemployed in the high school trigonometry
course to derive manyother trigonometric identity equations like the onein Equation 2.22.
Remark 2.8 (Using
Complex Number
Multiplication
to Perform a Vector Rotation)
It is rather a simple task to rotate a vector in a two-dimensional space using complex number mul-tiplication.
The rotation of a vector X1 = M1e
j?1 by an angle a can be accomplished by multiplying a
complex
number
ejatoX1
sothat
theresultant
vector
willbeM1e
j?1eja =M1e
j(?1+a).
Asshown
in
Figure 2.3, the multiplication of ej45 to X1 = 2ej30? will rotate the vector CCW (counter clockwise) by
45?to
X3 = 2ej30?ej45? = 2ej75?
(2.24)
The complex number employed to accomplish the rotation can befurther modifiedto a more general
form as Aeja. In addition to rotation, this more general linear operator can also change the magnitude
of resultant
from
vector. In other
X1to anywhere
words, by complex
in the two-dimensional
number
multiplication
the vector
The samerotation process can also be achieved using the conventional rotation
but the computation is more complicated:
X3 =
The complex
position
can be moved
space.
matrix multiplication,
?cos45?-sin45???v
? ?1 -1??v
?? ?
3
sin45? cos45?
number representation
1
1
= v2
3
1 1
of the vector X3 is 0.5176+
1
j1.9318,
0.5176
=
1.9318
(2.25)
which is equal to 2ej75?.
16
2
Linear Systems
Analysis I
Fig. 2.3: Usingcomplex number multiplicationto rotatethe vector X1 CCWby 45?to X3.
Fig. 2.4: Using complex number addition to change the vector position from X1to X3.
Remark 2.9 (Using
Complex Number
Addition to
Change Vector Position)
The rotation or position change of vectors also can be accomplished using complex number addi-tion.
If the initial vector position X1and the end vector position X3 are all given, the difference can be
computed
asX2=X3-X1,
and
then
obviously
wehave
X3=X1+X2.
Asdemonstrated
inFigure
2.4,
byadding
X2= -1.2144+
j0.9318
toX1=1.732+
j,theresultant
vector
willbe0.5176+
j1.9318,
whichis the sameasthe result obtainedbyrotating X1 CCWby 45?.
Remark 2.10 (Quaternionsa
Generalization
of Complex Numbers)
Inspired by the success of the complex numbers in two-dimensional space geometry applications,
William Rowan Hamilton determined to find a generalization of complex numbers that can be applied to
mechanicsin three-dimensional space. Since a complex number consists of only two numbers, onereal
and one imaginary, it is quite natural to believe that one more number is needed to add one moredimen-sion.
Initially,
Hamilton
spent
most of his effort in looking
was made,but he never found a meaningful wayto
As one of the greatest
out that four
numbers, instead
mathematicians
of three,
solution.
Some progress
multiply triples like the complex number multipli-cation.
of all time,
are required
for a triple-number
Hamilton
still
to solve this important
needed about 16 years to find
problem.
The quaternion
2.3
a generalization
of complex
numbers
to 3D space,
on October 16, 1843. The quaternions
aerospace
engineering,
In addition
variables
robotics,
to their impact
numbers
of the
by
Laplace
William
Transform
Rowan
17
Hamilton
have been widely applied to 3D mechanics, 3D graphics,
3D animation
enriched the
video, and so on.
and
mathematical
tools
mechanics, the complex
and
methodologies
numbers
and
used in almost
all
The concepts, tools, and methodologies sprouted from the theory of complex
and variables include
the stability
Review
was discovered
on geometry, trigonometry,
have fundamentally
engineering disciplines.
A Brief
the Laplace transform,
theory, the frequency
the transfer
response, and anything
function,
the
poles and zeros of sys-tems,
related to the frequency-domain
analysis
and design, which are almost the entire contents of a typical systems and control book.
In the following
subsection,
we will briefly review the fundamentals
a short list of Laplace transform
partial fractional
equations
expansion,
and understanding
pairs that are frequently
and inverse
Laplace transforms,
the frequency-domain
of Laplace transform,
used, basic Laplace transform theorems,
which are essential in solving
properties
includ-ing
differential
of systems.
2.3 A Brief Reviewof the Laplace Transform
Oncea renowned mathematician said, Mathematics is atransformation that transforms a difficult prob-lem
into an easy one. Although some students may not agree on the remark in general, most of the
engineering students do realize that the Laplace transform indeed transforms a differential equation
problem into
an algebraic
equation
problem,
which is
much easier to solve. In fact, the
Laplace
trans-form
performs two transformations
in the same act: one is from differential to algebraic and the
other is the transformation from the time domain to the frequency domain. The transformation
to the
frequency
domain is a big step forward
C. Maxwells
1868 historical
from the time-domain
not fully
developed
paper, On
in the development
Governors,
of systems
theory.
James
was believed the first to address the stability
perspective. However, the frequency-domain
until the Laplace transform
and control
became
issue
analysis and design approach was
widely known in engineering
community
after
1940.
2.3.1 Laplace Transform
Pairs
The Laplace transform is an indispensable tool in the design and analysis of almost every engineering
problem. In addition to the capability of transforming
a differential equation into an algebraic
one, the
Laplace transform
serves as a bridge
connecting
the two
worlds: the time-domain
and the
frequency-domain
worlds. In the time domain, the signals and systems are represented asfunctions and
differential equations in terms of time t, whichis an understandable real variable. Onthe other hand,the
signals and systems in the frequency
seems to
be a mystery.
the tremendous
Although
domain
this
benefits brought forth
are described in terms
complex
variable
of s, which is a complex
s appears intangible,
by the frequency-domain
functions
variable that
we soon
will
witness
and equations.
Although the time-domain and frequency-domain attributes of signals and systems are closely re-lated,
they provide quite different aspects that are both essential in system analysis and design. To
achieve
the
both time-domain
Laplace transform
and frequency-domain
bridge
back and forth
performance,
it is necessary
between the time
and frequency
to be able to cross
domains.
In the fol-lowing,
we will derive the Laplace transform pairs of the unit step function, the unit impulse function,
the ramp function, and the exponential function based on the defined Laplace transform integral. Then
the Laplace transform
pair of the exponential
function,
together
to derive
more Laplace transform
transform
pairs will be employed in the study of time-domain
this and the next chapters
with Eulers formula,
pairs for damped and undamped
sinusoidal
functions.
can be employed
These Laplace
and frequency-domain responses later in
18
2
Definition
Linear Systems
Analysis I
2.11 (Laplace
Transform)
The Laplace transform
of f(t) is defined as
F(s)
=L[
f(t)]
=
?8
e-st f(t)dt
(2.26)
0-
where
s =s+j?isacomplex
variable,
0- =lim(0-e)
isthe
instant
right
before
t =0,andf(t)is
e?0
piecewise
continuous
when
t =0.The
lower
integration
limitissetatt =0-toresolve
theissue
caused
by some f(t)
with discontinuity at t = 0. Thus,the initial
condition of f(t)
and its derivatives are meant
tobetheir
values
att =0-.
Example 2.12 (Laplace
Transform
of the Unit Step Function us(t))
Recallthat the unit step function is defined as
f(t)
= us(t)
?
=
0
t
<0
1
t >0
(2.27)
The Laplace transform of us(t) is
?8 ?8 ?-1 ?8-1 e-8-e-0e-stdt =
L[1] =
e-stdt =
0Example 2.13 (Laplace
0-
Transform
e-st
s
of the Impulse
=
0-
Function
s
?
? =
1
(2.28)
s
d(t))
Recallthat the unit impulse function is defined as
f(t)
= d(t) =
?
0
t
<0
0
t
>0
?8
d(t)dt = 1
and
(2.29)
-8
The Laplacetransform of d(t) is
L[d(t)]
?8
e-std(t)dt =
=
Example 2.14 (Laplace
Transform
=1
(2.30)
is defined as
f(t)
of t is
d(t)dt
of the Ramp Function t)
Recall that the unit ramp function
The Laplace transform
?8
0-
0-
=t,
t
>0
(2.31)
?8 ?8
te-stdt :=
?8
0- 000wherethe formula ofintegrationby partsis employed withu and dv, chosenasu =t, dv = e-stdt;
L[t] =
u dv = [uv]8
v du
hence,
v=-e-st?
? ?8?8 ? ?8
s, du
(2.32
= dt. Therefore,
L[t] = t
-e-st
s
0-
-
-e-stdt
0-
s
=
-e-st
s2
0-
1
=
s2
(2.33)
2.3
A Brief
Review
of the
Laplace
Transform
19
Example 2.15(Laplace Transform ofthe Exponential Function e-at)
The Laplacetransform of e-at is
?8 ?8 ?-(s+a)?8
Le-at?
?
e-ste-atdt
=
=
0-
e-(s+a)tdt
e-(s+a)t
=
1
=
0-
s+a
0-
(2.34)
Remark 2.16(Laplace Transform of the Complex Exponential Function e-(a-j?)t )
Notethat Laplacetransform
fore
we have the following
? s+ais
1
also valid whenais a complex number, andthere-(2.35)
pair e-at
Laplace transform
pair:
e-(a-j?)t
?
1
s+a- j?
Remark 2.17(Laplace Transform of DampedSinusoidal Functions e-at (cos?t
The complex
Laplace transform
e-ate j?t
pair in Equation
WithEulers formula ej?t
j?
=
s+a-j? (s+a-j?)(s+a+j?)
(2.36)
=cos?t +j sin?t, this Laplacetransform pair becomes
e-at (cos?t +j sin?t)
Separating the real and the imaginary
e-at cos?t
as follows:
s+a+
1
?
= e-(a-j?)t
2.35 can be rewritten
+j sin?t))
(s+a)+
j?
(s+a)2
+?2
partsto obtain two Laplacetransform
(s+a)
?
?
(s+a)2
+?2,
e-at sin?t
(2.37)
pairs,
?
?
(s+a)2
+?2
(2.38)
Remark 2.18(Laplace Transform of Undamped Sinusoidal Functions cos?t and sin?t)
In casethat a = 0, wehavethe following
cos?t
?
Laplacetransform pairs for undamped sinusoidal functions:
?
s
s2 +?2, sin?t ? s2
(2.39)
+?2
Exercise2.19(Laplace Transform of Undamped Sinusoidal Functions with Phase Angle ?)
Provethe complex Laplacetransform pair:ej(?t+?)
sinusoidal Laplace transform pairs:
cos(?t
+?)
?
cos?s-?sin?
, sin(?t
s2 +?2
? ej?
s-j? anduse
it to derive
thefollowing
+?)
?
sin?s+?cos?
s2 +?2
(2.40
20
2
2.3.2
Linear Systems
Laplace
Analysis I
Transform
Properties
So far in this section we have learned how to compute the Laplace transform
functions
and have built a list of common
function
f(t)
Laplace transform
pairs together
Laplace transform
general engineering.
pairs, by which
we can transform
a time-domain
onthe list to its counterpart F(s) in frequency domain, and vice versa. Thislist of
with the
next section to solve linear time-invariant
the
Laplace transform
of basic time-domain
properties that
partial fractional
expansion
method
will be employed
in the
differential equations. In the rest of the section, we will review
are essential
The proofs of the theorems
and relevant to
our study in systems,
are also given except the trivial
controls,
ones, which are left
and
as
exercises.
Theorem 2.20 (Linearity)
L[a1f1(t)+a2
f2(t)]
= a1L[ f1(t)]+a2L[
f2(t)]
(2.41)
Proof: Left as an exercise.
Theorem 2.21 (Frequency
If
Proof: Since f(t)
? F(s)
e-at f(t)
Shift Theorem)
f(t)
? F(s), then
e-at f(t)
e-ste-at f(t)dt
=
2.22 (Time
(2.42)
?8
e-st
f(t)dt,
the
Laplace
transform
pair
for
e-at
f(t)
should
be
?8
?8
= 0-
?
0-
Theorem
? F(s+a)
e-(s+a)tf(t)dt
= F(s+a)
Delay Theorem)
If f(t) ? F(s),
then
f(t -T)us(t-T) ?e-sT
F(s)
Proof:
(2.43)
0-
(2.44)
?8
e-st
f(t)dt,
the
Laplace
transform
pair
for
f(t-T)
us(t
-T)
should
be
f(t -T)us(t-T) ? ?8
e-st
f(t -T)us(t-T)dt= ?8
e-st
f(t -T)dt
Sincef(t)
? F(s) = 0-
(2.45)
0-
T
Let
t -T =t and
change
theintegration
variable
fromdttodt,thenwehave
f(t -T)us(t-T) ?
Theorem 2.23 (Scaling
?8
e-s(T+t) f(t)
dt
= e-sT
0
?8
e-st f(t)
dt
= e-sTF(s)
(2.46)
0
Theorem)
If f(t)
Proof: Left as an exercise.
? F(s), then f
?t/a
?aF(as)
?
(2.47
2.3
Theorem
2.24 (Convolution
A Brief
Review
of the
Laplace
Transform
21
Theorem)
I f f1(t)
? F1(s) and f2(t)
? F2(s), then
?t
(2.48)
f1(t)* f2(t) = 0-f1(t)f2(t-t)dt ? F1(s)F2(s)
Proof:
Since
both
f1(t)and
f2(t)are
zero
fort <0,f2(t-t) =0when
t >t,and
thus
their
convolution
is
0-f1(t)f2(t-t)dt=0?t
?8
f1(t)
f2(t
-t)dt.
According
tothe
definition
ofthe
Laplace
transform,
L[f1(t)*f2(t)]=
?8
??8
f1(t)f2(t-t)dt?
e-stdt
0-
(2.49)
0-
Let
t -t =?and
change
theintegration
variable
fromdttod?,then
we
have
??8 ?
?8
??8 ?
?8
L[f1(t)*f2(t)]= 0=
Theorem 2.25 (Differentiation
0- f1(t) f2(?)dt
f2(?)e-s?d?
0- f1(t)e-stdt
0-
e-ste-s?d?
(2.50)
= F1(s)F2(s)
Formula)
df(t)? dt ?sF(s)-f(0-)
I f f(t)
d2f(t)?dt2 ?s2F(s)-s
f(0-)-?f(0-)
d3 f(t) ? dt3 ?s3F(s)-s2
f(0-)-s?f(0-)-f(0-)
? F(s), then
(2.51)
...
Proof:
L?f?(t)? =
?8
d f(t)
?8
e-stdt :=
u dv =[uv]8
-
?8
v du
(2.52)
00- dt
00wheretheformula ofintegrationby partsis employedwithu and dv,chosenas dv = df(t), u = e-st;
hence,
v =f(t),du=-se-stdt.
Therefore,
we
have
?8
f(t)e-stdt
=-f(0-)+sF(s)
L?f?(t) ? =[e-st f(t)]8
0- +s0-
?
d
=s[sF(s)-f(0-)]-?f(0-)=s2F(s)-s
f(0-)-?f(0-)
L?...
f(t) =Ldt
=sLf(t) -f(0-) =ss2F(s)-s
f(0-)-?f(0-)-f(0-)
?df(t)=s3F(s)-s2
f(0-)-s?f(0-)-f(0-)
? =Ldt
?
?
?
?
Lf(t)
f?(t)=
sLf?(t)-?f(0-)
?
Theorem 2.26 (Integration
?
?
?
?
?
Formula)
If f(t)
? F(s), then
?t
f(t)dt
0Proof:
(2.53)
?
F(s)
s
(2.54
22
2
Linear Systems
Analysis I
??t ? ?8?t ?8 ?8
?t
e-stdt;
hence,
v=-e-st?
-e-st
??t ??-e-st?t ?8?8
f(t)dt
L
f(t)dt
=
0-
e-stdt :=
u dv =[uv]8
0- 0-
0-
0-
-
v du
wherethe formula ofintegration by partsis employed with u and dv, chosenas u = 0- f(t)dt,
s, du = f(t)dt.
f(t)dt
L
=
Theorem 2.27 (Final-Value
If f(t)
f(t)
dv =
Therefore, wehave
f(t)dt
s
0-
(2.55)
0-
0-
0-
-
0-
s
f(t)
dt =
F(s)
s
(2.56)
Theorem)
? F(s) andthe real part of all poles ofsF(s) are strictly negative,then thefinal value of
can be computed in the frequency domain asfollows:
lim f(t)
=limsF(s)
t?8
This theorem is important since mostofcontrol
where f(t) is not available
Proof:
Recall that
L
(2.57)
s?0
system designs are carried out in the frequency domain
before the design is completed.
? ??8e-stdt=sF(s)-f(0-)fromTheorem
2.25.
Assapproaches
to
d f(t)
= 0-
dt
d f(t)
dt
zero, the equation becomes
?8
sF(s)-f(0-)= 0s?0
lim
Therefore,
d f(t)
dt =[ f(t)]8
dt
(2.58)
we have
f(8)
=lim f(t)
= limsF(s)
t?8
Theorem 2.28 (Initial-Value
If f(t)
0- =f(8)- f(0-)
Theorem)
? F(s) and lim sF(s) exists, then the initial
frequency-domain
(2.59)
s?0
s?8
as follows:
f(0+)
value of f(t)
at t
= 0 can be computed i
= lim sF(s)
(2.60)
s?8
Proof:
From Theorem 2.25,
L
? ??8e-stdt=sF(s)-f(0-).Assapproaches
toinfinity,
the
d f(t)
= 0-
dt
term e-st become zero; hence, lim
d f(t)
dt
sF(s)=f(0-) =f(0+)if f(t) hasnodiscontinuity
att =0.Inthe
s?8
case
thatf(t)hasadiscontinuity
jumpat =0fromf(0-)tof(0+),df(t)? dtcontains
animpulse
[f(0+)-f(0-)]d(t).
Now
we
have
?8
d f(t)
dt
sF(s)-f(0-) =lim
0s?8
s?8
lim
?0+
?0+
e-stdt = 0-
d f(t)
dt
dt
=0-[f(0+)-f(0-)]d(t)
dt=f(0+)-f(0-)
(2.61)
2.4
2.4 Time-Domain
Time-Domain
Response of Typical
Response of Typical First-Order
First-Order
Dynamic
Systems
23
Dynamic Systems
As discussed in Section 2.1, alarge group of first-order systems share mathematical equivalency and can
be characterized by the typical first-order differential equation described by Equation 2.4. The equation,
for ease of reference, is repeated asfollows:
t?x(t)+x(t)
In the equation,
the
x(t) is the system output
= xssu(t)
or the variable
mechanical system in Figure 2.1(a), the voltage e(t)
(2.62)
of interest,
which
on the capacitor
may be the velocity
of the electric
v(t)
of
circuit in Figure
2.1(b), or the angular velocity ?L(t) of the electromechanical system in Figure 2.1(c). The variable u(t)
is the control input by which the behavior of the system can be altered or controlled. Thetime constant
t and the step response steady-state value xssare determined by system component values.
Given
theinitial
state
x(0)=x0and
theinput
u(t)fort =0,we
can
solve
thedifferential
equation
to
findx(t)fort =0.After
taking
theLaplace
transform,
Equation
2.62
becomes
thefollowing
algebraic
equation,
(2.63)
t[sX(s)-x0]+X(s)
=xssU(s)
where
X(s)= L[x(t)],
U(s)= L[u(t)],
and
sX(s)-x0= L[
?x(t)].
The
algebraic
equation
canbe
rearranged to give a solution for X(s) asfollows:
t
X(s)
=
xss
x0 +
ts+1
ts+1
U(s):= Xi(s)+Xu(s)
(2.64)
Notethat X(s) consists of two parts: Xi(s), the response dueto the initial state x0, and Xu(s), the response
due to the control input U(s). Therefore, the complete response of the typical first-order system is
x(t)
= xi(t)+xu(t)
(2.65)
=L-1[Xi(s)]+L-1[Xu(s)]
2.4.1 The Response of the Typical First-Order
System Dueto Initial
Condition
The initial state response xi(t), which is the response of the first-order dynamic system due to x0, can
be computed using the exponential Laplace transform pair of Equation 2.34 andthe Laplace transform
linearity property of Theorem 2.20.
xi(t)
=L-1
? ? ? ?
t
ts+1
The initial state responses of the first-order
t = 0.5s,1s,2s are shown in Figure 2.5(a).
x0
=L-1
1
s+1/t
system with initial
x0
= x0e-t/t
(2.66)
state x0 = 1 and the time constant
Notethat the initial andfinal values of xi(t) are xi(0) = x0e0 = x0 and xi(8) = x0e-8 = 0,respec-tively,
and xi(t) = x0e-1 = 0.368x0, which means
the time constantt is the time whenthe exponential
term e-t/t
equals to e-1. In other words, the initial
state response graph is an exponential curve line
connectingthe following three points: x(0) = x0,x(t) = 0.368x0,andlim xi(t) = 0.
t?8
Example 2.29 (The
For the
mass-friction
and the initial
equation
Response of the
System Dueto Initial
Velocity v0 = 1 m/s)
system in Figure 2.1(a), if there is no external force other than the friction
velocity of the
of the system
Mass-Friction
will b
mass Mis assumed v(0)
force
= v0 = 1 m/s,then the governing differential
24
2
Linear Systems
Analysis I
t?v(t)+v(t)
= 0,
v(0) =v0 = 1 m/s,
t = M/B
and the solution v(t) can be found using the Laplacetransform as follows:
1
t(sV(s)-1)+V(s)
=0 ? V(s)
=s+(1/t)
? v(t)=e-t/t,t =0
Then we have
B=1Ns/m,M=1kg ? t =1s ? v(t)=e-t,t =0
t =0
B=1Ns/m,
M=0.5
kg ? t =0.5
s ? v(t)=e-2t,
t =0
B=1Ns/m,
M=2kg ? t =2s ? v(t)=e-0.5t,
Themass
velocity
v(t)willgodown
exponentially
from
v0tozero
ast ?8with
time
constant
t =M/B.
Assuming the friction
coefficient
is constant, the velocity
Note that the response curve due to the initial
will go down faster if the
velocity v(0)
mass Mis smaller.
= v0 = 1 m/sis the exponential
passingthrough the following three points: v(0) =v0 = 1 m/s,v(t)
= e-1 = 0.368 m/s,and v(8)
0.
Fig. 2.5: Theresponses of the first-order system dueto initial
condition or unit step input.
The time response plots in Figure 2.5 are generated using the following
%
CSD
clear,
x0=1,
Fig2.5a
first-order
t=linspace(0,6,61);
tau1=0.5,
initial
state
tau2=1,
CSD
x_ss=2,
Fig2.5b
tau1=0.5,
x3=x0*exp(-(1/tau3)*t);
grid
first-order
t=linspace(0,6,61);
x1=x_ss*(1-exp(-(1/tau1)*t));
plots
tau3=2,
x2=x0*exp(-(1/tau2)*t);
plot(t,x1,'b-',t,x2,'m-.',t,x3,'r--'),
clear
step
response
on,
grid
minor,
on,
grid
mino
plots
figure(12)
tau2=1,
tau3=2,
x2=x_ss*(1-exp(-(1/tau2)*t));
x3=x_ss*(1-exp(-(1/tau3)*t));
plot(t,x1,'b-',t,x2,'m-.',t,x3,'r--'),
MATLAB code:
figure(11),
x1=x0*exp(-(1/tau1)*t);
%
response
grid
curve
=
2.4.2
2.4
Time-Domain
The Response of the Typical
First-Order
Response of Typical
Systems
Due to
First-Order
Dynamic
25
Systems
Unit Step Input
The step response xu(t) is the response of the first-order dynamic system dueto the input u(t) = us(t),
where us(t) is the unit step function defined by Equation 2.27. According to Equation 2.64 and the
Laplace transform of the unit step function,
L[us(t)] = 1/s, we have
Xu(s) =
To find
xu(t),
partial fraction
follows:
which is the inverse
expansion
xss
ts+1
1
s
=
Laplace transform
xss/t
s(s+1/t)
(2.67)
of Xu(s), a common
practice is to employ
the
methodto break down the right-hand side of Equation 2.67 into two parts as
Xu(s)
=
xss/t
s(s+1/t)
A1
=
s
A2
+
(2.68)
s+1/t
There are three common ways to compute the residue constants A1 and A2. The first is the
residue approach by which the constant Ai is evaluated asthe residue at its corresponding pole asfol-lows:
0A2
sXu(s)|s=0
? (0+1/t)
xss/t= A1+0+1/t
? A1=xss
(s+1/t)Xu(s)|s=-1/t
? xss/t
-1/t =0A1
s
(2.69)
? A2=-xss
+A2
The second is the polynomial substitution approach, which also uses pole value substitution like the
residue approach but it is carried out on a polynomial equation. The first step is to multiply both sides
of Equation 2.68 by the least common denominator to obtain the corresponding polynomial equation as
follows:
xss/t = A1(s+1/t)+A2s
(2.70)
Then, by the substitution
of pole values,
we have the following:
s =0 ? xss/t=A1(0+1/t)? A1=xss
(2.71)
s =-1/t ? xss/t=A2(-1/t)? A2=-xss
The third approach is the coefficient comparison approach that sets up equations to
of each term
of the polynomial.
Equation 2.70 can be rearranged
matchthe coeffi-cients
asthe following:
xss/t = (A1 +A2)s+(1/t)A1
By comparing
the coefficients
(2.72)
of the s terms and the constant terms,
we have the following:
A1+A2
=0 and(1/t)A1
=xss/t? A1=xss
=-A2
(2.73)
Now, with A1and A2 obtained, Equation 2.68 becomes,
Xu(s) =
xss/t
s(s+1/t)
xss
=
s
Using two basic Laplace transform pairs, L[1] = 1/s and
the first-order system xu(t) is found as follows:
+
-xss
s+1/t
L[e-at]
= 1/(s+a),
xu(t)=L-1[Xu(s)]
=xss(1-e-t/t),
when
t =0
(2.74)
the step response of
(2.75
26
2
Linear Systems
Analysis I
The step responses
of the first-order
system
with the final
steady-state
value xss = 2 and the time
constant t = 0.5s,1s,2s are shownin Figure 2.5(b). Notethat the initial andfinal values of xu(t) are
xu(0)
=xss(1e0)=0and
xu(8)=xss(1e-8)=xss,
respectively,
and
xu(t)=xss(1e-1)=
0.632xss,
whichmeans
thetimeconstantt is thetime when
the exponential
terme-t/t equalse-1.In
other words, the step response graph is an exponential
points: x(0) = 0, x(t)
= 0.632xss,and lim xu(t)
t?8
curve line connecting the following
= xss.
Forthe mass-friction system in Figure 2.1(a), if the initial
unit step force is applied to the system
at t
three
= 0, then the
velocity of the mass Mis assumed 0 and a
mass velocity
would go up exponentially
from
0to0.632xss
when
t =t =M/B
and
continue
torise
tothesteady
state
xss
as
t ?8,where
BandM
are the friction
coefficient
and the
mass of the system, respectively.
Note that
when the time
constant is
smaller, the velocity of the system rises faster.
In case that the system is subjected to both of nonzero initial condition and unit step input, the
complete response of the system will be the sum of the initial state response and the step response, as
shown in the following
equation:
x(t)=xi(t)+xu(t)=x0e-t/t
+xss(1-e-t/t)
(2.76)
=x0+(xss
-x0)(1-e-t/t)
Fig. 2.6: The responses of the first-order system dueto both of initial condition and unit step input.
The complete
responses
of the first-order
system
with initial
with the final state value xss = 2 and the time constant t
is clear that the complete response graph is the superposition
Figure 2.5(b) since x(t)
= xi(t)+xu(t).
state x0 = 1 and the unit step input,
= 0.5s,1s,2s,
are shown in Figure 2.6. It
of the two graphs in Figure 2.5(a) and
It also can be seen that the complete
response
can be rewritten
asx(t)=x0+(xss
-x0)(1-e-t/t)
fromEquation
2.76.
Hence,
theinitialand
finalvalues
ofx(t)are
x(0)=x0and
x(8)=xss,
respectively,
and
x(t) =x0+(xss-x0)(1-e-1)
=x0+0.632(xss
-x0),
which means
the time constantt is the time whenthe exponentialterm e-t/t equalse-1. In other words,
the step response graph is an exponential curve line connecting the following three points : x(0)
x(t) = xss.
= x0,
x(t)=x0+0.632(xss
-x0),
and
lim
t?8
Example 2.30 (The
Response of the
Mass-Friction
System Dueto Unit Step Input)
For the mass-friction system in Figure 2.1(a), if the applied force input is fa(t) = us(t)N and the
initial velocity is assumed v(0) = 0, then the governing differential equation of the system will b
2.5
t?v(t)+v(t)
Frequency-Domain
=xssus(t),
Properties
v(0) = 0,
of Typical
First-Order
27
Systems
t = M/B, xss =1/B
and the solution v(t) can be found using the Laplacetransform as follows.
s + -xss
s ? V(s)
=s(s+1/t)
=xss
t =0
? v(t)=xss(1-e-t/t),
tsV(s)+V(s)
xss/t
= xss
s+1/
Then we have
B=1Ns/m,M=1kg ? t =1s ? v(t)=1-e-t,t =0
t =0
B=1Ns/m,M=
0.5
kg ? t =0.5
s ? v(t)=1-e-2t,
t =0
B=1Ns/m,M=
2kg ? t =2s ? v(t)=1-e-0.5t,
Themass
velocity
v(t)willgoupexponentially
from
v(0)=0m/s
tov(t)=1m/s
ast ?8with
time
constant t
= M/B. Assumingthe friction coefficient is constant,the velocity will go up faster if the
mass Mis smaller.
Note that
the step response
curve is the
exponential
curve
passing through
the
following
three
points:
v(0)=0m/s,
v(t)=1-e-1=0.632
m/s,
and
v(8)=1m/s.
2.5 Frequency-Domain Properties of Typical First-Order Systems
In the study of the typical first-order dynamic systems, up to now our focus has beenin time-domain
properties and analysis. In this section, we will study the frequency-domain properties of the system in-cluding
the transfer functions, the poles and zeros, the characteristic equations, the frequency responses,
and the Bode plot.
2.5.1 Transfer
Functions and Differential
Equations
As described in Section 2.4, the Laplace transform of the system output, X(s) =L[x(t)] is related to
the initial state x(0) = x0 and the control input U(s) =L[u(t)]
according to Equation 2.64, which is
rewritten
in terms
of G0(s) and
X(s) =
It is noted that the initial
G(s) as follows:
t
ts+1
x0 +
xss
U(s) :=
ts+1
state x0 and the control input
U(s) affect the output X(s) via the two chan-nels
G0(s)and G(s),respectively. Thefunction G(s) = xss/(ts+1),
relationship
described
between
U(s) and
by the differential
X(s) in frequency
equation
Equation
the early development
whichspecifiesthe input-output
domain, is called the transfer
2.62. In addition
representing systems, the transfer function allows the investigation
and facilitates
of the classical
(2.77)
G0(s)x0 +G(s)U(s)
to
function
of the system
being another convenient
of the frequency-domain
way of
properties
control theory.
Since the transfer function G(s) only specifies the relationship between the control input U(s) and
the output X(s), it has nothing to do with the initial condition x0. But the assumption of the initial
condition
being zero in the definition
of the transfer
G(s)
=
X(s)
function
xss
?
?
U(s)
=
?
?
x0=0
ts+1
(2.78)
maycause some confusion. In fact, this zero initial condition assumption is madeonly for the computa-tion
of the transfer function.
Asfor the computation
ofX(s)
or in the process of solving a differential
28
2
equation,
Linear Systems
the effect
Analysis I
of the initial
condition
should
not be discounted.
From this discussion, we know how to obtain the transfer function from a given differential equation
by using the Laplace transform. Conversely, the same procedure can beimplemented reversely to find
the corresponding differential equation. Considerthe system described bythe following transfer function
representation:
b1s+b0
X(s) =
s2 +a1s+a0
This equation can be rewritten asthe following
(2.79)
algebraic equation:
s2X(s)+a1sX(s)+a0X(s)
Fig. 2.7: The frequency-domain
U(s)
= b1sU(s)+b0U(s)
and time-domain
representations
(2.80)
of differentiators
and integrators.
According the Laplace transform theory regarding the frequency-domain andtime-domain represen-tations
of the differentiation and integration operators as shown in Figure 2.7, the complex variable s
here actually represents a differentiation operator, with which sX(s) and s2X(s) mean?x(t) and
x(t) in time domain, respectively. Therefore, the differential equation corresponding to Equation 2.80
can be easily obtained as
x(t)+a1
2.5.2
Characteristic
Observing the initial
Equation
?x(t)+a0x(t)
and System
state response in
(2.81)
= b1
?u(t)+b0u(t)
Poles
Equation
2.66 or the step response in
Equation
2.75,
we can see
that the transient responseis mainly determined by the exponentialterm e-t/t. This exponentialterm
clearly is associated with the denominator polynomial ts+1 of the transfer function in Equation 2.77.
This polynomial is named the characteristic polynomial of the system, and its associated polynomial
equation, ts + 1 = 0, is called the characteristic equation of the system. In general, if the transfer
function
of a system is represented
by
G(s) =
N(s)
(2.82)
D(s)
then the characteristic equation of the system will be
D(s) = 0
and the
roots
of the characteristic
equation
are
defined
(2.83)
as the
poles of the system.
Similarly
the
roots of the equation, N(s) = 0, are defined asthe zerosof the system. Forthe system withtransfer
function
shown
inEquation
2.77,
there
isnozero,
but
there
isone
pole
ats =-1/t,where
tisthetim
2.5
constant and the pole is located
s = 1/t,
Frequency-Domain
on the left-hand
Properties
of Typical
side of the complex
First-Order
plane. If this
pole
Systems
29
were located
at
whichis on the right-hand side of the complex plane,the corresponding exponential term
would be et/t
and
would increase
without
bound
as time increases.
In this case, the system is said to
be unstable. Theformal definition and more detailed discussion of system stability
will be given in later
chapters.
For
thesystem
with
transfer
function
defined
in Equation
2.79,
there
isone
zero
ats =-b0/b1,
which is the root of the equation b1s +b0 = 0. The characteristic equation is s2 +a1s +a0 = 0, and
the system
complex
poles are the two
numbers
by the location
roots of the quadratic
or both real numbers.
equation.
The two roots can be a pair of conjugate
We will seethat the behavior
of the system poles on the complex
plane.
of the system is
Detailed discussion
mainly determined
will be given in the next
chapter regarding the typical second-order systems.
2.5.3 The Responses of the Typical First-Order
Systems Dueto Sinusoidal Inputs
In Section 2.4, weinvestigated the time-domain response of the typical first-order system using the step
function as the input testing signal and observed that the transient responses dueto the initial state x0
and the unit step input are related to the time constant t. Later,the discussions of the transfer functions,
the characteristic
equations,
and the system poles in Section 2.5.1 and 2.5.2 confirmed
that the transient
response of the system is mainly determined by the location of the system poles. In this section, we
will investigate how the typical first-order system will respond to sinusoidal input signals with different
frequencies.
Before considering
the
more general problem
posted on Figure 2.8 at which the sinusoidal
?is a variable, we will consider a special case of the problem ? = 1rad/s in the following
Example 2.31 (Response of a First-Order
frequency
example.
System Dueto Sinusoidal Input)
Consider the system described in Figure 2.8.
Assume the initial
condition
of the system is zero, the
parametersin the transfer function G(s) are xss = 2 and t = 2s, andthe control input is u(t) = cost,
which means
the frequency ofthe signalis ? = 1rad/s. Findthe outputresponsey(t) that includes both
transient response ytr(t) and steady-state response yss(t). Thetransient response will die out asthe time
increases,
and the steady-state
response is the output response after the system reaches the steady state.
Fig. 2.8: Output response dueto sinusoidal input.
The output Y(s) is obtained asfollows:
Y(s) = G(s)U(s)
The output response y(t) is the inverse
first
=
2
2s+1
s
s2 +1
Laplace transform
s
=
(s+0.5)(sj)(s+j)
ofY(s).
To find the inverse
(2.84)
Laplace transform,
weshould decompose the rational function into the sum of several simpler parts using the partial
fractional
decomposition
i
30
2
Linear Systems
Analysis I
Y(s)
=
s
A1
(s+0.5)(s-j)(s+j)
=
s+0.5
c*
c
+
+
s- j
(2.85)
s+ j
whereA1is areal numberandcis a complexnumberto bedetermined.Thereis no needto computec*
since it is the complex
conjugate
of c. These partial fraction
expansion residue
constants can be evaluated
using the residue approach asfollows:
A1 = lim
-0.5
= (-0.5)2+1
= -0.5
1.25=-0.4
(s+0.5)Y(s)
s?-0.5
c =lim
s?j
(2.86)
j
1
(s- j)Y(s)=(j+0.5)(2
j) =1+j2=
1
j163.4?=1v5e-j63.4?
j arctan2
= v5e
v1+22e
?v?
?v?
y(t) =-0.4e-0.5t
+?v?
1/
?
=-0.4e-0.5t
+?v??
1/
=-0.4e-0.5t
+?v?
2/ 5 cos(t-63.4?)
Sincec* is the complex conjugate of c, we havec* = 1/
. Now,
5 ej63.4?
and c* into Equation2.85, and usethe Laplacetransform pair 1/(s+a)
response in
5 e-j63.4?ejt
5
+ 1/
plug the values
of A1, c,
? e-at; wehavethe output
5 ej63.4? e-jt
ej(t-63.4?) +e-j(t-63.4?)
(2.87)
=-0.4e-0.5t
+0.8944cos(t
-63.4?)
=ytr(t)+yss(t)
where
thefirsttermofthesolution,
ytr(t)= -0.4e-0.5t,
isthetransient
response
part
thatwilldecay
quickly.
After a few seconds, the output response
sinusoidal function
will reach the following
with the same oscillation frequency,
steady state,
? = 1 rad/s,
which is a sus-tained
but the amplitude has
changed
from1to0.8944
and
thephase
dropped
from0?to -63.4?.
Thatmeans
thesteady-state
re-sponse
part of the response is
yss(t)=0.8944cos(t
-63.4?)
(2.88)
Fig. 2.9: Efficient way of computing the sinusoidal steady-state response yss(t).
Since only the steady-state response term, yss(t), is of interest in almost all the applications of
the sinusoidal analysis and design and there exists a much easier way to compute yss(t), usually it
is not required to gothrough the inverse Laplace transform procedure to compute the complete response.
A much more efficient
way of computing
Theorem 2.32 (Steady-State
ej(?t+f)
yss(t) is given in the following
theorem.
Complex Response)
Assumethe system G(s) shown in Figure 2.9is stable, which meansit has no poles on the imaginary
axis or in the right half of the complex plane. If the input is u(t)
response ofthe system will be
yss(t) = A(?)Bej(?t+f+?(?))
= Bej(?t+f), then the steady-state
(2.89
2.5
Frequency-Domain
Properties
of Typical
First-Order
Systems
3
whereA(?) and ?(?) arethe magnitudeandthe phaseofG( j?), respectively.
Proof:
(s-j?),theoutput
Y(s)= L[y(t)]
consists
ofthetransient
re-sponse
Since
U(s)
=L[u(t)]
=Bejf?
part Ytr(s) and the steady-state response part, Yss(s), we have
Y(s) = G(s)U(s)
Bejf
= G(s)
=Ytr(s)+Yss(s)
s-j?
Based on the assumption that G(s) has no poles on the imaginary
(2.90)
axis or in the right halfofthe
complex
plane,the only polethat wouldcontributeto the steady-state
responseis s =j?, and,hence,
Then usethe Laplace transform
Yss(s) asfollows:
A(?)e j?(?)Bejf
Bejf
Yss(s) = G(j?)
s-j?
pair ej?t
=
A(?)Bej(f+?(?))
=
s-j?
(2.91)
s-j?
?
?1(s-j?)toobtain
the
inverse
Laplace
transform
of
yss(t) =L-1 [Yss(s)] = A(?)Bej(f+?(?))e j?t = A(?)Bej(?t+f+?(?))
Corollary 2.33 (Steady-State
Sinusoidal
(2.92)
Response)
Assumethe system G(s) shown in Figure 2.9is stable, which meansit has no poles on the imaginary
axis orin the right halfofthe complexplane(RHP) . Ifthe input is Bcos(?t +f) or Bsin(?t +f), then
the steady-state
yss(t)
response
ofthe system
= A(?)Bcos(?t
+f
will be
+?(?))
or
yss(t)
= A(?)Bsin(?t
+f
+?(?))
(2.93)
whereA(?) and ?(?) arethe magnitudeandthe phaseofG( j?), respectively.
Proof:
UseEulers formula ej(?t+f)
Now the efficient
approach
considered in Example
= cos(?t
+f)+ j sin(?t
will be employed
+f)
to solve the steady-state
sinusoidal
response
problem
2.31.
Example 2.34 (Use Corollary 2.33 to Compute the Steady-State Sinusoidal
Response)
Giventhe transfer function,
G(s) =
xss
ts+1
2
=
(2.94)
2s+1
Find the steady-state sinusoidal response of the system driven bythe control-input signal u(t)
=cost.
Solution:
Sincethe frequency of the input signalis ? = 1 rad/s, the magnitudeand phaseof G(j?)
= G(j1)
are computed as follows:
G(j?)
= G(j1)
=
2
1+ j2
2
= Aej?
= ve-j
tan-1
2=0.8944e-j63.4?
5
(2.95)
Hence,
yss(t)
=Acos(?t
+?)=0.8944cos(t
-63.4?)
which is exactly the same as Equation 2.88, the result of Example 2.31.
(2.96)
32
2
2.5.4
Linear Systems
Frequency
Analysis I
Responses and the
Bode Plot
Forthe samesystem G(s)in Figure 2.9,if the frequency ? ofthe control-input signal u(t) = cos?t is a
variable,then the magnitudeA(?) andthe phase ?(?) of G(j?) will also befunctions of?. Therefore,
the amplitudeandphaseof yss(t) will changewith ? accordingly.Theresponseof yss(t,?), dueto the
change of the frequency
?, is called the frequency response of the system. Thefrequency
response of a
systemincludes the magnitudefrequency response A(?) and the phasefrequency response ?(?).
Thefrequency response of the system G(s) = ts+1
xss = 2s+1
2 =s+0.5is
1
computedasfollows:
G(j?)
where
=
1
=|G(j?)|?G(
j?) =A(?)e
j?(?)
j?+0.5
(2.97)
20log102
=6.02
dB when??0.5
?
???
|G(j?)|dB=20log10v
1
?=?G(
j?)=-tan-1
?? ???
=
20log10
v2
= 3.01 dB
when
? = 0.5
?2+0.52
0.5
(2.98)
-20log10?
dB when?? 0.5
?
Fig. 2.10: The Bode plot of G(s) = 2/(2s+1).
Thefollowing
%
CSD
num=1;
Fig2.10
den=[1
magb=20*log10(mag);
MATLAB program is employed to obtain the Bode plot shown in Figure 2.10:
Bode
0.5];
plot
of
the
w=logspace(-2,1);
figure(1)
system
G(s)=1/(s+0.5)
[mag,phase]=bode(num,den,w);
2.5
Frequency-Domain
semilogx(w,magb),title('Magnitude
figure(2),
response
semilogx(w,phase),title('Phase
Properties
in
of Typical
dB'),
First-Order
Systems
33
grid,
response
in
deg'),
grid
The frequency response plots shown in Figure 2.10 are called the Bode plot, which consist of the
magnitude
response
plotand
thephase
response
plot.Themagnitude
response
plot
is a|G(j?)|dB
versus
?plot,where
|G(j?)|dB
isin dBscale
asdefined
in Equation
2.98
and
thefrequency
? (rad/s) is in log scale. The actual magnitude response is plotted in blue, which can be approxi-mated
by its two asymptote lines (in red) when ? is either very small or very large compared to the
corner frequency ?c. The corner frequency ?c, also called the 3 dB frequency, is the ?c that sat-isfies
|G(j?c)|=20log10
|G(j0)|-3dB.The
asymptote
forthefrequency
? <?cisahorizontal
linewith
|G(j?)|dB=20log10
|G(j0)|.The
other
asymptote
for?>?c
isthestraight
linewith
slope
-20dB/decade
thatintersects
thehorizontal
asymptote
atthecorner
frequency
?c.Note
thattheasymp-tote
approximation deviates the actual response by 3 dB at the corner frequency.
Example 2.35 (Use the Bode Plot to Compute the Steady-State Sinusoidal
From the Bode plot in Figure 2.10, at the corner frequency
?c = 0.5 rad/s
Responses)
weobserve that
20log10
|G(j0.5)|=|G(j0.5)|dB
=6dB-3dB=3dB
so that
|G(j0.5)|=103/20
=1.414
and
? =?G(
j0.5)=-45?
Hencethe steady-state response dueto u(t)
= cos0.5t is
yss(t)
=|G(j0.5)|cos(0.5t
+?)=1.414cos(0.5t
-45?)
Fig. 2.11: Outputresponsey(t) of G(s) = 1/(s+0.5)
dueto u(t) = cos0.5t.
The sinusoidal response in Figure 2.11 is obtained using the following
MATLAB code
(2.99)
34
2
%
CSD
Linear Systems
Fig2.11
clear,
Analysis I
Sinusoidal
response
t=linspace(0,30,301);
u=cos(0.5*t)
u=cos(0.5*t);
y=-exp(-0.5*t)+1.414*cos(0.5*t-45*pi/180);
figure(11),
plot(t,u,'b--',t,y,'r-'),
grid
on,
xlabel('t,sec'),
grid
ylabel('y'),
minor,
grid
figure(12),
on,
grid
The graphs of the input u(t)
ylabel('u(t)
plot(u,y,'b-'),
and
y(t)'),
xlabel('u'),
minor,
= cos0.5t and its corresponding output response are shown in Figure
2.11.
Note
that
thetransient
response
part
ofy(t),ytr(t)=-e-0.5t,
has
decayed
to0.0027
byt =10s
and
thereafter
y(t)=yss(t)
=1.414cos(0.5t
-45?).
The
steady-state
response
yss(t)
has
exactly
thesame
frequency ? = 0.5rad/s or period T = 2p/0.5 = 12.57sasthe input u(t) =cos0.5t, but with different
amplitude
and
different
phase.
The
amplitude
is1.414
instead
of1,and
thephase
is -45?
instead
of0?.
It also can be seenthat the lag time of the output with respect to the input is 1.57s. Thelag time 1.57s is
equivalent
tothephase
shift
? =-360?(1.57/12.57)
=-45?,
which
isexactly
thesame
as
that
shown
in
Equation
2.99.
When
???c,
say?=10rad/s,
we
observe
|G(j10)|dB
=-20
dBsothat
|G(j10)|=10-20/20
=0.1and? =?G(
j10)=-87?
Therefore, the steady-state response dueto u(t)
= cos10t is
yss(t)=Acos(10t
+?)=0.1cos(10t
-87?)
Fig. 2.12: Output response y(t) of G(s) = 1/(s+0.5)
The graphs of the input u(t)
dueto u(t)
(2.100)
= cos10t.
=cos10t andits corresponding output response y(t) are shown in Figure
2.12.
The
transient
response
part
ofy(t),ytr(t)=-0.0052e-0.5t,
has
decayed
toless
than
0.002
by
t =2s
and
thereafter
theoutput
response
isalmost
atthesteady
state
asyss(t)
=Acos(10t+?)
=0.1cos(10t87?). Thesteady-stateresponse yss(t) has exactly the samefrequency ? = 10 rad/s or period T =
2p/10 = 0.628s, but with different amplitude and different phase. The amplitudeis 0.1 instead of 1,
and
thephase
is -87?
instead
of0?.
Italso
can
beseen
that
thelagtimeoftheoutput
with
respect
tothe
input
is0.152s.
The
lagtime0.152s
isequivalent
tothephase
shift? =-360?(0.152/0.628)
=-87?,
which is exactly the same as that shown in
Equation
2.100
2.6
Remark
2.36 (Low-Pass
From the
Exercise
Problems
35
Filter)
Bode plot in Figure 2.10, and the sinusoidal
time responses to the input signals
with low
and
high frequenciesshownin Figure 2.11and Figure 2.12, weobservethat the system G(s) = 1/(s+0.5),
or the typical first-order system G(s) = xss/(ts +1) in general, allows the low-frequency signals to
pass while rejecting the high-frequency signals. The ability of selectively rejecting signals in certain
frequency ranges is called frequency filtering.
Thosethat are particularly designed to reject the high-frequency
noises, like some RC circuits, are called low-pass filters. The low-pass filtering is widely
used in practice to filter out unwanted high frequency noises; however, not all low-pass filtering
properties are desirable. Since ?c = 1/t, as shown in the Bode plot of Figure 2.10, if the corner
frequency ?c decreases,the bandwidth of the system will shrink and the time constant willincrease
to slow down the step response, as demonstrated in Figure 2.5 and Figure 2.6.
2.6 Exercise Problems
P2.1: Convert X = a+ jb to polar form X = ?ej?.
P2.2: Convert X = ?ej? to rectangular form X = a+ jb.
P2.3: Convert X =
v3+ j to
polar form.
P2.4: Convert X = 2ej3p/4 to rectangular form.
P2.5: Compute X1 = 2ejp/6
+ 2ej3p/4.
P2.6:
Compute
X2=2e
jp/6-2e
j3p/4.
P2.7: Compute X3 = (
v3+j)(-1+j).
v3+j
P2.8:
Compute
X4=-1+j
.
P2.9: Showthat sin(a+)
= sinacos
+sin cosa using Eulersformula.
P2.10:
Show
thatcos(a+)=cosacos
-sinasinusing
Eulers
formula.
P2.11: Express cos2? and sin2? in terms of cos? and sin? using Eulers formula.
P2.12: Express cos3? and sin3? in terms of cos? and sin? using Eulers formula.
P2.13: Computecos18? basedonthe fact of ej90? = cos90? +j sin90? = j andthe formulas in P2.11
and
P2.12.
(Hint:Use
ej5?=ej3?e
j2?=(cos3?
+j sin3?)
(cos2?
+jsin2?)= =j)
P2.14: Consider a vector in atwo-dimensional
wiseby 90? wewillobtaina newvectorX2 =
P2.15:
As described in Section (2.2.3),
number,
and vector rotation
?v
?T
3 -1?T
Find
a22 rotation
matrix
Rso
thatX2=RX1.
?v
space X1 = 1
3
a vector in a two-dimensional
can be accomplished
X1in P2.14 can berepresented as X1 = 1+ j
by complex
. After arotation of the vector clock-.
space can be represented
number
multiplication.
v
on a complex plane.
v3=
2e
j60?
. Find the
clockwise
by90?
toobtain
thevector
X2= 3-j =2e-j30?
complex
by a com-plex
The vector
Wecan rotate the vector
number
X3 so that th
36
2
new vector
Linear Systems
Analysis I
X2 is the product
of X3 and X1(i.e.,
X2 = X3X1). Comment
complex number multiplication approach andthe rotation
P2.16a:
Solve the following
differential
equation
on the relationship
matrix approach in P2.14.
using Laplace transforms.
x?(t)+2x(t)
between the
Assume zero initial
condi-tions.
= 4us(t)
P2.16b: Plot the solution x(t) as a function of t. Specifythe time constant t andthe steady-statestep
response value xss on the graph.
P2.16c:
Repeat
Problem
P2.16a
with
initial
condition
x(0)=-1.
P2.16d:
Repeat
Problem
P2.16b
with
initial
condition
x(0)=-1.
P2.17: Solvethe differential equation
x?(t)+2x(t)
=4cos2t
us(t)
using the Laplace transform/complex number approach (see Example 2.31). Assume zero initial condi-tions.
Specify the steady-state response part and the transient response partin your solution x(t).
P2.18a: Find the transfer function
G(s) = X(s)/U(s)
of the system described bythe differential equa-tion,
x?(t)+2x(t)
= u(t)
where X(s) and U(s) are the Laplacetransforms of the output x(t) and the input u(t), respectively.
P2.18b:
Let
u(t)=4cos2t
us(t).
Find
thesteady-state
response
ofthesystem
xss(t)
using
thesteady-state
sinusoidal
response
approach
(as shown in
Corollary
2.33 and Example
2.34).
P2.18c: Nowyou havethe sinusoidalsteady-stateresponseofthis form xss(t) =Acos(?t+?).
Thetran-sient
response of the system should be xtr(t) = Be-t/t, wheret is the time constant of the system and B
is a constant to be determined. Assumethe initial condition is zero, then we have x(0) = 0 =B+Acos?
by whichthe constant B can be computed.
P2.18d: Comparethe solution x(t)
from Problem P2.17.
= xtr(t)+xss(t)
from P2.18b and P2.18c with the solution x(t) ob-tained
P2.18e: Plot the transient response xtr(t) in a red, dashline andthe steady-state response xss(t) in a blue
solid line on the same graph.
P2.18f: Plot the complete solution x(t)
= xtr(t)+xss(t)
on a separate graph.
Fig. 2.13: Cascadeconnection of two subsystems G1(s) and G2(s)
2.6
Exercise
Problems
37
P2.19: The differential equations of the subsystems G1(s) and G2(s) shown in Figure 2.13 are given
as 2
?w(t) +3w(t) = 4u(t) and 5
?y(t) +y(t) = w(t), respectively. Find the differential equation of the
overall system that relates the output y(t) andthe input u(t).
P2.20a: Consider a first-order linear time-invariant
system with the differential equation
x?(t)+2x(t)
where x(t)
and u(t) are the output and the input
pole of the system, and explain the physical
= u(t)
of the system.
Find the transfer
function
G(s) and the
meaning of the pole.
P2.20b: Consider the same system. Drawthe Bode plot of the system with asymptotes onthe magnitude
response graph and the corner frequency ?c on both of the magnituderesponse andthe phase response
graphs.
P2.20c: Findthe magnitudeand phaseof the systemat the following frequencies: ? = 0.1?c, ? = ?c,
and ? = 10?c, respectively.
P2.20d: Find and plot the steady-state response xss(t) of the system dueto each of the following inputs:
(i) u(t) =sin(0.1?ct), (ii) u(t) =sin(?ct), and(iii) u(t) =sin(10?ct), respectively.
P2.20e:
Comment
on the frequency-domain
behavior
obtained in P2.20a, P2.20b, P2.20c, and P2.20d
of the simple first-order
system based on the results
3
Linear Systems AnalysisII
M
ANY practical systems are stand-alone second-order systems, whose behavior is governed
by second-order differential equations. Furthermore, complicated higher-order systems are
fundamentally
composed of several second-order and first-order
systems as building
blocks. For example, in the study of the aircraft flight dynamics and control, the aircraft usually is
considered a rigid body with six degrees of freedom including three rotational motions andthree transla-tional
motions. The mathematical modeling of the motion of each degree of freedom requires a second-order
differential
equation;
hence the flight
dynamics
model
would include
differential equations. Therefore, the study of the second-order
and design of dynamic systems and control.
six coupled
second-order
system is essential in the analysis
Many of the fundamental concepts welearned from Chapter 2 like the mathematical equivalency,
typical equation characterization, time-domain analysis and frequency-domain properties, will be ex-tended
to the study of the typical second-order dynamic systems in this chapter. Mostof the extensions
are quite straightforward,
although the computations
involved
may be slightly
more complicated.
The
basic difference betweenthese two typical systems is that the behavior of the first-order system is mainly
determined
by the time constant
while that of the second-order
system is characterized
by both the damp-ing
ratio and the natural frequency.
From the study in Chapter 2, we have a pretty clear idea on how the time constant would affect
the time-domain and frequency-domain responses of the typical first-order systems. Similarly, we will
learn how to select the damping ratio and the natural frequency to achieve desired performances for the
typical
second-order
be employed
systems.
to simplify
For the cases that involve
the computations
just
sinusoidal
manipulations,
as we did for the sinusoidal
Eulers formula
response
computations
will
in
Chapter 2.
3.1 Typical Second-Order
As discussed in
of the
Dynamic Systems
Chapter 2 regarding
mathematical
equivalency
the typical
first-order
among systems
systems,
and the typical
we will continue to take advantage
equation
characterization
to enhance
our learning experience. Thethree typical second-order physical systems to be considered in this section
are shown in Figure 3.1: the massdamperspring system in (a) is a mechanical system, the RLC circuit
system in (b) is an electrical
feedback
control
system,
system
and the
DC
motor position
control
system in (c) is an electrome-chanical
40
3
Linear Systems
Analysis II
Fig. 3.1: Mathematically equivalent systems: (a) massdamperspring
and (c)
3.1.1
DC motor position
Mathematical
The governing
Equivalency
dynamic
(MBK) system, (b) RLC circuit,
control system.
Among
differential
Typical
equation
Second-Order
for the
Dynamic
massdamperspring
Systems
(MBK)
system in
Figure
3.1(a) can be derived from Newtons law of motion. The effective force fef f(t) applying to the mass will
causethe mass Mto move with accelerationy(t)
= fe f f(t)/M,
where y(t) is the displacement variable
ofthesystem.
The
friction
force
ffri(t)=-B
?y(t)
isproportional
tothemagnitude
ofthevelocity,
but
itsdirection
isalways
against
themotion
ofthemass.
The
spring
force
fspr(t)=-Ky(t)
isproportional
to the
magnitude of the displacement,
but its direction
is in the opposite
direction
of the displacement.
Hence,
wehave
feff(t) =fa(t)-B
?y(t)-Ky(t),
which
leads
tothefollowing
equation:
y(t)+B
M
?y(t)+Ky(t)
= fa(t)
(3.1)
Notethat y(t), the displacement of the mass M,is the output or the variable of interest in the system,
and fa(t), the applied force, is the control input by which the motion of the system can be altered or
controlled.
The RLC circuit in Figure 3.1(b) consists of a voltage source ea(t), a resistor
and a capacitor
C. Let i(t)
be the current flowing
of the voltage source, ea(t), through
clockwise
R, L, C, and back to the negative terminal
of ea(t).
the characteristicsof the resistor, eR(t) = RiR(t),the inductor, eL(t) = LdiL(t)/dt,
iC(t) = ?
CeC(t), and based on Kirchhoffs
around the loop should be zero:
di(t)
=?q(t), or q(t)
= ? i(t)dt,
L
According
to
andthe capacitor,
voltage law (KVL), the algebraic sum of the voltage drops
-ea(t)+Ri(t)+L
dt
Sincei(t)
R, an inductor
around the loop from the positive terminal
+(1/C)
?
i(t)dt
=0
(3.2)
this equation can be rewritten asfollows,
Lq(t)+R
?q(t)+(1/C)q(t)
= ea(t)
(3.3
3.1
The DC
motor position
control
Typical
Second-Order
Dynamic
Systems
41
system in Figure 3.1(c) consists of a DC motor with transfer
func-tion
G(s) = b/(s2 +as) that relates the control input U(s) andthe angular displacement output T(s), a
proportional controller K with E(s) and U(s) asits input and output, respectively. The error signal E(s)
is E(s)= TR(s)-T(s),
where
TR(s)
isthereference
orcommand
input.
The
objective
ofthecontrol
systemis to designthe controller Kso that the output ?(t) can follow ?R(t) as closely as possible. We
will seethat the choice of K will affect the performance of the control system. To find the differential
equationthat relatesthe referenceinput ?R(t) andthe output ?(t), it wouldbeeasierto firstly determine
the transferfunction between TR(s)and T(s) andthen covertit into the correspondingdifferential
equation.
Since
T(s)= KG(s)E(s)
andE(s)= TR(s)-T(s),
we
have
T(s)=KG(s)
[TR(s)-T(s)]
=KG(s)TR(s)-KG(s)T(s)
(3.4)
which yields the following transfer function:
T(s)
KG(s)
TR(s)
=
bK
=
1+KG(s)
(3.5)
s2 +as+bK
Hence, we havethe differential equation for the DC motor position control system:
s2T(s)+asT(s)+bKT(s)
=bKTR(s)
? ?(t)+a
3.1.2 Characterization
of Typical Second-Order
Now we have derived the
mathematical
??(t)+bK?(t)
= bK?R(t)
(3.6
Dynamic Systems
models for the three systems in Figure 3.1,
which are the three
differential equations given in Equation 3.1, Equation 3.3, and Equation 3.6, respectively. Thesethree
systems are very different physically: one mechanical, one electrical, andthe last an electromechanical
position
control
system.
However, these three systems
are mathematically
equivalent,
and indeed they
all can berepresented in the same form of the typical second-order system differential equation:
x(t)+2??n
?x(t)+?2 nx(t)
= xss?2
nr(t)
(3.7)
where x(t) represents the output or the variable of interest in the system and r(t) is the input, which
can be the control input for an open-loop system or a reference input for a feedback control system.
Thetypical second-order system is characterized by three parameters: the damping ratio ?, the
natural frequency ?n, and the steady-state step response xss. Thephysical meaningof these three
parameters will be clearly described later in this section.
After rewriting the massdamperspring (MBK) system governing equation Equation 3.1 in terms
of the typical second-order differential equation in Equation 3.7, we obtain the new MBK equation
y(t)+(B/M)
?y(t)+(K/M)y(t)
wherethe output variableis x(t) = y(t), theinput is r(t)
the damping ratio is ? = 0.5B/
The RLC circuit
system
v MK, andthe
fa(t)
(3.8)
=fa(t), the naturalfrequencyis ?n =
vK/M,
steady-state step response is xss = 1/K.
equation in Equation
q(t)+(R/L)
= (1/K)(K/M)
3.3 can be rewritten
?q(t)+(1/LC)q(t)
in the form
=C(1/LC)ea(t)
of Equation
3.7,
(3.9)
v
where
the
output
variable
isx(vt)
=q(t),
the
input
isr(t)=ea(t),
the
natural
frequency
is?n=1/LC,
the damping ratio is ? = 0.5R C/L, andthe steady-state step response is xss =C.
42
3
Linear Systems
Analysis II
The DC motor position feedback
control
system
governing
equation,
Equation
3.6, is already in the
form of Equation 3.7.
?(t)+a
??(t)+bK?(t)
= bK?R(t)
(3.10)
wherethe outputvariableis x(t) = ?(t),theinputis r(t) =?R(t),the naturalfrequencyis ?n = vbK,
the dampingratio is ? = 0.5a/ vbK, andthe steady-statestepresponseis xss = 1.
These discussions have clearly revealed the mathematical equivalency among the above three sys-tems.
For this reason, the study of the large group of all typical second-order systems is boiled
downto simply investigating the three parameters ?, ?n,and xssin Equation (3.7).
3.2 Transfer Function,
As discussed in
Section
Characteristic
2.5.1, the transfer
Equation, and System Poles
function
of a given system can be derived from the corre-sponding
differential equation using the Laplace transform.
associated
with the differential
equation,
Equation
The frequency-domain
3.7, of the typical
second-order
algebraic equation
system can be found
as
s2X(s)+2??nsX(s)+?2
nX(s) = xss?2
nR(s)
(3.11)
which yields the following transfer function between the input R(s) and the output X(s):
xss?2
X(s)
=
R(s)
n
(3.12
s2 +2??ns+?2=G(s)
n
The behavior ofthe first-order system is mainly determined bythe root of its characteristic equation,
as described in Section 2.5.2. It is also true for the typical second-order system, with the exception
that the second-order system hastwo roots instead of one. The characteristic polynomial of the system
in Equation3.12 is s2 +2??ns +?2
n, which is the denominator polynomial of the transfer function.
Therefore, the characteristic
equation
is
s2 +2??ns+?2 n = 0
(3.13)
and its roots are the systems poles. Since the natural frequency ?n and the damping ratio ? are non-negative
real numbers, the roots of the second-order algebraic equation will either betwo real numbers
or one pair of conjugate complex numbers, and no roots would be on the right half of the complex plane.
We will learn
in this chapter.
more about the physical
By comparing
Equation
meaning of the damping ratio and the natural frequency
3.8, the governing
equation
of the
massdamperspring
later
(MBK)
system, to the typical second-order equation, Equation 3.7, we can easily see that the damping ratio
? is proportional to the friction coefficient B. That means morefriction will cause a higher damping
ratio. Similarly for the RLC electric circuit, the damping ratio will go up if the resistance R of the
resistor increases. Accordingto the values of the dampingratio ?,the system behaviorsare classified
into four cases:
Case A: ? > 1. The system is overdamped, and Equation 3.13 hastwo distinct negative real roots.
s1
s2
Example
A: s2 + 2??ns
?
=-??n?n ?2-1= -a1
-a2
+?2n=s2 + 5s + 4=(s
+ 1)(s + 4)=0. This characteristic
(3.14)
equation has
twodistinct
roots
at -1and-4,and
thedamping
ratioandnatural
frequency
are? =1.25
and
?n = 2rad/s, respectively.
3.3
Time-Domain
Response of Typical
Second-Order
Dynamic
Systems
4
Case B: ? = 1. Thesystem is critically damped,and Equation3.13 hasdoublereal roots at
s1
=
s2
Example B: s2 +2??ns+?2 n = s2 +4s+4
-?n
(3.15)
-?n
= (s+2)2
= 0. This characteristic
equation
has two
identical
roots
at-2,and
thedamping
ratioand
thenatural
frequency
are? =1and?n=2rad/s,
respectively.
Case C: 0 < ? < 1. The system is underdamped, and Equation 3.13 has a pair of complex conjugate
?
roots at
s1
s2
Example C: s2 +2??ns+?2
:= -a j?
=-??n j?n 1-?2
v
v3)
(3.16)
= 0. This characteristic
n=s2+2s+4
=(s+1-jv3)(s+1+
j
3, and the damping ratio and the natural
equation
has
apair
ofcomplex
conjugate
roots
at-1 j
frequency are ? = 0.5 and ?n = 2rad/s, respectively.
Case D: ? = 0. Thesystem is undamped,and Equation3.13hasa pair of complex conjugateroots on
the imaginary
axis at
s1
s2
Example D: s2 +2??ns+?2
= j?n
(3.17)
n =s2+4=(s-j2)(s+j2) =0.This
characteristic
equation
has
a
pair
ofcomplex
conjugate
roots
atj2,and
thedamping
ratio
and
thenatural
frequency
are
? =0
and ?n = 2rad/s, respectively.
3.3 Time-Domain Responseof Typical Second-Order Dynamic Systems
As discussedin Section 3.1, alarge group of second-order systems share mathematical equivalency and
can be characterized by the typical second-order differential equation described by Equation 3.7, which
is repeated in the following for ease of reference:
x(t)+2??n
?x(t)+?2 nx(t)
= xss?2
nr(t)
(3.18)
In the equation, x(t) is the system output orthe variable of interest, which may be the displacement y(t)
of the mechanical system in Figure 3.1(a), the electric charge q(t) on the capacitor of the electric circuit
in Figure3.1(b),orthe angulardisplacement?(t) ofthe electromechanical
systemin Figure3.1(c).The
variable r(t) is the control input or the command by which the behavior of the system can be altered
or controlled. The damping ratio ?, the natural frequency ?n, and the steady-state step response xss are
determined
by system component
values.
Given
initial
conditions
x(0)and
?x(0)
and
input
r(t)fort =0,we
can
solve
thedifferential
equation
tofindx(t)fort =0.After
taking
theLaplace
transform
ofEquation
3.18
and
some
straightforward
algebraic manipulations, we havethe solution
X(s) =[G0(s)x(0)+G1(s)
X(s)
=L[x(t)]
as follows:
?x(0)]+G(s)R(s)
:=
XI(s)+XR(s)
where
G0(s) =
s+2??n
D(s)
,
G1(s) =
1
D(s)
,
G(s) =
xss?2
n,
D(s)
D(s) =s2 +2??ns+?2n
(3.19)
44
3
Linear Systems
Analysis II
IntheLaplace
transform
manipulations,
theLaplace
transform
pairs
?x(t)?sX(s)-x(0)
and
x(t) ?
s2X(s)-sx(0)?x(0)
fromTheorem
2.25
were
employed.
Note
thatX(s)
consists
oftwoparts:
XI(s),
the response due to the initial conditions x(0) and?x(0), and XR(s), the response due to the control or
command input R(s). Therefore, the complete response x(t) of the typical second-order system, de-scribed
by Equation 3.18, is the inverse Laplace transform of X(s):
x(t)
=L-1 [X(s)]
=L-1 [XI(s)]+L-1
3.3.1 The Response of the Typical Second-Order
[XR(s)]
=xI(t)+xR(t)
Systems Dueto Initial
(3.20)
Conditions
The initial state response xI(t) is the response of the second-order dynamic system due to the initial
conditions x(0) and?x(0). It can be computed using the inverse Laplacetransform in the following:
xI(t)
=L-1 [XI(s)]
?
x(0)s+2??nx(0)+
?x(0)
s2 +2??ns+?2 n
=L-1
?
(3.21)
Notethat the two poles of the function X(s) are the roots of the characteristic equation s2 +2??ns+
?2
n = 0. As discussedin Section 3.2, the behaviors of the typical second-order system can be classified
into the following four cases according to the values of the damping ratio ?.
?
Case A: ? > 1. Thesystem is overdamped,andthe characteristic equation hastwo distinct negative
XI(s)
can
bedecomposed
intotwo
terms
real
roots:
-??n?n ?2-1or-a1and-a2.Hence,
asfollows using the partial fraction expansion:
XI(s)
=
x(0)s+2??nx(0)+
?x(0)
A1
=
(s+a1)(s+a2)
A2
+
(s+a1)
(3.22)
(s+a2)
Then the initial state response xI(t) for Case Ais
xI(t)
=L-1 [XI(s)]
Case B: ? = 1. The system is critically
= A1e-a1t +A2e-a2t
(3.23)
damped, andthe characteristic equation hastwo identical neg-ative
real
roots
at-?n.Hence,
XI(s)
can
bedecomposed
intotwo
terms
asfollows
using
thepartial
fraction expansion:
XI(s)
Then the initial
=
x(0)s+2?nx(0)+
state response xI(t)
xI(t)
?x(0)
for
A3
=
(s+?n)2
(s+?n)
A4
+
(3.24)
(s+?n)2
Case Bis
=L-1 [XI(s)]
= A3e-?nt +A4te-?nt
(3.25)
?
Case C: 0 < ? < 1. The system is underdamped, and the characteristic equation has two complex
conjugate
roots
at-??nj?n 1-?2or-a j?. Hence,
XI(s)
canbedecomposed
intotwo
terms as follows
using the partial fraction expansion:
XI(s) =
x(0)s+2??nx(0)+
?x(0)
Aej?
=
(s+a-j?)(s+a+j?) (s+a-j?)
+
Ae-j?
(s+a+
j?)
(3.26
3.3
Time-Domain
Response of Typical
Second-Order
Dynamic
Systems
45
Then the initial state response xI(t) for Case Cis
xI(t)
=L-1 [XI(s)]
= Aej?e-(a-j?)t
?
= Ae-at ej(?t+?)
+e-j(?t+?)
+Ae-j?e-(a+j?)t
?
(3.27)
= 2Ae-at cos(?t
+?)
Case D: ? = 0. Thesystem is undamped, andthe characteristic equation hastwo complex conjugate
roots
atj?n.Hence,
XI(s)
can
bedecomposed
intotwoterms
asfollows
using
thepartial
fraction
expansion:
XI(s)
=
A5e-jf
A5ejf
x(0)s+ ?x(0)
=
(s-j?n)(s+j?n) (s-j?n)
+
(3.28)
(s+ j?n)
Then the initial state response xI(t) for Case Dis
xI(t) =L-1 [XI(s)] = A5e
jfe j?nt +A5e-jfe-j?nt
?
= A5 ej(?nt+f)
+e-j(?nt+f)
?
(3.29)
= 2A5cos(?nt +f)
In the following example, the massdamperspring (MBK) system as shown in Figure 3.1(a) will be
employed to demonstrate the behavior of the system in four cases.
Example 3.1 (The
Responses of an MBK System Dueto Initial
Conditions)
Consider a massdamperspring (MBK) system as shown in Figure 3.1(a), wherethe massandthe
spring constant are chosen to be M = 1 kg and K = 4 N/m, respectively. The friction coefficient B of
the damper is kept as afree design parameter at this moment. Assumethere is neither applied force nor
initial velocity (both fa(t) and?y(0) are zero), but the initial massdisplacement is a little bit away from
the equilibrium, say, y(0) = y0 = 0.1m. The objective is to find the trajectory of the massdisplacement,
y(t), after it is released at t = 0.
Withthese assumptions, the governing differential equation now is
y(t)+B
Take the Laplace transform
?y(t)+4y(t)
= 0,
y(0)
= y0, y?(0)
=0
to obtain
y0(s+B)
s2Y(s)-sy0
+B[sY(s)-y0]+4Y(s)
=0 ? Y(s)=s2
+Bs+4
and then the solution y(t) can be found by computing the inverse Laplace transform of Y(s). Before
doing that, we will see how to classify the behavior of the system into four classes according to the value
of the friction
with that
coefficient
of the typical
B of the damper.
second-order
Comparing the characteristic
polynomial
of the
MBK system
system,
s2 +Bs+4
? s2 +2??ns+?2n
we have ?n = 2 rad/s and ? = B/4. Therefore,if
M = 1 kg and K = 4 N/mare chosenfor the MBK
system, the behavior of the MBK system can be classified into the following four cases according to the
values of the friction coefficient B:
Case
A: B>4Ns/m,?>1, ?n=2rad/s,s1,
s2=-a1,-a2,overdamped
Case
B: B=4Ns/m,? =1, ?n=2rad/s,s1,s2=-2,-2, critically
damped
Case
C: 0<B<4Ns/m,
0<?<1, ?n=2rad/s,s1,
s2=-a j?, underdamped
Case
D: B=0Ns/m,?=0, ?n=2rad/s,s1,
s2=j2, undampe
46
3
Linear Systems
Analysis II
Based on this investigation
and analysis,
we should
be able to envision
the trajectory
of the
motion
even before solving for y(t).
Example 3.2 (Case A Overdamped
Response of the
MBK System Dueto Initial
Conditions)
For the MBK system considered in Example 3.1, with the assumptions M = 1 kg, K = 4 N/m,
fa(t) = 0,?y(0) = 0, and y(0) = y0 = 0.1 m, we have the solution in s-domain as a function of the
friction coefficient B as follows:
L[y(t)]
To investigate
Case A overdamped
initial
=Y(s)
=
0.1(s+B)
s2
+Bs+4
state response, let
B = 5 Ns/m so that the damping
ratio
and
natural
frequency
are? =1.25
and?n=2rad/s,
respectively,
and
the
tworealpoles
are
s1=-1
and
s2=-4.Now,
tofindtheinverse
Laplace
transform
ofY(s),
we
decompose
Y(s)
intotwoparts
using the partial fraction
expansion:
0.1(s+5)
s2 +5s+4
A1
=
s+1
A2
+
? 0.1s+0.5
=A1(s+4)+A2(s+1)
s+4
This
polynomial
equation
isvalid
forany
realorcomplex
value
ofs;thus,
thesubstitutions
ofsby-1
and-4are
thebest
choices,
leading
toeasy
solutions
forA1andA2.
s =-1 ? -0.1+0.5
=A1(-1+4)? A1=0.4/3
=0.1333
s =-4 ? -0.4+0.5
=A2(-4+1)? A2=0.1/-3=-0.0333
Taking the inverse Laplace transform,
we have the following initial state response of the
MBK sys-y(t)
tem:
=0.1333e-t
-0.0333e-4t,
t =0
Notethat the initial stateresponsey(t) hastwo exponentialterms, y1(t) = 0.1333e-t and y2(t) =
-0.0333e-4t,
which
are
associated
with
thesystem
poles
s =-1and
s =-4,respectively.
Asshown
from Figure 3.2, y2(t) decays much morequickly than y1(t), and therefore the pole closerto the
imaginary
axis plays a more dominant role than the other pole.
Fig. 3.2: Case Ainitial
Example 3.3 (Case B Critically
state response
Damped Initial
of the overdamped
State Response of the
MBK system.
MBK System)
The MBK system considered in the following is identical to the onein Example 3.2, except the value
of the friction
coefficient
Bin the following
s-domain
trajectory
solution
3.3
Time-Domain
Response of Typical
L[y(t)]
To investigate
Case B critically
=Y(s)
damped initial
=
Second-Order
Dynamic
Systems
47
0.1(s+B)
s2 +Bs+4
state response, let B = 4 Ns/m so that the damping
ratio and naturalfrequency are ? = 1.0and ?n =2rad/s,respectively,andthe two real poles areidentical
s1=s2= -2.Now,
tofindtheinverse
Laplace
transform
ofY(s),
we
decompose
Y(s)
intotwoparts
using the partial fraction expansion:
0.1(s+4)
B1
=
s2 +4s+4
As mentioned previously,
we can substitute
s+2
B2
+
(s+2)2? 0.1s+0.4
=B1(s+2)+B2
this polynomial
equation is valid for any real or complex
s by any two values to obtain two linearly
independent
algebraic
value of s; thus,
equations from
which
the two unknowns B1and B2 can be solved. As alternative, the coefficient comparison approach can be
employed
to set up two linearly
Taking the inverse
MBK system:
independent
equations to solve for
? B1=0.1
?B2=0.4-0.2
=0.2
s1term :
0.1 = B1
s0term :
0.4 = 2B1 +B2
Laplace transform
B1 and B2 as follows:
of Y(s), we have the following
initial state response of the
y(t)=0.1e-2t
+0.2te-2t
, t =0
Fig. 3.3: Case B critically
The Case B critically
MATLAB code:
%
CSD
clear,
Fig3.3
Case
t=linspace(0,6,61);
y=y1+y2;
grid
damped initial
figure(31),
minor,
xlabel('Time,
B
damped initial
state response
of the
MBK system.
state response in Figure 3.3 is obtained
critically
damped
y1=0.1*exp(-2*t);
initial
state
ylabel('Init
response
y2=0.2*t.*exp(-2*t);
plot(t,y,'k-',t,y1,'b--',t,y2,'r--'),
sec'),
using the following
grid,
Response'),
legend('y','y1','y2')
Notethat the initial stateresponsey(t) hastwo terms, y1(t) = 0.1e-2t and y2(t) = 0.2te-2t, which
are
theresults
ofthedouble
system
poles
ats =-2.The
system
considered
inExample
3.2
and
theon
48
3
Linear Systems
Analysis II
in this example are basicallythe sameexceptthe value ofthe dampingratiothe former with ? = 1.25,
overdamped,andthe latter with ? = 1, critically damped. By comparingthe initial stateresponsesin
Figure 3.2 and Figure 3.3,
equilibrium
we can observe that the critically
state response. The critically
overshoot and oscillations.
Example
The
damped step response
faster and reaches the steady state at the equilibrium
3.4 (Case
MBK system considered
Response of the
in the following
MBK System
is almost identical
Due to Initial
to the
s-domain trajectory
in
coefficient B, which is
solution:
L[y(t)]
Case C underdamped
Conditions)
previous two examples
Example 3.2 and Example 3.3. The only difference is in the value of the friction
To investigate
the
initial
damped step response is the least damped one that still can avoid an
C Underdamped
smaller in the following
moves towards
earlier than the overdamped
=Y(s)
initial
=
0.1(s+B)
s2
+Bs+4
state response, let
B = 2 Ns/m so that the damping ratio
and naturalfrequency are ? = 0.5 and ?n = 2rad/s, respectively,andthe two complex conjugate poles
are
s1,s2=-1 j
v3.
Now,to find the inverse Laplacetransform ofY(s),
parts using the partial fraction
0.1(s+2)
s2
c
=
+2s+4
we decompose Y(s) into two
expansion:
s+1-j v3 +s+1+
v
c*
j
v
j 3)
v3 ? 0.1(s+2)
=c(s+1+
j 3)+c*(s+1-
Notethat the two complex numbers c and c* are conjugate, which meansif either oneis known
we will have the other one immediately.
-1+
v3, the
polynomial
equation
Thus, only one needs to be computed.
v
v3+1+
0.1(-1+
j 3+2)=c(-1+j
. Taking
hence,c* = 0.1v
3ej30?
the
By substituting s by
becomes
the inverse
j
v
0.1(1+ j
3) ? c=
Laplace transform
ofY(s),
j2
v3
v3)
0.1
=
we have the initial
v3 e-j30?
state response
of
MBK system:
y(t)
=
v
0.1
v3 e-j30? e(-1+j 3)t
which can be rewritten
in terms
y(t)
=
+
0.1
v3 ej30? e(-1-j
v
3)t =
0.1
v3 e-t
?
ej(
v
v
?
3t-30?) +e-j( 3t-30?)
of cosine or sine functions:
0.2
v3e-t
cos(
v
0.2
3t-30?)
= v3 e-t sin(
v3t
+60?)
v3rad/s
v3 rad/s is the imagi-nary
It canbeseenthattheinitial stateresponsey(t) is asinusoidalfunction withfrequency ?=
and exponentially decaying amplitude. Notethat oscillation frequency ? =
part of the system pole, and the decaying rate in e-t is determined by the real part of the
pole. Furthermore, the phase of the sine function is 60?, which revealsthat the damping ratio is
? =cos60? =0.5. Asshownin Figure3.4,theunderdamped
initial stateresponse
has much
fasterrise
time thanthe overdampedor critically dampedcases. Butthe 18% overshootdueto the choice of ? = 0.5
is not a good design. A new selection of Bto increasethe dampingratio to about ? = 0.9 will greatly
improve
the performance.
The Case C underdamped initial
code:
state response in Figure 3.4 is obtained using the following
MAT-LAB
3.3
%
CSD
clear;
Fig3.4
Case
C
Time-Domain
underdamped
t=linspace(0,6,61);
figure(32),
initial
state
Second-Order
grid,
sec'),
Redesign the
49
Systems
response
grid
ylabel('Init
minor,
Response')
Fig. 3.4: Case Cunderdamped initial state response of the
Exercise 3.5 (Redo
Dynamic
y=0.1155*exp(-t).*sin(sqrt(3)*t+pi/3);
plot(t,y,'k-'),
xlabel('Time,
Response of Typical
Example
MBK system
of this particular
3.4 by Selecting
by selecting
underdamped
B = 3.6
B = 3.6
design
MBK system.
Ns/m)
Ns/m and repeat
with the critically
Example
damped
3.4 to compare the per-formance
design in
Example
3.3 and
give your comments.
Example 3.6 (Case D Undamped Response of the
MBK System Dueto Initial
Conditions)
The MBK system considered in the following is almost identical to the previous three examples in
Example
3.2, Example
3.3, and Example
which is now zero, in the following
3.4. The difference is in the value of the friction
s-domain
L[y(t)]
coefficient
B,
solution:
=Y(s)
=
0.1(s+B)
s2 +Bs+4
With B = 0 Ns/m,the damping ratio and natural frequency are ? = 0 and ?n = 2 rad/s, respectively,
and
thetwocomplex
conjugate
poles
are
s1,s2=j2. Now,
Y(s)
has
become
Y(s) =
0.1s
s2 +22
Theinitial state response y(t) is the inverse Laplace transform ofY(s),
y(t)
= 0.1cos2t
whichis a sinusoidalfunction withfrequency ? = 2rad/s and undampedamplitude asshownin Figure
3.5. Notethat the imaginary part of the system pole, ? = 2 rad/s, is the oscillation frequency, and
the zero real part of the pole meansthat the decaying rate in e0tis zero
50
3
Linear Systems
Analysis II
Fig. 3.5: Case D undamped initial
3.3.2
The Response of the Typical
state response
Second-Order
Systems
of the
Due to
MBK system.
Unit Step Input
As shown in Equations 3.19 and 3.20, the output response x(t) of the typical second-order system con-sists
of two parts: xI(t), the response due to the initial conditions, and xR(t), the response dueto the
reference input or control input rR(t). The response due to the initial conditions has been discussed in
the previous subsection,
responses
Section 3.3.1.
due to initial
The initial
conditions
state responses
will decay exponentially
are basically
transient
and eventually
responsesall
die out
unless the
system has poles on the imaginary
axis or in the right half of the complex plane. Onthe other hand,
the responses due to the input r(t)
will include both transient and steady-state responses. The type of
input
signals
can be an impulse
or anirregular
function,
piecewise
a step function,
continuous
function.
a sinusoidal
function
However, the
of a wide range
of fre-quencies,
most common testing signals
in practice are the unit step function us(t) andthe sinusoidal function cos?t or sin?t
used
with a widerange
of frequencies.
The sinusoidal
response
of the second-order
system
will be discussed in the next subsection.
subsection, we will investigate the effect of the damping ratio ? and the natural frequency
transient
behavior
of the typical
second-order
Recall that in Section 2.4.2 welearned
In this
?n on the
system.
how to find the step response
of the typical first-order
Asshownin Figure 2.6,the procedureoffinding the stepresponseof the system t?x(t)+x(t)
system.
= xssu(t)
isfairly
straightforward
since
thesystem
has
only
one
pole,
s =-1/t,inwhich
t isthetime
constant
of
the system.
The step response
graph can be easily sketched
equation since the equation hasrevealed the information
without the need of solving
the differential
ofthe time constant t andthe steady-state step
response xss, which provide enough information to construct the step response.
Although the step response of the typical second-order system is not as simple asthat of the typical
first-order system,it still can beeffectively characterizedby onlythree parameters:the dampingratio ?,
the naturalfrequency ?n,andthe steady-statestepresponsexss.ThestepresponsexR(t) is the response
of the second-order dynamic system dueto the unit step input us(t), and can be computed asthe inverse
Laplace transform of XR(s):
xR(t) =L-1 [XR(s)] =L-1
?
xss?2
n
s2 +2??ns+?2
1
n
s
where XR(s) was given in Equation 3.19. In the following,
typical
second-order
response study
system for each case of Cases Ato
? ?
=L-1
xss?2
n
s(s2 +2??ns+?2
n)
?
(3.30)
we will investigate the step response of the
D as we did in Section 3.3.1 for the initial
state
3.3
Time-Domain
?
Response of Typical
Second-Order
Dynamic
Systems
51
Case A: ? > 1. Thesystem is overdamped,andthe characteristic equation hastwo distinct negative
XR(s)
can
bedecomposed
into
three
terms
real
roots:
-??n?n ?2-1or-a1and-a2.Hence,
as follows
using the partial fraction
XR(s) =
expansion:
xss?2
n
A1
xss
=
s(s+a1)(s+a2)
+
s
A2
+
(s+a1)
(s+a2)
(3.31)
Then the step response xR(t) for Case Ais
xR(t)
=L-1 [XR(s)] = xss +A1e-a1t +A2e-a2t
Example 3.7 (Case A Step Response Example
(3.32)
with ? = 1.25, ?n = 2 rad/s and xss = 1)
Considerthe following system with ? = 1.25, ?n = 2rad/s, xss = 1, andthe roots ofthe character-istic
equation
are
at-1and-4:
XR(s)
R(s)
= G(s) =
xss?2
n
4
s2 +2??ns+?2
=
n
4
=
s2 +5s+4
(s+1)(s+4)
(3.33)
Withthe unit stepinput R(s) = 1/s, we havethe step response XR(s) with its partial fraction expan-sion,
4
XR(s)
where A0, A1, and
=
A2 are the real
These constants can be evaluated
A0
=
s(s+1)(s+4)
partial fraction
using the residue
A0 =lim sXR(s) = (1)(4)
4
= 1,
s?-4
A1
+
s+1
expansion
approach
A2
+
s+4
residue
4
= (-4)(-3)
(3.34)
constants to be determined.
as follows:
A1 = lim (s+1)XR(s)
s?0
A2 = lim (s+4)XR(s)
s
s?-1
4
= (-1)(3)=
3,
-4
(3.35)
=1
3
Fig. 3.6: Case A: Overdamped step response.
Now, plug the values of A0, A1, and A2into
1/(s+a)
Equation 3.34, and use the Laplace transform
pair
? e-at, and wehavethe stepresponsein the following:
:= 1+x1(t)+x2(t)
xR(t)=L-1
[XR(s)]
=1-(4/3)e-t
+(1/3)e-4t
(3.36
52
3
Linear Systems
Analysis II
Note
thatthestep
response
xR(t)
has
twoexponential
terms,
x1(t)=-1.33e-t
andx2(t)=
0.33e-4t,
which
areassociated
with
thesystem
poles
s =-1and
s =-4,respectively.
Asshown
from Figure 3.6, x2(t) decays much more quickly than x1(t), and therefore the pole closer to
the imaginary axis plays more dominant role than the other pole.
Case B: ? = 1. Thesystemis critically damped,andthe characteristicequationhastwo identical nega-tive
real
roots
at-?n.Hence,
XR(s)
can
bedecomposed
intothree
terms
asfollows
using
thepartial
fraction
expansion:
XR(s) =
xss
xss?2
n
s(s+?n)2
A3
+
= s
A4
+
(s+?n)
(3.37)
(s+?n)2
Then the step response xR(t) for Case Bis
xR(t)
=L-1 [XR(s)] = xss +A3e-?nt +A4te-?nt
(3.38)
Example 3.8(Case B Step ResponseExample with ? = 1, ?n = 2rad/s and xss = 1)
Consider the following
system with ? = 1, ?n = 2rad/s, xss = 1, andthe two roots of the charac-teristic
equation
areboth-2:
XR(s)
= G(s)
=
R(s)
xss?2
n
4
s2 +2??ns+?2
4
=
s2
n
(3.39)
=
+4s+4
(s+2)2
Withthe unit stepinput R(s) = 1/s, wehavethe output XR(s) andits partialfraction expansion,
XR(s) =
4
A0
s(s+2)2
=
s
A3
+
A4
+
s+2
(3.40)
(s+2)2
where A0, A3, and A4 are the real partial fraction expansion residue constants to be determined.
These constants can be evaluated using the residue approach as follows:
A0 = lim sXR(s)
= 4
s?0
d
A3 = lim
s?-2
ds
22
= 1,
-2 =-2,
A4 = lim (s+2)2XR(s)
= 4
s?-2
? (s+2)2XR(s)
? = lim
s?-2
d?4
ds
s
? = lim
-4
s?-2 s2
=
(3.41)
-4
(-2)2=-1
The partial fraction expansion residue constants can also be found using the polynomial
substitution approach and the coefficient comparison approach as described in Remark 3.9.
Now, plug the values of A0, A3, and A4into Equation 3.40, and use the Laplace transform pairs
1/(s+a)
? e-at and 1/(s+a)2
? te-at,and wehave
thestepresponse
in thefollowing:
:= 1+x1(t)+x2(t)
xR(t)=L-1[XR(s)]
=1-e-2t-2te-2t
The Case B critically
code:
%
CSD
Fig3.7
clear,
ylabel('Step
Case
B
critically
t=linspace(0,6,61);
xR=1+x1+x2;
grid,
damped step response in Figure 3.7 is obtained using the following
grid
figure(33),
minor,
damped
x1=-exp(-2*t);
plot(t,xR,'k-',t,x1,'b--',t,x2,'r--'),
xlabel('Time,
Response'),
sec'),
legend('xR','x1','x2'
step
response
x2=-2*t.*exp(-2*t);
(3.42)
MAT-LAB
3.3
Time-Domain
Response of Typical
Second-Order
Dynamic
Systems
53
Fig. 3.7: Case B: Critically damped step response.
Note that in addition
to the steady-state
response xss = 1, the step response xR(t) has two
transient
response
terms,
x1(t)=-e-2t
and
x2(t)=-2te-2t,
which
are
theresults
ofthedouble
poles
ats =-2.Ashown
s from
Figure
3.7,
ifthere
were
nox2(t)=-2te-2t
term,
thestep
response
would be the same as that
of the typical
first-order
system,
with time
constant
equal to 0.5 second,
orequivalently
with
thesingle
pole
ats =-2.The
additional
pole
atthesame
location
s = -2
leads
tothecreation
ofthetermx2(t)=-2te-2t
thatapparently
slows
down
thestep
response.
The systems
considered
in
Figure
3.6 and Figure
3.7 are both typical
second-order
systems
with
identical ?n and xss. The only differenceis in the value of the dampingratioone
with ? = 1.25,
overdamped,andthe other with ? = 1, critically damped. Bycomparingthe stepresponsesin Figure
3.6 and Figure 3.7, wecan observe that the critically
the steady state earlier than the overdamped
the least
damped
one that
Remark 3.9 (Alternative
can still
Ways to
damped step response rises faster and reaches
step response.
avoid an overshoot
Evaluate the
The critically
damped
step response is
and oscillations.
Partial
Fraction
Expansion
Residue Con-stants.)
As discussed in Section 2.4.2, the partial fraction expansion residue constants A0, A3, and A4 in
Equation 3.40 can also be evaluated using the polynomial substitution approach and the coefficient
comparison approach. Multiply both sides of Equation 3.40 by s(s +2)2 to obtain the following
polynomial equation:
4 = A0(s+2)2
+A3s(s+2)+A4s
(3.43)
Sincethis polynomial equation is valid for any value of s, any set of three distinct numbers can be
selected to substitute s to obtain three linearly independent equations so that the three constants can
be uniquely determined. However, some particular numbers may be chosen to makethe equations
easierto solve. For this example, if s is selected as 0 then Equation 3.43 will become 4 =A0(2)2,
which
leads
toA0=1.Similarly,
substituting
sby-2in Equation
3.43
willgive
thesolution
A4=-2.After
plugging
these
twoknown
residue
constants
intoEquation
3.43,
there
isonly
one
unknown residue constant A3left in the equation:
4 =(s+2)2+A3s(s+2)-2s
Then there
are several
ways to
determine
the value of A3in
Equation
(3.44)
3.44: (1)
By comparing
thes2term,we
have
0=s2+A3s2,
which
gives
A3=-1;(2)Bysubstituting
sby1,Equation
3.44
becomes
4=32+3A3
-2and
then
A3=-1;or(3)Bymoving
-2sfrom
theright-hand
sid
54
3
Linear Systems
Analysis II
to the left-hand side, Equation 3.44 becomes 2(s+2)
= (s+2)2
+A3s(s+2),
whichis simplified
to0 =s+A3s;
hence,
A3=-1.
?
Case C: 0 < ? < 1. The system is underdamped, and the characteristic equation has two complex
conjugate
roots
at-??nj?n 1-?2or-a j?. Hence,
XR(s)
can
bedecomposed
intothree
terms as follows
using the partial fraction expansion:
XR(s) =
xss?2
n
xss
=
s(s+a-j?)(s+a+j?)
Usingthe Laplacetransform pair 1/(s+a)
Ae-jf
Aejf
+
s
+
s+a-j?
s+a+
(3.45)
j?
? e-at, wehavethe stepresponsexR(t) for CaseCas
follows:
xR(t)
=L-1 [XR(s)]
= xss +Aejfe-(a-j?)t
+Aejfe-(a+j?)t
(3.46)
= xss +Ae-at(ej(?t+f)
The residue
constant
Aejf in
Equation
+e-j(?t+f))
= xss +2Ae-at cos(?t
3.45 can be found (See Equation
+f)
3.64) in terms
of the damp-ing
ratio ? and the steady-state step response xss as
Aejf =
? ??
1-?2?
ej(?+p/2),
0.5xss
where ?= cos-1?
(3.47)
Hence, Equation 3.46 can be rewritten as
???
1-?2?
?1- ???
?
=xss
1 1-?2?
xR(t) =xss +2Ae-at cos(?t
+f)
=xss +
e-at cos(?t
xss
e-at sin(?t
+? +p/2)
(3.48)
+?)
Example 3.10(Case C Step ResponseExample with ? = 0.5, ?n = 2 rad/s and xss = 1)
Considerthe following
equation
are-1 j
XR(s)
R(s)
system with ? = 0.5, ?n =2rad/s, xss = 1,and the roots of the characteristic
v3:
= G(s) =
xss?2
n
s2 +2??ns+?2
Withthe unit step input R(s) = 1/s,
XR(s) =
=
n
4
s2
+2s+4
=
4
(s+1-j
v3)(s+1+
j
v3)
(3.49)
we havethe step response XR(s) andits partial fraction expan-sion,
4
s(s2 +2s+4)
A0
=
s
c
+
s+1-j
v3 +s+1+
c*
j
v3
(3.50)
where A0is areal constant and cis a complex constant to be determined. Thereis no needto compute
c* since it is the conjugate of c. Theseconstantscan be evaluated usingthe residue approachas
follows:
v?
lim
v =3+j-2v3 =1v 3ej150?
v ? s+1-j 3 XR(s) = (-1+j v4
3)( j2 3)
s?-1+j 3
,
A0 =lim sXR(s) = 4/4 = 1
c* = 1/ 3 e-j150?
s?0
c =
?v?
(3.51
Now,plugthe values of A0,c, and c*into Equation 3.50,and usethe Laplacetransform pair 1/(s+
a) ?e-at,then wehavethe stepresponse
in thefollowing:
3.3
Time-Domain
Response of Typical
Second-Order
?v
?
?v
?
?
?v? ? 3t +150?)=1- 2/ 3e-t
sin(
?v?
?v?
The
twocomplex
poles
are-??nj?n?
1-?2=-1 j v
v
= 1+ 1/ 3 ej150?e-(1-j 3)t + 1/
v
v
+e-j( 3t+150?)
3 e-t ej( 3t+150?)
xR(t) =L-1[XR(s)]
= 1+ 1/
= 1+ 2/
3 e-t cos(
Dynamic
Systems
3 e-j150?e-(1+j
v
v3t
55
v3)t
(3.52)
+60?)
3in rectangularform, or ?nej(p-?)
thattheunderdamped
stepresponse
in
= 2ej120?in polarform, where? =cos-1? =60?.Note
Figure 3.8 rises more quickly than the critically damped step response in Figure 3.7, but it has over-shoot
and oscillations. The oscillation frequency is determined by ?, the imaginary
part of the
complex pole, and the amplitude of the oscillation is decreasing exponentially
with decreasing
rate
determined
by-a,therealpart
ofthecomplex
pole.
Fig. 3.8:
Case C: Underdamped
step response.
The Case C underdamped step response in Figure 3.8is obtained using the followingMATLAB
code:
%
CSD
clear,
xR=1+x1;
xlabel('Time,
Fig3.8
Case
C
underdamped
t=linspace(0,6,61);
step
response
x1=-1.16*exp(-t).*sin(sqrt(3)*t+pi/3);
figure(34),
plot(t,xR,'k-',t,x1,'b--'),
sec'),
ylabel('Step
grid,
Response'),
grid
minor,
legend('xR','x1')
CaseD: ? =0. Thesystemis undamped,andhasa pairof complexconjugaterootsontheimaginary
axis
ofthecomplex
plane
at j?n.Hence,
XR(s)
canbedecomposed
intothree
terms
asfollows
using the partial fraction expansion:
XR(s) =
xss?2
n
xss
s(s-j?n)(s+
j?n)
Usingthe Laplacetransformpair 1/(s+a)
=
s
Aejf
+
s-j?n
+
Ae-jf
s+ j?n
(3.53)
? e-at, wehavethe stepresponsexR(t)for CaseDas
follows:
xR(t) =L-1 [XR(s)] = xss +Aejfej?nt +Aejfe-j?nt
(3.54)
= xss +A(ej(?nt+f)
+e-j(?nt+f))
= xss +2Acos(?nt +f)
The residue Aejf in Equation 3.53 can be found as 0.5xssejp, and therefore the step response xR(t)
for
Case D can be rewritten
a
56
3
Linear Systems
Analysis II
xR(t)
=xss+2Acos(?nt
+f)=xss
(1-cos?nt)
(3.55)
Example 3.11(Case D Example with ? = 0, ?n = 2rad/s and xss = 1)
Considerthe following
system with ? = 0, ?n = 2rad/s, xss = 1, andthe roots of the characteristic
equation
j2:
XR(s)
R(s)
= G(s) =
xss?2
n
4
s2 +2??ns+?2
=
n
4
s2 +4
=
(s-j2)(s+j2)
(3.56)
Withthe unit stepinput R(s) = 1/s, wehavethe stepresponse XR(s)andits partial fraction expan-sion,
4
XR(s) =
A0
s(s2 +4)
=
s
c
+
s-j2
+
c*
(3.57)
s+ j2
where A0is areal constant and cis a complex constant to be determined. Thereis no needto compute
c* sinceit is the conjugateof c. Theseconstantscanbeevaluatedusingthe residueapproachas
follows:
4 =-0.5=0.5e
(s-j2)XR(s)
=j2(j2+j2) = -8
jp
sXR(s) = 4/4 = 1
c* =-0.5=0.5e-jp,
A0=lim
s?0
4
c = lim
s?j2
(3.58)
Fig. 3.9: Case D: Undamped step response.
Now,plugthe values of A0,c, and c* into Equation 3.57. Thenby taking its inverse Laplacetrans-form,
we have the step response in the following:
xR(t)=L-1[XR(s)]
=1-0.5e
j2t-0.5e-j2t
=1-cos2t
(3.59)
?
The
twocomplex
poles
inthisexample
are-??nj?n 1-?2= j2 inrectangular
form,
or
?nej(p-?) in polarform, where? = cos-1? = 90?. Notethat the undampedstepresponsexR(t)
in
Figure 3.9 rises
more quickly
than the
underdamped
step response
in
Figure
3.8, but it
has
100% overshoot and undamped oscillations. The oscillation frequency is ? = ?n, which is the
imaginary part of the complex pole, and the amplitude of the oscillation is sustained without
damping since the real part of the complex pole is zero
3.4
3.4 Characterization
Characterization
of the
Underdamped
of the Underdamped Second-Order
Second-Order
Systems
5
Systems
For
atypical
first-order
system
described
byEquation
2.62,
itsstep
response
isx(t)=xss(1-e-t/t),
which can be easily characterized
by the initial
value x(0)
and plotted,
as in Figure 2.5(b).
= 0, the steady-state
value x(8)
The step response is basically
= xss, and the value of x(t)
deter-mined
at the time
constant
x(t) =xss(1-e-1)
=0.632xss.
It has
neither
overshoot
noroscillations;
itjustsimply
rises
exponentially with a decreasingslope?x(t) = (xss/t)e-t/t
x(8)
to approachto the steady-statefinal value
=xss.
Although the typical second-order system can be characterized by the damping ratio ?, the natural
frequency ?n, and the step response steady-state value xss,the plotting of the second-order system step
response according to the three characterization
system.
As discussed in the previous two
parameters is not astrivial
sections,
Sections
as that of the typical first-order
3.2 and 3.3, according
to the pole
locations of the system, the system can be categorized into four cases:the overdamped case with ? > 1,
the critically damped case with ? = 1, the underdamped case with 0 < ? < 1, and the undamped case
with ? = 0. Among these four cases, the most common and interesting case that occurs in practice
is the underdamped case with 0 < ? < 1. For this reason, in this section we will focus on the study
of the underdamped case to gain much more clear understanding about the geometry of the poles
on the complex plane and the relationship between the geometry and the time-domain response.
3.4.1
Geometry
of Conjugate
System
Poles on the
Complex
Plane
When0 < ? < 1, the system poles, which arethe roots of the characteristic equation, Equation 3.13, are
a pair of complex
conjugate
roots at
?
p
where
a=??n,
?=?n 1-?2,? =cos-1
?
p* =-a j? =?nej(p-?)
The
pair
ofcomplex
numbers
are
either
represented
inrectangular
form
as-a j? orinpolar
form
?nej(p-?).
(3.60)
Bothrepresentationsareimportant in revealing the physical meaningof the mathematical
results and in serving
as efficient
computation
tools in the analysis and design of dynamic
systems.
?
The
realpartofthecomplex
number,
-a =-??n,
determines
how
fastthesystem
response
part, j? = j?n
1-?2,manifests
theos-cillation
would converge to the steady state, and the imaginary
frequency of the system response. In polar form, the complex numbers are represented in
terms of their magnitude and phase angle. It can be seen that the magnitude and phase angle of the
complex numbers are
esting to note that algebraically the relationship between the angle ? and the damping ratio ? can be
v
a2+?2=?nand(p-?),respectively,
where
? =cos-1?.
Itisinter-?
represented in
manydifferent
=tan-1
ways by trigonometric
functions as follows:
?1-?2 ?1-?2=cos-1?
?
=sin-1
p
and
2
?
?
-? =tan-1
(3.61)
1-?2
The algebraic relationship between a, ?, ? and ?, ?nseems complicated; however, it is rather straight-forward
graphically,
3.4.2
Step
as shown in Figure 3.10.
Response of the
The step response
Underdamped
of the system, represented
Second-Order
System
by Equation 3.7, is the solution
due to the unit step input assuming zero initial conditions: x(0)
briefly
described
a procedure
of the differential
= 0 and?x(0)
to obtain the step response for the underdamped
equation
= 0. In Section 3.3.2, we
case in
Equation
3.48.
58
3
Linear Systems
Analysis II
Fig. 3.10: Geometrical relationship between a, ?, ?, and ?, ?n.
In the following,
the second-order system poles geometry shown in Figure 3.10 will be employed to
derive and explain the physical
Laplace transform
of Equation
XR(s)
meaning of the step response solution in
3.7, the transfer function
= G(s) =
Equation
3.48.
After taking the
of the system can be found as follows:
xss?2
xss?2
n
n
=
(3.62)
(s+a-j?)(s+a+j?)
n = 0. Withthe
Note
that-a j? are
thesolutions
ofthecharacteristic
equation,
s2+2??ns+?2
R(s)
s2 +2??ns+?2
n
unit
step input R(s) = 1/s, we havethe step response XR(s) andits partial fraction expansion,
XR(s)
=
xss?2
n
A0
s(s2 +2??ns+?2
=
n)
s
c*
c
+
s+a-j?
+
s+a+
(3.63)
j?
where A0is a real number and c is a complex number to be determined. Since c* is the conjugate of c,
only the two residue
c =
constants
A0 and c need to be evaluated:
xss?2n
xss?2n
(s+a- j?)XR(s)
=(-a+j?)(
j2?)=?nej(p-?)2?n v1-?2ejp/2
s?-a+j?
lim
v
n/?2n
A0 =lim sXR(s) = xss?2
c* = 0.5xsse-j(?+p/2),
1-?2
ej(?+p/2
v1-?2
= 0.5xss
= xss
s?0
(3.64)
?
Intheabove
complex
number
manipulations,
weused
thegeometrical
facts
that-a j? =
1-?2,
j =ejp/2,and
j =e-j3p/2
=ejp/2.Now,
plug
thevalues
ofA0,
c,and
?nej(p-?), ? =?n
c*into Equation 3.63, and withthe Laplacetransform pair 1/(s+a)?e-at,
wehavethe following:
x(t) =L-1 [XR(s)] =xss+ 0.5xss
e-j(p/2+?)e(-a-j?)t
v1-?2
v ej(p/2+?)e(-a+j?)t + 0.5xss
?v ?
1-?2
= xss 1+ 0.5e-at ej(p/2+?t+?)
+e-j(p/2+?t+?)
1-?2
??? v
= xss
1+
e-at
1-?2cos(p/2+?t
?
+?)
(3.65)
Notethat the Eulers formula (Theorem 2.1) and Corollary 2.2 have been employed to convert the sum of
two complex functions into one real cosine function. Furthermore, with the basic trigonometric formula
cos(p/2+?)
=-sin?,
Equation
3.65
can
besimplified
tothe
following:
x(t)
= xss
?1-?
1-?2
e-at
sin(?t
?
+?) ,
where a=??n,
? = ?n
?1-?2,? =cos-1?
(3.66)
3.4.3
Graphical
Interpretation
In the following,
3.4
Characterization
of the
Underdamped
of the
Underdamped
Second-Order
Second-Order
Step
Systems
5
Response
we will plot the step response according to Equation 3.66, and characterize the step
?1e-at??
1-?2?
response in terms of the final steady-state value xss,the rise time tr, the peak time tp, the envelope
functions xss
, the maximumovershoot OS,the oscillation frequency ? and period
T, and the settling time ts. All of these characterizations are determined by the damping ratio ?, the
natural frequency ?n, and the step response steady-state value xss. The numerical example to be used
for demonstration is based on ? = 0.2, ?n = 3 rad/s, and xss = 5.
1-?2and
t =0,we
have
theinitial
value
1.
Initial
and
Final
Steady-State
Values:
With
?
=
sin-1?
?
1-sin(?)??
1-?2 =0.Italso
can
beseen
that
thefinalsteady-?
x(t)=xss
(1-0)=xss
if a>0.
??
1-?2,which
? ??
?1-e-at??
1-?2?
1-?2?
from Equation 3.66, x(0)
= xss
state value is lim
t?8
2. A Pair of Envelope Functions: Sincethe amplitude of the sinusoidalterm is e-at
is decreasing
with time, the step response
xss
waveform is confined
1+e-at
and
between
a pair of envelope func-tions:
(3.67)
xss
As shown in Figure 3.11, these two envelopes (in blue dotted lines) are symmetrical
the x(t) = xss horizontal line.
with respect to
3. Sinusoidal Oscillation Frequency and Period: The step response curve will start from x(0) = 0,
and reaches x(8) = xss = 5 at steady state. The curve will swing up and down with decreasing
oscillationamplitudeatfrequency ? = ?n
theoscillation
period
is
?1-?2=2.939rad/s;hence,
T = 2p/?
4. Valleys and Peaks of the Decaying Sinusoidal
= 2.138s
(3.68)
Curve: The valleys and peaks (local
minimums and
maximums)
ofthestep
response
willoccur
att =0.5kT,
k=0,2,4,and
t =0.5kT,
k=1,3,5,,
respectively.
The
valley
values
arex(0.5kT)
=xss(1
-e-0.5kaT)
while
thepeak
values
are
x(0.5kT)
=xss(1
+e-0.5kaT).
Hence,
thevalleys
att =0,T,2T,3T,
arex(0)=0,x(T)=
x(2.138)
=3.61,
x(2T)=x(4.276)
=4.62,
x(3T)=x(6.414)
=4.89,, respectively
and
the
peaks
att =0.5T,
1.5T,
2.5T,
3.5T,are
x(0.5T)
=x(1.069)
=7.63,
x(1.5T)
=x(3.207)
=5.73,
x(2.5T)
=x(5.345)
=5.2,
x(3.5T)
=x(7.483)
=5.06,, respectively.
These
valleys
and
peaks
are marked as pink dots onthe graph. Notethat the valleys and peaksin general are not on the pair
of envelopes defined by Equation 3.67.
5. Peak Timetp: Peaktime is the time whenthe maximumovershootoccurs. Althoughthe amplitude
of the sinusoidal
term is decreasing, the local
period T = 2p/?.
global
maximum,
Since the first
maximums
minimums still
= 0, the first
occur according to the
maximum, which is also the
will occur at
tp = T/2 = p/?
The peak time
and
minimum occurs at t
can also be found
by searching
= 1.069s
the time
at
(3.69)
which the first
derivative
of x(t)
with
respect to t is zero. By setting?x(t) equal to zero, wehave
? ?? ??
x?(t)= -xss 1-?2
d
dt
e-at sin(?t
+?)
?
=0
(3.70)
which
will
lead
to ?cos(?t
+?)-asin(?t
+?)=0,and
therefore
we
have
thefollowing:
60
3
Linear Systems
Analysis II
tan(?t
+?) =
?1-?2
?
=
a
?
=tan?
(3.71)
This
equations
holds
only
at?t=kp,k =0,1,2,. The
firstpeak
occurs
atk =1;hence,
thepeak
time is tp = p/?,
whichis exactlythe sameas Equation 3.69.
Fig. 3.11: Graphical interpretation
The typical
following
%
CSD
second-order
system
of the typical second-order system step response.
step response
shown
in
Figure
3.11 is
obtained
using the
MATLAB code:
Fig3.11
Typical
x_ss=5,
ze=0.2,
den=[1
2*ze*wn
wn=3,
wn2],
2nd-order
system
alpha=ze*wn,
step
response
w=wn*sqrt(1-ze2),
t=linspace(0,8,800+1);
num=x_ss*wn2,
x=step(num,den,t);
x_envelope_down=5*(1-(1/sqrt(1-ze2))*exp(-ze*wn.*t));
x_envelope_up=5*(1+(1/sqrt(1-ze2))*exp(-ze*wn.*t));
figure(1),
plot(t,x,'r',t,x_envelope_down,'b--',t,x_envelope_up,'b--'),
grid
minor,
xlabel('Time,
grid,
sec'),
ylabel('Step
Response'),
legend('x(t)','Envelope')
?
6. Maximum
Overshoot
OS:Since
sin(p+?)=-sin?=- 1-?2,
thevalue
ofx(t)atpwillbe
x(tp)
= xss
?1-?
1-?2
e-atp
That meansthe step response overshoots
sin(p+?)
?
=xss? 1+e-??ntp ? = 7.633
over the steady-state
(3.72)
value xss = 5 by 2.633. The maximum
overshoot usually is measuredas percentage of its steady-state value, xss;therefore, the maximum
overshoot
can be computed
from
either one of the following
two formulas
3.5
OS =e-?p/
Analysis and
v1-?2 = 0.5266
Design of a MassDamperSpring
System
61
= 52.66%
(3.73)
?xss
?5 =0.5266
OS=(xR(tp)-xss)
=(7.633-5)
=52.66%
It can be seen from this equation that the maximum overshoot is solely determined by the damping
ratio ?. In many applications like system identification or control system design, we may need to
compute the damping ratio ? based on a given information of the maximum overshoot. It can be
obtained as follows:
?
?
1-?2 ? tan?= -ln(OS) ? ?=cos
-ln(OS)
ln(OS)
=?
?1-?2?
? -ln(0.5266)
?
p
-p?
tan-1
Notethat the geometrical relationships between ? and ?, tan?
=
p
?, and cos?
(3.74)
= ?, were
employed in the proof of Equation 3.74. For verification, the damping ratio corresponding to OS =
0.5266 is computed as follows:
p
? =cos tan-1
=cos78.46? = 0.2
(3.75)
7. Rise Time tr: Therise time usually is defined asthe time a step response needsto rise from 10% to
90% of its desired steady-state value, xss. However,for the underdamped case,it is more meaningful
to define the rise time asthe time required to rise from 0to xss(i.e., the smallest t so that x(t) = xss).
leads
to?n?
1-?2tr+?=p.
?1-?2tr+?)=0,which
p-cos-1
?
1-?2=0.603
tr= ? s
Basedonthis definition, wehavesin(?n
Hence,the rise time is
(3.76)
?n
8.Settling
Time
ts:The
settling
time
tsisdefined
asthetime
atwhich
thestep
response
is within
2%
??
1-?2=0.02,
which
leads
to
? ?1-?2?
of the steady-state value xss.It can be computed via the amplitude of the sinusoidal term in Equation
3.66, e-??nts
ts =
-ln
0.02
= 6.55 s
??n
(3.77)
Ifapproximation
isallowed,
this
can
be
ts-ln0.02
(??n)
4(??n)
=6.52
s?.
In
cthat
ase
the
?
?
2%error
isrelaxed
to 5%,thesettling
timewould
bets -ln0.05? (??n) 3 (??n)=5.0s.
3.5 Analysis and Designof a MassDamperSpring
System
In this section,
we will analyze how the mass, damper, and spring affect the behavior of the massdamperspring
(MBK) system in Figure 3.1(a) and then consider how to choose appropriate values of
mass,friction coefficient, and spring constant, so that the system has a desired performance. For ease of
reference, the schematic diagram of the massdamperspring system is repeated in Figure 3.12.
The transfer function
corresponding to the governing differential equation of the
massdamperspring
system in Equation 3.8 is
(1/K)(K/M)
Y(s)
Fa(s)
=
s2 +(B/M)s+(K/M)
(3.78
62
3
Linear Systems
Analysis II
Fig. 3.12:
Comparing this transfer function
A massdamperspring
system.
with the typical second-order system transfer function in Equation
3.12, we havethe dampingratio ?,the natural frequency ?n, andthe steady-statestep responseyssas
follows:
? = 0.5B/
v MK,
?
?n =
K/M,
(3.79)
yss = 1/K
In the following,
we would like to know how the friction coefficient B,the spring constant K, and
the mass M,individually affect the step response of the system. First, we would let M = 1 kg and
K = 1 N/m and observe how the step response of the system varies with the change of the damper
friction coefficient B = 3, 2, 1.414, 1, 0.5 Ns/m. As it can be seen from Equation 3.79 and Fig-ure
3.13,the variation of the friction coefficient B does not affect the natural frequency ?n, whichis
?n = K/M= 1rad/s. However,the dampingratio ? will change with B = 3, 2, 1.414, 1, 0.5 Ns/m,
v
to ? =1.5,1, 0.707,0.5, 0.25,respectively.WhenB =3 Ns/m,the polesare onthe negativereal axis
of
thecomplex
plane
ats =-2.618
and
s =-0.382,
and
thecorresponding
step
response
isoverdamped
with ? = 1.5 and ?n = 1rad/s. As B decreases,the two poles will movetoward each other and coincide
together
ats =-1onthecomplex
plane
when
B=2Ns/m
and
thecorresponding
step
response
is
critically damped with ? = 1 and ?n = 1rad/s.
The following
MATLAB
program
damping
shown in
Figure 3.13:
%
CSD
ratio
Fig3.13
clear,
Effect
t=0:.1:20;
of
is employed
varying
to plot the
damping
M=1,
B=0.5,
G1=tf(num1,den1),
M=1,
B=1,
G2=tf(num2,den2),
M=1,
B=1.414,
G3=tf(num3,den3),
M=1,
B=2,
K=1,num4=K/M,
G4=tf(num4,den4),
M=1,
B=3,
K=1,
G5=tf(num5,den5),
x1=step(G1,t);
x4=step(G4,t);
x5=step(G5,t);
K=1,
K=1,
MBK system step responses
ratio
on
num1=K/M,
num2=K/M,
K=1,
MBK
num3=K/M,
B/M
den5=[1
x2=step(G2,t);
K/M],
K/M],
den3=[1
den4=[1
responses
B/M
den2=[1
num5=K/M,
step
den1=[1
B/M
B/M
with varying
K/M],
K/M],
B/M
K/M],
x3=step(G3,t);
figure(35),
plot(t,x1,'b.',t,x2,'m-',t,x3,'g-',t,x4,'k-',t,x5,'r--'),
grid,
legend
grid
minor,
xlabel('Time,
sec'),
ylabel('Step
Response'),
('B=0.5','B=1','B=1.414','B=2','B=3')
Asthe friction coefficient Bfurther decreases, the two identical poles will split to become a pair
of complex conjugate poles moving along the circle with radius equal to ?n = 1 centering at the
, andthe corresponding
origin.
When
B=1.414
Ns/m,
thepoles
are
ats =-0.707
j0.707
=1ej135?
step responseis underdamped with ? = 0.707 and ?n = 1rad/s. Asthe friction coefficient drops to
which is
B=1Ns/m,
thepoles
areats =-0.5j0.866
=1ej120?
closer to the imaginary
and the corresponding step responseis moreunderdamped with ? = 0.5 and ?n = 1 rad/s.
, and the corresponding
B=0.5Ns/m,
thepoles
are
ast =-0.25j0.968
=1ej104.5?
axis,
When
step response is
oscillation
frequenc
?1-?2rad/s,thestepresponse
even moreunderdamped with ? = 0.25 and ?n = 1rad/s, which hasoscillations with high overshoot.
Sincethe oscillation frequency is ? = ?n
3.5
Analysis and
Design of a MassDamperSpring
System
63
Fig. 3.13: Theeffect ofthe dampingratio ? onthe stepresponseand polelocations ofthe massdamperspring
system whilethe naturalfrequencyremainsthe sameat ?n = 1rad/s.
will increase as the damping ratio decreases and, therefore, asshown in Figure 3.13, the oscillation
frequency
of the step responses
becomes higher as the damping ratio decreases.
Now wehaveobservedhowthe stepresponseand polelocations vary withthe dampingratio ? while
the natural frequency remains at ?n = 1 rad/s, as shown in Figure 3.13. Next, we would like to observe
how the natural frequency ?n affects the pole locations andthe step response while keeping the damping
ratio at ? = 0.707. As discussed, for the case with M = 1 kg, K = 1 N/m and B = 1.414 Ns/m, the
, andthe step response is underdamped with ? = 0.707 and
?n = 1 rad/s, as shown in both Figure 3.13 and Figure 3.14. If the massand the friction coefficient
poles
areast =-0.707
j0.707
=1ej135?
are changedto
M = 4 kg and B = 2.828 Ns/m, respectively,then the naturalfrequency will reduceto
while keeping the
?n=0.5
rad/s
and
thepole
location
willchange
to -0.354j0.354
=0.5ej135?
damping ratio at ? = 0.707.
Notethat the step response with the smaller natural frequency is shown in Figure 3.14in blue, which
has a
much slower rise time
?n = 1 rad/s.
and a slower
oscillation
frequency
than the step response
in
Onthe other hand, the massand the friction coefficient can be changed to
black
with
M = 0.25 kg
and B = 0.707 Ns/m, respectively,to increasethe natural frequencyto ?n = 2 rad/s and changethe
pole
location
to -1.414j1.414
=2ej135?
while keeping the damping ratio at ? = 0.707. The step
response corresponding to the increased natural frequency ?n = 2 rad/s is shown in Figure 3.14in
red, which has afaster rise time and a higher oscillation frequency than the step response in black
with ?n = 1 rad/s.
The following
MATLAB program is employed to plot the typical
with varying natural frequency shown in Figure 3.14:
%
CSD_Fig3.14
clear,
Effect
of
varying
natural
ze=0.707,
wn=0.5,
G1=tf(num1,den1),
ze=0.707,
wn=1,
G2=tf(num2,den2),
ze=0.707,
wn=2,
G3=tf(num3,den3),
x1=step(G1,t);
figure(36),
xlabel('Time,
legend
t=0:.1:20;
frequency
num1=wn2,
ylabel('Step
('\omega_n=0.5','\omega_n=1','\omega_n=2'
step
den1=[1
system step re-sponses
response
2*ze*wn
wn2],
num2=wn2,
den2=[1
2*ze*wn
wn2],
num3=wn2,
den3=[1
2*ze*wn
wn2],
x2=step(G2,t);
plot(t,x1,'b-',t,x2,'k-',t,x3,'r-'),
sec'),
on
second-order
x3=step(G3,t);
grid,
Response'),
grid
minor,
64
3
Fig. 3.14:
Linear Systems
The effect
Analysis II
of the
natural frequency
on the step response
and pole locations
of the
massdamperspring
system while the damping ratio remains the same at ? = 0.707.
3.6 Analysis and Designof a Simple DC Motor Position Control System
As described in the beginning
(MBK)
of Section
3.1, there is a mathematical
system, the resistorinductorcapacitor
(RLC)
equivalence
among the
system, and the
massdamperspring
DC motor posi-tion
control system. Like the MBK mechanical system and the RLC electrical system, the DC motor
electromechanical position control system is also governed bythe typical second-order system differen-tial
equation. In this section, we will design a simple proportional controller to achieve a desired
position control for a DC motor system. Considerthe block diagram of the DC motor position control
system shown in Figure 3.1(c), which for ease of reference is repeated here in Figure 3.15.
Fig. 3.15: Design of a proportional controller
Kfor the DC motor position control system.
As given in Equation 3.5, the transfer function of the closed-loop system is
T(s)
TR(s)
and therefore
the characteristic
KG(s)
=
1+KG(s)
bK
=
s2 +as+bK
(3.80
equation is
s2 +as+bK
=0
(3.81)
The behavior of the closed-loop system is mainly determined bythe closed-loop system poles, which
are the roots of this characteristic equationor,
more specifically, bythe corresponding damping ratio
and natural frequency. By comparing this characteristic equation to the typical second-order charac-teristic
equation, s2 +2??ns +?2
n = 0, it can be easily seenthat the damping ratio ? and natural
frequency ?n arefunctions ofthe proportional controller parameter K asfollows: ? = 0.5a
?v
bK
3.6
and ?n =
Analysis and
Design of a Simple
DC
Motor Position
Control
System
65
bK. Increasing K will enlarge the natural frequency ?n and reduce the damping ra-v
tio at the same time, as shown in Table 3.1. As explained and demonstrated in Sections 3.4 and 3.5,
the increase of ?n will bump upthe step response by increasing the oscillation frequency according to
? = ?n
damping
ratio
? may
also
lead
toafaster
step
response,
but
itcertainly
will
?1-?2.Asmaller
cause larger
overshoot
and oscillations.
Table 3.1: The effect of K on ?, ?n, andthe closed-loop system performance
K
?
?n,rad/s
0.5 1.414
1
-0.293,
-1.707
0.707
1
Overshoot ? = ?n
Poles
1
-1, -1
2 0.7071.414 -1 j = v2e
j135?
v3 = 2ej120?
4
0.5
2
-1 j
16
The following
responses
%
CSD
clear,
a=2,
?1-?2
overdamped
0
critically damped
5%
1 rad/s
16%
1.732 rad/s
45%
-1 j3.87=4e
j104.5?
3.75
MATLAB program is employed to plot the DC motor position control system step
with varying
Fig3.16
4
0.25
0
control
Effect
b=1,
of
parameter
varying
t=0:.05:10;
Kshown
K
on
K=0.5,
in
DC
Figure
motor
num1=b*K,
3.16:
position
den1=[1
control
a
b*K],
G1=tf(num1,den1),
x1=step(G1,t);
K=1,
num2=b*K,
den2=[1
a
b*K],
G2=tf(num2,den2),
x2=step(G2,t);
K=2,
num3=b*K,
den3=[1
a
b*K],
G3=tf(num3,den3),
x3=step(G3,t);
K=4,
num4=b*K,
den4=[1
a
G4=tf(num4,den4),
x4=step(G4,t);
K=16,
G5=tf(num5,den5),
x5=step(G5,t);
figure(37),
num5=b*K,
den5=[1
plot(t,x5,'b.',t,x4,'m-',t,x3,'g-',t,x2,'k-',t,x1,'r--'),
grid
minor,
xlabel('Time,
b*K],
a
b*K],
grid,
sec'),
ylabel('Step
Response'),
legend('K=16','K=4','K=2','K=1','K=0.5')
Fig. 3.16: Stepresponses and polelocations of the DC motor position control system withfive different
proportional controllers
66
3
Linear Systems
Without loss
Analysis II
of generality
we assume a = 2 and b = 1 in the following
for ease of demonstration.
Then the characteristic equation of the closed-loop system becomes s2 +2s+K
= 0, and the damping
ratio and natural frequency would be ? = 1/
K and ?n =
K. When K = 0.5, in Figure 3.16, wecan
v
see how the variations
the pole locations,
of the controller
gain
and the step response.
v
K would affect the damping ratio, the natural frequency,
Note that the results in Figure 3.16, associated
with
K = 0.5,
arecfrequency
olored
inred.
Itisclear
that
thetwopoles
are
at-1.707
and-0.293.
The
damping
ratio
and
nat-ural
are ? = 1.414 and ?n = 0.707 rad/s, respectively, and the step response is overdamped.
Similarly
the results
associated
with K = 1,2,4,16
are in black,
green, purple,
and blue, respectively.
When K = 1, ? = 1, the system is critically damped. The K = 2,4,16 cases are underdamped
0.707, 0.5, and 0.25 damping ratios, respectively.
with
The performance results associated withthe five simple controller designs are also tabulated in Table
3.1. It is clear that
overshoot.
both extremes
The better choice
are either too
of Kis between
overdamped
K = 1 and
or underdamped
with large
oscillation
and
K = 2 where the overshoot is less than 5% and
the rise time is acceptable.
3.7 Steady-State Sinusoidal Responseand Bode Plot of Typical Second-Order
Systems
As discussed in Section 2.5.3, particularly in Theorem 2.32 and Corollary 2.33, the steady-state sinu-soidal
response can be obtained simply
rather than solving
for the complete
by computing
solution
the
magnitude and the phase of a complex
of a differential
equation.
Although
presented in
number
Chapter
2 wherethe mainemphasis wasin the first-order systems, Theorem 2.32 and Corollary 2.33 are also
valid for higher-order systems as long as the system transfer function has no poles on the imagi-nary
axis or in the right half of the complex plane. To recap, Figure 2.9 withthe essence of Corollary
2.33 is shown in Figure 3.17.
Now,let G(s) bethe transfer function of the typical second-order system,
G(s) =
and the input be u(t)
=sin?t.
xss?2
n
s2 +2??ns+?2
(3.82)
n
Thenthe steady-state output response will be
yss(t) = A(?) sin (?t
+?(?))
where
A(?)=|G(j?)|and
?(?)=?G(
j?)are
themagnitude
and
thephase
ofG(
j?),respectively.
Notethat G(j?) is ajust a constantcomplex numberif the frequency ?is aconstantreal number. How-ever,
if the sinusoidal input frequency
? is changing, the amplitude and the phase of the steady-state
output response will change accordingly.
Withs replaced by j? in Equation 3.82, wehave G(j?) asfollows:
G(j?)
Let ?
=
xss
j?) =A(?)e
j?(?)
1-(?/?n)2
+j2?(?/?n)=|G(j?)|?G(
(3.83)
= ?/?n bethe normalized frequency, then Equation 3.83 becomes
G(j?)
=
xss
j?) =A(
?)ej?(
?)
1-?2 +j2?
? =|G(j?)|?G(
The magnitudeandthe phaseof G(j?)
will be
(3.84
3.7 Steady-State
Sinusoidal
Response and Bode Plot of Typical
Second-Order
Systems
67
Fig. 3.17: Efficient way of computing the sinusoidal steady-state response yss(t).
when
??1
?
??
A(
?) =|G(j?)| =xss
(1-?2)2+4?2
?2 =???xss
when
??1
???
?(
?)=?G(
j?)=-tan-1
?2??/(1-?2)
xss/2?
when ?
=1
(3.85
xss/?2
?
The Bode plot of G(j?)
with xss = 1 and six damping ratio values ? = 0.05, 0.1, 0.5, 0.7, 1, 1.5,
are shown in Figure 3.18. It can be seen that these plots are very different from the Bode plot of the
first-order systems.The maindifference occurs when
? = ?/?n = 1 orthe sinusoidalinput frequency
? equalsthe naturalfrequency ofthe system, ?n. The magnitudeof G(j?n) can becomevery large if
the damping ratio is approaching to zero. This phenomenon is called resonance.
Example 3.12 (Sinusoidal
Response Example with ? = 0.05, ?n = 4 rad/s and xss = 1)
Given the transfer function
G(s) =
xss?2
n
s2 +2??ns+?2
=
n
16
s2
(3.86)
+0.4s+16
Find the three steady-state sinusoidal responses of the system respectively driven bythe following
control-input signals: u(t) = sin0.4t, u(t) = sin4t, u(t) = sin40t.
three
Solution:
Thesystem shown in Equation 3.86is atypical second-order system withthe damping ratio ? = 0.05,
natural frequency ?n = 4rad/s, and steady-state step response gain xss = 1. The normalized magnitude
and phase response graphs (i.e., the Bode plot) of the system, A(
?) and ?(?),
are shown in Figure
3.18. These graphs reveal that the magnitude A(
?) and the phase ?(?) at the normalized frequencies
? = 0.1, ? = 1, and?
= 10 are:
A?0dB=1, ??0? when ? = 0.1 or ? = 0.4rad/s
A=20dB
=10,? =-90? when ? = 1 or ? = 4 rad/s
A?-40dB
=0.01,
? ?-180? when ? = 10 or ? = 40rad/s
Therefore, the steady-state
u(t)
= sin0.4t, u(t)
responses
of the system
=sin4t, and u(t)
yss(t) = sin0.4t
given in
Equation
3.86 due to the sinusoidal
(3.87)
inputs:
= sin40t, are found respectively in the following:
when
? = 0.4rad/s
yss(t)
=10sin(4t
-90?) when ?
yss(t)
=0.01sin(40t
-180?) when
= 4rad/s
? = 40 rad/s
(3.88)
68
3
Linear Systems
Analysis II
Fig. 3.18: The Bodeplot of the typical second-ordersystems with varying ?.
The following
MATLAB program is employed to plot the Bode plot of the typical
systems with varying ? shown in Figure 3.18:
%
CSD_Fig3.18
%
typical
Effect
of
2nd-order
varying
damping
ratio
on
Bode
plot
second-order
of
systems
w=logspace(-1,1,300);
num=1;
ze=0.05,
den=[1
2*ze
1];
[mag,p1]=bode(num,den,w);
m1=20*log10(mag);
ze=0.1,
den=[1
2*ze
1];
[mag,p2]=bode(num,den,w);
m2=20*log10(mag);
ze=0.5,
den=[1
2*ze
1];
[mag,p3]=bode(num,den,w);
m3=20*log10(mag);
ze=0.7,
den=[1
2*ze
[mag,p4]=bode(num,den,w);
m4=20*log10(mag);
ze=1,
[mag,p5]=bode(num,den,w);
m5=20*log10(mag);
ze=1.5,
[mag,p6]=bode(num,den,w);
m6=20*log10(mag);
figure(1)
den=[1
den=[1
2*ze
2*ze
1];
1];
1];
semilogx(w,m1,'b-',w,m2,'b.',w,m3,'m-',w,m4,'r-',w,m5,'k-',w,m6,'g.'),
grid,
grid
minor,
legend('\varsigma=0.05','\varsigma=0.1','\varsigma=0.5',...
'\varsigma=0.7','\varsigma=1','\varsigma=1.5'),
title('Magnitude
response
in
dB'),
figure(2)
semilogx(w,p1,'b-',w,p2,'b.',w,p3,'m-',w,p4,'r-',w,p5,'k-',w,p6,'g.'),
grid,
grid
minor,
legend('\varsigma=0.05','\varsigma=0.1','\varsigma=0.5',...
'\varsigma=0.7',
title('Phase
'\varsigma=1','\varsigma=1.5'),
response
in
deg')
Remark 3.13 (Resonance)
For a system with a small damping ratio like the onein the previous example, ? = 0.05, the vibration
amplitude of the system due to a disturbance input with frequency at the natural frequency ?n will be
10 times of the input amplitude. The vibrations near the natural frequency of a system with small
damping ratio can be very damaging; the collapse of the Tacoma Narrows Bridge in 1940 is a
notorious example
3.8
Exercise
Problems
6
3.8 Exercise Problems
P3.1a: Consider Equation 3.1, whichis the governing differential equation of the massdamperspring
system (the
MBK system)
shown in Figure 3.12 (or
Figure 3.1(a)).
spring constant is K = 1.6 N/m, and the damper friction
coefficient
Assume the
massis
M= 0.1 kg, the
B = 0.4 Ns/m. Then the governing
equation of the system is given by
y(t)+4
where y(t) and fa(t)
?y(t)+16y(t)
= 10 fa(t)
(3.89)
are the displacement (m) of the massand the external force (N) applied to the
mass,
respectively. Findthe transfer function Y(s)/Fa(s), the poles,the dampingratio ?,andthe natural
frequency ?n of the system.
P3.1b: Assumezero initial conditions (y(0)
0.1us(t)
= 0 m,?y(0) = 0 m/s) and let the input be fa(t)
N, where us(t) is the unit step function.
P3.1c: Use MATLAB command
Solve and plot y(t), the solution
of Equation
=
3.89.
step to plot the step response of the system and verify your result in
P3.1b.
P3.1d: Explain how the poles,the damping ratio ?, and the natural frequency ?n affect the step response
you obtained in P3.1b.
P3.2a: Considerthe same massdamperspring system (MBK system) shown in Problem P3.1a. Assume
the external applied force is zero, fa(t) = 0. Thenthe differential equation becomes,
y(t)+4
?y(t)+16y(t)
=0
Let
the
initial
conditions
bey(0)=0.01
m?y(0)
, =0 m/s.
Solve
and
plot
y(t)fort =0.
P3.2b: If the damper friction
equation will become
coefficient
Bis increased from 0.4 Ns/mto 1 Ns/m, then the differential
y(t)+10
?y(t)+16y(t)
=0
Findthe poles,the dampingratio ?, andthe naturalfrequency ?n ofthe system. Lettheinitial conditions
bey(0)=0.01
m?y(0)
, =0 m/s.
Solve
and
plot
y(t)fort =0.
P3.2c: Consider a special case with no damping, B = 0. In this case,the differential equation becomes,
y(t)+16y(t)
=0
Let
the
initial
conditions
bethesame:
y(0)=0.01
m
?y(0)
, =0 m/s.
Solve
and
plot
y(t)fort =0.
P3.3a: Consider a massdamperspring system with the same mass Mandidentical spring constant K as
those in P3.1a whilethe damper friction coefficient is changed to B = 0.08 Ns/m. Thenthe differential
equation of the system is,
y(t)+0.8 ?y(t)+16y(t)
= 10 fa(t)
Findthe transfer function Y(s)/Fa(s), the poles,the dampingratio ?, andthe naturalfrequency ?n of
the system.
P3.3b: Consider the same system asthe onein P3.3a. Drawthe Bode plot of the system with asymptotes
on the magnitude response graph and the corner frequency ?c on both of the magnitude response and
70
3
Linear Systems
the phase response
Analysis II
graphs.
P3.3c: Findthe magnitudeand phaseofthe systemY(s)/Fa(s), atthe following frequencies: ?= 0.1?c,
? = ?c, and ? = 10?c, respectively.
P3.3d: Find and plot the steady-stateresponseyss(t) of the system dueto eachof the following inputs:
(i) fa(t) =sin(0.1?ct), (ii) fa(t) =sin(?ct), and(iii) fa(t) =sin(10?ct), respectively.
P3.3e: Comment on the frequency-domain
behavior of the typical second-order system based on the
results obtained in P3.3d.
P3.4a: Consider a system with transfer function,
?2
n
Y(s)
G(s) =
U(s)
=
s2 +2??ns+?2n
where Y(s) and U(s) are the Laplace transforms of the output y(t) andthe input u(t), respectively. Let
the damping ratio and the natural frequency be ? = 0.6 and ?n = 1rad/s, respectively, and assumethe
input u(t)
= us(t), the unit step function.
Then Y(s) can be expressed as
Y(s) =
Find y(t), the inverse Laplacetransform
1
s(s2 +1.2s+1)
ofY(s),
and plot it using the
MATLAB command
plot.
P3.4b: Repeat Problem P3.4a for (i) ? = 1, (ii) ? = 1.4, and(iii) ? = 0.2, respectively, and compare the
rise time, overshoot, and settling time of the four time responses including the one from P3.4a.
P3.4c: Repeat Problem P3.4a for ?n = 2 rad/s, and compare the rise time, overshoot, and settling time
of the two time responses.
P3.5a: Consider a system with transfer function
G1(s) =
Y(s)
U(s)
=
?2
n
s2 +2??ns+?2n
Let ? = 0.6, ?n = 1rad/s. Usethe MATLAB command step to plot the stepresponseofthe system.
P3.5b:
Repeat
P3.5a
forthesystem
modified
byadding
aLHP
zero
ats =-1:
G2(s) =
Comment
on how the adding
Y(s)
U(s)
=
?2
n(s+1
s2 +2??ns+?2
LHP zero affects the rise time,
n
overshoot,
and settling time.
P3.5c: Repeat P3.5afor the system modified by adding a RHP zero at s = 1:
G3(s) =
Y(s)
U(s)
=
-?2
n(s-1)
s2 +2??ns+?2n
Comment on how the adding RHP zero affects the rise time, overshoot, and settling time.
3.8
P3.6a:
Consider a system
with transfer
Exercise
Problems
7
function
X(s)
R(s)
4
= G(s) =
s2 +3s+2
and let r(t) and x(t) bethe inverse Laplace transforms of R(s) and X(s), respectively. Find the differen-tial
equation corresponding to the transfer function G(s).
P3.6b: Find the characteristic equation and the poles of the system.
P3.6c: Find the damping ratio ?, the natural frequency
?n, andthe steady-state step response xssof the
system.
P3.6d: Is the system overdamped? critically damped? underdamped? or undamped?
P3.6e: Assumethe initial conditions x(0) = 0 and?x(0)
Find the output response x(t).
= 0, and let the input r(t) be a unit step function.
P3.6f: Plot the constant term, the two exponential terms of x(t) and the sum of these three terms versus
t on the same graph. Discuss how these two exponential terms relate to the system poles, and how the
constant term relates to xss.
P3.7a: Consider a system with transfer function
X(s)
R(s)
= G(s) =
20
s2
+2s+10
and let r(t) and x(t) bethe inverse Laplacetransform of R(s) and X(s), respectively. Find the differen-tial
equation corresponding to the transfer function G(s).
P3.7b: Find the characteristic equation and the poles of the system.
P3.7c: Find the damping ratio ?, the natural frequency
?n, andthe steady-state step response xssof the
system.
P3.7d: Is the system overdamped? critically damped? underdamped? or undamped?
P3.7e: Assumetheinitial conditions x(0) = 0 and?x(0) = 0,andlet theinput r(t) bea unit stepfunction.
Find the output response x(t).
P3.7f: Plot the constant term, the decaying sinusoidal term of x(t) andthe sum of these two terms versus
t onthe same graph. Discuss how the decaying sinusoidal term relates to the system poles, and how the
constant term relates to xss.
P3.8a: Consider a system with transfer function
X(s)
R(s)
= G(s) =
xss?2
n
s2 +2??ns+?2n
where0 < ? < 1. Letr(t) and x(t) bethe inverse Laplacetransforms of R(s) and X(s),respectively.
Assume r(t) is a unit step function, and the step response of the system can be found as,
72
3
x(t)
Linear Systems
= xss
Analysis II
? 1-?2
1-?
e-at
sin(?t
+?)
?
, where a=??n,
? = ?n
?1-?2,? =cos-1?
(3.90)
Let ? = 0.4, ?n = 1rad/s, xss = 1. Findthe differential equation correspondingto the transfer function
G(s).
P3.8b:
Find the characteristic
equation and the poles of the system.
P3.8c: Assumethe initial conditions x(0) = 0 and?x(0)
Find the output response x(t) using Equation 3.90.
= 0, and let the input r(t) be a unit step function.
P3.8d: Verifythe value x(0) = 0 and?x(0) = 0 usingthe result of P3.8c,lim x(t) and limx ?(t).
t?0
P3.8e: Verify the steady-state step response lim x(t),
which should be xss.
t?8
P3.8f: Computethe sinusoidal oscillation frequency
t?
? = ?n
step response x(t).
?1-?2andtheperiodT=2p/?ofthe
P3.8g: Computethe peaktime tp = p/? at whichthe maximumovershootoccurs. Notethat tp is half
of the period T.
) for
P3.8h: Compute
the valuesof x(ktp) =xss(1+e-aktp)fork =1,3,5
andx(ktp)=xss(1-e-aktp
k = 2,4,6. These values provide the information
step response x(t).
P3.8i: Usethe above information
graph x(t) versus t.
regarding the locations of the peaks and valleys of the
obtained from the results of P3.8d, e, f, g, hto construct the step re-sponse
P3.8j: Verify the step response graph using
P3.8c.
MATLAB command
plot
to plot the x(t) obtained in
P3.8k: Usethe above step response graph x(t) to compute the maximum overshoot according to the
following
formula:
OS=(x(tp)-xss)/xss.
P3.8l: Usethe formula OS = e-?p/
result of P3.8k.
v1-?2 to
compute the
maximum
overshoot,
and verify it
with the
4
Modelingof MechanicalSystems
I
Nthe previous two chapters wehave employed some simple physical systems: mass-friction system,
RClow-pass filter, DC motor, massdamperspring system, resisterinductorcapacitor
circuit, and
motor position control system, to demonstrate fundamental systems and control concept, analy-sis,
and design. It is clear that a meaningful control system design begins with a trustworthy dynamics
model of the physical system to be controlled. In this chapter, we will focus on the dynamics model-ing
of fundamental translational and rotational mechanical systems using the Newtonian approach and
the
Lagrange
approach.
Both approaches
eventually
will lead to exactly the same dynamic
model equa-tions.
However,they have completely different road mapsto reach the same destination. The Newtonian
approach involves the geometry and vector relationship of each component in the mechanical sys-tem.
On the other
the potential
hand, the
energy, the
Lagrange
power
approach
dissipation,
only
needs the information
and the external
forces,
of the
but it requires
kinetic
energy,
differential
cal-culus
computations.
Both have advantages and disadvantages,
but interestingly
they complement
each other perfectly: the
disadvantage of one is the advantage of the other, and vice versa. The Newtonian approach is moreintu-itive,
providing
clear picture
process can be tedious,
of the interactions
especially
for larger
among components
multi-body
systems.
inside the system,
but the
modeling
Conversely, the Lagrange approach is
moreelegant without the needto worry about the directions of vectors and detailed interconnection of
components, but it allows virtually no insight of the interactions among components within the system.
The Lagrange approach is not included in mostof the undergraduate control and dynamics textbooks.
Part of the reasons
approach together
more learning
may be due to its lack
with the
hours.
Newtonian
We will apply
of intuitiveness
approach
and insight.
Actually, learning
will be more effective,
both approaches to almost all
the
Lagrange
and will not necessarily
require
mechanical systems in this
chapter
so that the students can learn both approaches together and compare their results. It is easy, especially
for the
beginners, to
miss hidden reaction
It is also very possible to
forces
or commit
misssome energy and power
sign errors using the
dissipation
terms
Newtonian
or commit
differential
approach.
calculus
computation errors using the Lagrange approach. However, it is rare for these two approaches to
have same errors in their end results. Hence, employing both approaches for system modeling not
only enables the students to understand the system better from two different prospectives, but also
provides additional assurance on modeling correctness.
4.1 Translational
MechanicalSystems
To analyze the behavior or to obtain a mathematical model of a translational
first
step is to
understand the force-velocity
or the force-displacement
mechanical system, the
relationship
of the three fun
74
4
damental
Modeling of
Mechanical
translational
Systems
mechanical elementsdamper,
spring,
and
mass. As shown in
Figure
4.1, the
force-velocity
relationship
ofthedamper
isffri =-Bv=-B
?x,
in which
thenegative
signmeans
that
the viscous friction
f, velocity
and
force is always in the opposite
v, displacement
Ns/m, respectively.
x, and friction
direction
coefficient
The force-displacement
of the
velocity.
B are Newton (N),
relationship
The units for the force
meter/sec (m/s),
of the spring is described
meter (m),
by Hookes law,
fspr=-Kx,where
thespring
force
isproportional
tothemagnitude
ofthedisplacement
but
itsdirection
is in the opposite
direction
The force-velocity
of the displacement.
The unit of the spring
or the force-displacement
relationship
of the
constant
Kis
massis described
N/m.
by Newtons
sec-ond
lawof motion,
fnet= M
?v=M
x = -fine,
where
fnet
and
fine
are
thephysical
net
force
and
the
Mis kilogram (kg). Notethat the physical net
inertial force, respectively. The unit of the massor inertia
force needsto havethe same magnitude of the inertial force to keep the mass M moving with acceler-ation
x. The magnitude of the inertial force is proportional to the acceleration. Strictly speaking, the
inertial force is not a physical force. It is even called asfictitious force in some dynamics books, yet
it plays an essential role leading to the key concept of the famous dAlemberts principle, which will
be addressed in the next subsection. The characteristics of these three translational
elements
are summarized
in Figure 4.1.
Fig. 4.1: Basictranslational
4.1.1
dAlemberts
mechanical system
mechanical system elements.
Principle
Recall that in the beginning
of
Chapter
3, we considered
a simple
massdamperspring
mechanical
system, which is shown again here on the left-hand side of Figure 4.2. The mathematical model of this
simple mechanical system can be derived based on Newtons second law of motion, fnet = x,
M
where
fnetis the net physical force, Mis the massorinertia, andx is the acceleration, which welearned from
physics
courses in high school.
For the simple
MBK system shown in
Figure 4.2, fnet is the physical
net force applied to the mass M, whichis the algebraic sum of the applied force fa(t), the spring force
-Kx(t),and
thedamper
friction
force-B
?x(t),andtherefore
wehave
thefollowing:
fnet=fa +(-B
?x)+(-Kx)=xM
This is exactly the same governing
dynamics
equation
of the
MBK system
(4.1)
we had in
Equation
3.1
4.1
Fig. 4.2:
Asimple
The direct
simple
massdamperspring
Newtons
systems,
but
f
=
(MBK)
Ma method
will become
Translational
Mechanical
Systems
system and the associated free-body
of assembling
more confusing
and
the
equation
difficult
to
of
motion
manage for
75
diagram.
works
well for
complicated
multi-body
systems. In a multi-body system, its subsystems may be interconnected each other to con-strain
relative motions in a complicated way. This disadvantage can be eliminated by employing
dAlemberts
principle that allows the system to be systematically partitioned into several isolated free
bodies to
make it easier to assemble the equations
times acceleration,
of acceleration.
of
motion.
x,
M as simply another force called the inertial
Theorem 4.1 (dAlemberts
Assume all the inertial
The central idea is to consider the
mass
force that acts opposite to the direction
Principle)
forces
and physical forces in the system are considered.
Then the
algebraic
sum ofall forces acting at any point in the system is zero.
fsum =?
fi
=0
(4.2)
i
Notethat this principle can be applied in each admissible direction of motion. The significance of
dAlemberts
principle
is far greater than a simple
depth by Lanczos [Lanczos,
For example, the algebraic
Min the
sum of the forces
MBK system shown in
directions
restatement
of Newtons
second law, as explained
Figure
acting
onto a point between the spring
4.2 is zero since the two forces
on this
with equal magnitude of Kx. For another example, if the term
from the right-hand
in
1970].
side of the equation to the left-hand
point
K and the
xM in Equation 4.1 is
side, the equation
mass
are in opposite
moved
will become the following:
fa +(-B
?x)+(-Kx)+(-M
x) =0
(4.3)
This
equation
fulfills
dAlemberts
principle
iftheterm-M
x isconsidered
asaforce.
Equation
4.3 is
mathematically
equivalent
to
Equation
4.1,
which
was derived from the
well-known
Newtons second law of motion, but the significance of dAlemberts
principle is more than just a
restatement of Newtons laws of motion. It provides an important
dynamic equilibrium concept
that allows complicated
multi-body system dynamics to be broken down into several free-body
diagrams and makeit much easier to construct the dynamics modelfor mechanical systems.
Remark
4.2 (Dynamic
For any
zero there
will remain
Equilibrium)
mechanical system, if the algebraic
would be no acceleration
at rest if the initial
sum of all the physical
forces
acting
on the system is
for the system, and the system is said to be at equilibrium
condition
of the system is at rest.
Now, for the
since it
MBK system, the algebrai
76
4
Modeling of
sum of all the three
nonzero net force
equilibrium
Mechanical
Systems
physical forces,
which is the net force,
will keep the massin
acting
on the system is usually
motion with the acceleration x(t);
nonzero.
The
hence,the system is not at
in the general sense of static equilibrium.
However, according to Equation 4.3 or dAlemberts
principle, the algebraic sum of all the four
forces
acting
onthesystem
iszero
iftheinertial
force
term,-M
x(t),isconsidered
asaforce.
Then
the system is at an equilibrium,
a special
kind
of equilibrium
called
dynamic
equilibrium,
for time
t =0.Although
x(t)may
change,
thealgebraic
sum
ofthese
fourforces
remains
zero
allthetime.
The dynamic
equilibrium
complicated
multi-body
concept
has madeit easier to construct the equations
of motion, especially
for
systems.
Example 4.3 (Newtonian
Approach
Modeling of the Simple
MBK System)
Considerthe massdamperspring (MBK) system in Figure 4.2, wherefa(t) is the applied force and
the variable of interest is the displacement of massx(t). The origin reference (the position of x = 0)
is chosen at the equilibrium
condition
when the spring is
neither stretched
nor compressed,
reference of direction of motion is assumed positive if it points to the right.
and the
For example, a positive
displacement x(t) meansthe massis at the right-hand side of the equilibrium position and a negative ve-locity
?x(t) indicates that the massis movingtowards the left. Now,the objective is to find the dynamics
model of the system,
which is a differential
equation
describing
the relationship
and the output x(t), using the Newtonian approach and dAlemberts
between the input
fa(t)
principle.
Solution:
In general, for a complicated system, it
of
motion for the system.
The free-body
diagram is drawn
mayrequire several free-body diagrams to assemble the equa-tions
But for the simple
MBK system,
based on the schematic
diagram
only one free-body
of the
diagram is needed.
MBK system shown
on the left
of Figure 4.2. It can be seenthat there are three physical forces acting onthe mass M. Sincethe applied
force fa(t)
is assumed to point to the right
should be drawn on the free-body
when it is positive,
diagram. Note that
a right
directional
arrow
both the arrow direction
with label
fa(t)
and the label sign
arerelevant.
Aleftdirectional
arrow
together
with
anegative
sign
label-fa(t)can
also
represent
the
applied
forceonthefree-body
diagram.
Since
both
thefriction
force-B
?x(t)andthespring
force
-Kx(t)have
negative
signs,
theyshould
berepresented
byleftdirectional
arrows
with
labels
B
?x(t)
and Kx(t), respectively,
on the free-body
diagram.
Since
theinertial
force-M
x(t)isnotaphysical
force,
itisrepresented
byadashed-line
arrow
instead of a solid-line arrow on the free-body
therefore, its representation on the free-body
diagram. Theinertial force also has a negative sign;
diagram should be a left directional dashed-line ar-row
withlabel x(t).
M
Now, with the completed free-body diagram in Figure 4.2 and dAlemberts
sum of the four forces
on
M should
be zero, and
we have the following
principle, the algebraic
dynamics
model of the
massdamperspring
system:
x(t)+B
M
?x(t)+Kx(t)
Remark 4.4 (Tips to Avoid Errors in Free-Body
The key step in the
make mistakes on the
Newtonian
=fa(t)
Diagrams)
approach is to draw a correct free-body
magnitude of the forces,
(4.4)
but it takes
only one force
diagram.
direction
Students seldom
or sign error in the
free-body diagram to void the effort of constructing the dynamics model of the system. The first tip
to avoid confusion and mistakesis to follow conventional rules if they are available. For example, th
4.1
Cartesian coordinate
system
with the origin located
Translational
at an equilibrium
Mechanical
position
Systems
can be employed
77
to serve
as a reference for all vectors in the system sothat a positive horizontal vector is pointing to the right and
a positive
vertical
vector is in the up direction.
The second tip is not to have negative sign on any label so that the direction of each vector on the
free-body
diagram is purely
4.2, the positive
determined
motion direction
by its arrow
direction.
For the
is assumed to point to the right
MBK system shown in
and all labels
Figure
of the four force vectors,
fa(t), B
?x(t), Kx(t), and x(t)
M
are with no negative sign attached to them. Therefore, the vector arrow
of fa(t) points to the right, but the vector arrows of the other three forces arein the left direction.
This
mechanical system dynamics
model construction
approach is a Newtonian
approach,
which is
mainly based on detailed geometry and interactions among the component vectors. The Lagrange ap-proach,
on the other hand, is basically
a scalar function
approach involving
energy and power functions
without the need of knowing the detailed system configuration and interactions among the components.
As mentionedin the beginning of this chapter, these two approaches actually complement each other
very well. Learning these two approaches at the same time certainly will enhance students learn-ing
experience. Both Newtonian and Lagrange approaches will be employed to derive the equations of
motionfor almost every mechanical systems discussed in this chapter.
4.1.2 A Brief Introduction
of the Lagrange Approach
Unlike the Newtonian approach, the Lagrange approach is a scalar-and
energy-based approach
requiring the kinetic energy function, the potential energy functions, and the dissipated power
functions. It does not need free-body diagrams, but it requires differential calculus computations in-volving
the following Lagrange equation:
??
?L
d
??q
dt
?D
?L
-
+
?q
?q
= Q
(4.5)
In the equation, qand?q are generalizedcoordinate vector and generalizedvelocity vector,respectively.
L(?q,q) representsthe Lagrangian function, whichis defined as
L(
?q,q)=T(
?q,q)-U(q)
(4.6)
where T(?q,q) is the kinetic energy function of the system, and U(q) is the potential energy function.
D(
?q) is the dissipated power function of the system, and Q is the generalized external force vector
acting on the system.
Example 4.5 (Lagranges
Approach
Modeling of the Simple
MBK System)
Consider the same simple MBK dynamics modeling problem shown in Example 4.3 but find the
dynamics model of the system using the Lagrange approach.
Solution:
For the simple
MBK system shown
on the left-hand
side of Figure
4.2, the
generalized
coordinate
and
velocity are simply x and?x. The kinetic energy of the system is stored in the massas T = 0.5M?x2, the
potential
energy of the system is stored in the spring
as U = 0.5Kx2,
and the power
dissipated
in the
damper
is D=0.5B
?x2.
Hence,
theLagrangian
function
is L=T-U=0.5M
?x2-0.5Kx2.
Then
the
key differentials
in the
Lagrange equation can be found
?L
?x =
?L
?x =
)
?(0.5M
?x2-0.5Kx2
=M?x,
?x
)
?(0.5M
?x2-0.5Kx2
?x
as,
?D
?x =
?(0.5B?x2 )
??x
=-Kx, Q=fa
=B
?x
(4.7
78
4
Modeling of
Now, from
Equation
d
dt
which is exactly
Remark
Mechanical
4.5, the
Lagrange equation,
??
?L
??x
?L
-
working
?x
we have the dynamics
model equation:
=Q ? M
x(t)+Kx(t)+B
?x(t)
=fa(t)
obtained
using the
Newtonian
approach
shown in
(4.8)
Equation
4.4.
Between the Two Approaches)
on the same
Lagrange approaches,
?D
+
?x
the same as that
4.6 (Differences
After
Systems
modeling problem
of the simple
we have observed some differences
MBK system
between the two
using the
Newtonian
approaches.
and
The Newtonian
approach relies onthe free-body diagram that depicts the force interactions among the components and
the applied force.
Onthe other hand, the
of the system interact
Lagrange approach
with each other. Instead,
does not need to know how the components
it only requires
the information
of the
kinetic
energy
function, the potential energy function, the power dissipation function, and the external force. These
two approaches are very different, but amazingly the end results are identical.
4.1.3 A Quarter-Car
A simplified
of the
Suspension System
quarter-car
suspension
system is shown in Figure
mass of the car body, B2is the friction
coefficient
4.3. In the figure,
M2represents
of the shock absorber (damper),
quarter
K2is the spring
constant of the strut, M1represents the massof the wheel, and K1is the spring constant of the tire. The
displacements, x2(t), x1(t), and d(t) of the masses, M2, M1,and the tire, respectively, are measured
from their equilibrium conditions. The equilibrium conditions are assumed not affected by gravity,
and both springs K2 and K1 are neither stretched nor compressed. Thus,the gravity forces will be in-corporated
in the free-body diagram and appear on the dynamics model.In case that the equilibrium
positions are chosento offset according to the gravity effect, then the two gravity forces M1gand M2g
should not appear on the free-body diagram or the dynamics model.
The car is assumed moving to the right, and it may encounter uneven road conditions with dis-turbance
d(t). The objective of the study here is to find the dynamics model of the system that
describes how the disturbance input d(t) would affect the output variables x1(t) and x2(t). The
dynamics model also can be employed to choose the desired values of the strut spring constant K2
and the damper friction coefficient B2so that the suspension system has a desired performance.
Example 4.7 (Newtonian
Consider the
we follow
the
quarter-car
Approach
Modeling of the Quarter-Car
suspension
Cartesian coordinate
system shown
convention
Suspension System)
on the right-hand
to choose the up direction
side of Figure 4.3.
as the positive
Note that
direction
for all
vectors in the system and that the displacements x1(t), x2(t), and d(t) are measuredfrom their respective
equilibrium
positions.
Find the dynamics
model of the system,
describing the relationship between the input
Newtonian approach and dAlemberts
principle.
which is a set of differential
equa-tions
fa(t) and the outputs x1(t) and x2(t), using the
Solution:
Since there
free-body
are two
diagram
masses, M1 and
on the top and the
M2,the free-body
M1free-body
Figure 4.3. There are four forces acting on the
diagram is broken
diagram at bottom,
down into two
as shown
parts: the
on the left-hand
M2
side of
mass M2and five forces acting on the mass M1.
Thetwo gravitational forces ofM2 and M1always point down to the center of the Earth; hence,they
are represented
by the two
downward
arrows
with labels
M2g and
M1g, respectively,
on the top and th
4.1
bottom free-body
diagrams.
The inertia
forces
up direction, so they are represented
for
M2 and
A quarter-car
Based on dAlemberts
spring
K2 should
suspension
principle
have the same
M1 are opposite
by the two downward
M2
x2 and M1
x1 on the top and the bottom free-body
Fig. 4.3:
Translational
Mechanical
79
to the conventional
arrows, respectively,
pos-itive
with labels
diagrams.
system and the associated free-body
or Newtons third
Systems
law of
magnitude but in opposite
diagram.
motion, the two forces
directions.
at both ends of the
If x2 = x1, the spring is neither
stretched nor compressed, and the forces at both ends are zero. If x2 > x1, the spring will be stretched
but the recoil spring force will resist the elongation of the spring. In other words, the same spring
forceK2(x2
-x1)willpullM2
down
andat hesame
timepullupM1.
Similarly,
thesame
friction
force
B2(
?x2-?x1)
isresisting
themotions
ofM1
andM2
from
increasing
thegap
between
them.
For
the
spring force on K1,if x1 > d, the spring will stretch and the recoil spring force at the lower end of M1
willgodownward
with
the
labelK1(x1
-d).
Now, with the completed free-body diagrams onthe left-hand side of Figure 4.3, we will obtain two
differential
equations
by summing
up the four forces
on
M2and the five forces
on
M1,
M1
x-B2(
?x2-?x1)-K2(x2
-x1)+K1(x1
-d)+M1g
=0
M2
x+B2(
?x2
-?x1)+K2(x2
-x1)+M2g
=0
which can be rewritten in the following
(4.9)
matrix form:
? ????B2-B2????K1+K2-K2????? ? ?
M1 0
x1
0 M2 x2
+
Example 4.8 (Lagranges
Consider
the same
-B2 B2
Approach
quarter-car
4.7, but find the dynamics
x?1
x?2
+
-K2 K2
x1
x2
=
Modeling of the Quarter-Car
suspension
system
model of the system
dynamics
using the
K1
0
d+
-M1 g
-M2
(4.10)
Suspension System)
modeling
Lagrange
problem
shown in
Ex-ample
approach.
Solution:
Forthe quarter-car suspension system shown onthe right-hand side of Figure 4.3, the generalized coor-dinate
and velocity
are ? x1 x2
?T ?T
and ?x?1x?2
, respectively.
The kinetic
energy of the system is store
80
4
in the two
Modeling of
Mechanical
Systems
1 +M2
?x2
2), the potential energy of the system is stored in
masses M1and M2as T = 0.5(M1?x2
thesprings
K1,K2and
inthegravity
fieldasU=0.5[K2(x2
-x1)2
+K1(x1
-d)2]+M1gx1
+M2gx2,
and
thepower
dissipated
inthedamper
B2
is D=0.5B2
(?x1-?x2)2.
Hence,
theLagrangian
function
is
1 +M2
?x2
L= T-U =0.5(M1
?x2
2)-0.5[K2(x2
-x1)2+K1(x1
-d)2]-M1gx1
-M2gx2 (4.11)
Then the key differentials
in the Lagrange equation
can be found
as,
??B2(?x1-?x2)?
?? ? ??D??
??L??
?x1
?x1
-x1)-K1(x1
-d)-M1g
?
??K2(x2-K2(x2
??L??x1
-x1)-M2g
B2(
?x1-?x2)???
-x1)-K1(x1
-d)-M1g
? ??K2(x2-K2(x2
??-B2(
?x1-?x2)
-x1)-M2g
B2-B2??
? ????-B2B2
??K1-K2+K2-K2K2????? ?-M2?
?
?L
?x =
?L
?L
?x = ?L
? x2
?
?x2
M1
?x1
,
M2
?x2
=
?
?D
??x =
?D ??x2
=
-B2(
?x1-?x2)
,
=
(4.12
Q =0
Then from the Lagrange equation Equation 4.5 we have the dynamics model equation:
M1
x1
M2
x2
0
-
which can be rewritten in the following
x1
M2 x2
+
(4.13)
0
matrix form:
x?1
x?2
M1 0
0
=
+
x1
+
x2
K1
=
0
-M1 g
d+
(4.14)
This is exactly the same asthat obtained using the Newtonian approach shown in Equation 4.10.
Remark 4.9 (Equilibrium
Offset Dueto Gravity)
The differential equations in Equation 4.10 or Equation 4.14 can be considered as a special case
of the generalized
a constant
MBK system matrix equation of this form,
gravitational
force
vector
and Fd is a disturbance
x+B
M
?x+Kx
input
vector.
= Fd +Fg, where Fg is
Note that the equilibrium
occurs
atxe=K-1Fg.
Define
x =x-K-1Fg,
then
thegeneralized
MBK
matrix
equation
willbecome
x+B ?x+Kx = Fdandthe equilibrium occursatxe = 0.
M
4.2 Rotational MechanicalSystems
As engineers, we mayencounter morerotational systems than translational systems since almost all of
the translational motions are driven by rotational systems like wheels, pulleys, motors, and engines. Just
like the translational
mechanical system, the rotational mechanical system also has three fundamental
rotational mechanical elements: rotational damper, torsional spring, and moment of inertia. Asshown in
Figure
4.4,
the
torque-angular
velocity
relationship
oftherotational
damper
istfri=-B?=-B
??,in
which the negative sign
angular
velocity
coefficient
meansthat the viscous friction
torque is always in the opposite
direction
of the
??. The unitsfor the torque t, angular velocity ?, angular displacement ?, andfriction
B are Newton (N-m), radian/sec (rad/s), radian (rad), and Nms/rad,respectively. Thetorque-angular
displacement
relationship
ofthe
torsional
spring
isdescribed
byHookes
lawtspr=-K?,where
the spring torque is proportional to the magnitude of the angular displacement ? but its direction is in
the opposite rotation
direction
The unit of the torsional
of the angular
spring
displacement relationship
displacement.
coefficient
of the
Kis
Nm/rad. The torque-angular
velocity
or the torque-angular
moment of inertia is described by Newtons second law of
motion,
tnet=J??=J? =-tine,
where
tnet
and
tine
are
thenet
torque
and
the
inertial
torque,
respec-tively.
Note that the
net torque
needs to
have the same
magnitude
of the inertial
torque to
move the
? . The magnitude of the inertial torque is proportional to
the angular acceleration ? . The unit of the moment of inertia J is kgm2 or Nms2/rad. The characteristics
moment of inertia
J with angular acceleration
of these three rotational
mechanical system elements are summarized
in Figure 4.4.
4.2
Fig. 4.4: Basic rotational
4.2.1
dAlemberts
Principle
in
Rotational
Rotational
Mechanical
Systems
81
mechanical system elements.
Systems
It can be easily seen that there exists an analogy between the rotational system and the translational
system.
That meansthe concept and the dynamics
can be applied to rotational
pendulum
system in
systems
Figure 4.5 is
modeling approaches
with little
mathematically
modifications.
equivalent
welearned in the previous sec-tion
We will see that the single torsional
to the simple
MBK system in Figure
4.2.
In addition, these two systems are analogous to each other.
Fig. 4.5: Asingle torsional pendulum.
Recallthat the rotational version of Newtons Second Law of Motionis described by the equation,
tnet = Ja = J?, wheretnetis the net physical torque, J is the momentof inertia, and a =? is the
angularacceleration.Forthe singletorsional pendulumsystem shownin Figure 4.5, tnetis the physical
nettorque appliedto the momentofinertia J, whichis the algebraicsum ofthe appliedtorque ta(t), the
spring
torque
-K?(t),
and
thefriction
torque
-B
??(t),
and
therefore
we
have
thefollowing:
(4.15)
tnet=ta +(-B
??)+(-K?)=J?
This equation is analogous to the governing
dynamics
equation
of the
MBK system
we had in
Equation
3.1.
The direct (Newtons t = Ja) method of assembling the equation of motion works well for
simple systems, but will become more complicated and difficult to managefor large multi-bod
82
4
systems.
Modeling of
Mechanical
This disadvantage
Systems
can
be eliminated
it easier to assemble the equations
of
if the
dAlemberts
motion. The rotational
principle
is employed
version of dAlemberts
to
make
principle is
described in the following.
Theorem 4.10 (dAlemberts
Assume all the inertial
sum ofall torques
Principle for
torques
acting
Rotational Systems)
and physical torques in the system are considered.
at any point in a multi-body
Then, the algebraic
system is zero.
tsum =? ti = 0
(4.16)
i
Forthe singletorsional pendulumsystem,if theterm J? in Equation4.15is moved
fromthe right-hand
side of the equation to the left-hand side, the equation will become the following:
ta+(-B
??)+(-K?)+(-J
?) =0
(4.17)
This
equation
fulfills
dAlemberts
principle
iftheterm-J
? isconsidered
asatorque.
As in the translational
dynamic
equilibrium
down into
that
several
systems,
allows
dAlemberts
complicated
free-body
diagrams
principle
multi-body
and
provides
rotational
makes it
an important
system
concept
dynamics
much easier to construct
to
the
of
be bro-ken
dynamics
model.
Example 4.11 (Modeling
of the Single Torsional Pendulum System by Newtonian Approach)
The rotational mechanical system shown in Figure 4.5 is a single torsional pendulum. Thelink be-tween
the disk, which has moment of inertia J, and the fixture at top is a flexible shaft with torsional
spring constant
K.
Whenthe system is in
motion, there is a viscous friction
torque
with friction
coeffi-cient
B. The positive rotation direction is chosen according to the right-hand rule with the thumb
pointing down at the center of the disk, or it is in the clockwise direction viewing from the top.
The angular displacement ? of the disk J is measuredfrom the equilibrium position at which the
torsional
spring
torque
is zero. In this example,
will be employed to find the dynamics
the
Newtonian
approach
with dAlemberts
model of the system, which is a differential
prin-ciple
equation
describing the relationship betweenthe input ta(t) and the output ?(t).
Solution:
The free-body
diagram is shown on the right-hand
view of the disk.
The positive rotation
side of Figure 4.5. This diagram
direction
is in the clockwise
direction.
shows the same top
This free-body
diagram
should
consist
offour
torques:
theapplied
torque
ta,thefriction
torque
tfri=-B
??,
thespring
torque
tspr
=
-K?,
and
the
inertial
torque
tine
=-J
?
.
The
applied
torque
taobviously
isinthe
clockwise
direction, butthe otherthree torques are all withnegativesigns, asshown in Figure 4.4. Therefore, B
??,
K?, and J?
should all be in the counterclockwise
Weutilize the signs in the torque
The torque direction
direction.
equations to determine the torque
determination
can also be validated
directions
in the free-body
dia-gram.
by some simple virtual experiment,
for example,if ? is positive, which meansthe disk has beenrotated clockwise away from equilibrium.
Just imagine
torque
from
your
hand is holding
the torsional
spring
the
disk at this
attempting
displaced
position;
you
to go back to the equilibrium,
spring torque is indeed in the opposite direction
would feel the recoiled
which verifies the tor-sional
of ?. Similarly, the friction torque is always
in the oppositedirectionofthe angularvelocity;hence,B
??shouldbein the counterclockwise
direction
4.2
Rotational
Mechanical
Systems
83
The
negative
inertial
torque,
-tine
=J? and
itsdirection
onthefree-body
diagram
may
beconfusing.
The confusionis understandablesince J? actually is equalto the nettorque required to perform the
motion to move J with acceleration ? toward the same direction of ? according to Newtons
second law of motion. Then why,in the free-body diagram, is ? shown in the opposite direction of ??
rotational
This confusion can beresolved using the following
see that the following
equations
explanation of dAlemberts
principle. It is easy to
are correct:
tnet(t) = J?(t)
??(t)-K?(t)-J
?(t)=0
tnet
(t) =ta(t)-B
??(t)-K?(t) ? ta(t)-B
The above equation
equation
on the right
on the left is obviously
verifies
dAlemberts
the result
Principle
of
Newtons
second law
that the algebraic
of
(4.18)
motion, and the
sum of all torques
acting
on J
is zero.Thisequationclearlyshowsthat J? is in the samedirectionwith B
?? and K?. Therefore,
onthe free-body diagram, J? is alwaysin the oppositedirection of the assumedpositiverotation
direction. Forthis example,J? shouldbein the counterclockwise
direction.
Once the free-body
diagram is completed,
to see that the sum of the torques
as shown
on the right-hand
in the counterclockwise
direction
side of Figure 4.5, it is easy
should
be equal to the sum of the
torques in the clockwise direction:
J?(t)+B ??(t)+K?(t)
=ta(t)
(4.19)
which is the dynamics model equation for the single torsional pendulum system shown on the left-hand
side of Figure 4.5. Notethat the equation is mathematically equivalent and analogous to that of the
simple massdamperspring
system shown in Figure 4.2.
4.2.2 The Lagrange Approach for
Rotational
As discussed in the previous section, the
Systems
Newtonian
and the Lagrange approaches
have their
advantages
and disadvantages, but they complement each other very well. Students are urged to apply these two ap-proaches
to every
mechanical system
dynamics
modeling problem.
experience,
but also greatly reduces the probability
approaches
are very differentone
relies
of obtaining
The application
Lagrange
approach
system that involves
may be less complicated
geometrical
modeling results.
relationship
These two
and the other
Therefore, it is rare for these two approaches to
of the Lagrange approach to the rotational
For a more complicated
incorrect
on the vectors and their
is based on the scalar energy and power functions.
lead to same errors.
Doing so not only enhances learning
system is similar to the translational
both translational
than
the
Newtonian
and rotational
approach
since the
approach requires separate free-body diagrams for translational
and rotational
for the Lagrange approach the rotational energy functions and the translational
can be simply combined together.
sys-tem.
subsystems,
the
Newtonian
subsystems while
energy functions
The Lagrange equation is basically the same as follows:
d
dt
??
?L
??q
-
?L
?q
The only difference is that the generalized coordinate
will be rotational
system
variables
?D
+
?q
= Q
(4.20)
q vector and the generalized velocity ?q vector
84
4
Example
Modeling of
Mechanical
4.12 (Modeling
In this example,
Systems
of the Single
Torsional
we will employ the
Pendulum
Lagrange approach
by the
Lagrange
Approach)
to find the dynamics
model for the same
single torsional pendulum system considered in Example 4.11.
Solution:
For the single torsional pendulum system shown on the left-hand side of Figure 4.5, the generalized
coordinate and velocity are ? and ??, respectively. The kinetic energy of the system is stored in the mo-ment
ofinertia J as T =0.5J
??2,the potentialenergyofthe systemis storedin thetorsionalspring Kas
U = 0.5K?2, and the power dissipated with friction
coefficient Bis
D= 0.5B
??2. Hence,the Lagrangian
function
is L =T-U=0.5J
??2-0.5K?2.
The
key
differentials
in Lagranges
equation
can
befound
as,
?
=
?
?
J
?,
?D
?
=
?
?
0.5B
??2??
?L?=??0.5J
??2
-0.5K?2??
??=-K?, Q=ta
?L?=??0.5J
??2
-0.5K?2??
? ? =Q ? J?(t)+B??(t)+K?(t)
=ta(t)
?
?
?
?
?
??? = B
??
?
(4.21)
Then from Lagranges equation we havethe dynamics
d
dt
?L
?L
-
??
??
model equation in the following,
?D
+
(4.22)
???
which is exactly the same asthat obtained using the Newtonian approach shown in Equation 4.19.
4.2.3 A Two-Rotor/One-Shaft
Atwo-rotor/one-shaft
Rotational System
rotational
system is shown in Figure 4.6. In the figure,
J1 and J2 represent the two
rotors with moment of inertia J1 and J2, respectively. B1and B2 arethe rotational friction
coefficients of
the two rotors, respectively.
(the torsional
These two rotors
are connected
with a shaft
whose stiffness
spring constant) is K. The positive rotation direction is chosen according to the
with the thumb pointing from the left side of the figure at the center of the two
clockwise direction viewing from the left of the figure. The angular displacements
two rotors are measured respectively from the equilibrium
positions at which the
torque is zero.
Fig. 4.6: Atwo-rotor/one-shaft
rotational system
right-hand rule
rotors, or in the
?1 and ?2 of the
torsional spring
4.2
Example
4.13 (The
Two-Rotor/One-Shaft
Consider the two-rotor/one-shaft
Rotational
rotational
System
system shown
Rotational
by
Mechanical
Newtonian
Systems
Approach
85
)
on the left side of Figure 4.6. In this exam-ple,
the Newtonian approach with dAlemberts principle will be employed to find the dynamics model
of the system, whichis a set of differential equations that govern the relationship betweenthe input ta(t)
and the two outputs ?1(t) and ?2(t).
Solution:
The free-body
diagram is shown
on the right-hand
side of Figure
4.6.
This diagram
shows the same
left side view of the two rotors. The positive rotation direction for all vectors in the system is in
the clockwise direction viewed from the left side. Sincethere are two rotors, J1 and J2,the free-body
diagram is broken
down into two
free-body
on the right,
diagram
parts side by side: the
as shown
J1 free-body
on the right-hand
diagram
on the left
and the J2
side of Figure 4.6.
There are four torques acting on the rotor J1, and three torques on the rotor J2. Since the ap-plied
torque ta is in the chosen positive rotation direction when it is positive, it is represented by a
clockwise (CW) arrow withlabel ta(t) onthe J1 free-body diagram. Thefriction torques B1
??1 and
B2
??2are alwaysagainsttheir respectiverotor motion;hence,they arerepresentedby counterclock-wise
(CCW) arrows, respectively, on the J1 and the J2 free-body diagrams. For the same reason, the
inertial torquesJ1
?1 andJ2
?2 are drawnas CCWdashedarrowsonthe J1 andthe J2free-body
diagrams, respectively.
According to dAlemberts principle or Newtons third law of motion, the spring torques at both
ends of the torsional spring Kshould have the same magnitude but in opposite rotation directions.
These
twoare
labeled
thesame
with
K(?1
-?2),
buttheir
rotation
direction
areoppositethe
onewith
the J1 free-body
diagram is
CCW and the other
with J2 free-body
diagram is in
CW. The reason that
K(?1
-?2)
is CCW
ontheJ1rotor
isexplained
asfollows:
IfthelabelK(?1
-?2)
ispositive,
(i.e.,
?1 > ?2), the torsional spring Kis twisted clockwise, which will cause the spring to recoil. Hence,
the recoiled spring torque is CCW, in the opposite rotation direction. Notethat if the spring torques are
labeled
differently
asK(?2
-?1),
then
their
rotation
directions
onthefree-body
diagrams
need
tobe
reversed!
Now the free-body diagrams are completed, as shown on the right-hand side of Figure 4.6, and we
can obtain two
the three torques
differential
equations
by summing
on the J2 free-body
up the four torques
on the J1 free-body
diagram
J1
?1+B1
??1
+K(?1
-?2)=ta
(4.23)
J2
?2+B2
??2
-K(?1
-?2)=0
which can be rewritten
in the following
matrix form:
? ???? ????K -K?????
J1 0
0 J2
Example
4.14 (The
In this example,
two-rotor/one-shaft
?1
?2
B1 0
+
0
Two-Rotor/One-Shaft
we will employ
rotational
and
diagram, respectively.
the
B2
??1
??2
+
Rotational
-K K
System
?1
?2
by the
1
=
0
ta
Lagrange
Lagrange approach to find the dynamics
system considered in
Example
(4.24)
Approach)
model of the same
4.13.
Solution:
Forthe two-rotor/one-shaft
rotational system shown in the left-hand side of Figure 4.6, the generalize
86
4
coordinate
Modeling of
Mechanical
Systems
are ? ?1 ?2
and velocity
?T ?T
and ???1??2 , respectively.
The kinetic
energy
of the system is
storedin the moments
ofinertia, J1andJ2,as T = 0.5(J1
??2
1 +J2
??2
2 ), the potential energy of the system
isstored
inthe
torsional
spring
KasU=0.5K(?1
-?2)2,
and
thepower
dissipated
infrictions,
B1and
B2, is
1 +B2
??2
2 ).
D = 0.5(B1 ?? 2
Hence,the Lagrangian function is
1 +J2
??2
L=T-U =0.5(J1
??2
2)-0.5K(?1
-?2)2
Then the key differentials
in Lagranges
equation
J1
??1
??? = ?L/? ??2
?L
?? =
as,
? ?? ? ? ?? ?
? ??K(?1-?2)? ??
?L/? ??1
?L
can be found
(4.25)
=
?L/??1
?L/??2
J2
??2
,
?? = ?D/? ??2
-K(?1-?2),
=
B1
??1
?D/? ??1
?D
=
B2
??2
(4.26)
ta
0
Q =
Now, from Lagranges equation Equation 4.20 we havethe dynamics
model equation
-?2)?? ???
? ??-K(?1
K(?1
-?2)
J1
?1
J2
?2
which can be rewritten
-
in the following
B1
??1
B2
??2
+
general
=
ta
(4.27)
0
matrix form:
? ???? ????K -K?????
J1 0
0 J2
?1
?2
B1 0
+
0
B2
??1
??2
+
-K K
?1
?2
1
=
0
ta
(4.28)
This equation is exactly the same as that obtained using the Newtonian approach shown in Equa-tion
4.24.
Fig. 4.7:
Arotational
system
with a gear train.
4.3 A Rotational System with a Gear Train
A gear train is an important
the best power
mechanical device
matching between
a
widely used in almost any power driving
motor or an engine and the system to
system to pro-vide
be driven.
A simple
rotational system with a geartrain is shown in Figure 4.7. Anideal caseis considered at this moment,
assuming the inertia J1, J2 and friction B1, B2 of the gears are negligible and the power transferred
from the input gear to the output gear is lossless
4.3
A Rotational
System
with a Gear Train
87
Forthe lossless special case,the work done by the input gear,t1?1, is equalto that by the outpu
gear,t2?2.
t1?1 = t2?2
(4.29)
wheret1 and?1arethetorqueandthe angulardisplacement
associatedwiththeinput gear,andt2 and
?2 are associated withthe output gear.Sincethe arclength traveled bythe input gearshould bethe same
asthat traveled by the output gear, we have
r1?1 = r2?2
(4.30)
where r1 and r2 are the radii of the input gear and output gear, respectively.
With Equation 4.29, Equation 4.30, and the fact that gear ratio should be equal to the radius ratio,
the relationships between the input and output gear variables are summarized asfollows:
?2
The dynamics
easily obtained
model equation
using the
=
r2
=
N2
of the rotational
Newtonian
t1
N1
r1
?1
=
??2
=
t2
(4.31)
??1
system in
Figure 4.7 without the gear train
or the Lagrange approach
as follows,
JL
?2(t)+BL ??2(t)+KL?2(t) =t2(t)
which is
mathematically
equivalent
between the input
JL
Due to the angular
N1
??1, ?2
?1, ??2 =
N2
N2
=
velocity
and torque relation-ships
4.31, we have
N1
?1,
N2
t2 =
N2
t1
N1
(4.33)
? ?2 ??2 ??2
N1
?1(t)+BL
N2
This equation can be further
friction
4.19.
(4.32)
model equation of the rotational system in Figure 4.7 with the gear train can be
found as
rotational
Equation
N1
?2 =
Then the dynamics
to
and output gears shown in Equation
can be
coefficient
rewritten
N1
N2
??1(t)+KL
in terms
Be, and the equivalent
N1
?1(t) =t1(t)
N2
of the equivalent
torsional
spring
(4.34)
moment of inertia
constant
Je, the equivalent
Keas follows:
Je
?1(t)+Be ??1(t)+Ke?1(t) =t1(t)
where
Je =
(4.35)
??2 ??2 ??2
N1
N2
JL,
Be =
N1
N2
BL,
Ke =
N1
N2
KL
(4.36)
If the gears are chosen so that the gear ratio N2/N1 = 4, the torque required to drive the
system from the input gear side will be four times smaller. Furthermore, the moment of inertia,
the friction
coefficient,
all become sixteen times
Remark 4.15 (A
and the spring
Realistic Gear with
Consider the same rotational
B1, B2are not negligible. In this
can still
constant
of the load
observed
from
the input
gear side
will
smaller.
system
Moment of Inertia
with a gear train
and Friction)
whose
moments of inertia
J1, J2, and frictions
more general case,the dynamics modelequation of the rotational system
be described in the same form
of Equation 4.35,
88
4
Modeling of
Mechanical
Systems
Je
?1(t)+Be ??1(t)+Ke?1(t) =t1(t)
but with modified equivalent
Je = J1 +
moment of inertia
??2
N1
N2
(J2 +JL),
Be = B1 +
(4.37)
Je and equivalent friction
??2
N1
N2
(B2 +BL) ,
Ke =
coefficient
Beasfollows:
??2
N1
N2
KL
(4.38)
4.4 A Simple Inverted Pendulum
In this section, the Newtonian and Lagrange approaches will be employed to derive the nonlinear dy-namics
model of a simple inverted pendulum system in the form of nonlinear differential equation. By
defining the angular displacement and angular velocity of the pendulum as state variables, the nonlinear
differential equation can be converted into a state space modelthat consists of two first-order differential
equations, which can be packaged into a matrix form called the state equation.
Then the nonlinear state equation can be utilized to determine the equilibriums and the local linear
state-space model of the pendulum system at each equilibrium.
We will investigate the behavior of the
system at the vicinity of the equilibrium of interest and use the linear state-space models to design a
feedback controller to stabilize the system orto improve the performance of the closed-loop system.
Fig. 4.8:
4.4.1
Modeling the Simple Inverted
Newtonian
A simple inverted
pendulum
system.
Pendulum
Approach
Asshown in Figure 4.8,the angular displacement ?is assumed zero whenthe pendulum is atthe upright
position.
The length
of the stick is ?(m), but the
weight of the stick is assumed negligible
compared
with
the mass m(kg) of the black ball, which is attached to one end of the stick. The other end of the stick is
connected to the pivot. The rotational friction coefficient is assumed b(Nms/rad) and the moment
of inertia of the black ball is m?2. The black ball is subjected to the gravitational force mg and the
applied control force fa(N), which is perpendicular to the gravity. Since the angular displacemen
4.4
A Simple Inverted
Pendulum
89
? is a variable, the torque dueto the gravity is mg?sin? and the torque generated by the control
force is fa?cos?.
The dynamics model of the simple inverted pendulum system can be obtained using the Newtonian
approach with the free-body diagram in Figure 4.8 asfollows:
m?2
? +b
??-mg?sin?
=fa?cos?
Lagranges
(4.39)
Approach
Notethat the system is a nonlinear system. The dynamics model can also be derived using the Lagrange
approach. The kinetic energy, the potential energy, and the power dissipated in the system are T =
0.5m?2
??2,
U=-mg?(1-cos?),
and
D=0.5b
??2,
respectively.
Hence,
theLagrangian
is
L=T-U =0.5m?2
??2+mg?(1-cos?)
and the differentials
in Lagranges
?L
?? =
?L
?? =
equation can be found
as
?(0.5m?2
??2+mg?(1-cos?))
= m?2
??,
??
?D
??? =
?(0.5m?2
??2+mg?(1-cos?))
= mg?sin?,
??
Now, from Lagranges equation,
d
dt
??
?L
-
??
?L
??
?(0.5b
??2)
=b
??
???
Q = fa?cos
?D
+
??
= Q
we havethe dynamics model
m?2
? +b
??-mg?sin?
=fa?cos?
which is exactly the same as Equation 4.39, obtained by Newtonian approach.
4.4.2 Equilibriums
The dynamics
and Linearization
modelequation of the simple inverted pendulum can be slightly rearranged asfollows:
? +
b
g
1
m?2
?? - sin?= m?cos? fa
(4.40)
?
Nonlinear State-Space
In order to identify
of interest,
are infinite
Model
the equilibriums
of the system and to find
a linearized
model for each equilibrium
we will first convert the nonlinear differential equation into a state equation. Although there
many choices in
defining the state variables,
usually the state variables
are selected
based
on two considerations. Oneis to associate the state variables with physical variables in reality like
displacements, velocities, voltages, and currents. The other is to choose state variables so that the
state-space modelis in a special form for the purpose of analysis or design.
In the following, the angular
state variables. Let
displacement
? and the angular velocity ?? are chosen to be the
90
4
Modeling of
Mechanical
Systems
x1 = ?,
where u is used to represent
equation in
x2 =??,
fa since u is a common
and u = fa
notation for control input.
Then
we have the state
matrix form:
????sinx1-b
x?1
x?2
=
g
x2
m?
cosx1
u
m?2
x2 + 1
Assumeg = 9.8 m/s2,? = 1.089 m, m = 0.918 kg, and b = 0.551 Ns,
g
b
= 9,
?
Therefore, we havethe following
system:
x? =
Finding
1
m?2=0.6,
nonlinear state-space
m?
=1
modelto represent the simple inverted pendu-lum
???9sinx1-0.6x2+cosx1u?
x?1
x?2
x2
= f(x,u)
=
=
? ?
f1(x1,x2,u)
(4.41)
f2(x1,x2,u)
Equilibriums
Let the control input u be zero, and then the equilibriums of the system can be found by solving the state
equations with the derivative of the state variables set to zero. Now, we have
u = 0,
??=0
x?1
x?2
? ???? ??
x2 = 0
x*
1
x*
2
? ?sinx1
- m?2
bx2 =0
Thesystem hastwo equilibriums:
g
=
0
or
0
p
0
?T
?T
Oneis x*U = ? 0 0 , which represents the upright stick position
? = 0, and the other is x*D = ? p 0 , which represents the downward stick position ? = p.
Linearized State-Space
Model at the Unstable Equilibrium
Next, we will find alinearized state-space modelfor each of the two equilibriums.
?T
At the upright equi-librium
x*U = ? 0 0 , wehave the state-space model
x?(t)
where the
matrices AU and BU are computed
(see appendix
(4.42)
= AUx(t)+BUu(t)
via Jacobian
matrices Jx and Ju, respectively,
as follows
C):
AU = Jx =
BU = Ju =
???=?9cosx1-0.6???=?9 -0.6?
?? ?
?? ? ???=? ???=??
?f
?x
=
x*
?f
?u
=
x*
?f1/?x1
?f2/?x1
?f1/?x2
?f2/?x2
?f1/?u
?f2/?u
0
x*=
x*=
0
0
x*=
0
0
1
0
0
1
0
0
0
0
cosx1
x*=
1
0
Note
thattheeigenvalues
ofAU
are2.715
and-3.315,
which
shows
thattheequilibrium
isunstable.
4.4
Linearized
State-Space
Model at the Stable
A Simple Inverted
Pendulum
91
Equilibrium
, letx(t) =x(t)-x*
?T
Onthe other hand,at the downward equilibrium x*D = ? p 0
state-space model
x?(t)
D, and we havethe
= AD
x(t)+BDu(t)
(4.43)
wherethe matrices AD and BD are computed via Jacobian matricesJx and Ju,respectively, asfollows:
AD = Jx =
BD = Ju =
???=?9cosx1-0.6???=? ?
?? ?
?? ? ???=? ???=? ?
?f
?x
?f
?u
?f1/?x1
?f2/?x1
=
x*
?f1/?x2
?f2/?x2
0
x*=
?f1/?u
?f2/?u
=
x*
p
x*=
0
1
p
p
x*=
0
0
cosx1
0
0
1
-9 -0.6
0
p
x*=
-1
0
Note
thattheeigenvalues
ofADare-0.3j2.985,
which
verifies
thattheequilibrium
isstable.
These eigenvalues also reveal that the damping ratio andthe natural frequency of the system around the
downward equilibrium are ? = 0.1 and ?n = 3 rad/s, respectively.
Open-Loop Simulation
Using Nonlinear Simulink
Model
A Simulink computer simulation program built based on the nonlinear state-space model Equa-tion
4.41 will be employed to conduct simulations for the simple inverted pendulum. This Simulink
file name is SIPmodel.mdl,
which will be called by either CSD fig4p10
SIPmainOL.m
for
open-loop simulation or CSD fig4p11
SIPmainCL.m
for closed-loop simulation.
The block diagram
of the Simulink
model is shown in
Figure
4.9. Since the system
has two state
variables, there are two integrator s-1 blocks. The output of the integrator blocks are: x1, the angular
displacement, and x2, the angular velocity, respectively. The initial conditions x1(0) and x2(0) can be
set by clicking
Integrator2,
the integrator
respectively.
Fig. 4.9: The nonlinear
be employed
to conduct
block to type
x10 and x20 in the Initial
The values of x10 and x20
Simulink
model, SIPmodel.mdl,
both open-loop
condition
will be given in the
and closed-loop
boxes for Integrator1
of the simple inverted
simulations
and
main m-file program.
pendulum
system
will
92
4
Modeling of
Mechanical
Systems
Click the ToWorkplace
blocks: x1, x2, time, and cntrl, type x1, x2, t, and cntrl as Variable names,
and select Array in Save Format box. The simulation results are recorded simultaneously
on these arrays
x1, x2, t, and cntrl, and can be plotted after the simulation is completed. Click the Gain block, F*u, and
type F as the gain and select
nonlinear function,
Matrix(K*u)
in
Multiplication
box. In the Fcn block, you need to type the
9*sin(u(1))-0.6*u(2)+cos(u(1))*u(3)
where u(1), u(2), and u(3) represent the state variables x1, x2, and the control input u, respectively.
We will run the following
MATLAB program: CSD fig4p10
SIPmainOL.m,
which will auto-matically
call the Simulink model program SIPmodel.mdl,
to conduct the open-loop simulation.
The MATLAB code is listed asfollows:
%
CSD_Fig_4.10_SIPmainOL.m,
%
BC
%
MATLAB
%
Run
%
x1_dot=x2,
%
Initial
Chang,
R2015a
or
later
Inverted
on
=
Use
this
1;
Part
which
15,
x10=x10_deg*pi/180,
for
open-loop
-9
2
-0.6],
will
automatically
call
SIPmodel.mdl
x20=0,
response.
B=[0;
At
1],
=
Part3
the
eig(A),
down
F=[0
XD
equilibrium:
0];
Simulation
20
%
for
open('SIPmodel'),
open-loop
simulation
'current',
sim('SIPmodel',
Plot
the
figure(41),
simulation
[0,
sec'),
figure(42),
grid
rad
and
plot(t,cntrl,'r-'),
on,
grid
ylabel('cntrl'),
grid
minor,
Fig. 4.10: Time responses
of the
inverted
the pendulum
at x*D = ? p 0
with initial
on,
grid
grid
legend
minor,
('x1','x2'),
minor,
plot(x1,x2),
of the uncompensated
pendulum
system simulation
=0), the pendulum
despite
condition x(0)
nonlinear
simple
pendulum
equilibrium.
?T ?
expected, without control action (i.e., u(t)
stable equilibrium
rad/s'),
ylabel('x2')
and phase plane trajectory
open-loop
on,
x2
figure(43),
xlabel('x1'),
which converge into the stable downward
The result
'current');
sim_options);
results
ylabel('x1
xlabel('t,sec'),
'DstWorkspace',
sim_time],
plot(t,x1,'b-',t,x2,'r-'),
xlabel('t,
system,
Simulation
versions
sim_options=simset('SrcWorkspace',
grid
Open-loop
x2_dot=9*sinx1-0.6*x2+cosx1*u
sim_time
%
Pendulum
3/11/2020
Conditions
A=[0
%
University,
CSD_fig4p10_SIPmainOL.m,
x10_deg
%
Simple
Drexel
where the initial
= ? 0.26 0
is shown in
Figure
4.10.
As
will always spiral into the downward
condition is. Figure 4.10 shows that
would depart from the nearby
unstable
upward
4.4
A Simple Inverted
93
Pendulum
?T
equilibrium x*U = ? 0 0 , movetoward and overshoot passingthe downwardequilibrium x*D, oscillate
around several times, and eventually settle at the equilibrium after about 15 seconds. The high overshoot
oscillation is caused by the low damping property of the system with ? = 0.1 and ?n = 3 rad/s.
4.4.3
State Feedback
In the following,
Controller
Design to Stabilize
a Simple Inverted
a simple state-feedback controller
Pendulum
will be designed to stabilize the simple inverted
pendulum atthe upright position(i.e., to convertthe originally unstableequilibrium x*U = ? 0 0
stable one).
Moreover, the closed-loop
plane to obtain a desired closed-loop
at the upright equilibrium
performance.
obtained in
x?(t)
Since the controllability
system poles can be placed at desired locations
Equation
= AUx(t)+BUu(t)
For ease of reference,
4.42 is repeated
=
the linearized
?T
to a
on the complex
state-space
model
here:
?9 -0.6? ??
0
1
x(t)+
0
1
u(t
matrix
[BU
AUBU] =
?1 -0.6?
0
1
is nonsingular, there exists a state-feedback control strategy, u(t) = Fx(t), so that the closed-loop sys-tem
poles orthe eigenvalues of AU +BUF can be placed anywhere in the complex plane as long asthe
control-input constraints are satisfied. The definition of controllability
matrix and controllability theory
will be discussed in
Chapter 10.
Let the state-feedback
matrix be F = ? F1 F2 ?. Then the closed-loop system characteristic equation
will be
det
[sI-(AU+BUF)]
=s2+(0.6-F2)s-9-F1
=0
The desired closed-loop
characteristic
equation
(4.44)
can be chosen as
s2 +2??ns+?2n = 0
with damping ratio ? = 0.8 and ?n = 5rad/s,
which is
s2 +8s+25
=0
(4.45)
Comparing
thecoefficients
ofEquations
4.44
and
4.45,
we
have
F1=-34and
F2=-7.4.Hence,
the
state-feedback controller is designed as
u(t)=F1x1(t)+F2x2(t)
=-34x1(t)-7.4x2(t)
Now, withthe state-feedback controller, the closed-loop system state equation at the upright equilib-?
rium becomes
x?(t)
= (AU +BUF)x(t)
=
?
0
1
-25-8
x(t)
(4.46)
Note
thattheeigenvalues
ofAU+BUF
are-4 j3,which
verifies
thattheoriginally
unstable
up-right
equilibrium is now stable and has a desired transient
ratio ? = 0.8 and natural frequency ?n = 5rad/s.
response characterized
by its damping
94
4
Modeling of
Mechanical
Systems
Fig. 4.11: Withcontrol compensation, the originally unstable upright equilibrium is stabilized.
Closed-Loop Simulation
Using Nonlinear Simulink
Model
The same Simulink simulation program, SIPmodel.mdl,
will be employed to work together with
the closed-loop stabilization
program, CSD fig4p11
SIPmainCL.m,
to conduct the closed-loop
control simulation. Figure 4.11shows that the pendulum withinitial condition x(0) = ? 0.26
0
?T
would movetoward and settle at the upright equilibrium after just about one second without much
overshoot and oscillation. This desired transient response is the direct result of the choice of damp-ing
ratio and natural frequency as ? = 0.8 and ?n = 5 rad/s, respectively.
The simulation
results showed in
CSD fig4p11
SIPmodel.mdl,
SIPmainCL.m
shown in
4.9. The
filename:
CSD_fig4p11_SIPmainCL.m,
%
BC
Drexel
%
MATLAB
%
Run
%
SIPmodel.mdl
%
x1_dot=x2,
%
Initial
R2015a
Use
A=[0
on
later
=
1;
which
9
x10=x10_deg*pi/180,
for
closed-loop
-0.6],
will
model,
asfollows:
Inverted
Pendulum
aoutomatically
call
1],
0.6-2*ze*wn],
2
x20=0,
stabiliztion.
B=[0;
At
eig(A),
ze=0.8,
Acl=A+B*F,
%
Part3
up
XU
equilibrium:
wn=5,
eig(Acl),
for
closed-loop
simulation
'current',
sim('SIPmodel',
Plot
the
figure(41),
simulation
[0,
xlabel('t,
sec'),
figure(42),
grid
rad
plot(t,cntrl,'r-'),
minor,
Although the controller
state-feedback
on,
x2
rad/s'),
on,
grid
figure(43),
xlabel('x1'),
was designed
controller
?T
and
grid
ylabel('cntrl'),
grid
'current');
sim_options);
results
ylabel('x1
xlabel('t,sec'),
'DstWorkspace',
sim_time],
plot(t,x1,'b-',t,x2,'r-'),
on,
the
Simulation
open('SIPmodel'),
is capable
pendulum
pro-gram
versions
15,
sim_time=3
linear
inverted
MATLAB
8/11/2018
sim_options=simset('SrcWorkspace',
grid
nonlinear
code of is listed
the
x2_dot=9*sinx1-0.6*x2+cosx1*u
this
Part
%
running
Conditions
F=[-9-wn2
%
or
Simulink
MATLAB
Simple
University,
CSD_fig4p11_SIPmainCL.m,
x10_deg
%
with the
Figure
%
Chang,
Figure 4.11 are obtained from
from
legend
minor,
('x1','x2'),
minor,
plot(x1,x2),
ylabel('x2')
based on the linearized
not only performs satisfactorily
of bringing the pendulum
grid
a wide range
of initial
model at the upright
within the linearized
equilibrium,
region,
positions to the upright
this
it actually
equilibrium
x*U = ? 0 0 , as shown in Figure 4.12. The graph at the right shows the 14 phasetrajectories
at
of th
4.4
A Simple Inverted
Pendulum
95
closed-loop
system
with
theinitial
pendulum
positions
from-10?
to -70?
and
from10?
to70?.
Within
therange
of-70?
and70?,
thecontroller
only
requires
asmall
amount
ofeffort
toquickly
bring
the
pendulum to the upright
equilibrium.
Fig. 4.12: Phase portraits of the simple inverted pendulum with feedback control.
Since the control-input
force u(t) can only work in the horizontal direction, the effective control-input
torque is u?cos?, which getssmaller as ? movescloserto 90?. When? becomesgreaterthan
90?,the effectivecontrol-inputtorque u?cos? will changesign andjoin the gravityforceto pushthe
pendulum
ball down and swing it to the other side.
The pendulum
will continue
to swing
up and
may
change swing direction several times before reaching the top to swing back through almost one full
?T
revolutionto the uprightequilibriumx*U = ? 0 0 . Therefore, for the phase trajectories with the initial
position beyond 80?the controller wouldneedto make moreeffort andtravel alonger distanceto bring
the pendulum to the upright
equilibrium
as shown on the left
graph of the figure.
Now,
thestate-feedback
controller
u(t)=F1x1(t)+F2x2(t)
=-34x1(t)-7.4x2(t)
has
successfully
?T
? ? ??
converted the originally unstable upright equilibrium to a stable one.We may wantto know how the same
controller affect the downward equilibrium
state-space
model at this equilibrium
x?(t)
at x*D = ? p 0 . Recallthat state equation of the linearized
is
= AD
x(t)+BDu(t)
=
0
1
x(t)+
-9 -0.6
0
-1
u(t)
Hence,the closed-loop system state equation at the downward equilibrium is
x?(t) = (AD+BDF) x(t)
=
??
0 1
x(t)
25 6.8
(4.47)
Note
that
theeigenvalues
ofAD+BDF
are9.4465
and-2.6465,
which
verifies
that
theoriginally
stable
downward equilibrium is now unstable.
Remark 4.16 (Radian
is
and Degree)
The common units for angles and angular displacements areradian and degree. It seems that degree
moreintuitive to humans, but in engineering analysis and design we haveto use radian instead o
96
4
Modeling of
Mechanical
Systems
degree. It is not only because radian
reasons
has become a standard
why we should use radian instead
One of the practical
of degree in engineering
reasons is relevant to the linearization
model at the upward pendulum equilibrium.
convention;
process
For instance, sin?
there
are several
and scientific
we employed
practical
computations.
to obtain the linearized
= 0.2571 is approximately equal to ?
for ? = 0.26rad whenthe unit usedis radian. Butit is notthe casefor ? = 15? whenthe unit usedis
degree.
4.5 Cart-Inverted Pendulum System
The cart-inverted
pendulum
control system is one of the
most well-known
benchmark
control
problems
that has been extensively studied for decades. The problem is so intriguingit
is inherently unstable,
nonlinear, and under actuatedthat
almost all available control design approaches have been employed
to solve the problem.
The main objective of this section is to derive the nonlinear dynamics model of the cart-inverted
pendulum system using both the Newtonian and the Lagrange approaches. Oncethe nonlinear dy-namics
modelis derived, it can be employed to analyze the nonlinear system behavior around the
unstable equilibrium, to obtain a linearized dynamics model at the unstable equilibrium, and to
utilize the information
of the linearized
model to design a controller so that the unstable equilib-rium
can be converted to a stable one.
Fig. 4.13: A cart-inverted pendulum system.
A schematic diagram of atypical linear-rail inverted pendulum system is shown in Figure 4.13. The
cart has mass Mkilograms and is driven by a force fa Newtons, which can be generated by a DC motor
inside the cart. The displacement of the cart is represented by s meters measuringfrom the center of the
rail to the right.
A positive s
meansthe cart is at the right
side of the rail.
Onthe other hand, a negative s
indicates a position onthe other side of the rail. The direction of the force is defined accordingly so that
a positive force fa
left.
The friction
would
coefficient
move the cart forward
of the translational
to the right,
and a negative fa
motion is assumed a constant
would go reverse to the
Bs.
The stick with mass mkilograms andlength 2? metersis hinged to the cart with a pivot so that the
stick can rotate freely and make a whole 360-degree swing in either clockwise or counter clockwise
direction. The friction coefficient of the rotational motionis assumed as a constant B?. The angula
4.5
Cart-Inverted
Pendulum
System
97
displacement is represented by ? radians, measuringfrom the upward reference in the clockwise
direction. A positive ? meansthe stick is tilted to the right side of the pivot.
Fig. 4.14: Free-body
diagrams
of the cart-inverted
pendulum
system.
Fig. 4.15: More detailed free-body diagrams for the stick subsystem: (a) horizontal translational
of the stick, (b) vertical translational motion of the stick, (c) rotational motion of the stick.
4.5.1 The Newtonian
Translational
Approach
Modeling of the Cart-Inverted
motion
Pendulum System
Motion Equation of the Cart:
The cart-inverted pendulum system consists of the cart subsystem and the stick subsystem. Thefree-body
diagrams of the stick and the cart are shown in Figure 4.14. V and H are the vertical and horizontal
reaction forces at the pivot, respectively. There are four forces acting on the cart, and the algebraic
sum of these forces is zero according to dAlemberts principle,
M
s+Bs
?s+H-fa =0
Horizontal
Translational
Unlike the cart subsystem,
(4.48
Motion Equation of the Stick CG:
the stick subsystem
involves
two translational
motionshorizontal
and
vertical, and one rotational
motion. To makeit easier to establish the three equations of motion for
the stick subsystem, the stick free-body diagram needsto befurther
decomposed into three more
98
4
detailed
Modeling of
free-body
Mechanical
diagrams,
Systems
shown
in
Figure
4.15.
Figure 4.15(a) shows the free-body diagram for the horizontal translational
motion of the cen-ter
of gravity (CG) of the stick. Notethat the massof the stick is m,and the horizontal displacement
of the CGis s+?h, where?h = ?sin?. Atthe CG,there aretwo forces:the horizontalreaction force H
pointing to the right and the negative inertia force,
of s. According to dAlemberts
principle,
?
obtained as
H = m s+
Vertical Translational
m(
s+ ?h),
the horizontal
which is always in the opposite direction
translational
equation
of
motion for the
?=m
s+?cos?
?-?sin?
??2?
d2(?sin?)
?
dt2
CG is
(4.49
Motion Equation of the Stick CG:
Figure 4.15(b) shows the free-body diagram for the vertical translational
motion of the center of
gravity of the stick. Notethat the vertical displacement ofthe CGis ?v, where?v = ?cos?, and there are
three forces
acting
on the
CG. Apparently
the vertical reaction force
V is in the up direction,
while the
gravitational force mgis always pointing down. Since the up direction is assumed positive, the inertia
force, m
?v, shouldpoint downonthe free-bodydiagramsinceit is alwaysin the oppositedirectionof
the positive reference
direction.
According to dAlemberts
principle,
the vertical translational
equation
of motionfor the CGis obtained as
=-m?
sin?
?+cos?
??2?
d2(?cos?)
V-mg=m
Rotational
?
dt2
(4.50)
Motion Equation of the Stick:
Figure 4.15(c) shows the free-body diagram for the rotational
motion of the stick about the CG
of the stick. Notethat the torques about the CG dueto the vertical reaction force V and the horizontal
reaction force Hare ?Vsin? and ?Hcos?, respectively. Thetorque ?Vsin? is in the clockwise direction
while?Hcos?is in the opposite,
the CCWdirection.Thefrictiontorque B?
?? andtheinertialtorqueJ?
are also in the CCW direction.
Hence,the rotational equation of motionfor the CGis obtained as
?Vsin?
-?Hcos?
-B?
??-J
? =0
(4.51)
where B?is the rotational friction coefficient of the stick, which usually is very small or negligible and
J is the moment of inertia of the stick about the CG, which is
(4.52)
J = m?2/3
and ?is half of the length of the stick.
Nonlinear
Equation of Motion of the Cart-Inverted
After eliminating
have the following
the internal
reaction
forces
V and
Pendulum:
H from
Equations
4.48, 4.49, 4.50, and 4.51,
we
nonlinear governing differential equations for the cart-inverted pendulum system:
(M+m)
s+Bs
?s-m?sin?
??2+m?cos?
? =fa
(4.53)
(J+m?2)
? +B?
??-mg?sin?
+m?cos?
s =0
Note that the translational
dynamics
of the cart and the rotational/translational
dynamics
of the stick are
coupled in a nonlinear fashion. In the next subsection, the Lagrange approach will be employed to derive
the equation
of motion for the same cart-inverted
pendulum
system.
4.6
4.5.2
Lagranges
Approach
Modeling
of the
Cart-Inverted
Pendulum
Exercise
Problems
99
System
For the cart-inverted pendulum system shown in Figure 4.13, the generalized coordinate and velocity
are q = ? s ?
?T ?T
and
?q = ?s???
M and mand in the
, respectively.
moment of inertia
The kinetic energy
masses
J as
T = 0.5M?s2 +0.5m
?? ?2
+?sin?
??2?
s?+??? cos?
where the first, the second, and the third terms
of the cart, the translational
of the system is stored in the
+0.5J
??2
?
of the kinetic
(4.54)
energy are due to the translational
motion of the stick, and the rotational
motion
motion of the stick, respectively. In
thesecond
term,
?s+?
??cos?
isthehorizontal
translational
velocity
oftheCG
ofthestick
and-?
??sin?
is the vertical translational
velocity
of the
CG of the stick.
The potential
energy of the system is stored
in the gravitational field as
U = mg?cos?
(4.55)
where ?cos? is the vertical displacement of the CG of the stick. The power dissipated in the frictions
??2. Hence,
the Lagrangian
functionis
are D = 0.5Bss?2 +0.5B?
?? ?2
+0.5J
??2-mg?cos?
+?sin?
??2?
L=T-U=0.5M
?s2+0.5m
s?+?
??cos?
Then the key differentials
?L
??q =
?L
equation
can be found
as,
? ??D??
??L??
??
?? ?
?s
?s
? ??
??L??s
??-m??s??sin?+mg?sin?
?L
?q = ?L
Now, from
in Lagranges
(4.56)
?
Lagranges
?
???
?s+m(
M
?s+??? cos?)
,
m?cos?
?s+m?2?? +J??
=
0
?
??
?D
??q =
,
=
Q=
?
?D ? ?
Bs
?s
=
B?
??
(4.57
fa
0
equation,
d
dt
??
?L
??q
?L
-
?q
?D
+
?q
= Q
(4.58)
we havethe dynamics model equation
(M+m)
s+Bs
?s-m?sin?
??2+m?cos?
? ???
?(J+m?2)
? +B?
??-mg?sin?
+m?cos?
s
fa
=
which is identical to the dynamics
in Equation 4.53.
model equation
(4.59)
0
derived using the Newtonian approach shown
This nonlinear dynamics model described by the two coupled second-order differential equa-tions
will be employed in Section 7.6 for further investigation of the cart-inverted pendulum system
regarding nonlinear state-space model construction, equilibriums, linear state-space models, and
stabilization controller design and analysis.
4.6 Exercise Problems
P4.1a:
Consider the double
MBK (massdamperspring)
system shown in
Figure 4.16,
where M1 =
M2 =1 kg, K1 =K2 =2 N/m, B1 =B2 =0 Ns/m, and the initial conditions of the system are x1(0)
=x10,
100
4
Modeling of
Mechanical
Systems
x?1(0) = 0, x2(0) = x20, and?x2(0)
Newtonian approach.
= 0. Verify the following
governing equations of the system by
x1(t)+4x1(t)-2x2(t)
=0
-2x1(t)+
x2(t)+2x2(t)=0
And show that the Laplace transform of these equations are:
s2X1(s)-sx10
+4X1(s)-2X2(s)
=0
-2X1(s)+s2X2(s)-sx20
+2X2(s)
=0
P4.1b:
Repeat P4.1a using the Lagrange approach.
Fig. 4.16: A double
MBK system.
P4.1c: Solvefor X1(s) and X2(s) as
A1s
A3
C1s
C3s
X1(s) = s2+0.76393
+s2+5.23607
X2(s) = s2+0.76393+s2+5.23607
and find
A1, A3, C1, and C3 as functions
of the initial
conditions
x10 and x20.
P4.1d: Chooseinitial conditions x10 and x20so that A3 = 0 and C3 = 0, and show that the two
M1and M2 will exhibit in-phase harmonic motions at the frequency ?L = 0.76393rad/s.
v
masses
P4.1e: Underthe initial conditions given in P4.1d, find the inverse Laplacetransform of X1(s) and X2(s)
to obtain x1(t) and x2(t). Plot x1(t) and x2(t) and usethe graphs to verify the in-phase harmonic motions
and the oscillation frequency obtained in P4.1d.
P4.1f: Chooseinitial conditions x10 and x20sothat A1 = 0 andC1 = 0, and show that the two masses M1
5.23607rad/s.
and M2 will exhibit 180-degreeout-of-phaseharmonic motionsatthe frequency ?H =
v
P4.1g: Underthe initial conditions given in P4.1f, find the inverse Laplace transform of X1(s) and X2(s)
to obtain x1(t) and x2(t). Plot x1(t) and x2(t), and usethe graphs to verify the 180-degree out-of-phase
harmonic motions and the oscillation frequency obtained in P4.1f.
P4.1h:
Randomly
choose a set of the initial
conditions,
x10 and x20, and then find the inverse
Laplace
transform of X1(s) and X2(s) to obtain x1(t) and x2(t). Plot x1(t) and x2(t), and comment on the motion
of the system.
4.6
P4.2:
Consider the simple
pendulum
positioning
system shown in
Exercise
Figure 4.17.
Problems
101
Unlike the simple in-verted
pendulum control problem, the objective of this project is not to stabilize the pendulum at the
upright
lightly
unstable equilibrium.
damped pendulum
Instead, it is an angular
position tracking/regulation
control
problem for a
system.
As shown in the figure,
one end of the
and the other end attaching to a black ball
pendulum
with
stick is connected to the pivot under the ceiling,
mass m kg. The pendulum
sticks
length
is ? m, and its
massis assumed negligible. The gravity g = 9.81 m/s2, andthe control force is u N, whose direction is
assumed always
perpendicular
to the angular velocity
to the pendulum
with friction
stick.
The friction
torque
of the system is proportional
coefficient b Nm/rad/s.
Fig. 4.17:
A simple
pendulum
positioning
system.
P4.2a: Showthat the governing differential equation for the pendulum system is
m?2
?(t)+b ??(t)+mg?sin?
= u?
using the Newtonian approach.
P4.2b:
Verify the result
of P4.2a using the
Lagrange approach.
P4.2c: Definethe state variables x1 = ?, x2 =?? and then find the nonlinear state equation of the system.
P4.2d: Assume ?g = 16,
numerical values.
b
m?2=
0.6, m?
1 = 2, and then rewrite the nonlinear state equation with these
P4.2e: Showthat x*D = ? 0 0
pendulum
positions.
?T
is the static equilibrium
A static equilibrium
P4.2f: Find the linearized
is an equilibrium
of the system associated with the downward
of the system
u(t) is zero.
state equation associated with the downward equilibrium in the form of
x?(t) = ADx(t) +BDu(t) and show that the equilibrium x*D = ? 0 0
ues of AD.
when the input
?T
is stable based on the eigenval
102
4
Modeling of
Mechanical
P4.2g:
Find the characteristic
Systems
equation associated
with this equilibrium,
and useit to determine the pole
locations
-a j?,thedamping
ratio
?,and
thenatural
frequency
?n.Discuss
how
a,?,?,and?nwill
affect the time response
of the system around this equilibrium.
oscillation frequency based onthe information
P4.2h: Build a Simulink/MATLAB
and useit to conduct an initial
Measurethe
maximum
overshoot
x*D = ? 0 0
?T
with x1(0)
is stable,
= 0.26 rad and x2(0)
= 0 rad/s.
but the response is oscillatory
maximum overshoot and the oscillation frequency from the initial
graph obtained from the simulation
and the
program based on the nonlinear state equation obtained in P4.2d,
state response simulation
You will see that the downward equilibrium
large overshoot.
P4.2i:
Predict the
of a, ?, ?, and ?n.
in P4.2h, and compare the simulation
results
with
state response
with your predictions
in P4.2g.
P4.3: As welearnedfrom ProblemP4.2a Problem4.2i,the system nearx*D,the downward equilibrium,
is stable; however, the time response is oscillatory with large overshoot. The performance of the system
can be improved using feedback control. Consider the state equation,
x?(t)
Withthe state-feedback control, u(t)
= ADx(t)+BDu(t)
= Fx(t), the state equation of the closed-loop system will become
x?(t)
= (AD+BDF)x(t)
and the poles of the closed-loop system will bethe eigenvalues of AD+BDF.
P4.3a: Designa state-feedback controller u(t) = Fx(t) = F1x1(t)+F2x2(t)
sothat the closed-loop sys-tem
has damping ratio ? = 0.7 and natural frequency at ?n = 4 rad/s. After closing the loop, determine
the pole locations, and discuss how the polelocations, the damping ratio and the natural frequency
affect the time response
oscillation
frequency
of the system around this equilibrium.
based on the information
of the closed-loop
Predict the
maximum
overshoot
poles and their corresponding
will
and the
damping
ratio and natural frequency.
P4.3b:
controller
Modify the Simulink/MATLAB
program you usedin P4.2h to incorporate the state-feedback
you designed in P4.3a. Conduct simulations
to verify that the time response
performance
has
been greatly improved.
P4.3c: Measurethe maximum overshoot and the oscillation frequency from the initial state response
graph obtained from the simulation in P4.3b, and compare the simulation results with your predictions
in P4.3a.
P4.4: As mentionedin P4.2e, a static equilibrium is an equilibrium whenthe input u(t) is zero. In con-trast,
a dynamic equilibrium is an equilibrium whenthe input u(t) is a nonzero constant. Manypractical
systems
are required to
For example,
an aircraft
P4.4a: In this problem,
work at a dynamic
cruise level
equilibrium,
straight flight
or to switch
from
one equilibrium
maintains a constant speed
to another.
with a constant thrust.
we areinterested in investigating the behavior of the pendulum system at the
dynamic equilibrium x* =[0.26 0]T, which means
the angular displacementis ? = 15 degreeandthe
angular velocity ?? is zero. Usethe nonlinear state-space model obtained in P4.2d to determine the inpu
u* atthis equilibrium.
4.6
P4.4b:
Find a linearized
state-space
model of the pendulum
Exercise
system at the dynamic
in P4.4a. The linearized state-space modelshould be of the following
Problems
equilibrium
103
chosen
form:
x?(t)=A
x(t)+B
u(t), where
x(t) =x(t)-x* andu(t) =u(t)-u*
Notethat x(t) and u(t) arethe actualstate vectorandthe actualcontrol input, respectively; x* and u* are
the values of the state vector and the control input at the equilibrium; andx(t)
of the state vector and the control input,
respectively,
from the equilibrium
andu(t)
arethe deviations
values.
P4.4c: Find the characteristic equation associated with this equilibrium, and useit to determine the pole
locations
-a j?,thedamping
ratio
?,and
thenatural
frequency
?n.Discuss
how
a,?,?,and?nwill
affect the time response
of the system in the vicinity
of this equilibrium.
Predict the
maximum
overshoot
and the oscillation frequency.
P4.4d: Build a Simulink/MATLAB
state equation
obtained in
program, similar to
what you did in P4.2h, based on the nonlin-ear
P4.2a, and use it to conduct
an initial
state response
simulation
with
x1(0) = 0 rad, x2(0) = 0rad/s, and u(t) = u*. Observeif the time responseconvergesto the new dy-namic
equilibrium chosen in P4.4a.
P4.4e: Measurethe maximum overshoot and the oscillation frequency from the initial state response
graph obtained from the simulation in P4.4d, and compare the simulation results with your predictions
based on the pole locations, damping ratio, and natural frequency obtained in P4.4c.
P4.5: As welearned from Problem P4.4a Problem P4.4e,the system aroundthe (x*,u*), the dynamic
equilibrium chosen in P4.4a, is stable; however, the time response is oscillatory with large overshoot.
The performance of the system can beimproved using feedback control. Considerthe state equation,
x?(t)
Withthe state-feedback control, u(t)
= Fx(t),
x?(t)
and the
poles of the closed-loop
system
= x(t)+B
A
u(t)
the state equation of the closed-loop system will be
= (A+BF) x(t)
at the dynamic
equilibrium
will become the eigenvalues
of
A+BF.
P4.5a: Designa state-feedback controller u(t)
= Fx(t)
= F1
x1(t)+F2 x2(t) sothat the closed-loop sys-tem
has damping ratio ? = 0.7 and natural frequency at ?n = 4rad/s. After closing the loop, determine
the pole locations, and discuss how the polelocations, the damping ratio and the natural frequency will
affect the time response of the system in the vicinity of this equilibrium. Predictthe maximum overshoot
and the oscillation frequency.
P4.5b:
Modify the Simulink/MATLAB
program you usedin P4.4d to incorporate the state-feedback
controller
youdesigned
inP4.5a.
Since
x =x-x*and
u =u-u*,thestate-feedback
control
law
u(t) =F
x needs
tobereplaced
byu =F(x-x*)+u*
inthesimulation.
Conduct
simulation
tover-ify
if the time response performance is improved.
P4.5c:
Measure the
maximum
overshoot
graph obtained from the simulation
and the
oscillation
frequency
in P4.5b, and compare the simulation
from
the initial
results
state response
with your predictions
based on the pole locations, damping ratio, and natural frequency of the closed-loop system
5
Modelingof Electrical Systems
R
ECALL that in Chapter 2 and Chapter 3, westarted with afew simple mechanical, electrical, and
electromechanical systems and their associated mathematical model equations. These models
wereemployed in the analysis to understand how systems work and in a few simple design ex-amples
to achieve desired time-domain and frequency-domain performance. Since mathematical system
modeling is essential in the analysis and design of control systems, two chapters are allocated in the
book to address the fundamental modeling approaches for mechanical systems, electrical circuits, and
electromechanical
systems.
In Chapter 4, we havelearned how to employ the Newtonian approach, dAlemberts principle, and
the Lagrange approach to assemblethe governing dynamics equations as mathematical modelsfor me-chanical
systems. Electrical circuits or systems are alittle bit different, but there are close relationships
between the mechanical and electrical systems. For the MBK mechanical system andthe electrical RLC
circuit we discussed in the beginning of Chapter 3, their governing dynamics equations shown in the
following
clearly
reveal that the two systems
are
mathematically
equivalent
and their
physical
com-ponent
properties are analogous to each other:
y(t)+B
M
?y(t)+Ky(t)
= fa(t)
(5.1)
q(t)+R ?q(t)+(1/C)q(t)
L
= ea(t)
The mechanical system is driven by the applied force fa(t), which is analogous to the voltage
source ea(t). The voltage source was called electromotive force (emf) when the electricity era had
just started. This term is still used on some occasions since the voltage source serves as a means
to drive the electrical system. The resistance Rand the friction coefficient B are analogous to each
other because their dissipated powers are of the same form, with 0.5R?q2 and 0.5B?y2, respectively.
The inductor L and the inverse capacitance 1/C are analogous to M and K, respectively, since L
and M have the same form of kinetic energy, 0.5L?q2 and 0.5M?y2,respectively, and 1/C and K have
the form of potential energy (0.5/C)q2 and 0.5Ky2, respectively.
In Chapter 2 and Chapter 3 we briefly reviewed the characteristics of the three basic passive electric
elements, resistor, inductor, and capacitor, and usedthe component relations together with Kirchhoffs
current law and voltage law to derive the governing dynamics equations for a simple first-order RC cir-cuit
and atypical second-order RLC system. In this chapter, we will explain in detail how to assemble
the governing dynamic equations for electric circuits using several different approaches. Although elec-tric
circuits
straightforward
more limited
seem less tangible
than the
than
mechanical
mechanical
systems,
the electrical
system since the geometry
system
of the electric
modeling is
circuit
more
is, in a sense,
106
5
Modeling of Electrical
Systems
Fig. 5.1: Symbols and characteristics
of fundamental
two-terminal
electrical
elements.
5.1 Basic Electrical Circuit Elements and Circuit Conventions
To analyze the
behavior
or to
obtain
a
mathematical
model of an electrical
system, the first
step is
to understand the voltage-current relationships of the three fundamental passive electrical elements,
resistor, capacitor, inductorjust
like we did in Chapter 4 for mechanical systems about the force-velocity
relationships of the three mechanical elements, damper, spring, mass. As shown in Figure
5.1, the e-i relationship of the resistor is e = Ri, which is the well-known Ohms law. The units for the
voltage e, current i, charge q, and resistance R are volt, ampere, coulomb, and ohm, respectively.
The e-i relationship ofthe capacitoris i = Cde/dt, wherethe unit of the capacitance Cis Farad. The
e-i relationship of the inductor is e = Ldi/dt or dF = Ldi, wherethe units of the inductance L and
the magneticflux F are Henry and Weber,respectively.
Notethat the voltage polarity and the current flow direction need to be consistent: The current
always flows into the positive terminal of every passive element. Although in reality, especially in
a complicated circuit,
is completed,
interest
we may not know the actual polarity of the elements before the circuit analysis
it is a common
in the circuit.
practice to specify
For the simple
circuit in
voltage
Figure
and current
with polarity
5.2, the voltages
for each element
and currents
with polarity
of
and
flow direction are specified as shown. Sincethe specified current direction needs to be consistent with
the specified voltage polarity, only either one of them, voltage or current, needs to bespecified. Th
5.2
polarity
can be arbitrarily
Basic Time-Domain
Circuit
Modeling
chosen, but once the specified
Approaches
and
voltage and current
Kirchhoffs
variables
Laws
107
are selected they
should stay unchanged during the whole circuit analysis process.
Fig. 5.2: The convention of voltage polarity and current direction in an RLC circuit.
Fig. 5.3: (a) Thevenin and (b) Norton equivalent circuits.
The voltage of the voltage source and the current of the current source are prespecified, and the cur-rent
always flows out of the positive terminal of the source element. Behold, this is different from
the three passive electrical elements R, L, and C; with them the current always flows into the positive
terminal. The voltage and current sources are classified into two categories: independent and depen-dent
voltage/current sources. The independent ones represent the independent power sources whose
supplied voltages or currents do not depend on any parameters of the circuit. Onthe other hand, the
dependent sources represent some active devices like transistors or operational amplifiers whose sup-plied
voltages or currents are dependent on some parameters in the circuit. The voltage/current sources
listed in Figure 5.1 are ideal sources. Although there exists no ideal voltage/current source in the
real world, mostof the real signals and power sources can be modeled as a combination of an ideal
voltage/current source together with passive electrical components. For example, a signal source or
a battery can be modeled as a Thevenin or a Norton equivalent circuit, as shown in Figure 5.3.
5.2 Basic Time-Domain
As mentioned earlier, the
it
would be difficult
Circuit
mathematical
or even impossible
Modeling Approaches and Kirchhoffs
modeling of
to achieve
Laws
mechanical systems is essential, since
optimal
mechanical
without it
design and feedback
control
system design for performance enhancement, including precision, reliability, safety, and automation.
The mathematical modeling of electrical systems is even moreimportant since electronic devices and
electromechanical
systems
have been embedded in almost all systems that impact
start from the basic time-domain
circuit
modeling
approach.
(1) the 2k equations approach, (2) the nodetodatum
currents approach
This approach
(NTD)
our daily life.
includes
three
We will
variations:
voltages approach, and (3) the
mesh
108
5
5.2.1
Modeling of Electrical
Circuit
Modeling
Systems
Using the 2k Equations
For the electrical system or circuit
voltage
with k two-terminal
or current source, the dynamic system
relationship
Approach
and the interactions
elements that include at least one independent
model should
have the capability
among all the k elements.
of exhibiting
the cause-effect
Since there is one pair of voltage
and
current variables associated with each element, in total there are 2k variables, and therefore 2k inde-pendent
equations are required to solve these variables. Based on the e-i relationship of the passive
elements and the voltage/current information for the voltage/current sources, as described in Figure 5.1,
k equations can be easily set up. The other k equations must be obtained using Kirchhoffs voltage law
(KVL) and current law (KCL) according to the interconnection of the elements in the circuit.
Recall that voltage
was defined as the energy required to bring a positive
unit charge from
one point
to another in a conservative electric field. Based onthe conservation of energy law, the energy required
for a charge to travel around a closed path back to the starting point is zero, and the algebraic sum of
the voltages around any closed path is zero, whichis exactly the statement of Kirchhoffs voltage law
(KVL),
(5.2)
?ei = 0
i
On the other
hand,
Kirchhoffs
current
law
(KCL)
is derived
from
Conservation
of
Charge law.
At
any junction of elements, called node, charge can neither be destroyed nor created, and therefore the
algebraic sum of the currents entering any node is zero. Thus,
?i j
(5.3)
=0
j
Example 5.1 (2k Equations Approach)
Forthe electric circuit shown in Figure 5.2, there are k = 4 elements, including
L, capacitor
C,resistor
R, and voltage source es. Apparently,
we have the following
one each of inductor
k = 4 equations from
the e-i relationship of the L, C,and Relements and the value of the voltage source es:
e1 = L
As specified
on the circuit
di1
,
dt
i2
=C
de2
,
dt
e3 = R3i3,
and
e4 = es
elements in Figure 5.2, there are 2k variables:
(5.4)
i1,e1, i2,e2, i3,e3, i4,e4. It is
clear that k = 4 moreindependent equations are neededto determine the voltages and currents for all
the k = 4 elements.
interconnection
These k = 4 equations
of the k = 4 elements.
is a loop in the circuit that
obtained by applying
must be obtained
using
It can be seen that the circuit
does not contain any other loop,
KVL around the two
KVL and
KCL according
to the
has m = 2 meshes, where a mesh
and therefore
m = 2 mesh equations
can be
meshes:
(5.5)
-e4+e1+e2=0, and -e2+e3=0
There
are
n =3nodes
inthecircuit,
but
they
can
only
provide
n-1 =2independent
node
equations
by
applying
KCL to any 2 nodes:
i4 -i1=0 andi1 -i2-i3=0
The 2k equations
approach
described is conceptually
As a matter of fact, the system
straightforward;
model equations
example, in the circuit shown in Figure 5.2, if
(5.6)
however, it is tedious
do not need to include
computa-tionally.
all the 2k variables.
we know the voltage across the capacitor, e2,then
For
we
have
thevoltage
across
theresistor,
e3=e2,and
thevoltage
across
theinductor,
e1=es-e2,
simpl
5.2
by inspection.
Basic Time-Domain
With the knowledge
Circuit
Modeling
of the voltages for the three
i3 can be easily computed based on the simple e-i relationship
again,
employed
we have i4
= i1 and e4 = es. Hence, the system
Approaches
Kirchhoffs
Laws
109
L, C, R elements, the currents i1, i2, and
of each element, and finally, by inspec-tion
model needs only one equation that
can be
to solve for e2.
From this discussion, certainly there exists at least one electric circuit
moreefficient than the 2k equations approach.
5.2.2 The NodeToDatum
(NTD)
will serve as the common
of a voltage
or current
modeling approach that is
Voltages Approach
The first step of the nodetodatum
which
and
(NTD)
reference
voltages approach is to select one node as the datum,
point.
source is a good choice.
The selection is arbitrary,
but the negative terminal
The next step is to assign a nodetodatum
(NTD)
voltage variable for each node of the circuit except the positive terminal of voltage sources. Then
the final step is to apply KCL to all assigned nodes to set up equations with only the NTD voltages
as variables.
Fig. 5.4: (a)
Example 5.2 (NTD
The NTD voltages approach, (b)
The mesh currents approach.
Voltages Approach)
The circuit in Figure 5.4(a) is exactly the same as the onein Figure 5.2, which was considered in
Example
5.1. The difference
here is that by using the
NTD approach
only one voltage variable, instead of
2k variables, needsto be assigned. The node connecting to the negative terminal of the voltage source
is selected as the datum, and one NTD voltage variable eC is assigned at the node connecting to the
three elements L, C, and R. Thethird node is connecting to the positive terminal of the voltage source,
and its NTD voltage is obviously given as es. Therefore, only one NTD voltage variable is required to
be assigned. Now we apply Kirchhoffs current law to the assigned node to obtain the following KCL
equation:
deC eC
C
(5.7)
+(1/L)
+
dt
?(eC-es)dt=0
R
Note that the first term, the second term, and the third term of the equation are the currents flow-ing
out of the node through the capacitor, the resistor, and the inductor, respectively. This equation
describes the unique relationship
between the voltage source voltage
node, eC. The voltage eC will be determined
by solving
es and the voltage at the assigned
the equation if esis given, and once eCis known,
all the voltages and currents in the circuit can be easily determined.
Meanwhile, by taking
sCEC +
Laplace transform of Equation 5.7, we havethe following transfer function:
EC
R
+
EC-Es
sL
=0
?
EC(s)
Es(s)
R
=
RLCs2 +Ls+
110
5
5.2.3
Modeling of Electrical
The
Mesh Currents
Systems
Approach
The circuit in Figure 5.4(b) will be employed in the following
dynamic
equations
of the circuit
using the
mesh current
example to demonstrate how to assemble
approach.
This circuit
is exactly the same as
those of Figure 5.2 and Figure 5.4(a), considered in Example 5.1 and Example 5.2, respectively.
Example 5.3 (Mesh
Currents
Approach)
First we assign a meshcurrent variable to each mesh. There aretwo meshesin the circuit shown
in Figure 5.4(b). Thefirst meshis the loop consisting of the voltage source es,the inductor L, andthe
capacitor C,and the second meshis the loop composed of the capacitor Cand the resistor R. Thesetwo
mesh current
direction.
variables are denoted as i1 and i2, respectively,
Then
we apply
Kirchhoffs
equations:
di1
1
C
The left
side
of the first
+
mesh currents are in the clockwise
meshesto obtain the following
?(i1-i2)dt=0
1
-es+Ldt
drops around
and both
voltage law to the two
C
KVL equation
in
Equation
KVL
(5.8a)
?(i2-i1)dt+Ri2=0
mesh 1 clockwise, starting
two
(5.8b)
5.8a is the
algebraic
from the negative terminal
sum
of the voltage
of the voltage source.
The
first
term-es,
thesecond
termLdi1/dt,
and
thethird
term
(1/C)? (i1-i2)dt
are
thevoltage
drops
of
the voltage source, the inductor, and the capacitor, respectively. Similarly, the left side of the second
KVL equation, Equation 5.8b,is the algebraic sum of the voltage drops around mesh2 clockwise,
starting from the bottom terminal of the capacitor. The mesh current variables i1 and i2 can be
determined
by solving
the two
and currents in the circuit
For this particular
equations if es is given; and, once i1 and i2 are known,
all the voltages
can be easily determined.
circuit,
the
mesh currents approach
employs two
mesh current
variables requiring
two equations while the NTD voltages approach needs only one NTD voltage variable that only requires
one equation. It seems that the NTD voltages approach is simpler than the meshcurrents approach, but
this is not always the case. For some circuits,
than the
the
mesh currents
approach
may require
less equations
NTD voltages approach.
5.3 BasicImpedance Circuit Modeling Approaches
Recall that the e-i relationships
eR(t)
of the resistor, inductor,
= RiR(t),
eL(t)
=L
and capacitor,
diL(t)
dt
,
iC(t)
respectively
=C
are
deC(t)
dt
After the Laplace transform, the voltage-current relationships of the resistor, inductor,
respectively, will be described by the following algebraic equations:
ER(s) = RIR(s),
EL(s) = sLIL(s),
and IC(s)
=sCEC(s)
(5.9)
and capacitor,
(5.10)
These equations can be further rewritten in the form of E-I relationship, described by generalized
Ohms law :
ER(s) = ZR(s)IR(s),
EL(s) = ZL(s)IL(s),
EC(s) = ZC(s)IC(s)
(5.11
where
5.3
Basic Impedance
Fig. 5.5: E-I relationship of fundamental two-terminal
ZR(s) = R,
ZL(s) =sL,
Circuit
Modeling
Approaches
111
electrical elements in frequency domain.
ZC(s) = 1/sC
(5.12)
are called the impedance of the resistor, inductor, and capacitor. The frequency-domain E-I charac-teristics
of the fundamental two-terminal electrical elements are summarized in Figure 5.5.
The impedance
NTD voltages
circuit
and the
modeling approaches
mesh currents
that the impedance circuit
following
more versatile in
the generalized
described in the previous
domain.
which the inductors,
of the
The difference
is
The frequency-domain
the capacitors,
approach
and the resistors
with the impedance
are treated
the same
Ohms law.
Example 5.4 (Find the Transfer
Function EC(s)/Es(s)
Using Impedance
Consider the circuit in Figure 5.6(a). It can be seen that if
currents in the circuit
counterparts
section.
modeling approaches are carried out in the frequency domain while those in
the previous section are done in the time
concept is
are basically the frequency-domain
approaches
can be easily
derived. In the
Manipulations)
EC(s) is known, all the voltages and
next subsection,
we will use the impedance
NTD
voltages approach to solve for EC(s). But, here, we would like to show that the same objective can be
achieved
by applying
a couple of simple impedance
Fig. 5.6: (a) Impedance
manipulations.
NTD voltages approach, (b) Impedance
mesh currents approach
112
5
Modeling of Electrical
Systems
First combine the ZC and ZRinto
one impedance
as ZRC according
to the parallel combination
law of
impedances:
1
1
=
ZRC
1
1
+
ZR
=
ZC
R
+sC =
RCs+1
R
? ZRC
=RCs+1
R
(5.13
Then the voltage divider law can be applied to obtain the transfer function EC/Es asfollows:
EC(s)
Es(s)
Note that
=
ZRC +ZL
the transfer
system
R
RCs+1
ZRC
=
function
discussed in
R
RCs+1+sL
of the circuit
Chapter
=
R
is
R
=
sL(RCs+1)+R
mathematically
(5.14)
RLCs2 +Ls+R
equivalent
to the typical
second-order
3.
A Quiz Problem: Find a set of values for the inductor,
capacitor, and resistor so that the system
has a damping ratio ? = 0.7 with natural frequency ?n = 100rad/s.
5.3.1 The Impedance
The impedance
NTD Voltages Approach
NTD voltages approach
basically is the same asthe time-domain
NTD voltages
approach
welearned in Section 5.2.2. The only difference is that the impedance approach is mucheasier,requiring
only algebraic
computation
considered in
Example
the impedance
approach.
Example 5.5 (NTD
or differentiations.
example to demonstrate
The circuit
the advantage
of
Voltages Approach)
in
NTD voltages
identical,
with integrations
5.2 will be employed in the following
Consider the circuit
time-domain
without the need to deal
Figure 5.6(a),
approach in
but all variables
which is exactly
Example
5.2.
are in frequency
the same as the
The procedure
domain
one
we studied
of setting
as functions
using the
up equations
is
of s and all equations
al-most
are
algebraic.
As shown in Figure 5.6(a), the circuit has three nodes, and one of the three nodes, the negative
terminal
of the voltage source
Es, is chosen as the datum.
as Es since the node is the positive terminal
The voltage at the
of the voltage source
upper-left
Es. Hence, the
only
node is known
unknown
NTD
voltage is EC,the voltage at the node that connects to ZL, ZC and ZR.
Applying Kirchhoffs current law to the assigned NTD node, the algebraic sum of the three currents
leaving the node should be zero:
EC
EC
+
ZC
Note that
the currents
ZR
flowing
+
EC-Es
=0 ?
ZL
out of the
NTD
?
sC+
node into
1
R
1
+
sL
?
1
EC =
the capacitor
sL
Es
(5.15)
C, the resistor
R, and the
inductor
LareEC/ZC,
EC/ZR
and
(EC-Es)/ZL,
respectively,
according
tothegeneralized
Ohms
law. Plugging the impedances, ZC = 1/sC, ZR = R, and ZL = sL into the KCL equation and movingthe
Esterm to the right-hand
side of the equation,
we obtain the new equation
arrow sign. This new equation leads to the following
This transfer
function
is exactly
side of the
R
EC(s)
Es(s)
on the right-hand
transfer function:
=
RLCs2 +Ls+R
the same as the one obtained
by taking
Laplace
transform
of
Equation 5.7in Example 5.2. It is alsoidentical to the result obtained in Equation 5.14 based on simple
impedance
manipulations.
5.4
5.3.2
The Impedance
The impedance
Mesh Currents
meshcurrents
welearned in Section 5.2.3.
considered
advantage
in
approach
of the impedance
Example 5.6 (Impedance
Consider the circuit
time-domain
Circuit
basically is the same as the time-domain
The only difference
Example
Approach for
Modeling
113
Approach
only algebraic computation
The circuit
The Lagrange
is that
the impedance
meshcurrents
approach
is
without the need to deal with integrations
5.3 will be employed
in the following
approach
much easier, re-quiring
or differentiations.
example to
demonstrate
the
approach.
Mesh Currents Approach)
in
mesh currents
Figure 5.6(b),
approach in
which is exactly the same as the
Example
5.3. The procedure
one
of setting
we studied
using the
up equations is almost
identical, but all variables arein frequency domain as functions of s and all equations are algebraic.
Asshown in Figure 5.6(b), the circuit hastwo meshes;hence,two clockwise meshcurrents I1 and I2
are assigned. Thefirst meshis the loop consisting of the voltage source Es,the inductor impedance
ZL, and the capacitor
impedance
impedance
ZC and the
meshesto obtain two
ZC, and the second
resistor
impedance
mesh is the loop
ZR. Now
we apply
composed
Kirchhoffs
of the capacitor
voltage law to the two
KVL equations:
ZLI1+ZC(I1
-I2)-Es=0
(5.16)
ZC(I2
-I1)+ZRI2
=0
which can be rewritten in matrix form as
??? ? ?
?ZL+ZC-ZC??
-ZC ZC+ZR
To solve this
=
sL+
Es
0
1
sC
-1
sC
-1
1
sC
sC +R
???
??
I1
I2
=
Es
0
(5.17)
matrix equation for I1 andI2,
???
I1
I2
It is easy to verify that
5.14 and
I1
I2
Equation
sL+
=
-1
1
sC
-1
sC +R
sC
EC = I2ZR
= I2R.
5.15, obtained
?-1
??
1
Es
sC
1
=
0
RLCs2
+Ls+R
Hence, the result is consistent
using the impedance
? ?
RCs+1
1
Es
(5.18)
with those shown in
manipulations
Equation
and the impedance
NTD
voltages approach, respectively.
5.4 The Lagrange Approachfor Circuit Modeling
As discussed in
dynamic
Chapter 4,the
equations in the
Lagrange approach
mechanical system
provides
an elegant and efficient
modeling process.
The Lagrange
way of constructing
approach
can also be
applied to electrical systems in whichthe electric charge q andits derivative?q (the electric current) are
considered as configuration variables.
Recallthat the Lagrangian
L(?q,q) is defined as
L(
?q,q)=T(
?q,q)-U(q)
(5.19)
where T(?q,q) and U(q) are the total kinetic energy andthe total potential energy in the system, respec-tively.
For electrical systems, T(?q,q) is the total kinetic energy stored in inductors, and U(q) is the
total potential energy stored in capacitors. In addition to the energy stored in the inductors and capac-itors,
the energy delivered by the voltage/current sources and the dissipated power in resistors ar
114
5
Modeling of Electrical
Systems
represented by dW, whichis the total virtual work done by all forces action through dq,a variation
of the generalized coordinate vector:
? ?(dq)
dW=QTdqor Q=?(dW)
where Qis the generalized force vector. Withthe Lagrangian
Q defined, the Lagrange equation is given as
d ?L( ?q,q)
??q
dt
?L( ?q,q)
-
?q
(5.20)
L(?q,q) andthe generalized force vector
= Q
(5.21)
In the following, we will usethe same circuit in Figure 5.2 or Figure 5.4, whichis redrawn in Figure
5.7, to demonstrate how the Lagrange approach can be employed to derive dynamic equations in elec-trical
system modeling.
Fig. 5.7: Circuit modeling using the Lagrange approach.
As shown in
Figure
5.7, the electric
respectively, and their corresponding
charges traveling
around
mesh 1 and
mesh 2 are q1 and q2,
meshcurrents are denoted by?q1 and?q2. The generalized coor-dinate
?T
?T
and velocity vectors are q = ? q1 q2
and?q = ?q?1q?2 , respectively. The total kinetic en-1
ergy T(?q,q) = 0.5L?q2 is the kinetic energy stored in the inductor L, and the total potential energy
U(q)=(0.5/C)(q1
-q2)2
isthepotential
energy
stored
inthecapacitor
C.The
totalvirtualwork
through dq from the contribution ofthe voltagesourceandthe dissipationin the resistor will be
dW=esdq1
-R
?q2dq2
The generalized
force
vector is computed
according to
Q =
?(dW)
?(dq)
=
(5.22)
Equation 5.20 as
??
es
-R
?q2
(5.23)
Now the Lagrangian is
L(
?q,q)=T(
?q,q)-U(q)
=0.5L
?q2
1 -(0.5/C)(q1
-q2)2
(5.24)
Then according to the Lagrange equation in Equation 5.21, wehavethe following:
1 +(1/C)(q1
-q2)
?Lq-(1/C)(q1
??-R?q2?
-q2)
es
=
Since?q1 = i1 and?q2 =i2, Equation 5.25 can berewritten asthe following:
(5.25
5.4
Approach for
Circuit
Modeling
(i1-i2)dt=es
(i1-i2)dt=Ri2
Ldi1/dt
(1/C)
The Lagrange
+(1/C)
?
?
115
(5.26)
which is exactly the same as Equation 5.8, the dynamic equation obtained using the
meshcurrents
approach for the same circuit.
Remark 5.7 (An
Alternative
Note that the
Lagrangian
Way of Addressing the Dissipated Power in
used in the Lagrange
equation,
Equation
Lagranges
5.21, consists
Equation)
of only the total
kinetic energy and the total potential energy of the system. The consideration of the dissipated energy
is included
in the virtual
generalized
force vector
An alternative
work described in
Q on the right-hand
way of addressing
Equation
5.20 and appears in
Equation
5.21 as part of the
side of the equation.
this issue is to add one dissipated
power term
the total power dissipated in the system. Then the Lagrange equation will be slightly
d
dt
??
?L
?q
?L
-
?D
+
?q
??q
D to represent
modifiedto:
= Qs
where Qsis the generalized external force vector acting onthe system.
Example 5.8 (Use of the
Modified Lagrange Equation
With the Dissipated Power Term D)
Here we will employ the same circuit in Figure 5.7, to demonstrate how to utilize the modified La-grange
equation,
Equation
The generalized
tively.
5.27 to assemble the dynamics
coordinate
and velocity
system
vectors are still
model for electric
q = ? q1 q2
The Lagrangian is also the same:
circuits.
?T ?T
and?q = ?q?1q?2 , respec-(5.27
1 -(0.5/C)(q1
-q2)2
L(
?q,q)=T(
?q,q)-U(q)
=0.5L
?q2
The difference is the term ?D ? ??q added to the left-hand side of the equation, and the Qsterm on the
right-hand
including
side now only consists of the generalized
the forces that cause the dissipated
external force
vector acting on the system
without
power.
Thetotal dissipated power of the system is the power dissipated in the resistor R, D = 0.5R?q2
2. The
?T
? ?? ? ? ?? ?
? ??(1/C)(q1-q2)? ??
generalized external force vector is Qs = ? es 0 . Since the current?q1flows out of the positive terminal
of the voltage source, the positive voltage esis located at the first entry of the vector Qscorresponding
to q1,?q1. Now,with the following key Lagrange differentials,
?L
?q =
?L
?q =
?L/? ?q1
?L/? ?q2
=
?L/?q1
?L/?q2
=
Lq1
?
,
0
?D
?q =
?D/? ?q1
?D/? ?q2
-(1/C)(q1-q2) ,
Qs =
0
=
?q2
R
Then, from the Lagrange equation, Equation 5.27, we have the dynamics
circuit as follows:
1 +(1/C)(q1
-q2)???
?Lq-(1/C)(q1
-q2)+R
?q2
es
=
As expected, the result is identical to that shown in
regular Lagranges equation, Equation 5.21.
0
(5.28)
es
0
model equation for the
(5.29)
Equation 5.25, which was obtained based on the
116
5
Modeling of Electrical
In the following,
we will
Systems
work on two
more circuit
modeling examples
Thefirst is the circuit on Figure 5.8a, which consists of four nodes, two
resistor,
one inductor,
electrical
system. If
and two capacitors.
we choose to use the
to the negative terminal
two
meshes,one voltage source, one
many ways to assemble a dynamics
NTD voltages approach,
by which the two
are found, the voltage and current
current law
will be applied to these two
NTD voltage variables
can be solved.
of each element in the circuit
Fig. 5.8: More examples on circuit
model for the
we would select the node connecting
of the voltage source as the datum, and assign the two capacitor
NTD voltages. Then Kirchhoffs
KCL equations
There are
using the Lagrange approach.
voltages as the
NTD nodes to yield two
Once these two
NTD voltages
can be easily derived.
modeling using the Lagranges equation approach.
If the meshcurrents approach is selected, we would assign a meshcurrent variable for each mesh,
and apply
solving
Kirchhoffs
voltage law to assemble two
these two equations
to obtain the two
KVL equations in terms
mesh currents,
of these two variables.
all the voltages and currents
After
can be easily
computed.
Both of the NTD voltages approach and the meshcurrents approach are carried out based on
the circuit geometry and the interactions among the components, like the Newtonian approach for
mechanical
systems
does not require
does require
modeling.
detailed
differential
The Lagrange
information
approach
of the geometry
calculus computations,
is quite
different.
and interactions
It is energy
based that
among the components.
It
but they are quite straightforward.
Two versions of the Lagrange equations are included in this Chapter. Thesetwo versions are basically
identical. The only difference is in how the dissipated power and the generalized force vector Q are
included in the equation. The dissipated powerterm ?D ? ??q andthe generalizedforce vector Qare
explicitly included in Equation 5.27, while in Equation 5.21 the information of power dissipation and
the generalized force vector Qis included in the virtual work. Either version will do the job.
Example 5.9 (Apply the Lagrange Approach to the Circuit on Figure 5.8a)
There are two meshesin the circuit. Assume mesh1 is on the left and mesh2 is on the right. The
electric charges traveling around mesh1 and mesh2 are q1 and q2,respectively, and their correspond-ing
meshcurrents are denoted by?q1 and?q2. Both meshcurrents are assumed in clockwise direction.
Although the meshcurrent direction can be assigned arbitrarily, it is better to follow the convention
whenever possible to avoid confusions. Since the positive current direction is assumed flowing out of
the positive terminal of voltage source according to the convention rules shown in Figure 5.1, the mesh
1 current should be in the clockwise direction.
?T ?T
The generalized coordinate and velocity vectors are q = ? q1 q2
and?q = ?q?1q?2 , respectively.
Thetotal kinetic energy is stored in the inductor L, which is T = 0.5L?q2
2. Total potential energy of the
2 in whichthe first term is contributed by the capacito
system
isU=0.5(1/C1)(q1
-q2)2+0.5(1/C2)q2
5.4
The Lagrange
Approach for
Circuit
Modeling
117
C1 and the second term is by C2. Hence the Lagrangian is
L= T-U =0.5L
?q2
2 -0.5(1/C1)(q1
-q2)2-0.5(1/C2)q2
(5.30)
2
The total power dissipation is from the resistor R, which is D = 0.5R
?q2
1. The differentials and the
generalized force vector involved in the Lagrange equation can befound as
?L
?q =
?L
?q =
? ?? ? ? ?? ?
? ??(1/C1)(q1
? ??
-q2)-(1/C2)q2
?L/? ?q1
?L/? ?q2
0
=
?L/?q1
?L/?q2
?q2
L
?D
,
?q =
?q1
R
?D/? ?q1
?D/? ?q2
=
-(1/C1)(q1
-q2)
=
0
,
Qs =
0
Then, according to the Lagrange equation in Equation 5.27, we havethe following
the electric circuit:
dynamics modelfor
-q2)+R
?q1
?(1/C1)(q1
???
q2-(1/C1)(q1
L
-q2)+(1/C2)q2
(i1-i2)dt
???
? (i1-i2)dt
+Ldi2
es
=
(5.31)
es
(5.32)
0
This equation can be rewritten in terms of the meshcurrents i1 and i2 asfollows:
Ri1 +(1/C1)
es
?
dt +(1/C2)
-(1/C1)?
? i2dt
=
(5.33)
0
The next circuit to be considered is shown in Figure 5.8b, which consists of five nodes,two meshes,
one voltage source, one resistor, two inductors, andtwo capacitors. Sincethere arefive nodes, if the NTD
voltages approach werechosen to solve the circuit problem, then three NTD voltage variables needto be
assigned and three KCL equations would haveto be set up and solved. Onthe other hand, the circuit has
two
meshesthat requires
approach
only two
will be more efficient
KVL equations.
than the
Hence, for this
NTD voltages
particular
circuit,
the
mesh currents
approach.
Example 5.10 (Apply the Lagrange Approach to the Circuit on Figure 5.8b)
There are two
meshes in the circuit.
electric charges traveling
around
Assume
mesh 1 and
mesh 1 is on the left
and
mesh 2 is on the right.
mesh 2 are q1 and q2, respectively,
and their
The
corresponding
meshcurrents are denoted by?q1 and?q2. Mesh1 current direction is assigned to be in the clockwise
direction to matchthe polarity of the voltage source. Mesh2s current direction is also assumed to be in
the clockwise
direction.
The generalized
coordinate
and velocity
vectors are q = ? q1 q2
?T ?T
and?q = ?q?1q?2 , respectively.
1 + 0.5L2?q2
2. The total
The total kinetic energy stored in the two inductors L1 and L2 is T = 0.5L1?q2
potential energy of the system is U = 0.5(1/C1)q2 1 +0.5(1/C2)q2 2. Hence,the Lagrangian is
1 +0.5L2?q2
L=T-U=0.5L1
?q2
2 -0.5(1/C1)q2
1 -0.5(1/C2)q2
2
(5.34)
The total virtual work done by the generalized forces, including the applied power sources and the
internal onesto cause power dissipation, are
dW=esdq1
-R(
?q1
-?q2)(dq1
-dq2)
The differentials
and the generalized force
vector involved
in the Lagrange equation
(5.35)
can be found
a
118
5
Modeling of Electrical
?L
Systems
? ?? ? ? ?? ?
?q1
-?q2)
?
??es-R(
?
R(
?q1
-?q2)
?L/? ?q1
??q =
?L/? ?q2
=
Lq1
?
,
?q2
L
?q =
?(dW)/?(dq1)
?(dW)/?(dq2)
Q = ?(dW)
?(dq) =
?L/?q1
?L
-(1/C1)q1
=
?L/?q2
-(1/C2)q2
(5.36)
=
Then according to the Lagrange equation in Equation 5.21, we havethe following
for the electric circuit.
dynamics
?q1-?q2)
?
?
??esR(-R(
?q1
-?q2)
?L2di2/dt
???
L1di1/dt
+R(i1
-i2)+(1/C1)?
i1dt
-R(i1
-i2)+(1/C2)
L1
q1
L2
q2
This equation
can be rewritten
+(1/C1)q1
+(1/C2)q2
in terms of the
(5.37)
=
mesh currents i1 and i2 as follows,
es
?
=
i2dt
5.5 Circuit
model
(5.38)
0
Modeling Usingthe State-Space Approach
Dynamic systems can be described by differential equations, transfer functions, or state-space represen-tations.
Due to the recent rapid
theory,
more and
advancement
more sophisticated
control
of computing
systems
tools
are required
and the state-space
to
be analyzed
control
systems
and designed
using
state-space approaches, and therefore the state-space representation of systems has become increasingly
important. Although a state-space representation can be obtained from differential equations or transfer
functions, it is morefavorable if a state-space representation can be directly constructed in the modeling
process whenthe meaningful physical variables like displacements and velocities can be easily selected
as state variables.
Fig. 5.9: Circuit
modeling using the state-space
approach.
For electrical systems, the order of the system or the number of state variables in the system is
determined by the number of energy-storing elements in the circuit. As mentioned earlier, inductors
store kinetic energy T = 0.5Li2L in magneticfields, and capacitors store potential energy U = 0.5Ce2
Cin
electric field. Hence,it is natural to select the currents of inductors and the voltages of capacitors
as state
variables.
NTD voltages
The circuit
approach, the
shown in Figure 5.9 has been employed
mesh currents approach,
and the
several times to demonstrate
Lagrange approach.
the
We will also use this
circuit to demonstrate how to assemble state-space modelsfor electrical systems by directly using the
state-space
modeling approach
5.5
Example
5.11 (State-Space
Modeling of the
The circuit in Figure 5.9 has three
Circuit
Circuit
nodes,
Modeling
on Figure
marked as red dots
Using the
State-Space
Approach
119
5.9)
with numbers
1, 2, and 3 for ease of
reference. Thereis one inductor L between node 1 and node 2, a parallel combination of one capacitor
C,and one resistor Rconnected between nodes 2 and 3, andfinally one voltage source between nodes
1 and 3 with node 1 asthe positive terminal. The current iL of the inductor L and the voltage eC of
the capacitor C are chosen as the two state variables of the system. To construct the state equations
associated with the inductor current iL, Kirchhoffs voltage law is employed to obtain a KVL equation
around the loop 1-2-3-1, consisting of the inductor L. According to Kirchhoffs voltage law, the algebraic
sum of the voltage drops around the loop 1-2-3-1 is zero. That meansthe voltage drop from node 1 to
node 2, LdiL/dt, should be equal to the algebraic sum of the voltage drops along the other path 1-3-2,
which
ises-eC.
Hence,
we
have
theKVL
equation
inthefollowing,
diL(t)
L
=-eC(t)+es(t)
dt
Meanwhile, Kirchhoffs current law is applied to give a KCL equation at node 2, whichis connected
to the capacitorC. According to Kirchhoffs current law, the algebraic sum of the currents of all the three
branchesleaving node 2is zero. That means
the currentleaving node 2 via capacitorC, CdeC/dt, should
be equal to the algebraic
sum of the currents
entering
node 2 via the inductor
L and via the resistor
R,
which
isiL +(-eC)/R.
Hence,
we
have
theKCL
equation
inthefollowing:
deC(t)
C
dt
1
=iL(t)- ReC(t)
These KVL and KCL equations can be rewritten as follows in the standard
equation,
iL
1/L
diL/dt
es
+
=
eC
0
deC/dt
matrix form of state
0 -1/L ??
?? ?
? ??1/C-1/(RC)
In general, the state-space representation
or state-space
(5.39)
model of a system is described
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
by
(5.40)
wherethe first equation is called the state equation, andthe second is the output equation. The vectors
x(t) and u(t) are the state vector and the input vector of the system, respectively, and y(t) is the output
vector consisting of the variables of interest. Forthe circuit example considered above, the state vector,
input vector, the A and B matricesare
x(t)
=
??
iL(t)
eC(t)
,
If eC(t) is the variable of interest
u(t)
= es(t),
(i.e., y(t)
A =
approach
we will work on two
,
B =
1/L
0
(5.41)
= eC(t)) then the C and D matrices in the state-space
model will be C = ? 0 1 ? and D = 0. The state-space
casesis referred as (A,B,C,D) for convenience.
In the following,
0 -1/L ? ? ?
?1/C-1/(RC)
more circuit
model described
modeling examples
in
Equation
5.40 in
using the state-space
many
model-ing
120
5
Modeling of Electrical
Systems
Fig. 5.10: Morestate-space modeling approach examples.
Example 5.12 (State-Space
Modeling of the Circuit on Figure 5.10a)
The circuit in Figure 5.10(a) has four nodes,
marked as red dots with numbers 1, 2, 3, and 4 for ease
of reference. There is one inductor L between node 2 and node 3, and two capacitors with C1 between
node 2 and node 4, and C2 between
node 3 and node 4. There also exists one resistor
R connected
be-tween
nodes 1 and 2, and one voltage source between nodes 1 and 4 with node 1 as the positive terminal.
The current iL of the inductor L and the voltages eC1and eC2of the capacitorC1 andC2, respec-tively,
are chosen asthe three state variables of the system. To construct the state equations associated
with the inductor current iL, Kirchhoffs
loop
2-3-4-2,
Kirchhoffs
consisting
Voltagelaw is employed to obtain a KVL equation around the
of the inductor
Voltage law, the algebraic
L, the
capacitor
C2, and another
sum of the voltage
capacitor
drops around the loop
C1. According
to
2-3-4-2 is zero. That
meansthe voltage drop from node 2to node 3, LdiL/dt, should be equal to the algebraic sum of the
voltage
drops
along
theother
path
2-4-3,
which
iseC1-eC2.
Hence,
wehave
theKVL
equation
inthe
following:
L
Meanwhile,
Kirchhoffs
diL(t)
=eC1(t)-eC2(t)
dt
current law is applied to give two
KCL equations:
one at node 2,
which
connects to C1,and another at node 3, whichis connected to the capacitor C2. According to Kirchhoffs
current law, the algebraic
meansthe current leaving
sum of the
currents
of all the three
branches leaving
node 2 is zero.
That
node 2 via capacitor C1, C1deC1/dt, should be equal to the algebraic sum
ofthecurrents
entering
node
2viatheinductor
Land
viatheresistor
R,which
is -iL +(es
-eC1)/R.
Hence, we havethe KCL equation in the following:
C1
deC1(t)
dt
=-iL(t)+
es(t)-eC1(t)
R
Similarly, the algebraic sum of the currents of all the two branchesleaving node 3is zero. That means
the current leaving node 3 via capacitor C2, C2deC2/dt, should be equal to the current entering node 3
via the inductor L, whichis simply iL. Hence, we have the KCL equation in the following:
C2
deC2(t)
dt
The one KVL equation and two
matrix form:
= iL(t)
KCL equations can now be combined into one state equation i
5.5
? ??
? ?=?
diL/dt
deC1/dt
deC2/dt
The next circuit
Circuit
Modeling
-1/C1-1/(RC1)0
1/C2
to be considered
0
is shown
0
in
Figure
State-Space
Approach
121
?? ?? ?
?? ?+
? ?es
1/L -1/L
0
Using the
iL
eC1
eC2
5.10(b),
0
1/(RC1)
0
(5.42)
which consists
of five
nodes, two
meshes,one voltage source, one resistor, two inductors, andtwo capacitors.
Example
5.13 (State-Space
Modeling of the
The circuit in Figure 5.10(b)
ease of reference.
has five
There are two inductors
Circuit
nodes,
on Figure
5.10(b))
marked as red dots
with L1 between
with numbers
1, 2, 3, 4, and 5 for
node 1 and node 2, and L2 between
node 2
and node 3. Atthe bottom of the diagram, there are two capacitors with C1 between node 4 and node 5,
and C2between node 5 and node 3. There also exist one resistor Rconnected between nodes 2 and 5,
and one voltage source between
nodes 1 and 4 with node 1 as the positive terminal.
The two currents iL1 and iL2 of the inductors
L1 and L2, respectively, and the two voltages
eC1 and eC2 of the capacitor C1 and C2,respectively, are chosen asthe four state variables of the
system. To construct the state equation associated with the inductor current iL1, Kirchhoffs voltage
law is employed to obtain a KVL equation around the loop 1-2-5-4-1, consisting of the inductor
the resistor
algebraic
R, the capacitor
sum of the voltage
C1, and the voltage source
drops around the loop
es. According
1-2-5-4-1
to
is zero. That
Kirchhoffs
L1,
voltage law, the
meansthe voltage drop from
node 1to node 2, L1diL1/dt, should be equalto the algebraic sum of the voltage drops alongthe other
path
1-4-5-2,
which
ises-eC1
+(iL2-iL1)R.
Hence,
we
have
thefirstKVL
equation
inthefollowing:
L1
diL1(t)
=-RiL1(t)+RiL2(t)-eC1(t)+es(t)
dt
To construct
the state equation
associated
with the inductor
current iL2,
Kirchhoffs
voltage law is
employed to obtain a KVL equation around the loop 2-3-5-2, consisting of the inductor L2,the capacitor
C2, and the resistor R. According to Kirchhoffs voltage law, the algebraic sum of the voltage drops
around the loop 2-3-5-2 is zero. That meansthe voltage drop from node 2to node 3, L2diL2/dt, should
beequal
tothealgebraic
sum
ofthevoltage
drops
along
theother
path
2-5-3,
which
isR(iL1-iL2)-eC2.
Hence, we havethe second KVL equation in the following:
L2
diL2(t)
dt
=RiL1(t)-RiL2(t)-eC2(t)
Meanwhile, Kirchhoffs current law is applied to give two KCL equations: one at node 4, which
connects to C1,and another at node 3, whichis connected to the capacitor C2. According to Kirchhoffs
current law, the current entering node 4 via capacitor C1, C1deC1/dt, should be equal to the current
leaving
node 4 for the voltage source es, which is iL1. Similarly,
the current flowing
through
the capacitor
C2, C2deC2/dt, should be equal to the current through L2, which is iL2. Hence, we have the two
equations
KCL
in the following:
C1
Thetwo
KVL and two
deC1(t)
dt
= iL1(t),
C2
deC2(t)
dt
=iL2(t)
KCL equations can now be combined into onestate equation in matrix form
122
5
Modeling of Electrical
Systems
? ??R/L2-R/L2 ?? ?? ?
? ?=?1/C10 0???+??es
?
?
diL1/dt
diL2/dt
deC1/dt
deC2/dt
-R/L1 R/L1 -1/L1 0
0
?
?
?
?
0
1/C2
0
Remark 5.14 (Significance
of the State-Space
As mentioned early in the beginning
become increasingly important
control
regarding
systems theory.
the state-space
0
1/L1
?
?
?
?
0
iL1
iL2
eC1
eC2
?
?
?
?
1/L1
0
0
0
(5.43)
?
?
Model)
of this section, the state-space
representation
of systems
has
due to the recent rapid advancement of computing tools and the state-space
Although
concept,
we have not yet officially
state equation
solutions,
introduced
and state-space
the fundamental
control
theory
system analysis
and
design, we did start to employ the state-space modelsto conduct stability analysis, stabilizing controller
design, and computer simulations for the simple inverted pendulum system in the previous chapter,
Chapter 4. It is clear that the state-space model control system analysis and design is important
and will be one of the main emphases of the book.
5.6 Operational Amplifier Circuits
The operational
amplifier,
usually
called
op amp, is an almost ideal
electronic
amplifier
due to its
large voltage gain, high input impedance, and low output impedance. Theseproperties are significant
in two aspects. First, it virtually has noloading effect, which makesthe op amp circuit a perfect building
block in an interconnected
operating
interconnected
system.
The op amp circuit
system.
Secondly, these three
will preserve its performance
properties
in any normally
are the foundation
that lead to the
development of the virtual short concept approach for the op amp circuit analysis and design.
Fig. 5.11: Equivalent circuit of the op amp.
The virtual-short
concept
of op amp circuits,
approach
which otherwise
has made it possible to greatly
would be extremely
simplify
complicated.
the analysis
and de-sign
Dueto these two reasons,
the op amp has become one of the most versatile and widely used electronic devices. The op amp cir-cuits
can perform a variety of functions like signal addition, substraction, integration, detection, am-plification,
filtering,
disturbance
cancelation,
The symbol
one output,
binarytodecimal
conversion,
decimaltobinary
conversion,
and common-mode
and so on.
of the op amp in
Figure 5.11 shows that the op amp has two inputs,
eo. Note that the common
reference terminal
and the
DC power supply
eA and eB, and
do not show on the
symbol. Thereal circuit inside the op amp chip consists of numerous transistors and other components;
however, it is not a concern for users. A muchsimpler equivalent circuit is shown on the right-hand sid
5.6
of the figure.
The op amp is basically
afour-terminal
the ground. The voltages at the input terminals
input
impedance
between the two input
is assumed flowing
at the terminal
into
terminal
device.
One terminal
Amplifier
Circuits
is the common
123
reference,
A and B are represented by eA and eB,respectively.
terminals
A. The output
with respect to the ground.
Operational
is ri, and the
terminal
is
positive input
marked
Between the output terminal
current
with eo as the
ii
or
The
direction
output
voltage
and the ground there is a series
combination
ofadependent
voltage
source,
K(eB
-eA),
and
theoutput
resistor,
ro,where
Kisthe
voltage gain.
If
wefollow
the traditional
circuit
analysis
and design approach,
in Figure 5.11 as a model of the op amp and set up KVL and
mesh currents,
or the state-space
modeling approaches.
we would use the equivalent
KCL equations
The equations
using the
circuit
NTD voltages,
would be complicated
even for
asimple
circuit,
butthey
can
begreatly
simplified
byletting
K?8,ri ?8,and
ro ?0.The
virtual
but the difference is that it will carry out
short concept approach also utilizes the approximations,
approximations
before the equations are set up.
Now,
from
theequivalent
circuit
inFigure
5.11,
with
theaccompanied
approximations,
K?8,
ri ?8,and
ro ?0,wewillsee
thefollowing
intriguing
properties
oftheopamp.
Theorem 5.15 (Virtual
Short Concept)
From the equivalent circuit in Figure 5.11, it is easyto deduct the following
statement:
I f ri~= 8,ro~=0, K~= 8, then ii~=0 andeA
~=eB
If the input impedance
ri is virtually
infinity,
then the current flowing
meansthe circuit between the two input terminals
right side ofthe equivalent
circuit,
through
A and Bis virtually
the output impedance
ro is virtually
(5.44)
ri is virtually
open circuit.
zero,
which
Meanwhile, on the
zero; hence, the output voltage eo
willbevirtually
equal
toK(eB-eA).
Since
thevoltage
gainKisvirtually
infinity
and
theoutput
voltage
eoisfiniteatthesame
time,
theonlypossibility
isthat(eB-eA)has
tobevirtually
zero.
That
is,the
circuit betweenthe two input terminals A and Bis virtually short circuit. Therefore, the circuit between
the two input terminals A and Bis virtually open circuit and short circuit at the same time!
The virtual short concept seems to be a paradox becauseit is impossible for a circuit to be open
circuit and short circuit at the same time. However, it is true that the circuit between the two input
terminals A and B of the op amp is virtually open circuit and virtually short circuit at the same time.
Remark
5.16 (A
Subtle
but Big Difference
It is easy to get confused
Between
with these two terms,
Short
virtual
Circuit
and
short circuit
Virtual
Short
and short circuit.
Circuit)
Short circuit
meansthat the voltages of the two terminals A and B are equal and at the same time the resistance be-tween
these two terminals is zero. Onthe other hand, virtual short circuit simply meansthat the electric
potentials, or the voltages, eA and eB, are virtually identical, but the resistance between terminals
A
and Bis virtually infinite. The difference is big, although it seems subtle.
In this section, we will employ a few op amp circuits to demonstrate how to utilize the virtual short
concept to assemble equations for modeling op amp circuits. Thefirst example is the general inverting
amplifier circuit shown in Figure 5.12.
Example 5.17 (Virtual
Short Concept Approach for the Inverting
The op amp circuit in Figure 5.12 is a general inverting amplifier,
Amplifier
Circuit)
wherethe two impedances blocks
Z1(s) and Z2(s) can be any combination of capacitors, inductors, andresistors. Owingto the virtual shor
124
5
Modeling of Electrical
concept, the current
Kirchhoffs
Systems
iA flowing
into
terminal
A of the
op amp is virtually
zero; hence, according
current law, the algebraic sum of the two currents flowing into terminal
to
A should be zero.
Hence
I1(s)+I2(s)
(5.45)
=0
Fig. 5.12: General op amp inverting amplifier.
Based onthe virtual short concept again, the two voltages eA and eB are virtually identical, which
meansterminal Ais virtually connected to the ground as shown in the figure by a dotted ground
sign. Hence,the current I2 is Eo/Z2 according to the generalized Ohms law. Similarly, I1 = Es/Z1.
Therefore, we havethe following transfer function of the circuit:
Es(s)
Z1(s)
Eo(s)
+
Z2(s)
=0 ?
Eo(s)
Es(s)
=
-Z2(s)
(5.46)
Z1(s)
Wehave just experienced the effectiveness of the virtual short concept approach and weresurprised
by how ridiculously simple the problem had become. It seems that we had done something unlawful
to connect terminal A to the ground and ignored the existence of the op amp. Of course, the virtual
short concept has been proved and the procedure is legitimate. To be moresure about this, wecan
employ another approach to the same circuit problem to double check the result. In the next example, we
will employ the traditional equivalent circuit approach to assemble the equations and obtain the transfer
function
for the same circuit
Example 5.18 (Traditional
of Figure 5.12.
Approach for the Inverting
Amplifier
Circuit)
After replacing the op amp in the inverting amplifier circuit of Figure 5.12 by its equivalent circuit
shown on the right-hand
set up the equations
The circuit
side of Figure 5.11,
without applying
shown in
Figure 5.13 has three
assemble the equations for the circuit,
three
we have the circuit
approximations
diagram shown in Figure 5.13.
We will
until the last step.
meshes. If the
mesh current
we would have to set up three
KVL
approach
were selected to
mesh equations to solve for
meshcurrent variables. Onthe other hand, the circuit hasfive nodes: (1) The one at the bottom
of the diagram is selected
as the ground reference;
(2)
Two are the positive terminals
of the input
volt-age
source
Esand
thedependent
voltage
source
K(eB
-eA);
and
(3)There
areonly
twoNTD
voltage
variables needed to be assigned, and only two KCL NTD equations are required to solve for these
two unknown voltage variables. Thesetwo NTD voltages are Eo,the output, and EA,the voltage at
terminal A.
Applying Kirchhoffs current law at terminal
A will yield the following
KCL equation
5.6
Es-EA+ Eo-EA=
Z1
Similarly,
at the output terminal
Amplifier
Circuits
125
EA
Z2
O we have another
Operational
ri
KCL equation
as
Eo-EA Eo-K(0-EA)= 0
+
Z2
ro
Fig. 5.13: Equivalent circuit of the general op amp inverting amplifier.
Note that the terminal Bis connected to the ground; hence, EB has been replaced by zero in the
equation. After eliminating EA,the two equations will reduce to the following:
Es
Z1
Eo
+
Z2
?
1
=
Z1
1
+
1
Z2
+
ri
??
?
ro-KZ2
ro
+Z2
Eo
Now,
apply
approximations
letting
K?8,ri ?8,and
ro ?0.Then
we
have
Es(s)
Eo(s)
Eo(s)
= -Z2(s)
+
=0 ?
Z1(s)
Z2(s)
Es(s)
(5.47)
Z1(s)
As expected, the result matchesthe one obtained using the virtual short concept approach.
only difference is that the traditional approach is more tedious and complicated.
Example 5.19 (Inverting
Amplifier and Integrator
As mentioned earlier, the Z2 and Z1impedance
in Figure 5.12 can be any combination
of resistors,
The
Circuits)
blocks of the general inverting
capacitors,
and inductors.
amplifier
circuit shown
If the impedances
blocks
are chosen to be Z2 = R2and Z1 = R1,then the output voltage will be equal to
Eo(s) =
-R2Es(s)? eo(t)= -R2
es(t)
R1
R1
(5.48)
The circuit can serve as a voltage amplifier if R2is chosen to belarger than R1.
If the impedances blocks are chosen to be Z2 = 1/(sC)
relationship
and Z1 = R,then the input-output
voltage
will be
Eo(s) =
It is clear that the circuit
-(1/sC)Es(s)
R
=
-1
sRC
performs integration.
Es(s)? eo(t)
= -1
RC
?
es(t)dt
(5.49
126
5
Modeling of Electrical
The general inverting
Example 5.20 (PID
Systems
amplifier
circuit
Analog Controller
can also be employed to implement
a PID analog controller.
Circuits)
In Figure 5.12, assume the impedance block Z2 is a series combination
capacitor
C2. Meanwhile,
Z1is a parallel
combination
of a resistor
of a resistor
R1 and a capacitor
R2 and a
C1. Thus,
we
have the impedances Z2 and Z1 as follows:
Z2 = R2 +
R2C2s+1
1
=
Then combine the two equations together
Es
=
-Z2
Z1
=-
?
Z1
1
+
R1C2
R1C2s
?
Therefore,
the circuit
can be employed
+R2C1s
?
amplifier
to implement
?:=- ?KP+KI +KDs?
1
s
+KD
des(t)
the function
dt
(5.50)
?
of proportional,
integral,
and
control.
Note that for those circuits
change,
R1C1s+1
R1
R1
es(t)dt
eo(t)=- KPes(t)+KI
derivative
+sC1 =
with Equation 5.46 and we have
R1C1 +R2C2
That is,
1
=
sC2
sC2
Eo
1
and
which
constructed
may not be favorable.
using the inverting
This issue can be fixed
amplifier
structure, there always is a sign
by cascading the circuit
by another inverting
with unit gain. It is also possible to design a noninverting amplifier based onthe op amp circuit
configuration
shown in Figure 5.14.
Example 5.21 (General
Non-Inverting
Amplifier
Circuit)
The op amp circuit in Figure 5.14 is a general noninverting amplifier, where the two impedances
blocks Z1(s) and Z2(s) can be any combination of capacitors, inductors, and resistors.
Fig. 5.14:
Owing to the virtual
General op amp noninverting
short concept, the current iA flowing
amplifier.
into terminal
A of the op amp is virtually
zero; hence, according to Kirchhoffs current law, the current flowing into terminal A via Z2 should
be the same as the current leaving terminal A via Z1into the ground. Meanwhile, the virtual short
concept implies that the voltage at terminal Ais virtually equal to Es, which is connecting to the
terminal B. Thus, we hav
5.6
Eo-Es=
Z2
and the transfer
function from
Es
Eo
?
Z1
Z2
Es
=
Es
+
Z1
Operational
Z2
Z1 +Z2
=
Z1Z2
1+
Z2(s)
Fig. 5.15:
Amplifier
op amp summing
(5.51)
Z1(s)
The next op amp circuit to be considered is a summing amplifier,
and substraction of signals.
Consider the
127
Es
? ?
=
Es(s)
5.22 (Summing
Circuits
Es(s) to Eo(s) is
Eo(s)
Example
Amplifier
Op amp summing
which can perform the addition
amplifier.
Circuit)
amplifier
circuit
in
Figure 5.15.
The terminal
A of the
op amp is
connected to four resistors, R2, R1a, R1b,and R1c.In addition, it is also virtually connected to the ground
according to the virtual short concept condition. Since the current flowing into the terminal A of the op
amp is virtually zero, the algebraic sum of the four currents from the four resistors entering into
the terminal A node (the ground) is zero according to Kirchhoffs current law.
eo
R2
e1a
+
R1a
e1b
+
R1b
e1c
+
R1c
=0
Therefore, the output eo(t) is the algebraic sum of the three incoming
?
e1b(t)
e1a(t)
eo(t)=-R2 R1a
+
R1b
e1c(t)
+
R1c
?
weighted signals,
(5.52)
Thefollowing op amp circuit is a differential amplifier, which can be employed to eliminate the
common-mode disturbances during signal transmission.
There are two input signals, but only the
difference of the two signals will affect the output.
Example 5.23 (Differential
Amplifier
Circuit)
Consider the op amp differential amplifier circuit in Figure 5.16. Onthe upper half of the circuit
diagram, the terminal A of the op amp is connected to the first input e1 and the output eo via resistors
R1 and R2,respectively. Onthe bottom half of the circuit diagram, the terminal B of the op amp i
128
5
Modeling of Electrical
Systems
Fig. 5.16: Opamp differential amplifier.
connected
to the second input
e2 and the
ground
via another set of resistors
According to the virtual short concept, the current flowing into terminal
op amp is virtually
the terminal
zero.
Hence, the algebraic
sum
R2, respectively.
from
e1 and from
B of the
eo entering
A node is zero,
e1-eA+ eo-eA=0 ?
R1
R2
Similarly,
of the two currents
R1 and
A or out of terminal
the algebraic
sum of the two currents
? ?
e1
1
R1 -
from
R1
e2
R1
-
eA +
R2
e2 and from
B node is zero,
e2-eB+ -eB
R2 =0 ?
R1
1
+
eo
R2
the ground
=0
entering
? ?
1
R1
1
+
R2
eB = 0
From the two equations and the fact that the two voltages eA and eB are virtually
following differential amplifier input-output
relationship:
eo(t)
=
R2
R1
the terminal
equal, we have the
(e2(t)-e1(t))
(5.53)
5.7 DC Motor
Recall that in
dynamic
Figure
2.1, the
system examples.
DC motor system
Some students
was considered
as one of the three typical
might have been wondering
first-order
how it can be possible since the
DC motorseemsto be a complicated system, consisting of an electric circuit and arotational mechanical
system. Indeed, the internal structure andthe detailed theory explaining how it works may be alittle bit
complicated,
input
but in
control
most of the applications
voltage can be described
the relationship
by a first-order
between the
differential
angular
velocity
and the
equation.
A schematic diagram of a DC motor system is shown in Figure 5.17. Onthe right-hand side of
the diagram is a rotational
mechanical system including
the gear train
and the load that consists
of the
moment of inertia J? and the rotational damper B?. Onthe left-hand side of the diagram is the DC motor,
whose dynamic
behavior
will be determined
by a circuit
equation
involving
ea, La, Ra, and eb, and
a mechanicalequation of motioninvolving tm, Jm, and Bm. Before westartto assemblethe equations,
we will digress a little bit to refresh our memory on some relevant fundamental electromagnetism
learned from high school physics courses
we
5.7
Fig. 5.17:
5.7.1 Amperes
Force Law and Faradays
DC
Motor
129
DC motor system.
Law of Induction
Thefirst relevant electromagnetism law is Amperes force law regarding the interaction
between two
electric wires with currents. Should this sound familiar with the name Ampere? Yes,it is the unit of
electric current named after Andre-Marie Ampere, one of the greatest pioneers in electromagnetism.
Amperes force law was discovered in 1825, which revealed that the two parallel electric wires would
experience forces after being energized with electric currents. The forces would either attract the two
wiresto each other or do the opposite, expel each other, depending on the relative current directions. It
was a great discovery that manifested the possibility of converting electrical energy into mechani-cal
energy.
The DC motoris equipped with two
wound-up electric coils. Thefirst coil is called the field coil
that is fixed in the structure andis driven by a constantcurrent If to generatea constant magneticflux
Ff= kfIf. Thesecondcoil, calledthe armature coil, is woundaroundthe motorrotor, whichis designed
to be ableto rotate inside the magneticfield provided by Ff. Therotor will rotate whenthe armature
coil is energized. Therotor torque is proportional to both the magneticflux Ff and the armature
current ia(t),
tm(t) =c1Ffia(t) := Kmia(t)
?
Tm(s) = KmIa(s)
(5.54)
where Tm(s)andIa(s) arethe Laplacetransforms of tm(t) andia(t), respectively. This constant Kmis
called the torque
constant,
which usually is available
on the data sheet of the
DC motor.
The second relevant electromagnetism law we need to review is Michael Faradays law of elec-tromagnetic
induction,
which in the year of 1831 revealed how the variation of magnetic flux will
produce an electromotive force (EMF). This discovery is as significant as Amperes force law, if not
greater. Thelaw of induction is the underlying principle of the AC electric generators that convert
mechanical power like steam, hydraulic, or even nuclear power into electric power. In addition,
many other important
the inductors,
inventions
like
the induction
the
AC voltage
transformers,
motors, the solenoids,
the
and radio
AC electric
power
trans-missions,
wireless communications,
would not be possible without the knowledge of electromagnetic induction.
Now, back to the DC motor schematic diagram in Figure 5.17, there is a back EMF (electromotive
force) voltage eb developed across the two terminals of the DC motor. This back EMF is caused by the
variation of the magneticflux the armature coil has received. Although the field magneticflux Ff is
constant, the
magnetic flux
received
by the armature
is moving. Theback EMF voltageebis proportionalto
eb(t) = c2Ff?m(t) := Kb?m(t)
coil is a function
of time
since the
motor rotor
Ff andthe angularvelocity of the rotor, ?m(t),
?
Eb(s) = KbOm(s)
(5.55
130
5
Modeling of Electrical
Systems
whereEb(s)and Om(s)arethe Laplacetransforms of eb(t) and ?m(t), respectively. This constant Kbis
called the back EMF constant,
5.7.2 Assembling
In the schematic
whichis usually available on the data sheet of the DC motor.
Equations for the DC Motor System
diagram
of the
motor system in Figure 5.17, there is an electric circuit loop
on the left-hand
side of the diagram. Thisloop includes the control-input voltage source ea(t), the inductor La,the
resistor Ra,and the back EMF voltage eb(t). Applying Kirchhoffs voltage law, we havethe following
KVL equation:
La
Taking
Laplace transform,
dia(t)
+Raia(t)+eb(t)
dt
this equation
= ea(t)
becomes
sLaIa(s)+RaIa(s)+Eb(s)
Now, on the right-hand
by the
side of the diagram in
DC motor is employed
= Ea(s)
Figure
to drive a rotational
5.17,
we can see that the torque
mechanical system
discussion of gear train in Section 4.3, we havethe equivalent
(5.56)
via the gear train.
generated
Based on the
moment of inertia Je andthe equivalent
rotational friction coefficient Befrom the perspectiveof tm asfollows:
Je = Jm+
Then the equation for the rotational
??2
N1
J?,
N2
Be = Bm+
??2
N1
B?
N2
(5.57)
motion is given by
(Jes+Be)Om(s)
(5.58)
= Tm(s)
where Om(s)and Tm(s)arethe Laplacetransforms of ?m(t) andtm(t), respectively.
The electrical circuit equation, Equation 5.56, and the mechanical rotational
motion equation,
Equation 5.58, are coupled to each other via the torque equation, Equation 5.54 and the back EMF
equation, Equation 5.55. Thesefour equations are combined to yield the following:
sLaIa(s)+RaIa(s)+KbOm(s) = Ea(s)
(5.59)
(Jes+Be)Om(s)
In
many applications,
to the impedance
the impedance
of the resistor,
= KmIa(s)
of the inductor
R; hence, the first
Ia(s)
=
sLa is negligible
equation
since it is very small compared
of Equation
5.59 can be rewritten
Ea(s)-KbOm(s)
as
(5.60)
Ra
Substitute this expression for Ia(s) in the second equation of Equation 5.59to yield the following trans-fer
function between the control-input voltage source Ea(s) and the angular velocity of the DC motor
system, Om(s).
Om(s)
Km
b
?ss
(5.61)
=
=
=
RaJes+BeRa +KmKb
s+a
ts+1
Ea(s)
wherethe parameters a, b, time constant t, and the steady-state step response ?ss,are given asfollows:
a =
BeRa +KmKb
RaJe
Be
=
Je
+
KmKb
,
RaJe
b =
Km
,
RaJe
t
=
1
a
,
?ss =
b
a
(5.62
5.7
5.7.3
Torque-Speed
DC
Motor
131
Relationship
Some DC motor manufactures mayinclude the torque-speed relationships or graphs onthe datasheet of
the
DC motor. The torque-speed
graph
will look like the one shown in Figure 5.18.
Fig. 5.18: Torque-angular velocity graph for the DC motor.
From the torque
equation,
Equation
5.54,
we have Ia(s)
= Tm(s) ? Km. Substituting
it into
Equation
5.60 will yield the following torque-speed equation:
Tm(s)+
KmKb
Ra
Om(s) =
Km
Ra
Ea(s)
KmKb
? tm(t)+
Ra
?m(t) =
At the steady state with a constant voltage input,
we have the following
among the torque tm, angular velocity ?m, and the input voltage ea:
-KmKb
?m+
tm =
Km
ea(t)
(5.63)
Ra
algebraic relationship
Km
Ra
Ra
ea
(5.64
When ?m = 0, we have
tstall =
Km
ea
Ra
?
tstall
Km
Ra
=
(5.65)
ea
Onthe other hand, when tm = 0, we have
1
ea
?no-load
=Kbea ? Kb=?no-load
From the torque-speed relationship,
Equations 5.61 and 5.62.
5.7.4
A DC
Micromotor
we can compute the parameters
(5.66)
Km/Ra and Kb required in
Example
The DC micromotor to be considered is a Faulhaber 2230.012S micromotor with 1:14 gear head. First,
we will read the data sheet and associate the data sheet values
in
Equations
5.61 and 5.62.
A partial
data sheet
with the
with values of the
DC motor model equations
micromotor
shown
is given in Figure 5.19.
The data values shown here are fundamentally identical to those on the manufacturers data sheet. We
only slightly
revise the no-load speed from
9,500 rpm to 9,550 rpm and the stall torque
from
13.2
mNm
132
5
Modeling of Electrical
to 13.3
mNm in order to exactly
Systems
match the values of the torque
constant
Kb. The small discrepancies among the data sheet values mostlikely
Km and the back EMF constant
werecaused by rounding errors.
The manufacturers data sheet only shows rpm (revolutions per minute) asthe unit for rota-tional
speed. The unit rpm is moreintuitive for humans than the rad/s unit. However,in dynamics
system
analysis,
and errors.
design,
In
and computation,
Remark 4.16,
we have to use the
we had a brief
discussion
unit rad/s to avoid
on a similar
unit inconsisten-cies
issue regarding
degrees and
radians. On many occasions, we have no choice but to use both units, one for display to humans and
another for computation
angular velocity
in
machine. Hence, in Figure 5.19, we showed
both units, rpm and rad/s, for the
?.
Fig. 5.19: Partial data sheet values of Faulhaber
micromotor
2230.012S.
Intheanalysis,
computation,
and
design,
wewilluse,
forthefollowing
data
values
of?no-load
and
?maxin rad/s, and Kbin
mV/rad/s:
?no-load
=1,000
rad/s,?max
=1,152
rad/s,andKb=12mV/rad/s
instead of those in rpm or in
mV/rpm on the
manufacturers
data sheet:
?no-load
=9,550
rpm,?max
=11,000
rpm,andKb=1.25mV/rpm
The back EMF constant Kb was 1.25 in old mV/rpm unit, and now Kb equals to 12 in the new
mV/rad/s unit. Notethat now the back EMF constant, Kb = 12 mV/rad/s, and the torque constant,
Km = 12 mNm/A have the same value, 12. It is not a coincidence. Thereis a significant physical mean-ing
behind the equivalency of these two constants. Kb =Km meansthat the electric power ebiainto the
DC motor equals to the mechanical power tm?m out of the DC motor. That meansthe power transfer
is lossless. In mostcases, the loss of this electric power to mechanical power conversion is negligible.
Therefore, these two constants should be very close, if not equal. If the units used were not consistent,
the equivalency of these two constants would not berevealed.
Theinductance of the armature inductor, La = 430 H, andthe rotation frequency of the rotor is less
than ?max = 1,152 rad/s; hence, the magnitude of the impedance of the inductor
will beless than
|j?maxLa|
=0.495O,which
is negligible
compared
tothearmature
resistance
Ra=10.8O.This
verifies the underlying assumption in the previous subsection that the impedance of the inductor sLa i
5.7
DC
Motor
133
negligible.
The manufacturers
data sheet not only provides the values of torque constant Km,the back
EMF constant Kb, and the armature resistance Ra, but also gives the information
of the no-load
speed
?no-load
and
stall
torque
tstall
shown
in Figure
5.20.
We
willverify
thatallthese
data
values
are consistent.
Asit can be seen from the graph,
when the input
voltage is kept at 12 V, the relationship
betweenthe torque tm andthe angular velocity ?mis describedbythe straightline connectingthesetwo
points (1000,0) and (0,13.3). From Equations 5.65 and 5.66, we can compute the values of Kmand Kb
based on the information
Km =
tstall
ea
of the stall torque
13.3
Ra =
12
and no-load speed as follows:
10.8 = 12 mNm/A,
ea
Kb =
12
=
?no-load
1000
= 12 mV/rad/s
Fig. 5.20: Atorque-speed graph of Faulhaber micromotor 2230.012S.
Now,
we have all the parameters
inertia Jm and rotor friction
relevant to the
DC motor dynamics
coefficient Bm. The rotor friction
except the rotor
is negligible,
but the rotor
moment of
moment of
inertia
isJm=2.710-7
kgm2.
In the following,
we will consider the dynamics of the DC motor, only without the geartrain andthe
external load.
Example 5.24 (Analysis
of the DC Motor Without Load)
Thetransfer function of the DC motor system without the external load can be obtained from Equa-tions
5.61 and 5.62,
Om(s)
b
?s
Ea(s)
=
s+a
=
ts+1
wherethe parameters a and b are
a =
KmKb
RaJe
=
121210-6
Km
10.8
2.710-7
=49.38,
b=
RaJe
=
1210-3
10.8
2.7
10-7
=4115
andthe time constant t andthe steady-statestepresponsevalue ?ssare
t =
1
49.38
= 0.0203s,
?ss =
b
a
= 83.33rad/s
That is,
Om(s)
Ea(s)
4115
=
s+49.38
83.33
=
0.0203s+1
(5.67)
134
5
Modeling of Electrical
Systems
Fig. 5.21: Stepresponse of the DC micromotor without gear train or external load.
Rememberwhat welearnedin Chapter2?If the input ea(t)is a unitstepfunction (i.e., Ea(s) = 1/s),
we can solve for ?(t) by using the Laplace transform approach. Wecan even draw the response graph
and write down the solution without any computation just by inspection and reasoning. Thefirst-order
system step response graph is determined by only three points: the initial condition at t = 0, the final
steady-state
response
ast ?8,and
thetime
response
att =t.For
thisproblem,
theinitial
condition
is
zero
when
t =0and
thefinal
steady-state
response
is?ss,
which
is83.33
rad/s,
as
t ?8.The
response
curve
inbetween
isanexponentially
rising
curve,
which
has
tobeintheformA(1-e-t/t),
where
tis
the time constant, and Ais the steady-state value. Therefore, the step response of the system is
?m(t)
=?ss(1-e-t/t)
and
itsassociated
step
response
graph
isshown
inFigure
5.21.
Note
that?m(t)
=?ss(1-e-1)
=
0.632?ss. Thetime constant is t = 20.3 ms. Theresponseis very fast. If the input is a step function
with amplitude 12 (i.e., Ea(s) = 12/s), then the waveform shape will bethe same but the steady-state
value will be 12times aslarge; ?ss will be 1,000 rad/s.
Example 5.25 (Analysis
of the DC Motor with Gear Train and Load)
Consider the same DC micromotor,
The moment of inertia
and the rotational
but the
motor
friction
will drive an external
coefficient
load
via a 1:14 gear train.
of the external load are
J?=5.410-5
kgm2,
andB?=2mNm/rad/s
respectively.
Then the equivalent
Be are
Je = Jm+
moment of inertial
Je and the equivalent
rotational
friction
coefficient
??2
J?=5.45510-7,
Be=??2
B?=1.0210-5
1
1
14
14
Thetransfer function of the DC motorsystem with gear train and the external load can be obtained
from
Equations
5.61 and 5.6
Om(s)
Ea(s)
wherethe parameters a and b are
?ss
b
=
s+a
=
ts+1
5.7
a =
Be
Je
KmKb
+
RaJe
= 18.7+24.44
= 43.14,
b =
Km
RaJe
=
DC
Motor
135
1210-3
10.8
5.455
10-7
=2037
and the time constant t and the steady-state step response value ?ss are
t =
1
43.14
= 0.0232s,
?ss =
b
a
= 47.22rad/s
Thatis,
Om(s)
Ea(s)
=
2037
s+43.14
=
47.22
or
0.0232s+1
O?(s)
Ea(s)
=
145.5
s+43.14
=
3.373
(5.68)
0.0232s+1
Thestepresponseofthe systemfor ??(t), the angular velocity of the load can befound as
??(t)=3.373(1-e-t/0.0232)
rad/s=32.21(1-e-t/0.0232)
rpm
(5.69)
The time constant is t = 0.0232s, which is not much different from the no-load dynamics time
constant, t = 0.0203s. However, the steady-state angular speed has dropped tremendously, from
83.3rad/s to 3.373rad/s, which is about only 4% of its no-load angular speed. Thereduction of speed
mainly comes from the 1:14 gear ratio. The heavy load, whose moment of inertia is about 200 times of
that of the motorrotor, also contributes to slow down the rotation speed of the motor. Theloss of 96%
of the no-load speed seems terrible, but it is certainly
worthwhile for the gain in 14 times more
torque and ability to handle a heavy load with 200 times more moment of inertia than the motor
itself. Even at 4% of the no-load speed, the motor system still can movethe load up to the speed of
380 rpm, which is enough in many applications.
Fig. 5.22: Feedback speed control
of the
DC motor system
using integral
control.
In the following,
we will design a feedback speed control system based on the dynamics
obtained from Example 5.25.
Example 5.26 (Speed Control of the DC Motor System Using Integral
model
Controller)
The block diagram of the feedback control system is shown in Figure 5.22. Onthe left-hand side of
thediagram,
the
reference
input
signal
?r(t)
isfed
intoalittlecircle
with
+and-signs,
which
iscalled
the summer, a devicethat can add or subtract signals. The output signal ??(t), whichis the angular
velocity to be controlled, is fed back into the same summer with a negative sign sothat the output signal
ofthesummer
isthedifference
ofthetwoincoming
signals
?r(t)-??(t).
This
difference
signal
in
turn becomesthe input signal to the controller K(s). The controller will process the information,
make
decisions, and send out a control signal ea(t) to either increase or decreasethe speed of the motor until
??(t) is equalto ?r(t)
136
5
Modeling of Electrical
Systems
In general, the controller
K(s) is to be designed so that the closed-loop system is stable, the
steady-state error is zero if possible, and the transient error is as small as possible. Detailed dis-cussion
on stability
will be given in later chapters.
Roughly speaking,
stability
means that
no signals
withinthe system will grow without bound,and zerosteady-stateerrorimplies ??(t) will beequal
to ?r(t) at a steady state. A minimization of the tracking error representsa requirement to reachthe
goal as quickly as possible without much overshoot or oscillations.
tracking
controller
like this
one. In this example,
There are many ways to design a
we will use a very simple
design approach
without
much mathematics.
The first step is to find the overall transfer
function
of the closed-loop
system
between the reference
input Or(s) and the output O?(s). Therelationships among signals in frequency domain are all algebraic
and can be easily manipulated. Just by inspection of the diagram, we have following two equations:
O?(s)
=G(s)Ea(s)
andEa(s)
=K(s)
[Or(s)-O?(s)]
These two
equations
can be combined into
one by eliminating
the internal
variable
Ea(s).
Then,
with a
few moving around terms within the equation, we havethe following transfer function:
O?(s)
Or(s)
G(s)K(s)
=
(5.70)
1+G(s)K(s)
Plugging the expressions of K(s) = Ki/s and G(s) into the closed-loop transfer function,
145.5Ki
O?(s)
Or(s)
=
s(s+43.14)
145.5Ki
1+ s(s+43.14)
145.5Ki
=
=
s2 +43.14s+145.5Ki
we have
?2
n
s2 +2??ns+?2n
(5.71)
Notethat this closed-loop transfer function belongs to the category of the typical second-order system
westudied in Chapter 3. Recallthat the dynamic behavior of the typical second-order system is charac-terized
bythe dampingratio ? andthe naturalfrequency ?n.From Equation5.71, wehavethe following
two equationsthat relatethe dampingratio ? andthe naturalfrequency ?nto the integral constant Ki:
?2
n = 145.5Ki and
Fig. 5.23: Simulation
results
Dueto the restriction
to be determined,
of the feedback
of the integral
speed control
2??n = 43.14
of the
DC motor system using integral
control.
control structure that there is only one design parameter Ki
we only have one degree of freedom
in choosing
the damping
ratio and the natura
5.8
Exercise
Problems
137
frequency. If wechoosethe damping ratio to be 0.9, ? = 0.9, then the natural frequency will be
?n = 43.14/1.8 = 23.97rad/s and the integral gain should be Ki = 3.95.
Withthe integral control constant determined, the block diagram in Figure 5.22 can be employed
to build a simulation program and conduct simulations of the feedback speed control of the DC motor
system
using the integral
control
approach.
The simulation
results
are shown in Figure
5.23.
With the
damping ratio chosento be ? = 0.9,there will be, as expected,an almost undetectableovershootand
virtually
t
no oscillations. The step response reaches steady state with no steady-state error shortly after
= 0.2 s. The rise time
frequency
can be faster if there is another
in the design process.
control
structure
with two
We will revisit
design
this
degree of freedom
problems
parameters
later in
will be employed
to select a higher
Chapter
in
natural
6. A dual-loop
Example
feed-back
6.1 to achieve
a
better performance.
The control-input
plots are also shown in Figure 5.23. It can be seenthat whenthe angular speed of
the load, ??(t) reaches20 rad/s,the control input ea(t) is around 6 volts voltage. Similarly, when??(t)
follows anotherreferenceinput ?r(t) = 40 rad/sto get upto 40rad/s,it requires about 12 voltsfrom the
voltage source.
The control
inputs
always
have their
The system needs to work within the capability
perform poorly or become unstable.
limitations
in
magnitude
of the control inputs;
or in rate of change.
otherwise, the system
may
5.8 Exercise Problems
P5.1a: Findthe transfer function G1(s) = Eo(s)/Es(s) ofthe op amp circuit shown in Figure 5.24(a)
and explain the function and application of this circuit.
Fig. 5.24:
Op amp circuits.
P5.1b: Findthe transfer function G2(s) = Eo(s)/Es(s) ofthe op amp circuit shown in Figure 5.24(b).
Then compute the magnitude and phase of G2(j?)
system.
and explain the frequency-domain
behavior of the
P5.2: Assumethe DC motoris Faulhaber micromotor 2230.012S. You can find the data sheet values of
this DC motorfrom Figure 5.19. The DC motor system under consideration is shown in Figure 5.17.
Let
thegear
ratio
beN1/N2
=1/10,
and
J?=210-5kgm2,
B?=1mNm/rad/s.
Find
thetransfer
function
G(s) = O?(s)/Ea(s).
P5.3: Usethe transfer function G(s) = O?(s)/Ea(s) you obtained in Problem P5.2. Design an integral
controller K(s) = sK so that the closed-loopsystemshown in Figure 5.25 hasits characteristic equatio
138
5
Modeling of Electrical
Systems
with damping ratio ? = 0.8. Thenevaluatethe performanceof the closed-loopsystem by conducting
computer simulations withthe referenceinput ?r(t) = 40us(t) rad/s. Plotthe output response ??(t)
and the control input ea(t), and give your comments based on the simulation results.
Fig. 5.25:
A DC motor speed control
system.
P5.4: Consider the circuit shown in Figure 5.26, where eo(t) and es(t) are the output and the input,
respectively. Assign eC and iL as state variables, then find the state-space model of the electrical system.
Fig. 5.26: Find a state-space model of the RLC circuit.
P5.5: Consider the circuit shown in Figure 5.26, where eo(t) and es(t) are the output and the input,
respectively.
Let Eo(s) and
Es(s) be the
Laplace transforms
of eo(t)
and es(t), respectively.
Find the
transfer function Eo(s)/Es(s) Usingthe NTD voltagesapproach.
P5.6: Repeat Problem P5.5 using the meshcurrents approach.
P5.7: Repeat Problem P5.5 using the Lagrange approach.
P5.8: For the circuit shown in Figure 5.26, which hasjust been considered in Problems P5.5, P5.6,
and P5.7, assumethe component values of the resistors, the capacitor, and the inductor are: R1 = 1kO,
R2 = 10O, L = 2H, and C = 100F. Findthe transfer function Eo(s)/Es(s), andthe dampingratio ?
andthe naturalfrequency?nofthe system
6
Systems Representationsand Interconnected Systems
T
HEREis no question that mathematical dynamics models are essential in the study of dynamic
systems analysis, design, and control. Without a clear and truthful
description of a dynamic
system it would be very difficult or even impossible to understand how the system works,
how to fix problems if some engineering issues arise, or how to design and build a better system.
For this reason, we have to understand how physical dynamic systems are described by their mathe-matical
dynamic model equations. Onthe other hand, for every equation, engineering law, or scientific
theory
we learn,
we should
associate them
with some technical
reasoning,
experience,
lab
works, or
virtual experiment for verification.
As welearn from the previous chapters, although the dynamic system modelequations can be derived
using different approaches, the end results are essentially identical. However, dynamic system represen-tations
are not unique. For example, the simple RC circuit discussed in Chapter 2 can be represented by
afirst-order differential equation in the time domain or byits corresponding frequency-domain descrip-tion
as a transfer
function.
These two representations
the time domain, the differential
system,
different
equation reveals the charging
while the frequency-domain
property
provide
transfer
function
aspects of the same system. In
and discharging
helps us understand
activities of the
the low-pass
filtering
of the system.
It seems that the time-domain behavior and properties are moreintuitive than their counterparts in
the frequency domain; one would think the time-domain approaches would dominate the development
of the control system theory and technology in earlier years. But history showed the opposite: Mostof
the classical control
period
theory,
of almost a hundred
analysis,
and design tools
years since James
published in 1868 until late 1950s
Clerk
when the optimal
were developed in frequency
Maxwells
historic flyball
control theory
domain
stability
during the
analysis
and the state-space approach
paper
started
to emerge.
Due to the advancement
and the evolving
computer
representation
complicated
of the state-space control theory, the
technology,
has become
multivariable
control
the time-domain
more significant,
systems.
state-space
especially
matrix computing
approach
in its ability
The frequency-domain
algorithms,
based on the state-space
of handling
multivariable
the
control
more
approach
also madetremendous progress in 1970s and 1980s in addressing the robust stability and robust per-formance
issues
function
based on the transfer
matrices requires
matrix computations
The polynomial
function
polynomial
matrix representation.
matrix computations,
which is
The manipulation
more tedious
of the transfer
than the
constant
in the state-space approach.
matrix computation
transfer function
issue
was eliminated
in the late 1980s by converting
the un-derlying
matrix representation into the state-space representation and re-deriving th
140
6
Systems
optimal
robust
Representations
control
solution
and Interconnected
in state space.
is not a purely time-domain
approach
on the state-space
the frequency-domain
Systems
Hence, the state-space
control
approach. It is a merged time-domain
framework
for efficient
and time-domain
components
and reliable
systems
to-day
and frequency-domain
computation,
to be considered
approach
in the
yet allows
analysis
both of
and
design
process.
In this chapter,
we will present an overview
of the common
systems representations,
including
differential equations, the transfer functions, the state-space representations, the simulation
the block
diagrams,
representations,
interconnected
and the signal flow
and explain
graphs.
We will discuss the relationship
how to find the transfer
system of interest
function
or the state-space
that are usually required in analysis
the
diagrams,
among the system
representation
of an
or design.
Fig. 6.1: A simple block diagram with feedback structure.
6.1 Block Diagrams
Even before we officially define the term block diagram wehad usedit on two occasions: onein Figure
3.15 for DC motor position control, and another in Figure 5.22 for DC motorspeed control. Thesetwo
block diagrams look similar to the one shown in Figure 6.1. Assumethe plant, which meansthe system
to be controlled, is a typical first-order system G(s) representing the transfer function of a DC motor
whose output is the angular
integral) controller
velocity,
the input is a control
voltage signal,
and the PI (proportional
plus
K(s) are given in
G(s) =
b
s+a
and
K(s) = Kp +
Ki
s
respectively, whereb and a are given constants, and Kpand Ki arethe proportional andintegral constants
in the controller to be designed.
A block
diagram is an effective
interconnection relationship
graphical representation
of an interconnected
system that shows the
among the subsystems and components in the system. The block diagram
in Figure 6.1 shows that the output y(t) of the plant G(s)is the variable (the motorspeedin this case) to
be controlled. Thereference input r(t) represents the desired motor speed fed into the summer, which
isrepresented
byalittlecircle
with+and-signs
around
it foraddition
orsubstraction.
The
actual
speed variable y(t) is fed back into the negative terminal
of the summer so that the difference signal
e(t)=r(t)-y(t) willbesent
tothecontroller
K(s).
The
controller
K(s)willfollow
thecontroller
law
(the PI control law in this case) based on the difference input to
make decisions
and send out the control-input
signal to either increase or decreasethe speed of the motor system until the actual speed equals the
desired speed commanded by the reference input.
In the design and analysis of the feedback control system, we mayneedto find the closed-loop trans-fer
function Y(s)/R(s), or any other representationdescribingthe relationship betweenthe reference
input
r(t)
and the output y(t). If the transfer
function
of each block is available, the computation
of th
6.1
overall transfer
function
will be fairly
straightforward
since it only involves
this example, from the block diagram in Figure 6.1, we havethe following
Block
algebraic
Diagrams
141
manipulations.
For
two algebraic equations,
Y(s)= G(s)U(s)
and U(s)=K(s)
[R(s)-Y(s)]
Then these two equations can be combined into the following transfer function by eliminating the inter-nal
variable U(s):
Y(s)
R(s)
1+G(s)K(s)
Kps+Ki
b
G(s)K(s)
=
s+a
=
1+
s
b
s+a
Kps+Ki
s
bKps+bKi
=
s2 +(a+bKp)s+bKi
Fig. 6.2: Common basic block diagram connections: (a) cascade connection, (b) parallel connection, (c)
feedback connection.
Block diagram
connections
are pretty self-explanatory.
They consist
of several blocks that represent
subsystems or components. Each block is assumed to have an input terminal to receive an incoming
signal and an output terminal to send an outgoing signal. Signals are represented by arrows with ar-rowheads
on their tips showing
the signal flow
directions.
Signals
can be added or subtracted
to form
anew
signal
using
asummer,
which
isrepresented
byalittlecircle
with+and- signs
around
itto
determine whether the sign of the incoming
signal also can be tapped to create a new
or summer. Unlike the split of water flow
informationthe
signal magnitude wont
Three common
basic block
diagram
signal needs to be changed before entering the summer. A
branch to send exactly the same signal to another block
or electric current, the tapping here is just a sharing of
be affected.
connections
are shown in Figure 6.2. The block
diagram in (a)
is a cascade connection in which U is the input signal for the G2 block, whose output will be G2U.
Similarly, the output of the G1block is G1 multiplied by its input signal, G2U. Therefore, the output of
the G1block is G1G2U. The block diagram (b) is a parallel connection of G1and G2. The output Yis
the sum of the two incoming signals of the summer, which are G1U and G2U, respectively. Therefore,
Y = G1U+G2U = (G1 +G2)U. The block diagram (c) is a feedback connection with its forward path
gain
Ga1nd
theloop
gain-G1G2.
The
output
ofthesummer
isR-G2Y,
which
yieldsY
=G1(R-G2Y).
Bymoving
theterm-G1G2Y
totheother
side,
we
have
thefollowing:
G1
Y+G1G2Y
=G1R? Y=1+G1G2
142
6
Systems
Representations
and Interconnected
Fig. 6.3:
A dual-loop
Systems
feedback tracking
control
design.
In the following example, we will revisit the DC motorspeed control problem using a new controller
with dual-loop feedback structure. First, we will usethis opportunity to practice how to find the transfer
function
is a little
of the closed-loop
bit
system block diagram in a speed control
more complicated
than the previous integral
the new control system structure
control
design problem.
The block diagram
case considered in
provides two degrees of freedom in the controller
Example
5.26, but
design that allows
us
to choose the damping ratio and the natural frequency of the closed-loop dynamics independently.
Example 6.1 (Revisit the DC Motor Speed Control System Using Dual-Loop
Feedback)
The block diagram of the dual-loop feedback control system is shown in Figure 6.3. Onthe left-hand
side ofthe diagram, the reference input signal r(t) is fed into the summer. The output signal y(t), which
is the angular
speed of the
motor to be controlled,
is fed back into the same summer
with a negative
sign
sothat
theoutput
signal
ofthesummer
isthedifference,
r(t)-y(t),ofthetwoincoming
signals.
This difference signal in turn becomes the input signal to the integrator 1/s. There are two controller
parameters to be determinedone
is K2, whichis the integrator gain, andthe other is K1in the second
feedback loop.
Thefirst step is to find the overall transfer function of the closed-loop system between the reference
input R(s), whichis the Laplace transform of r(t), andthe output Y(s), the Laplace transform of y(t).
By inspection
of the diagram,
we have following
equation:
Y=GK1Y+GK2(1/s)(R-Y)
Then move all the Y terms to the left-hand
?1-GK1+ ?
GK2
Y =
s
which can be simplified
GK2
s
side and substitute Gby b/(s+a)
?
bK1
R ? 1- s+a1
bK2
+
s(s+a)
into the equation
?
Y =
bK2
s(s+a)
to
? Y=bK2R
s2+(a1
-bK1)s+bK2
?
and we have the closed-loop
transfer
function
Y
R
Using the
DC micromotor
Then the closed-loop
will be
dynamics
transfer
function
as follows:
bK2
=
(6.1)
s2+(a-bK1)s+bK2
data obtained in Equation
of the
dual-loop
feedback
5.68, we have b = 145.5 and a = 43.14.
DC micromotor
speed control
system
6.1
Y(s)
R(s)
Diagrams
143
?2
n
145.5K2
=
Block
=
s2 +2??ns+?2
s2+(43.14-145.5K1)s+145.5K2
(6.2)
n
Notethat this closed-loop transfer function belongs to the category of the typical second order system
we studied in
Chapter 3. Since the dynamic
behavior
of the typical
second-order
system is character-ized
by the damping ratio ? and the natural frequency ?n, from Equation 6.2, we have the following
two equations that relate the damping ratio ? = 0.9 and the natural frequency ?n = 50 rad/s to the
controller parameters K1and K2:
n =145.5K2
=502and2??n
=2(0.9)?n
=43.14-145.5K1
?2
Hence,
K1=-0.322
andK2=17.18.
Fig. 6.4: Simulation
result
of the tracking
control
design using dual-loop
feedback.
After the dual-loop feedback controller constants K1 and K2 are determined, the block diagram in
Figure 6.3 can be employed to build a simulation program to conduct simulations of the DC motorspeed
control system. The simulation results are shown on Figure 6.4. Withthe damping ratio chosen to be
? = 0.9,there are no oscillationsandthe overshootis almostinvisible. Thestepresponsereaches
steady
state
with zero steady-state
one with integral
error
shortly
after t
= 0.1s, which is two times faster
than
the
control shown in Figure 5.23. The faster response is due to the dual-loop feed-backs
ability to choosea higher natural frequency, ?n = 50 rad/s, compared to ?n = 23.97rad/s
in the integral
control
case.
The control-input plots are also shown in Figure 6.4. It can be seen that whenthe angular speed of
the load, y(t) reaches 20 rad/s, the control input u(t) is around 6 volts voltage. Similarly, when y(t)
follows the reference input r(t) = 40 rad/s to get up to 40 rad/s, it requires about 12.5 volts from the
voltage source,
only a slight increase than the integral
In this section,
function
would be if
we have experienced
approach in obtaining
the effectiveness
the overall
we had to find the differential
control
system
equation
case.
of the frequency-domain
model. Just imagine
how
of the overall system in time
block
diagram/trans-fer
much harder the
work
domain by combining
a group of differential equations. However,for a morecomplicated interconnected system, solving a set
of algebraic
equations
can still be time-consuming.
Some block
diagram reduction
but the graphic reduction procedure can still betedious. One of the mostefficient
overall transfer function (or gains) of a large system is Masons gain formula.
Although
Masons gain formula
can be applied to the block
diagram,
it is
techniques
may help,
waysto compute the
moreintuitive
for it to
work together with the signal flow graph. The signal flow graph is virtually equivalent to the block
diagram, but it looks neater and shows the signal flow path more clearly than its counterpart
144
6
Systems
Representations
and Interconnected
Systems
6.2 Signal Flow Graphs and Masons Gain Formula
6.2.1
Signal
Flow
Graphs
In signal flow graphs, signals are represented by circular dots, called nodes, and the transfer func-tions
or gains of systems between signals are represented by directional branches, with the ar-rowhead
in the middle of the branch showing the signal flow direction. Each node mayconnect with
several incoming and outgoing branches, butthe value of the signal at each node (except the input nodes)
is only determined by the signals via the incoming branches. Thatis, the signal at each node equals the
algebraic
sum of all incoming
outgoing
branches. Notethat the nodes with no incoming
the nodes
general
may still
signals
without outgoing
have outgoing
via the incoming
branches are considered
branches,
and it has nothing
to do with the
branches are regarded asinput nodes. Simi-larly,
output nodes, although
some output nodes in
branches.
Four basic signal flow graph connections are shown in Figure 6.5. The signal flow graph (a) is triv-ial
in which the node Y(s) has only one incoming branch, and the signal coming from node U(s) via
branch G(s) is the product, G(s)U(s). Forthe cascade connection in (b), it is easy to see that Y = G1E
and E = G2U; hence, we have Y = G1G2U. Unlike the parallel connection in the block diagram, we
do not need a summer in the signal flow
feeding
signals
via two
or moreincoming
graph (c).
To perform
branches into
an addition
operation, it
a node. For the parallel
only requires
connection
in (c), the
output node Y hastwo incoming branches, G1and G2,andthese two branches are connected to the same
signal source, U(s). Hence, Y = G1U+G2U,
Fig. 6.5:
(c) parallel
Basic signal flow
connection,
The feedback
which is Y =(G1
graph
connections:
(d) feedback
connection.
connection
consists of three
(a) two
nodes
nodes, the input
+G2)U.
one branch, (b) cascade
connection,
node R on the left, the output node Y on
the right, and the error signal node Ein the middle. The input node Rhas one outgoing branch to node E,
buthas
noincoming
branch.
The
error
signal
node
Ehas
twoincoming
branches,
1and-G2connected
to
R and Y, respectively.
Hence, the signal at node Eis
E=1 R-G2Y
= R-G2Y
The output node has one incoming
node i
signal from node E via branch G1; hence, the signal at the output
6.2
Signal
Flow
Graphs and
Masons
Gain Formula
145
Y = G1E
Combine the two equations by eliminating the intermediate
function
from
Rto
variable E to yield the following
Y
Y=G1(R-G2Y)
? Y+G1G2Y
=G1R?
6.2.2
transfer
Y:
R
G1
=
1+G1G2
Masons Gain Formula
As mentioned a while ago, Masons gain formula is one of the most efficient waysto compute the
overall transfer function (or overall gain) of alarge system. Theformula will be given in the following.
The general description of the formula mayseemlong and complicated. However, almost all the students
will find the formula is straightforward and easy after working on a few examples.
Theorem 6.2 (Masons
Gain Formula)
Consider a system represented by a signal flow graph, in which the relationships among signals and
systems are all algebraic. Thenthe transfer function from the input Uto the output Yis
Y
U
N
= ?
n=1
Fn?n
(6.3)
?
where
N =total number offorward
Fn =the n-th forward
paths
path gain
M= total number ofloops
?i = the i-th loop gain
? =1-
M
i,j ?i?
j -?NT
i,j,k?i?
j?k+
? ?i
+?NT
i
The superscript
NT means nontouching.
?NT
i,j ?i?j isthesumoftheproducts
oftwonontouching
loopgains.
?NT
i, j,k ?i?j?k is the sum ofthe
products
ofthree
nontouching
loop
gains.
?n: same asthe ?, but only the loops that do NOTtouch the n-th forward path are considered.
Example 6.3 (Masons
Consider the feedback
Gain Formula for the Basic Feedback System)
connection
signal flow
graph in Figure
6.5(d).
The transfer
function
of the
closed-loopsystem Y/R can be obtainedjust by inspection using Masonsgainformula.
The signal flow graph has one forward path and the forward path gain is F1 = G1. There is oneloop
with
loopgain
?1=-G1G2,
which
istouching
theforward
path.
Hence,
we
have
?=1-?1=1+G1G2,
?1=1, F1=G1 ?
F1?1
Y
R
=
?
G
=
1+G1G2
146
6
Example
Systems
Representations
6.4 (Masons
The signal flow
and Interconnected
Gain Formula
for the
Systems
Dual-Loop
graph shown in Figure 6.6 is equivalent
Feedback
Control
System)
to the block diagram shown in Figure
6.3,
where a dual-loop feedback controller wasemployed to achieve a desirable DC motor speed control. In
Example 6.1, algebraic equations werefirst set up according to the interrelationships among the subsys-tems
and signals within the system, and then some algebraic manipulations were performed to obtain
the overall transfer function of the system. Now, we will apply Masons gain formula to this signal flow
graph to determine the transfer function from Rto Y.
The signal flow
graph from
R to Y has one forward
path and two loops.
The forward
path gain and
the two loop gains are
F1=1(1/s)K2G =K2G/s
and
?1=GK1,?2=-1(1/s)K2G =-K2G/s
Hence,the ? and ?1 will be
?=1-(?1+?2)
=1-GK1
+K2G/s,
?1=1
since both loops are touching the forward path. Therefore, the transfer function is
bK2
Y
R
F1?1
=
?
K2G/s
=
bK2
s(s+a)
=
=
s+a + bK2
1-GK1
+K2G/s1-bK1
s(s+a)
s2+(a-bK1)s+bK2
Fig. 6.6: Signal flow graph for the DC motor dual-loop feedback speed control system.
Fig. 6.7: Signal flow
Example
6.5 (Masons
The signal flow
Gain Formula
graph shown in
for
graph of a four-loop
a System
with Four
Figure 6.7 has one forward
system.
Loops)
path and four loops.
This example is
employedto demonstratehow to computethe ? terms whenthere are multiple nontouching pairs of
loops. Theforward path gain and the four loop gains are
F1 = b0b1b2b3b4b
6.2
Signal
Flow
Graphs and
Masons
Gain Formula
147
and
?1 = b1a1, ?2 = b2a2, ?3 = b3a3, ?4 = b4a4
Hence,the ? and ?1 are
? =1-(?1+?2
+?3+?4)+(?1?3
+?1?4
+?2?4),
?1=1
Notethat there arethree nontouching pairs of loops for the third term in the ? equation andthere exists
no nontouchingtrio ofloops in the signalflow graph. ?1is 1 becauseall the four loopstouch the forward
path. Therefore,the transfer function Y/R is
Y
R
b0b1b2b3b4b5
F1?1
=
=
?
1-b1a1
-b2a2
-b3a3
-b4a4
+b1a1b3a3
+b1a1b4a4
+b2a2b4a4
If the conventional algebraic approach wereemployed to solve the problem, six equations would have to
be set up, and the five intermediate
the transfer
function
variables
E1, E2, E3, E4, and E5 would need to be eliminated
before
can be found.
Fig. 6.8: Signal flow graph of a state diagram.
Example 6.6 (Masons
The signal flow
Gain Formula for a State Diagram)
graph shown in
Figure 6.8
will serve for two
purposes.
One is to demonstrate
the
construction of ?n whenthere exists aloop that does not touch the n-th forward path. In this example, if
X2is considered asthe output, we will seethat the signal flow graph hastwo forward paths:
F1=1 (1/s)1(1/s)=1/s2,F2=2/s
and two loops:
?1=-a1(1/s)=-a1/s,?2=-a2(1/s)1(1/s)=-a2/s2
Notethat the two loops aretouching eachother;i.e., there is no nontouchingpair ofloops. Hence,the ?
is
a1
? =1-(?1+?2)
=1+s
a2
+
s2 +a1s+a2
s2 =
s2
Sincethese two loops are also touching the forward path F1,they are not counted in the construction
of ?1. However,the loop ?1should be countedin the construction of ?2 becauseit is nottouching the
forward path F2. Thus, ?1, ?2are:
s+a1
?1=1, ?2=1-?1=
148
6
Systems
Representations
Therefore, the transfer function
X2
R
and Interconnected
from
Rto X2 is
F1?1 +F2?2
=
?
1
s2
=
?
Systems
s2 +a1s+a2
The second purpose of this signal flow
2
s2 +
s+a1
s
s
?
2s+a12
=
s2
graph is to show that signal flow
+1
+a1s+a2
graph can also be employed
to represent a graphical representation of a state-space representation, called the state diagram.
We
can see that two of the nodes are labeled as x1 and x2, which are the state variables of the system.
Furthermore, we can obtain the state equation from this diagram asfollows:
??? ?????
x?1
x?2
-a1-a2
=
1
0
1
x1
x2
+
2
6.3 State-Space Model
In Section
4.4.2,
we informally
introduced
the state-space
approach to convert the
nonlinear
coupled
second-order differential equation of the simple inverted pendulum system into a state equation that
consists of two first-order differential equations. This conversion has madeit much easier to con-duct
analysis, linearization,
controller design, and simulations. Astate-space representation, called a
state-space model, can always be obtained by converting an existing dynamic system differential equa-tion
or transfer function. It also can be constructed directly from the dynamic system modeling process
as described in Section 5.5, where weusually assign inductor current and capacitor voltage as state vari-ables
for electrical systems. Similarly, displacement and velocity are usually assigned as state variables
for mechanical systems.
Thetransfer function frequency-domain approach played an essential role in early control systems
theory development due to its ability to simplify systems manipulations and to conduct frequency-domain
analysis and design. However,its application is restricted to linear time-invariant systems with
single input and single output. Onthe other hand, the state-space approach can handle multi-input/multi-output
systems that are nonlinear and time-varying. In fact, the state-space modelis a more complete
description
of dynamic
systems, since it not only
reveals the information
of the internal
in state space involve
those involve
only constant
complicated
polynomial
states.
describes
the input-output
Furthermore,
the analysis,
relationship,
design,
but also
and simulation
matrices, which are more efficient on digital computers than
manipulations.
The state-space model usually is represented in the compact matrix form
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
(6.4)
where
thevectors
x ?Rn,
u?Rm,
and
y ?Rp
are
then1 state
vector,
them1input
vector,
and
thep1 output
vector,
respectively.
Thematrices
A?Rnn,
B?Rnm,
C?Rpn,
andD?Rpm
are
thenn Amatrix,
thenm Bmatrix,
thepn Cmatrix,
and
thepm Dmatrix,
respectively.
The
state-space modelin expanded matrix form
will look like the following:
a12 a1n
b11
b12 b1m
???a11
?
?
?
?
???
a21
a22 a2n
b21
b22 b2m
?...??an1an2 ann????bn1bn2 bnm???
x?1
x?2
x1
x2
?
?
?
?
?
?
x?n
u1
?
=
?
?
... ... ... ...
?
?
?
?
?
?
u2
?
...
xn
?
?
?
+
?
?
... ... ... ...
?
?
?
?
?
?
?
...
um
?
?
(6.5a)
6.3
State-Space
Model
d12 d1m
c12 c1n
???
????d11
???c11
d21
d22 d2m
c21
c22 c2n
???cp1cp2 cpn????dp1dp2 dpm???
?
?
?
?
?
u1
u2
x1
x2
y1
y2
...
?
?
=
?
?
...
...
...
...
?
?
?
?
?
?
?
?
...
?
+
?
?
?
...
...
...
...
?
?
?
?
?
?
?
(6.5b)
?
...
?
um
xn
yp
149
The number of the state variables or the dimension of the A matrixis the order of the state-space
model.In general, the order of the state-space modelis the same asthat of its associated differential
equation or transfer function. For example, the equation of motion for the simple inverted pendulum
system discussed in Section 4.4.2 is a second-order differential equation; hence, the order of its state-space
modelis n = 2.
Example
6.7 (Find
a State-Space
Model for the
Simple
MBK System)
Assume we have a simple massdamperspring
mechanical system, shown in Figure 6.9; whatin-formation
do we need at any time instant, t =t0, in order to predict the motion of the system for t >t0?
We would need to know both of the displacement and velocity values at t =t0 and the applied external
force
fa(t)fort =t0.Hence,
thedisplacement
xand
velocity
?xareperfect
candidates
toserve
asstate
variables. Letthe state variables be
x1(t)
= x(t)
and
x2(t)
=?x(t)
Fig. 6.9: Find a state-space modelfor the simple massdamperspring
From the definition of the state variables, weimmediately
x?1(t)
(MBK) system.
havethe first state equation as
= x2(t)
Then from the free-body diagram in Figure 6.9, we have
-Bx2(t)+
x1(t)+
M
?x2(t)+Bx2(t)+Kx1(t)
=fa(t) ??x2(t)
=-K
M
M
1
fa(t)
M
Combine the two state equations into one state equation in matrix form and choose the displacement
variable asthe output; then wehavethe following state-space modelfor the massdamperspring system:
? ?? ???? ?
??
x?1(t)
x?2(t)
y(t)
0
=
=? 1
1
-K/M -B/M
x1(t)
0?
x2(t)
x1(t)
x2(t)
+
0
1/M
fa(t)
(6.6
150
6
Systems
Representations
The state diagram
associated
and Interconnected
Systems
with the state-space
model of the simple
MBK system given by Equa-tion
6.6 is shown in Figure 6.10, which is a graphical representation of the state-space model. The
graphicaltool can be either a signal flow graph or a block diagram.In the diagram,s-1 represents an
integrator,
and the constants
of the system, respectively.
M, B, and
The signal
K are the
mass,the friction
at each node (blue
coefficient,
and the spring
dot) can be represented
constant
by a time-domain
variable in lower-case blue or by a frequency-domain variable in capital-case red. That meansthe same
state diagram can be employed for analysis in both time domain and frequency domain. But caution
needs to betaken to not mix up the time-domain and the frequency-domain
signals. For example,
the initial state for the state variable x1(t) to be usedin a simulation (time-domain) should be just
the constant x10instead of x10/s,the integral of x10. Onthe other hand,to find the transfer function
from x10to X1(s), we would need to compute the transfer function from
we have
1
X1(s) = G(s)X10(s) = G(s) x10
X10(s) to X1(s) first.
Then
s
Fig. 6.10: The state diagram associated withthe state-space model of the simple
(MBK) system.
Example 6.8 (Simulation
of the Simple
massdamperspring
MBK System Using the State Diagram)
In this example we will conduct a simulation using a simulation diagram based onthe state diagram
shown in Figure 6.10. In this simulation,
the applied force input
fa(t) is assumed zero. The initial
veloc-ity
ofthe system?x1(0) = x2(0) = x20is also assumedzero. Butthe initial displacement of the system
x1(0) = x10is assumedto be x10 = 0.5 m. Thisinitial condition can beimplemented in the lab. Assume
the left end of the spring and damper are fixed to a wall asthe schematic suggested in Figure 6.9. Imag-ine
you are holding the massand gradually moving it away from its equilibrium toward the right. You
will feel that both of the friction and spring reaction forces are against the motion. The spring reaction
force, especially, will get stronger as you movethe massfurther away from the equilibrium. Asthe mass
reaches the 50 cm mark, you will stop moving but continue to hold the massat position. Then you may
wantto know how the mass will moveafter the massis released.
We will conduct the simulation two times using different sets of mass,spring, and damper.
In the first run, the mass,spring, and damper are chosen so that M = 0.1 kg, K = 0.1 N/m, and
B = 0.05 Ns/m. Since this simple MBK system is a typical second-order system, the following two
characteristic equations should be equivalent, as discussed in Section 3.5:
s2 +(B/M)
s+(K/M)
=0
?
s2 +2??ns+?2 n = 0
Hence,the damping ratio and natural frequency of the system corresponding to the chosen MBK values
will b
6.3
v MK
? = 0.5B/
= (0.5)(0.05)/0.1
= 0.25,
?n =
with high overshoot
and oscillations
? = ?n
which is equivalent to oscillation
Fig. 6.11: Simulation
Model
151
? ?
K/M=
Withthis low damping ratio ? = 0.25 and slow natural frequency
will be underdamped
State-Space
0.1/0.1 = 1rad/s
?n = 1rad/s,
weexpect the response
with frequency,
?1-?2=0.968rad/s
period = 6.49 sec.
using the state-space
model of the simple
massdamperspring
(MBK)
system.
Thesimulationresultsareshownin Figure6.11. Thefirst roundsimulationresults(? = 0.25)are
shown in blue lines, and the second round responses are in red (?
= 1.5). The left graph records both of
the blue and the red displacements x1(t), and the middle graph shows the velocities x2(t) for both blue
and red cases. For the blue (lightly damped with ? = 0.25) case,the mass movesfrom the initial 0.5 m
position to the left
overshooting
by
more than 50%, then oscillates
several times
before converging
to
the equilibrium shortly after 15 seconds. Both the displacement curve x1(t) and the velocity curve x2(t)
show the same oscillation period approximately equals 6.5 sec. Onthe right-hand side of the figure is
the phase plane trajectory
coordinates
to eventually
on the x2 versus x1 plot spiraling
converge to the equilibrium
from the initial
(0.5,0)
point
on the (x1,x2)
(0,0).
In the second round of the simulation, the values of the mass M and the spring constant K
remain the same, but the friction coefficient Bis changed from B = 0.05 Ns/m to B = 0.3 Ns/m.
This increase of the friction coefficient does not affect the natural frequency, but it changes the
damping ratio from ? = 0.25 to ? = 1.5. Hence,the initial state response will become overdamped
with no overshoot or oscillation. The x2 versus x1 plot on the right-hand side of the figure showed that
the phaseplane trajectory just
The simulation
% CSD
Fig6.11
monotonically
MBK
A=[0
x1=X(:,1);
initial
response
X0=[0.5
1;
-K/M
-B/M],
x2=X(:,2);
subplot(1,3,2),
0]'
K=0.1,
c=[1
0],
plot(t,x1,'b-'),
grid,
on,
x1=X(:,1);
grid,
subplot(1,3,3),
x=[0.5
MATLAB code:
b=[0;1/M],d=0,
G=ss(A,b,c,d),[y,t,X]=initial(G,X0,t);
hold
plot(t,x1,'r-'),
to
M=0.1,
subplot(1,3,1),
[y,t,X]=initial(G,X0,t);
on,
due
0]',
plot(t,x2,'b-'),
plot(x1,x2,'b-'),grid,
hold
without muchspiral
results in Figure 6.11 are generated using the following
t=linspace(0,20,201);
B=0.05,
movestoward the equilibrium
hold
on,
plot(x1,x2,'r-'),
B=0.3,
grid,
hold
on,
A=[0
1;
x2=X(:,2);
hold
on,
subplot(1,3,3),
-K/M
-B/M],
G=ss(A,b,c,d),
subplot(1,3,1),
subplot(1,3,2),
plot(t,x2,'r-'),
gri
grid,
motion.
152
6
Systems
Representations
6.4 State Transition
and Interconnected
Systems
Matrix
The state-space modelinclude a state equation and an output equation as shown in Equation 6.4. For
ease of reference, these equations arerepeated:
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
Recallthat x(t), u(t), and y(t) are the state vector, the input vector, andthe output vector, respectively,
and A, B, C, and Din general are all real constant matrices. The output equation is just an algebraic
equation. If the state vector x(t) is known, the output vector will become known accordingly. Hence,to
investigate the behavior of the system described by the state-space modelthe maintask is solving the
state equation.
The state equation is
a first-order
differential
equation.
The procedures
required
to solve a first-order
differential equation are less complicated than those involved in high-order differential equations.
However, the state equations in general require matrix manipulations and computations. Owing to the
advancement of matrix computing algorithms and computer technology in the last six decades,the digi-tal
tools for matrix computations and manipulations are now efficient andreliable.
Just like
other
differential
equations,
the solution
to the
state equation
includes
two
parts: the
homogeneous solution and the particular solution.
We will find the homogeneous solution first and
consider the particular solution later. The homogeneous solution does not depend on the input vector
u(t). Hence, only the following state equation with initial conditions is considered for now:
x?(t) = Ax(t),
x(0)
To get a clue on how to solve this equation in general
= x0
matrix form,
(6.7)
we will first
consider the scalar case
of this equation
x?(t)
which
we already
= ax(t),
x(0)
= x0
have a solution,
x(t)
= eatx0
Then, let us try to extend the scalar solution to the general matrix case and assumethe solution can
be written as follows:
x(t) = eAtx0
(6.8)
Note that this notation eAtis created by just replacing the ain eat with A. At this
moment, westill
donotknow
what
thisnotation
eAtwould
represent
except
thatitisannn matrix
and
x(t)=eAtx0
must satisfy Equation 6.7. Now, by substituting x(t) = eAtx0into Equation 6.7, we have the following
equation in question:
deAtx0
x?(t) =
? =? AeAtx0 = Ax(t)
dt
6.4.1 Theorems and Properties of the State Transition
Matrix
Intheprevious
discussion
thenotation
eAt
canbereplaced
byannn matrix
F(t),and
x(t)=F(t)x0
is a solution of Equation 6.7if and only if dF(t)
dt = AF(t). Thisfact is summarizedin the following
theorem and this
Equation
6.7.
matrix F(t) or eAt is called the state transition
matrix of the system described by
6.4
Definition
6.9 (State
Transition
State Transition
Matrix
153
Matrix)
Thenn matrix
F(t)thatsatisfies
thefollowing
differential
equation
iscalled
thestate
transition
matrix associated with the A matrix.
dF(t)
= AF(t)
dt
Theorem
IfF(t)
6.10 (State
Equation
Homogeneous
(6.9)
Solution)
is the statetransition matrixfor the systemdescribedby Equation6.7,then x(t) = F(t)x0 is
a solution ofEquation
6.7.
Proof:
IfF(t)
satisfies Equation6.9,then wehave
x?(t)
=
dF(t)x0
dt
= AF(t)x0
= Ax(t)
which completes the proof.
Theorem
6.11 (State
Transition
Matrix
Expression)
Thestatetransition matrix F(t) can beexpressedasfollows,
F(t) =eAt = I+At
+
A2t2
2!
A3t3
+
3!
A4t4
+
4!
A5t5
+
+
5!
(6.10)
Proof:
It is straightforward
to show that
deAt
dt
Theorem
A3t2
= 0+A+A2t
6.12 (Properties
+
2!
A4t3
+
3!
of State Transition
?
A2t2
+ =A I+At+ 2!
A3t3
+
3!
+
?=AeAt
Matrix)
Someproperties ofthe statetransition matrix F(t) are summarizedin the following:
F(0)
=I
F-1(t)= F(-t)
F(t1)F(t2) = F(t1 +t2) = F(t2)F(t1)
F(t2-t1)F(t1
-t0)=F(t2-t0)
Proof:
Use F(t) = eAt
154
6
Remark
Systems
Representations
6.13 (Significance
and Interconnected
Systems
of the State Transition
Matrix)
Forx(t) = F(t)x0 to be asolution ofthe stateequation
x?(t) = Ax(t),
the matrix F(t) hasto be the state transition
be rewritten in the general form
x(0)
= x0
matrixthat satisfies Equation 6.9. Notethat the solution can
x(t2)=F(t2-t1)x(t1)
which depicts the transition of the state of the system from x(t1) to x(t2), wheret1 andt2 can be any time
instant.
Theorem
6.14 (State
Transition
Thestate transition
Matrix
Expression
in
Frequency
Domain)
matrix F(t) is the inverse Laplace transform
of
F(s)=(sI-A)-1
(6.11)
Proof:
Let X(s) be the Laplace transform
yield the following,
of x(t), then taking the Laplace transform
of Equation 6.7 will
sX(s)-x0
=AX(s)? (sI-A)X(s)
=x0
? ?
Then we have
X(s)=(sI-A)-1x0 ? x(t)=L-1(sI-A)-1x0
6.4.2 Computing the State Transition
To compute the state transition
matrix, wecan use either one of the following
8
F(t)
Matrix
?
=eAt =
k=0
1
k!
Aktk= I+At +
or
F(t)
The first
analytic
formula,
Equation
work due to its infinite
this infinite-series
work.
solutions
formula
The second formula,
via
matrix inversion
6.12,
works
2!
well for
can be converted
A3t3
+
3!
A4t4
+
4!
(6.12
+
5!
(6.13)
machine computing,
but it is not suitable
However, based on the Cayley-Hamilton
into
a closed-form
Equation 6.13, is a Laplace transform
and inverse
A5t5
+
?(sI-A)-1?
series expression.
will first introduce the Cayley-Hamilton
the procedure.
=L-1
A2t2
two formulas:
Laplace transform
expression
convenient
approach that
manipulations.
provides
for
theo-rem,
for
an-alytic
closed-form
In the following,
we
approach and then use second-order examples to demonstrate
Shortly after that, the same examples
will be employed to illustrate
the Laplace transform
approach.
Theorem 6.15 (The
Cayley-Hamilton
Theorem)
Consider
annn matrix
A.Assume
?1,?2, , ?nare
theeigenvalues
ofthe
Amatrix,
which
means
they are the roots
ofthe following
characteristic
equation:
6.4
State Transition
Matrix
155
f(?):=det(?I-A)
=0
Then
f(?i)
= 0, i = 1, 2,..., n
i f and onlyi f
Based onthe Cayley-Hamilton theorem, the statetransition
asfollows:
f(A)
=0
matrix of Equation 6.12 can be rewritten
n-1
eAt= ?ai(t)Ai=a0(t)I+a1(t)A+a2(t)A2
+ +an-1(t)An-1
(6.14
i=0
Notethat the state transition
matrix now has only nterms, but the coefficients,
i =0,1,2,...,n-1
ai(t),
are required to be determined.
These time-dependent
e??t =
coefficients
can be solved from
n-1
? ai(t)?i?,
(6.15)
? = 1, 2, ..., n
i=0
where ??, ? = 1, 2,..., n arethe eigenvalues of the A matrix.
To makeit moreclear regarding the coefficient equations of Equation 6.15, without loss of generality
we will usean n = 3 caseto illustrate how to solve for the coefficients in these sets of equations. Assume
the A matrix has three distinct eigenvalues ?1, ?2, and ?3. The case of multiple eigenvalues will be
addressed shortly after this. Thenthe third-order version of Equation 6.15 will be
e?1t = a0(t)+a1(t)?1
+a2(t)?2
1
e?2t = a0(t)+a1(t)?2 +a2(t)?2
2
(6.16)
e?3t = a0(t)+a1(t)?3 +a2(t)?2
3
Since these three eigenvalues are distinct, the three equations are linearly independent; hence,
the three coefficients a0(t), a1(t), and a2(t) can be uniquely determined. Therefore, the state tran-sition
matrix of the 3rd-order system will be
3-1
eAt= ? ai(t)Ai = a0(t)I+a1(t)A+a2(t)A2
(6.17)
i=0
Now,let usconsiderthe case with multiple eigenvalues. Assume?2 = ?1;thenthe secondandthe first
equations of Equation 6.16 are identical.
Wehave to remove the second equation
and replace it by
a new equation obtained by taking derivative of a copy of the first equation with respect to ?1.In
other
words, Equation
6.16 is replaced
by the following
new coefficient
e?1t = a0(t)+a1(t)?1
+a2(t)?21
te?1t = a1(t)+2a2(t)?1
e?3t = a0(t)+a1(t)?3
equations:
(6.18)
+a2(t)?2
3
which are nowlinearly independent; hence, the three coefficients a0(t), a1(t), and a2(t) can be uniquely
determined.
156
6
Example
Systems
Representations
6.16 (Find
and Interconnected
a State Transition
Considerthe matrix A =
Matrix
Systems
Using the
Cayley-Hamilton
Approach)
?-2-3?
, whose
eigenvalues
can
befound,
?1=-1and
?2=-2.Based
0
1
on the Cayley-Hamilton theorem, the state transition
as
matrix associated with this A matrix can be written
eAt = a0(t)I+a1(t)A
From Equation
6.15 or Equation
6.16 we have the coefficient
equations:
e?1t = a0(t)+a1(t)?1
e?2t = a0(t)+a1(t)?2
Substitute
?1=-1and
?2=-2into
these
twoequations
toget
e-t=a0(t)-a1(t)
e-2t=a0(t)-2a1(t)
and solve these two equations to yield the two coefficients
a0(t)
=2e-t-e-2t
a1(t)=e-t-e-2t
Then we have the state transition
matrix as follows:
?
?
2e-t-e-2t e-t-e-2t
eAt=(2e-t-e-2t
)I+(e-t-e-2t
)A= -2e-t+2e-2t
-e-t +2e-2t
Example 6.17 (Cayley-Hamilton
Consider the
matrix A =
Approach for the Case with
Multiple Eigenvalues)
? ?, whose
twoeigenvalues
are
identical
as?1=-1and
?2=-1.
0
1
-1 -2
Basedon the Cayley-Hamilton theorem, the statetransition
matrix associated with this A matrix can be
written a
eAt = a0(t)I+a1(t)A
Sincethe two eigenvalues are identical,
equation in the following:
Equation 6.15 or Equation 6.16 can provide only one coefficient
e?1t = a0(t)+a1(t)?1
One more coefficient equation is required to solve for the two coefficient variables. Following the same
procedure as we did to obtain Equation 6.18, we can obtain the supplemental coefficient equation by
taking derivative ofthe aboveequation withrespectiveto ?1. Now wehavethe supplementalcoefficient
equation
te-t
= a1(t)
Together with the regular coefficient equation
e-t=a0(t)-a1(t)
we obtain the two coefficients
a0(t) = e-t +te-t
a1(t) =te-t
Then we have the state transition
matrix as follows:
eAt = (e-t
In the following
transform
approach
+te-t )I+te-tA
e-t +te-t
State Transition
te-t
?
Matrix
157
e-t-te-t
-te-t
we will employ the same example
matrices to illustrate
the procedure
of the Laplace
using the formula
F(t)
Example 6.18 (Find a State Transition
Consider the
?
=
6.4
matrix
=L-1
?(sI-A)-1?
Matrix Using the Laplace Transform
Approach)
? ?, whose
eigenvalues
can
befound,
?1=-1and
?2=-2.In
0
A =
1
-2 -3
order
toapply
theformula
shown,
thefirststep
istocompute
theinverse
ofthematrix
sI -Aand
perform
partial fraction
expansion
of each entry of the
? ?-1 ? ?
??
?
-1
(sI-A)-1= s2 s+3
Example 6.19 (Laplace
Consider the
matrix
1
s+1 + s+2
-1 s+1
+ s+2-1
-2 + s+2
-1 + 2
s+1
2 s+1
s+2
?
matrix as
2e-t-e-2t e-t-e-2t?
?(sI-A)-1??-2e-t+2e-2t
-e-t +2e-2t
=L-1
=
Transform
A =
=
s
(s+1)(s+2)
(s+1)(s+2)
Then we have the state transition
2
(s+1)(s+2)
-2
1
-2 s
s2+3s+2
1
(s+1)(s+2)
=
s+3
1
=
s+3
F(t)
matrix in the following:
Approach for the Case with Multiple Eigenvalues)
?-1 -2?, whose
twoeigenvalues
are
identical
as?1=-1and
?2=
0
1
-1.Thesame
Laplace
transform
approach
shown
in Example
6.18
canbeemployed
forthecasewith
multiple
eigenvalues
tocompute
theinverse
ofthematrix
sI-Aand
perform
partial
fraction
expansion
of each entry of the matrix before conducting the inverse Laplace transform.
? ?-1 ? ?
? ??
?
? ?=?
?(sI-A)-1??-te-t e-t-te-t?
-1
(sI-A)-1= s1 s+2
s+2
=
(s+1)2
-1
(s+1)2
Then we have the state transition
1
(s+1)2
=
s+2
1
1
-1 s
s2+2s+1
1
s+1 +
1
1
(s+1)2
s
-1
(s+1)2
(s+1)2
(s+1)2
1
s+1 +
matrix as
F(t) =L-1
=
e-t +te-t
te-t
-1
(s+1)2
158
6
Systems
Representations
and Interconnected
Systems
6.5 Solution of the State Equation
Considerthe state equation,
x?(t)
(6.19)
= Ax(t)+Bu(t)
where
A,Bx, (0)=x0,and
u(t),
t =0aregiven.
The
problem
istofindx(t)fort =0.
Let
X(s) and
U(s) be the
Laplace transforms
respectively. Taking the Laplace transform
of the state vector x(t)
and the input
vector
u(t),
on the state equation yields the following frequency-domain
equation:
sX(s)-x0=AX(s)+BU(s)
Since the objective is to find the state response, the state vector terms
are moved to the left-hand
side of
the equation while keeping or moving the input andthe initial state vector to the right-hand side:
(sI-A)X(s)
=x0+BU(s)
Multiplying
(sI-A)-1
toboth
sides
oftheequation
from
theleft,weobtain
thesolution
ofthestate
equation in the frequency
domain:
X(s)=(sI-A)-1x0
+(sI-A)-1BU(s)
Then we havethe state response x(t),
which is the inverse Laplace transform
of X(s),
?
?(sI-A)-1? ?(sI-A)-1BU(s)
Recall
that
theinverse
Laplace
transform
of(sI-A)-1isthestate
transition
matrix
F(t):
?(sI-A)-1?
x(t)
x0 +L-1
=L-1
6.21 can be rewritten
in terms
x(t) = F(t)x0 +
(6.21)
= F(t)
L-1
Thus, Equation
(6.20)
of the state transition
matrix as follows:
?tF(t-t)Bu(t)dt:=xh(t)+xp(t)
(6.22)
0
Fig. 6.12: State diagram for the simple
MBK system time response simulation.
The convolution theorem (Theorem 2.24) was employed in the inverse Laplace transform manipula-tions,
and the second term of the above state response is a convolution integral. Also note that the firs
6.5
Solution
of the
State Equation
159
term of the response xh(t) is the homogeneous solution due to the initial state vector x0 and the
second term of the response xp(t) is the particular solution due to the input u(t).
In the following, we will employ the simple MBKsystem again to illustrate how to compute the time
responsesin both time and frequency domains and verify the computational results with the simulation
results.
The Simulink
6.13 is constructed
simulation
diagram,
CSD fig6p13
response
shown
based on the state diagram in Figure 6.12. This simulation
shown in Figure 6.10. The difference is that this time
due to the initial
StateModel.mdl,
condition
and the input
due to the initial
B = 0.3 Ns/m, and
condition.
The
diagram is similar to that
we will consider overall time responses of all states
while in the previous simulation
mass, damper,
K = 0.2 N/m. This change
on the top of Figure
and spring
causes the
we only considered the output
are are chosen
damping
ratio
as
and natural
M = 0.1 kg,
frequency
of
the system to become ? = 1.06 and ?n = 1.414 rad/s, respectively. The response will be faster and
less damped than the previous overdamped case simulation in Example 6.8.
The simulation will be conducted three times, one for the homogeneous solution xh(t), the response
due to the initial state x0; the second one for the particular solution xp(t), the response dueto the input
u(t); and finally the complete solution x(t), the total response dueto the combined effort of x0 and u(t).
Theinitial conditions for the displacement state and the velocity state are assumed x1(0) = 0.5 m and
x2(0)
=0m/s,
respectively.
The
applied
force
input
isassumed
u(t)=0.05
Nfort =0.
Wewill run the following
Simulink
is listed
Run
CSD_fig6p13.m
this
% Nominal
to conduct
which will automatically
the simulation.
The
call the
MATLAB
code
program
and
4,
2020
to
automatically
plot_1Blue.m,
initial
&
step
call
plot_2Red.m,
response
the
Simulink
file:
plot_3Black.m,
plant
M=0.1,
K=0.2,
B=0.3,
Simulation
u=0,
May
CSD_fig6p13.m
% CSDfig6p13SM.mdl
%%
CSD fig6p13.m,
CSDfig6p13SM.mdl,
as follows:
% Filename:
%
MATLAB program:
model program
1
x10=0.5,
sim_time=10,
sim_options=simset('SrcWorkspace',
open('CSDfig6p13SM'),
'current',
'DstWorkspace
sim('CSDfig6p13SM',
[0,
',
sim_time],
'current');
sim_options);
run('plot_1Blue')
%%
Simulation
u=0.05,
2
x10=0,
sim_time=10,
sim_options=simset('SrcWorkspace',
open('CSDfig6p13SM'),
'current',
'DstWorkspace',
sim('CSDfig6p13SM',
[0,
'current');
sim_time],
sim_options);
run('plot_2Red')
%%
Simulation
u=0.05,
3
x10=0.5,
sim_time=10,
sim_options=simset('SrcWorkspace',
open('CSDfig6p13SM'),
'current',
'DstWorkspace',
sim('CSDfig6p13SM',
[0,
'current');
sim_time],
sim_options);
run('plot_3Black')
Before running this program, make sure to include the Simulink file CSDfig6p13SM.mdl,
and the three plotting files: plot
1Blue.m,
plot
2Red.m, and plot
3black.m
in the same
folder. Only one of the plot files is given. The other two are the same except the color.
% filename:
plot_1Blue.m
plot
after
running
simulink
file:
figure(1)
subplot(1,3,1),
plot(t,x1,'b-'),
grid
on,
grid
minor,
title('x1'),
hold
on,
subplot(1,3,2),
plot(t,x2,'b-'),
grid
on,
grid
minor,
title('x2'),
hold
on,
subplot(1,3,3),
plot(x1,x2,'b-'),
hold
on
grid
on,
grid
minor,
title('x2
vs
x1'),
160
6
Systems
Representations
The simulation
results
and Interconnected
are shown in Figure
Systems
6.13. The response
graph for x1 on the left shows three
displacement response curves: (1) x1h:the displacement response dueto the initial state x0 only, (2) x1p:
the displacement response due to the input u(t) only, and (3) x1: the displacement response dueto both
x0 and u(t) acting at the same time. It is observed that x1(t) = x1h(t)+x1p(t).
Fig. 6.13: Simple MBK system time response simulation results.
Similarly, the response graph for x2 in the
middle shows three velocity response curves: (1) x2h:
the velocity response due to the initial state x0 only, (2) x2p: the velocity response due to the input
u(t) only, and (3) x2: the response due to both x0 and u(t) acting at the same time. It is observed that
x2(t) = x2h(t)+x2p(t).
three
The response
graph
responses
displayed
on the
right
on the left
of Figure
and the
6.13 exhibits
middle graphs.
the
phase
blue) would movethe state from (0.5,0) towards the origin (0,0) if there
hand, the particular
solution
motion (in red)
would
plane trajectories
The homogeneous
move the state from
solution
was no input.
of the
motion (in
Onthe other
the origin (0,0) towards
its
intended destination (0.25,0) if the state conditions were zero. Actually the real motion (in black)
is the combination of both actions that together movethe state from (0.5,0) to (0.25,0).
Example 6.20 (Computation
in Time Domain)
of the
Homogeneous and Particular
Solutions of the State Equation
In the state diagram of the simple MBK(massdamperspring)
system shown in Figure 6.12, there is
oneintegrator betweenthe x1 node andthe x2 node,implying ?x1 =x2. By summing upthe three incoming
signals
atthe
?x2
node,
we
have
another
first-order
differential
equation
?x2=-2x1-3x2+10u.
Hence,
we havethe state equation of the system:
? ?? ?????
x?1(t)
x?2(t)
wherethe initial
input
with
0
=
1
-2 -3
x1(t)
x2(t)
0
+
10
u(t),
x(0)
= x0 =
??
x10
0
displacement state is assumed x10 = 0.5 m,and the applied force input u(t) is a step
magnitude 0.05N.
The objective
of this example is to find the homogeneous
solution
particular solution of the state equation. Thesesolutions can befound using the time-domain
via Equation
6.22,
which are
and the
approach
6.5
xh(t) = F(t)x0,
and xp(t) =
Solution
of the
State Equation
161
?tF(t-t)Bu(t)d
0
The state transition matrix F(t) in the above two equations can be obtained using the Cayley-Hamilton
approach or the Laplacetransform approach described in Section 6.4.2.
In
Example
state transition
6.16
we had applied the
Cayley-Hamilton
approach to the same A matrix to obtain its
matrix, which is
F(t) =
Thus, we have the homogeneous
xh(t) = F(t)x0 =
2e-t-e-2t e-t-e-2t?
?-2e-t
+2e-2t
-e-t+2e-2t
solution
as follows:
2e-t-e-2t e-t-e-2t??
e-t-0.5e-2t
?-2e-t+2e-2t
?
?
?
-e-t +2e-2t
-e-t +e-2t
0.5
0
The computation of the particular solution xp(t) is a little
convolution
=
bit more complicated since it involves
integral.
xp(t) =
?tF(t-t)Bu(t)dt= ?tF(t-t) ??
0
0
0.05us(t)dt
10
0
Notethat the variable t, instead of t, is the integration
variable. Whencarrying out the integration,
t should betemporarily treated as a constant parameter. According to the convolution theorem, the
convolution
integral
can be rewritten
asthe following:
xp(t)
=
?t ??us(t-t)dt
F(t)
0
0
0.5
Since
us(t-t)is1when
t <t and0elsewhere,
thevalue
ofus(t-t)is1inside
theintegration
range; hence,the above integral can further be simplified asfollows:
xp(t) =
?t ?? ?t
?e-t-e-2t? ?-e-te-t+0.5e-2t
?t
-e-2t
0 -e-t +2e-2t
0
dt = 0.5
0.5
F(t)
0
0
Then we have the particular
xp(t)
= 0.5
dt = 0.5
solution of the state equation as
-(-1+0.5)
-0.5e-t
+0.25e-2t
+0.25
?-e-t+0.5e-2t
?
?
?
e-t-e-2t-(1-1)
0.5e-t
-0.5e-2t
=
Next, we will apply the Laplace transform approach to compute the homogeneous and particular
solutions
of the same state equation.
Example 6.21 (Computation
of the Homogeneous and Particular
Using the Laplace Transform Approach)
Solutions of the State Equation
The state equation to be considered is repeated here for convenience.
? ?? ?????
x?1(t)
x?2(t)
0
=
1
-2 -3
x1(t)
x2(t)
0
+
10
u(t),
x(0)
= x0 =
??
x10
0
162
6
Systems
Representations
and Interconnected
Systems
wherethe initial displacement state is assumed x10 = 0.5 m,and the applied force input u(t) is a step
input with magnitude 0.05 N.
The state transition matrix associated with the A matrix had been computed in Example 6.18 using
the Laplace transform asthe following:
F(t)
2e-t-e-2t e-t-e-2t?
?(sI-A)-1??-2e-t+2e-2t
-e-t +2e-2t
2e-t-e-2t e-t-e-2t??
e-t-0.5e-2t
?-2e-t+2e-2t
?
?
?
-e-t +e-2t
-e-t +2e-2t
=L-1
=
Thus,the homogeneous solution is
xh(t)
= F(t)x0
=
0.5
=
0
Since
theLaplace
transform
oftheparticular
solution
is(sI-A)-1BU(s),
we
have
Xp(s)
=(sI-A)-1BU(s)
=?
s+3
(s+1)(s+2)
-2
solution
of the state equation
0
10
s
(s+1)(s+2)
Then the particular
??? ? ?
1
(s+1)(s+2)
0.5
s(s+1)(s+2)
0.05
(s+1)(s+2)
=
s
0.5s
s(s+1)(s+2)
can be obtained
by finding
the inverse
Laplace trans-form
of the previous matrix,
xp(t)
=L-1 [Xp(s)]
6.6 State-Space
?0.25 ??0.25-0.5e-t
+0.25e-2t
?
0.5e-t
-0.5e-2t
s
=L-1
+ s+1
-0.5 +0.25
s+2
0.5
s+1
+s+2-0.5
=
Models and Transfer Functions
Systems can be represented by differential equations, transfer functions, state-space models, or
state diagrams. In manyanalysis and design cases, we mayneed to convert from onerepresentation to
another. In this section, we will discuss how to do it correctly and efficiently. Since we have just utilized
the Laplace transform approach to solve a state equation in the previous section, we will first discuss
how to find
6.6.1
the transfer
Find Transfer
Considerthe following
function
Function
of a system from
from
State-Space
a given state-space
model of the system.
Model
state-space model:
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
Let X(s), U(s), and Y(s) be the Laplace transforms of the state vector x(t), the input vector u(t), and
the output vector y(t), respectively. Recallthat the solution X(s) of the transformed state equation was
given by Equation 6.20 as
X(s)=(sI-A)-1x0
+(sI-A)-1BU(s)
Plugging this solution
into the transformed
output equation
Y(s)
=CX(s)+DU(s
of the state-space
model,
6.6
we will have the following
State-Space
Models and Transfer
Functions
output response:
?
?
U(s)
Y(s)=C(sI-A)-1x0
+ C(sI-A)-1B+D
Note that the first term is the output response
response due to the input
Y(s) is
163
due to the initial
U(s). Therefore, the transfer
Y(s)
U(s)
(6.23)
state x0, and the second term is the output
function
from the input
U(s) to the output
:= G(s)=C(sI-A)-1B+D
(6.24)
Example
6.22
(State-Space
Model
?Transfer
Function)
For the simple MBK system considered in Example 6.7, if the system parameters are chosen to be
M= 0.1 kg, B = 0.3 Ns/m, and K = 0.2 N/m, the system dynamics can be described by the state-space
model.
x?(t) = Ax(t)+Bu(t)
y(t)
where
A =
The transfer
function
=Cx(t)+Du(t)
? ? ??
0
1
-2 -3
0
, B =
10
, C = ? 1 0 ?,
of the system can be obtained using
Equation
D =0
6.24 as follows:
? ?-1
??
-1
G(s)=C(sI-A)-1B+D
=? 1 0 ? s
2 s+3
Since
?s -1?-1? ?
1
=
2 s+3
we have the transfer function
G(s) =
?
asfollows:
1
?
? 1 0?
s+3
1
-2 s
where
? = s(s+3)+2
? ??
?
s+3
1
-2 s
0
10
=
0
10
10
?
=
10
s2
+3s+2
Fig. 6.14: Usingthe state diagram of a state-space modelto find the transfer function
164
6
Systems
Representations
and Interconnected
Systems
Example
6.23
(State-Space
Model
?State
Diagram
?Transfer
Function)
In this example,
we will demonstrate
how to use the state diagram to find the transfer function.
The
state diagram associated with the state-space model considered in Example 6.22 can be constructed as
shown in Figure 6.14. Since the state diagram is a signal flow graph, Masons gain formula can be
employed to find the transfer function from Uto Y. By inspection, we can see there are two loops
and one forward pathin the diagram. Thetwo loop gains and oneforward path gain are
?1=-3s-1,?2=-2s-2,andF1=10s-2
so that ? and ?1 are
?=1-(?1+?2)
=1+3s-1
+2s-2,?1=1
Therefore, the transfer function
G(s) is
Y(s)
G(s):=
U(s)
10s-2
F1?1
=
=
?
Given a state-space model (A,B,C,D),
10
1+3s-1 +2s-2 =s2
+3s+2
we can find the transfer function using the direct formula
C(sI-A)-1B+D
orutilizing
thestate
diagram
together
withMasons
gain
formula.
Itseems
that
the
state diagram approach requires
system is higher, the computation
direct formula
more steps. But in
involved
many cases, especially
in the state diagram
approach
when the order of the
is easier than that in the
approach.
6.6.2 Construct a State-Space
Modelin
Companion Form Using Direct Realization
As weknow, wecannot havetwo different transfer functions for a given system, butthere exist infinitely
many state-space models for the same system since state variables can be defined in infinitely
many
differently
ways. For this
despite using different
transfer function.
reason,
approaches.
give a state-space
Conversely
model,
we will obtain the same transfer
we may not get the same state-space
model from
function
a given
Thetransfer function of a system is unique, but the state-space modelis not.
There are many waysto assign state variables. Sometimes we would like the state variables to asso-ciate
with the physical variables in the real world, like the displacements, the velocities in mechanical
systems
and the inductor
currents, the capacitor
voltages in electric
analysis and design cases we may need a state-space
circuits.
Onthe other hand, in
model in some special form.
used forms of state-space modelsis called companion form,
form, orthe phase variable canonical form.
many
One of the commonly
whichis also named the controller canoni-cal
Withoutloss of generality, we assume the transfer function is a third-order strictly proper rational
function
as shown:
Y(s)
U(s)
:=
G(s) =
b2s2 +b1s+b0
(6.25)
s3 +a2s2 +a1s+a0
Here,strictly proper meansthat the degree of the numerator polynomial is less than that of the denomi-nator
polynomial. If both degrees are equal, a division needs to be performed to convert the transfer
function into two terms: one strictly proper rational function and the other a constant. Recallthat
the corresponding
differential
equation
of the transfer
...
y(t)+a2 y(t)+a1 ?y(t)+a0y(t)
function
is
= b2
u(t)+b1 ?u(t)+b0u(t)
...
In the equation, y andy represents the third and second derivatives of y, respectively.
Laplace transform
s variable
can be regarded
as a differentiation
Note that the
operator in the conversion
betwee
6.6
transfer function
and differential
State-Space
Models and Transfer
Functions
165
equation.
Thefirst step toward the construction
new variable X(s) to both the numerator
6.25 asfollows:
Y(s)
U(s)
of the companion form state-space modelis to multiply a
and the denominator of the transfer function in Equation
b2s2
=
+b1s+b0
X(s)
s3 +a2s2 +a1s+a0
X(s)
Split the numerator and denominator to get the following two equations:
Y(s) = b2s2X(s)+b1sX(s)+b0X(s)
? y = b2
x+b1 ?x+b0x
U(s) =s3X(s)+a2s2X(s)+a1sX(s)+a0X(s)
(6.26)
...
?
u = x +a2
x+a1 ?x+a0x
The second equation can be rearranged to the following,
...
x =-a2
x-a1?x-a0x+u
(6.27)
Including the input u,the output y, the variable x andits three derivative variables, six nodes are required
...
...
in the state diagram. They are labeled as u, x, x, ?x, x, and y. Based on Equation 6.27, the node x
shall receive the four incoming branches from nodes u,x, ?x, and x. Similarly according to the first
equation of Equation 6.26, three incoming branches from nodesx, ?x, and x will feed to the output
node y. Thus, we havethe state diagram shown in Figure 6.15.
Now, weare ready to assign state variables. Therule is to assign a state variables at eachintegrators
output. Although no specific order of the state variables is required, we would follow the companion
form convention to assign node x as x1, node?x as x2, and nodex as x3, as shown in the state diagram.
Then, according to the state diagram, wecan easily write down the state equation andthe output equation
in the following:
? ?? ?? ???
??u(t),
? ?=?-a0-a1-a2?? ?+
x?1(t)
x?2(t)
x?3(t)
This state-space
three features:
(1)
0
1
0
0
0
1
0
x1(t)
x2(t)
x3(t)
0
y(t)
1
modelis said to bein the companion form.
The entries right
above the
main diagonal
row arethe denominator coefficients of the transfer function
matrix entries
are all 0s.
The column
vector
= ? b0 b1 b2 ?
??
??
x1(t)
x2(t)
x3(t)
(6.28)
Notethat the A matrix hasthe following
are all 1s; (2) the
entries at the bottom
with sign change; and (3) the rest of the A
B has all 0s except the bottom
entry,
which is 1. The row
Direct
Realization)
vector Cis constructed using the numerator coefficients.
Example
6.24 (Construct
a Companion-Form
State-Space
Model Using
In this example, we will simply follow the companion form construction procedure to draw a state
diagram like
Figure 6.15, and then
form as shown in Equation
Consider the transfer
usethis state diagram to find the state-space
6.28.
function
G(s) =
Thefirst step is to multiply a new variable
transfer function as follows:
Y(s)
U(s)
Split the numerator
model in the companion
and denominator
s+3
s2 +3s+2
X(s) to both the numerator
=
s+3
s2 +3s+2
to get the following
X(s)
X(s)
two equations
and the denominator
of the
166
6
Systems
Representations
Fig. 6.15:
Construction
and Interconnected
of a state-space
Y(s) = sX(s)+3X(s)
?
Systems
model in companion
form
via a state diagram.
y =?x+3x
U(s)=s2X(s)+3sX(s)+2X(s)
? u=x+3?x+2x?x =-3
?x-2x+u
Including the input u, the output y, the variable x, and its two derivative variables, five nodes are required
in the state diagram. They are labeled as u,x, ?x, x, and y, asshown in Figure 6.16.
According
tothesecond
ofthetwoequations,
x =-3
?x-2x+u,
thenode
x shall
receive
thethree
incoming
branches
fromnodes
u,
?x,and
x with
gains
1,-3,and-2,respectively.
Similarly,
based
on
the other equation, y =?x+3x, the two incoming branches from nodes?x, and x with gains 1 and 3 feed
to the output node y. Thus, we havethe state diagram shown in Figure 6.16.
Fig. 6.16:
Construct a companion
Now, we are ready to assign state
of each integrator.
form
state-space
variables.
model from
a transfer function.
The rule is to assign state
variables
Although no specific order of the state variables is required,
at the output
we would follow the
companion form convention to assign node x asx1, and node?x as x2, as shown in the state diagram. Then
we havethe state equation and the output equation asfollows:
? ?? ?????
x?1(t)
x?2(t)
6.6.3
Construct
In the following,
state-space
=
a State-Space
0
1
-2 -3
x1(t)
x2(t)
Model from
+
0
1
u(t),
Interconnected
we will usethree examples to illustrate
models, respectively,
from
y(t)
=?3 1?
?
x1(t)
x2(t)
Systems
how to construct state diagrams, and conse-quently
a cascade connection
of two transfer
functions,
a parallel
6.6
connection
of two transfer
functions,
and a feedback
State-Space
connection
Models and Transfer
Functions
with plant and controller
transfer
167
func-tions.
In the following example, we will consider the construction of a state-space modelfrom a cascade
connection of two transfer functions. Thatis, the input-output relationship of the system is described
by
Y(s)/U(s)
= G1(s)G2(s)
where G1(s) and G2(s) can be from a decomposition of a given transfer function or they can be from
two different subsystems.
Example 6.25 (Construct
a State-Space
Model from
Cascaded Subsystems)
Consider a system, whichis composed of two subsystems G1(s) and G2(s) withthe following trans-2
fer functions:
Y1(s)
U1(s)
= G1(s) =
s+3
s+1
= 1+
,
s+1
Y2(s)
= G2(s) =
U2(s)
1
s+2
The relationship of the overall system G(s) and its subsystems is described by G(s) = G1(s)G2(s). Of
course, we could combine the two rational functions into one, and then construct a state-space model
based onthe combined system. However, for practical reasons, weshould keep the state variables of
the subsystems if possible.
Fig. 6.17: Construct a state-space
model from
a cascade connection
of two subsystems.
Hence, we will construct state-space models separately for all subsystems and then combine
them together to form a state-space modelfor the overall system. The state-space models of the two
subsystems
can be constructed
Y1
U1
2
= 1+
s+1
X1
X1
as follows:
?
x?1=-x1+u1
,
and
y1 = 2x1 +u1
Y2
U2
=
1
X2
s+2
X2
?
x?2=-2x2
+u2
y2 = x2
where x1 and x2 arethe state variables of the subsystems G1and G2,respectively.
Let y(t) and u(t) be the output and input of the combined cascade system, respectively. Accord-ing
to the cascade connection, we have y(t) = y1(t), y2(t) = u1(t), and u(t) = u2(t). Based on the
state-space models of subsystems and the above connection information, the state diagram for the
overall system can be completed as shown in Figure 6.17.
?T
The subsystems state variables x1(t) and x2(t) are of course the state variable of the combined
system. The state vector of the overall system is defined as x = ? x1 x2 . Then according to the stat
diagram in Figure 6.17, we have the following state-space modelfor the overall system:
168
6
Systems
Representations
and Interconnected
Systems
? ??0 -2?????
x?1(t)
x?2(t)
x1(t)
x2(t)
-1 1
=
0
1
+
u(t),
y(t)
= ?2 1 ?
??
x1(t)
x2(t)
In the following example, we will consider the construction of a state-space modelfrom a parallel
connection of two subsystems. Thatis, the input-output relationship of the system is described by
Y(s)/U(s)
where G1(s) and
G2(s) can be obtained from
= G1(s)+G2(s)
a decomposition
of a given transfer
function,
or they can
come from two different systems.
Example 6.26 (Construct
a State-Space
Model from a Parallel
Connection of Two Subsystems)
Assumethe relationship of the overall system G(s) andits subsystems G1(s) and G2(s)is described
by
Y(s)
U(s)
=
Y1(s)
U1(s)
+
Y2(s)
U2(s)
Fig. 6.18: Construct a state-space
= G1(s)+G2(s)
model from
a parallel
=
2
+
s+1
connection
-1
s+2
of two subsystems.
Withthe inputs and outputs of the subsystems G1and G2being denoted by U1, U2, and Y1, Y2,re-spectively,
and the given transfer functions of the subsystems, wecan construct the state-space models
for each of the two subsystems as follows:
Y1
U1
2
=
s+1
X1
?
X1
x?1=-x1+u1
,
and
y1 = 2x1
Y2
U2
=
-1
s+2
X2
X2
?
x?2=-2x2
+u2
y2=-x2
Based on the above state-space models of the two subsystems, and the fact of u1 = u2 = u and
y = y1 +y2 due to the parallel connection, the combined state diagram can be completed as shown
in Figure 6.18. Notethat the G1subsystem state diagram is drawn onthe top side andthe G2subsystem
state diagram is at the bottom side.
Both subsystems
the overall system is the sum of the two subsystems
are driven by the same input
?T
? ??0 -2?????
Now, assign the outputs
of the integrators
u(t), and the output of
outputs.
as state variables
x1 and x2 as shown in the state diagram.
Let x = ? x1 x2
bethe state vector of the overall system, where x1 and x2 arethe state variables of the
subsystems G1and G2,respectively. Then according to the state diagram in Figure 6.18, we havethe
following state-space modelfor the overall system:
x?1(t)
x?2(t)
=
-1 0
x1(t)
x2(t)
+
1
u(t),
1
y(t)=? 2 -1?
?
x1(t)
x2(t)
6.6
In the following
example,
State-Space
we will consider the construction
Models and Transfer
of a state-space
Functions
model from
169
a feedback
connection with both the plant andthe controller transfer functions. Thatis, the input-output relationship
of the system is described
by
Y(s)
R(s)
G(s)K(s)
=
1+G(s)K(s)
where G(s) is the plant, whichis the system to be controlled, and K(s) is the controller.
Fig. 6.19: Construct a state-space
Example 6.27 (Construct
a State-Space
model from
a feedback
connection.
Model from a Feedback Connection)
Assumethe feedback connection is depicted bythe block diagram shown on the top of Figure 6.19,
which
meansthe feedback
control system is described
by the following
three equations:
Y(s)=G(s)U(s),
U(s)=K(s)E(s),
andE(s)=R(s)-Y(s)
Here, G(s) represents the transfer function of the DC motor westudied in Example 5.25, and K(s) is
an integral controller to be designed. We will first build an individual state diagram for each of the
plant and the controller transfer functions asfollows:
Y
U
145.5
=
X1
s+43.14
?
X1
x?1=-43.14x1
+u
,
and
y1 = 145.5x1
U
E
=
KI
X2
s
X2
?
x?2 = 0x2 +e
u = KIx2
The G(s) and K(s) subsystem state diagrams are built and shown on the right-hand and the left-hand
sides, respectively, in Figure 6.19, and then are connected together based on the feedback connection
e =r -ytocomplete
thefeedback
system
state
diagram.
Now,
after
assigning
theoutputs
ofthe
integrators
as state variables
x1 and x2, we have the state equation
? ?? ?????
x?1(t)
x?2(t)
Once the state-space
=
-43.14
KI
-145.40
x1(t)
x2(t)
0
1
+
u(t),
and the output equation as follows:
y(t)
= ? 145.4 0 ?
model or the state diagram is obtained for the feedback
??
x1(t)
x2(t)
control system,
we can use
them for simulation, for analysis, and for design.
The characteristic equation of the closed-loop system can be computed from the A matrix of the
state-space
model as follows:
?
|sI-A| =
?
?
?
s+43.14
-KI
?=s2
+43.14s+145.4
=
?
?
145.4
s
?
170
6
Systems
Representations
and Interconnected
Systems
Recallthat the second-order characteristic equationcan be writtenin terms ofthe dampingratio ? and
the naturalfrequency ?nasfollows:
s2 +2??ns+?2 n = 0
Hence, we can relate the integral
following
constant KI to the damping ratio and the natural frequency by the
equations:
2??n = 43.14,
and
?2
n = 145.4KI
If the damping ratio is chosen to be ? = 0.9, the system response is underdamped with a very small
overshoot
and almost
undetectable
oscillations.
Then the natural frequency
and the integral
constant
will
be ?n = 23.97 rad/s and KI = 3.95, respectively.
Wecan conduct a computer simulation
diagram
or state-space
closed-loop
transfer
function
the state diagram together
of the DC motor speed control system based on this state
model, and the results
will be the same as those
can be computed
with
either using the
Masons gain formula.
shown in
Laplace transform
The Laplace transform
?
formula
approach
?
s
3.95
?
??
0
1
=
-145.4
s+43.14
The
or by using
will give
?-1
??
-3.95
Gclosedloop(s)
=C(sI-A)-1B
=? 145.4
0? s+43.14
145.4
s
= 1 ? 145.4 0 ?
?
Figure 5.23.
0
1
574.7
s2+43.14s+574.7
If the state diagram is employed together with Masons gain formula to compute the closed-loop
transfer function, we will find two loops and oneforward path in the state diagram. The loop gains and
the forward path gains are
?1=-43.14s-1,
?2=-574.7s-2,
andF1=574.7s-2
which leads to
?=1-(?1+?2)
=1+43.14s-1
+574.7s-2
and?1=1
Therefore, the closed-loop transfer function is
Gclosedloop(s)
=
F?1
?
=
574.7s-2
1+43.14s-1 +574.7s-2 = s2
574.7
+43.14s+574.7
6.7 Exercise Problems
P6.1a:
Consider the
PI (proportional
and integral)
Kp = 0 (i.e., the controller is an integral controller
closed-loop system has damping ratio ? = 0.707.
feedback
control
K(s) = Ki/s).
system shown in
Figure 6.20. Let
Determine the value of Ki so that the
Fig. 6.20: PIfeedback control system
6.7
Exercise
Problems
171
P6.1b: Withthe controller K(s) chosenin P6.1a,find the naturalfrequency ?nandthe transfer functions
Y(s)/R(s)
and U(s)/R(s)
of the closed-loop system.
P6.1c: Plotthe output response y(t) and control input u(t) dueto the reference input r(t) = 20us(t) on
separate graphs using the two transfer functions obtained in P6.1b and the MATLAB step command.
P6.2a: Consider the PI (proportional
andintegral) feedback control system shown in Figure 6.20. Let
Ki = 0(i.e., the controller is a proportional controller
closed-loop
P6.2b:
system has steady-state
Withthe controller
K(s)
error smaller than
=Kp). Determine the value of Kpsothat the
10%.
K(s) chosen in P6.2a, find the time constant t and the transfer functions
Y(s)/R(s) and U(s)/R(s) of the closed-loopsystem.
P6.2c: Plotthe output response y(t) and control input u(t) dueto the reference input r(t) = 20us(t) on
separate graphs using the two transfer functions obtained in P6.2b and the MATLAB step command.
Comment onthe limitation of the proportional control system.
P6.3a: Consider the feedback control system shown in Figure 6.20, wherethe controller is a PI con-troller
(proportional and integral controller) K(s) = Kp +(Ki/s).
Determinethe values of Kpand Ki so
that the closed-loopsystem has dampingratio ? = 0.707 and naturalfrequency ?n = 50rad/s.
P6.3b: Withthe controller K(s) chosen in P6.3a,find the transfer functions
of the closed-loop system.
Y(s)/R(s)
and U(s)/R(s)
P6.3c: Plotthe output response y(t) and control input u(t) dueto the reference input r(t)
separate graphs using the two transfer
functions
obtained in P6.3b and the
MATLAB
step
= 20us(t) on
command.
P6.4a: Considerthe dual-loop feedback control system shown in Figure 6.21. Determinethe values ofK1
and K2so that the closed-loop system has damping ratio ? = 0.707 and natural frequency ?n =50 rad/s.
Fig. 6.21: A dual-loop feedback control system.
P6.4b: Withthe values of K1 and K2 chosenin P6.4a,find the transfer functions Y(s)/R(s) and
U(s)/R(s) ofthe closed-loop system.
P6.4c: Plot the output response y(t) and control input
u(t) due to the reference input r(t)
= 20us(t)
on separategraphsusingthe MATLAB step command andthetwo transfer functions Y(s)/R(s) and
U(s)/R(s) obtainedin P6.4b
172
P6.5:
6
Systems
Representations
Compare the closed-loop
and Interconnected
system
Systems
performances
of the above four
controllers
obtained in
P6.1,
P6.2, P6.3, and P6.4, respectively, and give your comments. Remember that the objective of the feed-back
control is to find
reference input
a controller
so that the closed-loop
and the output is as small
as possible,
P6.6a: Consider a system described by the following
system is stable and the difference
subject to the control-input
between the
magnitude constrains.
state equation:
? ?? ???? ?
x?1(t)
x?2(t)
Is the system
=
stable? Explain the reasoning
-2 -2
2
3
x1(t)
x2(t)
+
-1
1
u(t)
behind your answer.
P6.6b: Design a state-feedback controller u(t)
= ? F1 F2 ?
??
x1(t)
x2(t)
so that the closed-loop
ble, andthe characteristic equation ofthe closed-loopsystemis s2+2??ns+?2
system is sta-n
= 0 with damping ratio
? = 0.866 and natural frequency ?n = 4 rad/s.
P6.6c: Build a Simulink programto conductthe closed-loopsystemsimulation withinitial statex1(0) =
2, x2(0) = 0. Plot x1(t), x2(t) on the same graph and u(t) on a separate graph using
command.
MATLAB plot
P6.7a: Consider the three state diagrams shown in Figure 6.22(a), Figure 6.22(b), and Figure 6.22(c),
respectively.
Find the state-space
model associated
with each of the three state diagrams.
P6.7b: Use Masonsgainformula to find the transfer function Y(s)/U(s)
associated with each of the
three state diagrams shown in Figure 6.22(a), Figure 6.22(b), and Figure 6.22(c), respectively.
Fig. 6.22: State diagrams
PartII
Linear Control System Desig
7
Fundamentalsof Feedback Control Systems
I
N Chapters 2 and 3, welearned that the frequency-domain approach not only transforms the differ-ential
equations into the algebraic onesto makethem easier to solve, but also reveals the valuable
frequency-domain properties and provides convenient, easy-to-use tools for the analysis and design
of control systems. Notethat the frequency-domain and the time-domain properties are closely related
via an easily crossed bilateral bridge: the Laplace transform. Asshown in the typical second-order sys-tem
step response graph in Figure 3.11, the time-domain attributes, like the maximum overshoot, the
oscillation
frequency,
the frequency-domain
the rise time,
and settling time
of the step response,
are
mainly characterized
by
poles.
Despite the success of the frequency-domain approach based on transformed transfer functions, it
only applies to linear, time-invariant SISO (single-input/single-output)
systems. Recallthat in Chapter
4, we obtained a dynamics model of the simple inverted pendulum system, which was described by a
nonlinear differential equation. In order to investigate the properties of the system, we would have to
introduce
the concept
of interest.
of equilibriums,
and explain
how to find
alinear local
model for each equilibrium
Although there exists other ad hoc ways of locating the equilibriums andfinding local linear
models, the
most systematic
and efficient
approach is to carry out the linearization
process in state space.
Forthis purpose, in Section 4.4.2, weintroduced the state-space analysis approach the first time in
the book. First, the original nonlinear differential equation wasconverted into a nonlinear state equation,
in which the state variables were chosen to bethe angular displacement and the angular velocity of the
pendulum. The nonlinear state equation wasthen employed to systematically determine the equilibrium
points,
and to assemble
state-space
plane,
a linearized
model associated
state-space
with the upright
which verifies the instability
local
model at the
equilibrium
equilibrium
has a pole in the right
of interest.
The linear
half of the complex
of the upright equilibrium.
Furthermore, in Section 4.4.3, students were guided to utilize the local linear state-space modelto
design a state-feedback controller based on a simple pole placement concept sothat the closed-loop sys-tems
poles are in the left half of the complex plane. A computer simulation was conducted to validate
that the linear
state-feedback
the originally
unstable
controller,
nonlinear
system.
designed based on the linear
This state-feedback
controller
model, was capable
design turned
of stabilizing
out to be the first
feedback control system design project the students would learn from the book.
In addition to this state-feedback controller design experience, students had also learned how to de-sign
anintegral feedback controller and a dual-loop feedback controller in Section 5.7.4 and Section 6.1,
respectively. In both of these design examples, the frequency-domain approach based on transformed
transfer
functions
was employed
to place the closed-loop
system damping ratio and natural frequencies
system poles according
to the desired closed-loop
176
7
Fundamentals
of Feedback
Feedback controllers
example,
past three
Systems
can be designed in state space or in frequency
its strength and limitations.
in the
Control
domain.
Either approach
Fortunately, the advancement of control theory and digital computing tech-nology
decades has
the frequency-domain
blended into the cost function
made it possible to integrate
MIMO (multi-input/multi-output)
(or performance
index)
these two
approaches
system robustness
of the state-space
together.
H2 and H8 optimization
formu-lation
with the
design approach to achieve a best possible performance.
The main objective
control
For
measure can be
process. Onthe other hand, the state-space LQR or LQG approaches can work together
root locus
has
of this chapter
systems in later
which include
chapters.
the advantages
is to provide
We will first
the fundamentals
address
and the limitations.
basic features
Then, after
for the discussion
of feedback
a brief review
of feedback
control
of system
systems,
represen-tations
and commonly used terminologies,
we will introduce stability, similarity transformations,
Routh-Hurwitz stability criterion, and basic pole-placement control systems designs.
7.1 Features of Feedback Control
The discovery
of feedback
control
seems not as well recognized
as the great inventions
of the steam
engine and aircraft. The steam engine wasthe main power force behind the Industrial Revolution at the
turn of the 19th century. About a hundred years later, the invention of aircraft by the Wright Brothers
created the aerospaceindustry and opened up a new era of aviation transportation that shrinks the world
and makesthe people all over the world more connected than before. However, it was James Watts
flyball governor control system that transformed the inefficient, unsafe, sometimes even explosive
steam engines into controllable, reliable, and efficient ones sothat the steam engine could be widely
employed and become the main power source for the Industrial
Revolution.
Forthe flight control of aircraft, Wilber Wrightenvisioned that the age of flying
will have arrived
when this one feature (the ability to balance and steer) had been worked out, for all other difficul-ties
are of minorimportance. The age of flying certainly has arrived, andthe implementation of feedback
control theory has greatly improved the ability to balance and steer and has enhanced the safety, qual-ity,
and performance of the flight. Feedback control technology have been embedded in almost all
aerospace vehicles, ships, manufacturing processes, power grids systems, automobiles, and a vari-ety
of machines/devices that require automation, optimization, precision, stability, safety, reliabil-ity,
and performance enhancement. In the following,
we will briefly discussthe features of feedback
control and its limitations.
7.1.1
A Demonstrative
Feedback
Control
System
Instead of just listing the features of the feedback control, we would associate these features with some
tangible feedback control system that we can experience in the lab or in computer simulations. The
simple
features
motor speed control system
of feedback
control.
we designed in
The block
diagram
Example
of the
6.1 will be employed
motor speed control
to demonstrate
system
of Figure
some
6.3 is
redrawn in Figure 7.1. Weadd the disturbance input d(t) to the block diagram in order to study the
system response due to the disturbance. The plant G(s) will be altered so that the effect of the plant
perturbations
on the system can be observed and analyzed.
In Figure 7.1, G(s) is the mathematical model of the plant, including the Faulhaber 2230.012S
the
function
1:14 gear head, and the external load
J? and B?, described in
Example
mi-cromotor,
5.25. The transfer
G(s) is
G(s) =
xss
ts+1
(7.1
7.1
Features of Feedback
Control
177
wheret = 0.02318sec and xss = 3.3727rad/s as givenin Equation 5.68. Recallthat the objective of
the speed control system is to design a controller
stable and the steady-state tracking
K(s) or K1 and K2 sothat the closed-loop system is
e(t) = 0), andthe transient error response is
error is zero (i.e., lim
t?8
as small as possible.
Fig. 7.1:
A speed tracking
feedback
control
system.
The
controller
was
designed
in Example
6.1,K1=-0.322
andK2=17.18.
With
theadded
distur-bance
input, the closed-loopsystem now hastwo inputs: the referenceinput r(t) =L-1 [R(s)], together
withthe disturbanceinput d(t) =L-1 [D(s)], and oneoutput y(t) =L-1 [Y(s)], whichrepresentsthe
angular speed of the load
moment of inertia,
J?. Applying
Figure 7.1, we can find the transfer functions
respectively,
GR(s) =
Masons gain formula
to the block diagram
GR(s) from R(s) to Y(s), and GD(s) from
K2xss
,
D(s) to Y(s),
ts2 +s
GD(s) =
of
(7.2)
ts2+(1-K1xss)s+K2xssts2+(1-K1xss)s+K2xss
and therefore, the output Y(s) hastwo parts: one dueto the reference input R(s) andthe other dueto the
disturbance input D(s).
Y(s) = GR(s)R(s)+GD(s)D(s)
(7.3)
:=YR(s)+YD(s)
Assume the reference input r(t) and the disturbance input d(t) are step functions with arbitrary
magnitude. Thatis, R(s) = R/s and D(s) = D/s, where R and Dare arbitrary constants. According to
Theorem
2.27, which is the final-value
lim yR(t)
t?8
theorem,
= lim sYR(s) =limGR(s)R
s?0
s?0
= R,
and Equations
lim yD(t)
t?8
7.2 and 7.3,
we have
=lim sYD(s) =limGD(s)D
s?0
=0
(7.4)
s?0
Thesefinal values of yD(t) and yR(t) reveal that at steady state the effect of the disturbance d(t) on
the output y(t) will be zero, and the output y(t) will perfectly follow the reference input r(t). Notethat
the steady-state response analysis is conducted completely in frequency domain. Similarly, the transient
response of the system also can be characterized without an explicit time-domain expression of y(t).
It can be seen from
Equation
7.2 that the characteristic
equation
of the closed-loop
ts2+(1-K1xss)s+K2xss
=0
system is
(7.5)
Withthe plant parameters t, xssand the controller constants K1, K2,the characteristic equation become
178
7
Fundamentals
of Feedback
Control
Systems
s2 +90s+2500 =s2 +2??ns+?2n = 0
(7.6)
Hence,the damping ratio, the natural frequency, andthe pole locations of the closed-loop system are
? = 0.9,
p1
?n = 50 rad/s,
=-a j? =-45 j21.795
p2
(7.7)
Thetransfer function GR(s)in Equation7.2 andthe dampingratio ? = 0.9revealthat the closed-loop
system is an underdamped typical second-order system. Thus, the formulas listed in Section 3.4.3
can be employed here to compute the rise time, the settling time, the peak time, the maximum over-shoot,
and other local peak and valley points of the step response. Since the oscillation frequency is
? = 21.795rad/s,the period of the decayingsinusoidal step responseis T = 2p/? = 0.288s. Hence,
the peaktime tp =T/2 = 0.144s, andy(tp) = R(1+e-45tp) = R(1+0.0015) imply that the maximum
overshoot is OS = 0.15%. Notethat the maximum overshoot is so small that the maximum peak point is
?
already
within
2%ofthesteady-state
value,
which
implies
thesettling
time
isless
than
thepeak
time
tp.Itisverified
bythesettling
time
formula,
ts =-ln(0.02
1-?2)/?=0.105
s.
Therefore, the feedback control not only achieves perfect disturbance rejection and reference track-ing
at steady state, it also provides excellent transient response, allowing fast, smooth convergence to
the steady state.
7.1.2 Performance
Verification
by Time-Domain
Simulation
The ability to envision the time-domain response based onthe frequency-domain attributes like transfer
function, pole-zero pattern, damping ratio, and natural frequency, is essential, especially in the phase of
feedback control design since the explicit expression of the time-domain response is not available until
the design is completed. However, after a controller is designed on the drafting board, it is a necessity
for the controller to undergo athorough time-domain simulation verification to determine if the control
system has fulfilled
all the required
closed-loop
system performance.
Fig. 7.2: Effect of the disturbance d(t) on y(t) and u(t).
A Simulink simulation based onthe speed tracking feedback control system block diagram shown
in
Figure 7.1
tracking
was conducted
performance
to verify the
of the system.
disturbance
The simulation
response
results
reduction
capability
and the reference
are shown in Figure 7.2. The reference input
r(t) = 20us(t) (in green color) is a step function intending to command the motorto increase its angular
velocity y(t) from 0 to 20 rad/s. The output y(t) (in blue color), following the command, would ris
7.1
Features of Feedback
Control
179
quickly
and
smoothly
to2098%=19.6
rad/s
in0.105
s(thesettling
time),
topass
thedesired
speed
20 rad/s at t
= 0.123 s, and reach its
maximum speed 20.03 rad/s (0.15%
maximum overshoot) at the
peak time tp = 0.144 s. The continued sinusoidal fluctuations of y(t) around the reference speed 20 rad/s
will decay quickly to negligible
shortly
after the peak time.
To demonstrate how the feedback control system reacts to a disturbance, a disturbance pulse d(t)
(in red color) is assumed to occur between t
= 0.2 s and t
= 0.3 s. It can be observed that the feedback
control system takes immediate action to quickly reduce the influence of the disturbance d(t) on y(t) to
about zero in 0.06 s.
Onthe right-hand side of Figure 7.2 is the graph of the control input u(t) versustime t. It can be seen
from the motortransfer function G(s) that to sustain the desired motor speed 20 rad/s at steady state,
the control-input voltage u(t) needsto be u(t) = 5.93 V at steady state. The graph in Figure 7.2 shows
exactly what weexpect after the system reaches the steady state shortly after t = 0.144 s. The graph also
shows that upon the sudden step change of the reference input r(t) from 0 to 20 rad/s at t = 0 s, the
control input u(t) immediately shoots up from 0 Vto about 6 Vin 0.05 s, then a little overshoot, and
converges to the new equilibrium at u(t) = 5.93 V.
Whenthe disturbance d(t) jumps from 0to 2rad/s at t = 0.2 s,the control input responds right away
to drop about 0.7 V and continues to adjust its value to counter the effect of the disturbance and bring
the motor speed back to the equilibrium 20 rad/s. Att = 0.3 s whenthe disturbance recedes to 0, the
control input u(t) again reacts accordingly to reduce the effect of the disturbance to zero in about 0.05 s,
and brings the system back to the equilibrium, y(t) = 20 rad/s and u(t) = 5.93 V, at the steady state.
The next simulation examines the robustness of the motor speed tracking feedback control system
against the uncertainties in the plant G(s). The mathematical model we employ in the design process
usually is not exactly the same asthe real physical system it represents. The discrepancy may be caused
by the
modeling inaccuracies
conditions
and the variations
like temperature,
capable of accommodating
of the system
pressure, and gravity.
the unmodeled
dynamics
parameters that
are functions
Hence, it is important
or the perturbations
of envi-ronmental
for a system to be
of the system parameters.
The graphs in Figure 7.3 record the results of three motorspeed tracking simulations. Thereference
input is assumedto be r(t) = 20us(t) rad/s (in green color), and the objective of the feedback control
system is to makethe output y(t), the motor speed,to follow the reference input r(t) as closely as pos-sible,
subject to the control-input constraints.
The first simulation is for the nominal case whenthe plant parameters t and xss are the same as
the nominal values t = 0.02318 s and xss = 3.3727rad/s, respectively. Thetracking responsey(t) is
shown in red onthe left graph of Figure 7.3. The corresponding control-input response u(t), also in red,
is shown on the right graph of the figure. These nominal responses y(t) and u(t) are the same asthose
shown in Figure 7.2.
In the second simulation, the plant parameter t remains the same but the parameter xssis reduced
20% to xss = (0.8)(3.3727) rad. The characteristic equation of the perturbed closed-loop system will
become s2 +80.6216s+1999.76
= 0, and the damping ratio andthe natural frequency will change to
? = 0.901 and ?n = 44.72rad/s, respectively. Thetracking response y(t) is shown in black onthe left
graph of Figure 7.3, slightly below the red nominal response. Thetransient response is a little bit slower
than the nominal one, but the steady-state response is identical to the nominal one, still at the desired
20 rad/s. The corresponding control-input response u(t), also in black, is shown on the right graph of
the figure. Notethat the control input u(t) is very different from the nominal one: it has to increase
by 25% (since
parameter
xss
1/0.8=1.25)
to 7.41
V at steady state to accommodate
the 20%
decrease
of the
plant
180
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.3: Effect ofthe perturbationsof t and xsson y(t) and u(t).
In the third simulation, the plant parameterxssremainsthe same, butthe parameter t is increased
20% to t = (1.2)(0.02318) s. The characteristic equationof the perturbedclosed-loopsystem will be-come
s2 +75s+2083.33 = 0, andthe dampingratio andthe naturalfrequency will changeto ? = 0.822
and ?n = 45.64rad/s,respectively. Thetracking responsey(t) is shownin blue ontheleft graphof Fig-ure
7.3, slightly abovethe red nominal response. The transient response has a little bit more overshoot
than the nominal one, but the steady-state response is identical to the nominal one, at the desired
20 rad/s. The corresponding control-input response u(t), also in blue color, is shown on the right graph
of the figure. Note that u(t) is different from the nominal one since it hasto adjust to minimize the
effect on y(t) due to the change of the plant parameter t.
In summary,
the feedback
control
mechanism in the
motor speed tracking
feedback
control
system
shown in Figure 7.1 provides excellent robustness against the parameter variations in the plant G(s). The
steady-state
tracking
the transient
tracking
performance
is perfectly
performance
robust,
and the effect of the plant parameter
is very smallalmost
negligible.
The feedback
variations
control
on
mechanism
is also capable of performing perfect steady-state step command tracking and arbitrary constant distur-bance
rejection to achieve fast convergent transient response.
7.1.3 Advantages and Limitations
From the
analysis
and simulation
of Feedback Control
verification
of the
motor speed tracking
control
system
discussed
in Sections 7.1.1 and 7.1.2, we have witnessedthe advantages brought forth by the feedback control
theory in dealing withthe important systems and control issues including disturbance rejection, precision
command tracking, robust stability, and performance against plant uncertainties.
We will first summarize the advantages of feedback control, andthen briefly discussthe limitations
and caution items in feedback
control
applications.
Figure 7.1 will continue to be employed
The basic feedback
in interpreting
the functions
G(s) is not restricted to be the DC motor, andthe controller
strategy.
control
block diagram shown in
of feedback
K(s) mayrepresent
control,
but the plant
more general control
Advantages of Feedback Control
1. Automatic
Control
Feedback control is a naturally
perfect
mechanism for automatic
control.
better way to achieve automatic control than using feedback control.
block
diagram shown in Figure 7.1 has been employed
There probably
exists no
The basic feedback control
to achieve automatic
speed control
of a D
7.1
motor. The same feedback
control
principle
Features of Feedback
can be applied to design the automobile
Control
181
cruise control
system or can be extended to address the 3D trajectory tracking issues relevant to aircraft guid-ance/navigation,
missile defense and satellite
positioning
systems. In a feedback
control
system, the
control input u(t) is not obtained by any complicated micro management process. Instead, it is au-tomatically
generated in real time by the controller according to the control strategy K(s) and the
real-time values of the reference input r(t) and the output y(t). The control strategy is predesigned
to achieve perfect reference tracking
to control-input
2.
Ability to
at steady state, and optimize the closed-loop
performance
sub-ject
constraints.
Modify the Dynamic Characteristics
via Pole Placement
The dynamics behavior of a system is mainly determined byits pole locations; therefore, through
feedback control the closed-loop system poles can be placed at desired locations to improve the
performance of the system. Recall that the poles of the unstable uncompensated simple inverted
pendulum
system
inSection
4.4.2
were
at2.715
and-3.315,
which
are
theeigenvalues
oftheAU
matrix.
With
astate-feedback
controller,
thecompensated
systems
poles
areplaced
at-4 j3,
which are the eigenvalues
of AU
+BUF, as shown in
Equation
4.46. The new poles are all in the
left half of the complex planeto ensure the stability ofthe compensated cart-inverted pendulum. For
another example, the
motor speed tracking
control system in
Figure 7.1, the
damping ratio
and the
natural frequency of the closed-loop system were purposely selected as ? = 0.9 and ?n = 50 rad/s,
respectively,
3. Steady-State
to achieve a desired performance
Disturbance
with small
maximum overshoot
and fast rise time.
Response Rejection
A feedback controller can be designed to achieve steady-state disturbance response rejection for
certain types of disturbances, including step disturbances with arbitrary magnitude, ramp distur-bances
with arbitrary slope, and sinusoidal disturbances with arbitrary amplitude and phase. Forthe
motor speed tracking control system example, the controller was designed to achieve steady-state
disturbance response rejection for the step disturbances with arbitrary magnitude. It wasshown in
Equation 7.4that the steady-state disturbance response is zero, lim yD(t) = 0. The steady-state dis-turbance
response rejection
4. Steady-State
Reference
Input
t?8
wasalso verified in the simulation shown in Figure 7.2.
Tracking
Steady-state reference input tracking
and steady-state
disturbance
response rejection
are mathemat-ically
the same problem. A feedback controller can be designed to achieve steady-state reference
input tracking for certain types of reference inputs, including step references with arbitrary magni-tude,
ramp references with arbitrary slope, and sinusoidal references with arbitrary amplitude and
phase. For the motor speedtracking control system example, the controller was designedto achieve
steady-state reference input tracking for the step reference inputs with arbitrary magnitude. It was
shown in Equation 7.4 that the steady-state output response is the same asthe reference input at
steady state, lim yR(t) = R. The steady-state reference input tracking wasalso verified in the simu-lations
t?8
shown in Figure 7.2 and Figure 7.3.
5.
Robust Performance
Against
Plant
Uncertainties
For an uncompensated system, whatever variations occurs in the plant dynamics
directly
affect the
output response.
Feedback control
provides
an effective
will pass on to
mechanism to tremen-dously
reduce the effect of the plant uncertainties and variations on the output. For the motor speed
tracking control system example, it can be seen from the simulation results on Figure 7.3 that the
20% variations of the plant parameters t and xss have no effect on the output y(t) at steady stat
182
7
Fundamentals
of Feedback
except a slight influence
Control
Systems
on the transient
response. It also can be seen from the right-hand
side of the
figure that the control input u(t) responds to the plant variations to adjust accordingly to
their effect on the output y(t).
Limitations
minimize
of Feedback Control
Despitethe overwhelmingly
may do
convincing advantages, feedback control hasits limitations
more harm than help if caution is not taken to address the issues
delay, actuator saturation,
and dramatic
andit occasion-ally
of sensor accuracy, time
change of plant dynamics.
1. Sensing Errors
For a speed tracking or position tracking control problem, the accuracy of tracking control accuracy
can only be as good as that of the sensor measurement. The sensor error will directly passto the
tracking
error.
wrongly
For example, if in an automobile
cruise control
system a speed of 70
measured as the set speed 65 miles/hours, then the feedback
miles/hour tracking error without knowing it.
filtering
can be addressed
system
miles/hour is
will have a 5
More serious measurement mistakeslike an orienta-tion
sensing error can cause a positive feedback
error and inaccuracy
control
and destabilize
by employing
diversified
the system.
redundant
The issues
of sensing
sensor suits and fault-tolerant
algorithms.
2. Time Delay
For a feedback control system to work, it requires continuous repetitive cycles of operations in-cluding
detection, information processing, decision, actuation, and control. Since it needs time to
perform tasks in each cycle of operation, more or less there is atime latency ortime delay for each
control cycle. If the time delay is too long, bythe time the information is received bythe controller
it is already out of date, not reflecting the current status of the system any more. Using out-of-date
information for decision and control maylead to poor performance or even cause the system to
become unstable. To alleviate the detrimental effect of time delay, the detailed dynamics model,in-cluding
dynamics
in the control
3.
Actuator
Actuators
of the plant, actuators, sensors, and data transport
delays, need to be incorporated
system design.
Saturation
are designated to implement
may not be able to execute
all faithfully
the
control
due to
actions
physical
determined
constraints.
by the controller,
but they
For example, the range
and
capabilities of the power supplies, the engines, and the control surfaces of the airplane are all physi-cally
constrained. Normally the control-input signals are within the control-input constrained limits,
but under some emergencies the controller may haveto issue stronger control signals in attempting
to mitigatethe crisis. If the control signal exceeds the actuator limits, the actuator will saturate at its
extreme value, and the actual control input delivered to the plant will be the saturated oneinstead
of the intended one by the controller. Unless a special contingency control strategy is in place to
address the actuator saturation issue, the saturation may cause a miscommunication between the
plant and the controller and drive the controller output even higher that would continue keeping the
input of the plant at the sustained saturation value. This wind-up phenomena and sustained actuator
saturation
may cause the feedback
control system to depart from its intended
and become unstable.
4.
Dramatic
Change of Plant
Dynamic
operating
equilibrium
7.2
Afeedback
control system is usually
System
Representations
designed based on the dynamics
and Properties
model representing
(the system to be controlled) in an operating region around a desired equilibrium.
controller
thus designed is supposed to
become inadequate
a violent
work well in this particular
if the plant dynamics
disturbance.
suffers
A remedy for this failure
a dramatic
condition
and replace it by a contingency controller specifically
operating
region;
183
the plant
The nominal
however, it
change due to a component
is to disengage the inadequate
will
failure
or
controller
designedto address the failure condition.
7.2 System Representations and Properties
A system is an operator that specifies the cause/effect relationship between the input u(t) andthe output
y(t): y(t) = G[u(t)], as shown in Figure 7.4.
Definition
7.1 (Linear
A system
superposition
Systems)
or an operator
principle
G, as shown in Figure
7.4, is said to be linear
if and only if the following
is satisfied:
G[c1u1(t)+c2u2(t)]
where c1 and c2 are arbitrary
real
or complex
= c1G[u1(t)]+c2G
[u2(t)]
(7.8)
numbers.
Fig. 7.4: An operator that specifies the relationship between the input u(t) and the output y(t).
Definition 7.2 (Time-Invariant
Systems)
A system Gis said to betime-invariant
ifand
only if
G
[u(t-t)] =y(t -t)
(7.9)
for any u(t) and any delay time t.
Definition
7.3 (Dynamic
A system
Systems)
Gis said to be dynamic if its output depends on past and present values ofthe input.
Example 7.4 (Linear
All the systems
Time-Invariant
described
Dynamic Systems)
by differential
equations
are dynamic
systems
since the variables
of the
differential equation at every instant always keeptheir results, dueto the past input, before updating to
the new values. The system described by the algebraic equation
184
7
Fundamentals
of Feedback
Control
Systems
y(t)
= au(t)+b
is not a dynamic system. The system has no memory; its output y(t) only depends on the current value
of input u(t). This algebraic equation looks simple, but it is not alinear system since the output due to
an input c1u(t) is not equal to c1y(t) unless b = 0.
The dynamic
system described
by the differential
dy(t)
equation
+ty(t)
dt
= u(t)
apparently is not time-invariant since one coefficient in the equation is function of time.
time-varying system is linear since
dy1
dt
+ty1 = u1
dy2
dt
+ty2
?
d (c1y1 +c2y2)
dt
= u2
For another example, the simple
pendulum
system
However,this
+t (c1y1 +c2y2) = c1u1 +c2u
described
by the differential
equation
J?(t)+B ??(t)+mg?sin?(t) =t(t)
is a nonlinear system because of the sin?(t) term. Butit can be considered alinear system if the operating
range of ?(t) is small so that sin?(t) is approximately equal to ?(t).
Alinear time-invariant
function,
an impulse
a signal flow
system (or subsystem) can be represented by a differential equation, atransfer
response,
a state-space
graph. For some particular
model, or a graphical representation
objective
of analysis
or design,
like
a block diagram
or
we may choose one representa-tion
of the system over the others, but it is quite easy to convert from one representation to another, as
described in Section 6.6.
7.2.1 Transfer
Function and Differential
Equation
As shown in Figure 7.5, alinear time-invariant
time
system can be represented by a differential equation in
domain:
y(n)(t)+an-1y(n-1)(t)+
+a1?y(t)+a0y(t)
=bmu(m)(t)+bm-1
u(m-1)(t)++b1
?u(t)+b0
u(t)
(7.10
or by a corresponding transfer function in frequency domain:
Y(s)
U(s)
= G(s) =
bmsm+bm-1sm-1
+bm-2sm-2
+ +b1s+b0:=
sn+an-1sn-1
+an-2sn-2
+ +a1s+a0
N(s)
D(s)
(7.11)
where
mand
nare
positive
integers.
The
system
issaid
tobeproper
orstrictly
proper
if m=norm<n,
respectively.
All the practical
systems to be considered
are proper or strictly
proper, and the controller
to
be designedis required to be proper or strictly proper since a non-proper controller is not implementable.
Note that the differential equation in Equation 7.10 is the time-domain description of the system,
while the transfer function in Equation 7.11 is the frequency-domain description of the same system.
The bridge
between the differential
The Laplace transform
equation
of the differential
and the transfer
equation
function
of Equation
7.10
in Equation 7.11. Onthe other hand, the inverse Laplace transform
D(s)Y(s)
= N(s)U(s)
will yield the same differential equation.
is simply
the Laplace transform.
will lead to the transfer
function
of the transfer function equation
7.2
Fig. 7.5: Representations of linear time-invariant
Although the initial
conditions
should
System
Representations
and Properties
185
system: differential equation and transfer function.
be considered in the process of solving
for the output response
y(t), they are irrelevant to system representations, andtherefore they should beignored in the represen-tation
conversion
process. To facilitate
the conversion
process, the variable s in the transfer
be considered as the Laplace transform of the differentiation
following differentiation operator conversions:
d/dt
Example 7.5 (Conversion
? s,
d2/dt2
? s2,
Between the Differential
Considerthe system withthe following
=L[y(t)]
and U(s)
Solution:
Using the differentiation
=L[u(t)],
can
,
dn/dtn
? sn
(7.12)
Equation and the Transfer Function)
differential equation,
y(t)+3 ?y(t)+2y(t)
Let Y(s)
function
operator d/dt. In summary, we havethe
=4
?u(t)+12u(t)
(7.13
find the transfer function of the system.
operator conversion
shown in
Equation
7.12, by inspection
we have
the conversion,
y(t)+3 ?y(t)+2y(t)
=4
?u(t)+12u(t)
? (s2 +3s+2)Y(s)
= (4s+12)U(s)
(7.14)
which leads to the transfer function
G(s) =
Y(s)
U(s)
Note that the conversion procedure is bidirectional.
transfer function as easily asthe other way around.
7.2.2 Impulse
4s+12
=
(7.15)
s2 +3s+2
A differential equation can be obtained from the
Response and Transfer Function
Asillustrated in Figure 7.6, let the input to the system
output will be
Y(s) = G(s)L[d(t)]
= G(s)
?
G(s) be a unit impulse function
y(t) = g(t)
d(t), then the
=L-1 [G(s)]
which meansthe impulse response g(t) is the time-domain counterpart of the transfer function G(s)the
inverse Laplacetransform ofG(s). Hence,the impulse response can be considered asa representation
of the system, and the output y(t) can be computed via the following convolution integral:
y(t)
=
?tg(t-t)u(t)dt
-8
(7.16)
186
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.6: Impulse
Example 7.6 (Impulse
response and transfer
Response and Convolution Integral)
Consider the system with transfer function
the impulse
response
function.
G(s) = b/(s
+a), where a and b are constants. Find
of the system and compute the step response
of the system using the convolution
integral.
Solution:
Theimpulse response is
g(t)
=L-1 [G(s)]
= be-at,
t =0
Then the step response (i.e., the output of the system due to the unit step input, us(t)) is
y(t)
-8g(t-t)us(t)dt=?t
-8be-a(t-t)us(t)dt
=0be-a(t-t)dt
?t
?t
=be-at
?t ae-ateat
? ae-at(eat-e0)=ba(1-e-at)
?tg(t)us(t-t)d
=
0eatdt = b
?t0 = b
Notethat the other form of the convolutional integral
y(t)
=
-8
will yield the same result.
7.2.3 State-Space
Model and Transfer
As mentioned in previous
has become increasingly
state-space
control
Function
chapters, the state-space
important
systems theory.
representation
due to the recent rapid
The state-space
or the state-space
advancement
model of systems
of computing
model usually is represented
tools
in the compact
and the
matrix
form:
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
(7.17)
where
thevectors
x?Rn,
u?Rm,
and
y ?Rpare
then1 state
vector,
them1input
vector,
and
the
p1 output
vector,
respectively.
Thematrices
A?Rnn,B?Rnm,
C?Rpn,
andD?Rpm
are
the
nn Amatrix,
thenm Bmatrix,
thepn Cmatrix,
and
thepm Dmatrix,
respectively.
As discussed in the previous
chapters,
particularly
of a system is not unique since the state variables
in Section 6.6, the state-space
can be chosen arbitrarily.
model representa-tion
However, the transfer
function, the characteristic equation, and the poles and zeros of all the state-space modelsrepresenting
the same system are invariant.
A state-space model can be obtained directly in the system modeling process, as demonstrated in
Chapters 4 and 5, where the physical
flowing
through
inductors,
variables
of interest, like the displacements,
and the voltages across capacitors,
are usually
the velocities,
the cur-rents
chosen as state variables.
It also can be assembled from a transfer function, a differential equation, or an interconnected system,
which is composed
of several subsystems
and components,
as described in
Chapter 6.
7.2
System
Representations
and Properties
187
Onthe other hand,the transfer function of the system represented by a state-space model (A,B,C,D)
can be computed
using the following
formula:
Y(s)
U(s)
:= G(s)=C(sI-A)-1B+D
Thetransfer function also can be obtained by applying
with the state-space
7.2.4 Characteristic
U(s)
Equation,
Poles, and Zeros
= G(s) =
transfer function (Equation 7.11),
bmsm+bm-1sm-1
+bm-2sm-2
+ +b1s+b0
:=
sn +an-1sn-1
+an-2sn-2
+ +a1s+a0
Under the assumption that the numerical polynomial
no common
Masons gain formula to a state diagram associ-ated
model, as shown in Section 6.6.1.
Considerthe system represented by the following
Y(s)
(7.18)
factor, the denominator
polynomial
N(s)
D(s)
N(s) and the denominator polynomial
D(s) is defined as the characteristic
system, and the equation D(s) = 0is the characteristic
D(s) have
polynomial
of the
equation:
D(s)=sn+an-1sn-1
+an-2sn-2
+ +a1s+a0=0
(7.19)
The poles of the system are defined as the roots of the characteristic equation, and the zeros of the
system are defined asthe roots of the equation N(s) = 0:
(7.20)
N(s)=bmsm+bm-1sm-1
+bm-2sm-2
+ +b1s+b0=0
Note that the dynamic
system in the complex
behavior
of the system is
plane, as welearned
from
mainly determined
have an effect onthe system response, asillustrated in the following
Example 7.7 (Characteristic
Equation,
12
s2 +3s+2
,
of the poles of the
The location
of the zeros also
example.
Poles, Zeros, and Their Effect on Dynamic Behavior)
Considerthe following three systems withtransfer functions
Ga(s) =
by the location
Sections 3.2 and 3.3.
Gb(s) =
4s+12
s2 +3s+2
,
Ga(s), Gb(s), and Gc(s), respectively:
Gc(s) =
-4s+12
The three transfer functions have exactly the same characteristic equation, s2 +3s+2
v2
has
tworealpoles
ats =-1and
s = -2.Their
damping
ratio
is? =3/2
overdamped.
However, these three systems
have different
zero structure.
(7.21)
s2 +3s+2
= 0, which
= 1.06, and they
Ga(s) has no zero,
are all
Gb(s) has a
zero
inthelefthalfofcomplex
plane
ats = -3,but
thesystem
Gc(s)
has
one
zero
intherighthalfof
the complex plane at s = 3.
The step responses of these three systems are plotted on Figure 7.7. The step response of the sys-tem
Ga(s),
which
has
nozero,
isya(t)=6-12e-t+6e-2t.
The
derivative
ofya(t)canbefound
as
y?a(t)=12e-t-12e-2t,
which
yields
?ya(0)
=0,azero
slope
att =0.Hence,
ya(t)rises
upslowly
in
the beginning.
t
The
step
response
ofthesystem
Gb(s),
which
has
one
zero
ats =-3,isyb(t)= 6-8e-t+2e-2t.
The
derivative
ofyb(t)can
befound
as
?yb(t)
=8e-t-4e-2t,
which
yields
?yb(0)
=4,aslope
of4at
= 0. Hence, yb(t) rises more quickly in the beginning and has afaster response than ya(t)
188
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.7: Effect of zeros on dynamic
behavior
of the system.
The
step
response
ofthesystem
Gc(s),
which
has
one
zero
ats =3,isyc(t)=6-16e-t
+10e-2t.
The
derivative
ofyc(t)
can
befound
as
?yc(t)
=16e-t
-20e-2t,
which
yields
?yc(0)
=-4,aslope
of-4att =0.
Hence,
yc(t)
atfirstwas
going
down
toward
theopposite
direction
until
t =0.223s
atyc(0.223)
=-0.4,
where it reverses its course back to the correct upward direction. Hence, yc(t) has a slower response
than ya(t). Thisinitial opposite direction movementis atypical property of the systems with zerosin the
right half of the complex plane. The systems with this property are called nonminimum phase systems.
7.3 Stability of Linear Systems
Although the definition of stability has not yet been defined, weroughly grasped a little physical sense
of it from the step response of the first-order system as discussed in Section 2.4.2. In Equation 2.75, the
exponential
term
e-t/tofthestep
response
would
become
unbounded
if -1/t >0,which
means
the
characteristic
equation,
ts+1=0,otfhe
first-order
system
would
have
apole
s =-1/t>0
inthe
right
half of the complex plane. Wealso learned from Section 4.4.3 that the simple inverted pendulum system
is unstable because some of its poles are in the right half of the complex plane. We were even more
convinced by the fact that the simple inverted pendulum system was stabilized after a state feedback
placed all the closed-loop
system poles in the left
half of the complex
plane.
Basically there are two stability definitions for linear time-invariant systems. Oneis called BIBO
(bounded-input/bounded-output)
stability, and the other is internal stability. Since these two defi-nitions
are relevant to how the input and the internal initial conditions, respectively, affect the output
response. Before giving the two definitions, we will review the zero-state and zero-input responses in
the following subsection.
7.3.1
Zero-State
Response and
Zero-Input
Response
The total response of a linear system can be decomposed into two parts, as shown in Figure 7.8: the
zero-state response yU(t),
whichis the response dueto the input u(t) only, and the zero-input respons
7.3
Stability
of Linear
Systems
189
yI(t), which is the response due to the initial conditions only. In the following, a simple second-order
linear system is employed to illustrate how the zero-state response andthe zero-input response relate to
the characteristic
equation
or the pole locations
Fig. 7.8: Total response y(t)
of the system.
= zero-state response yU(t)+
zero-input response yI(t).
Example 7.8 (Effects of System Poles on Zero-State Response and Zero-Input
Considerthe system withthe following
y(t)+3
Assume Y(s)
respectively,
=L[y(t)],
U(s)
of the system.
differential equation,
?y(t)+2y(t)
=L[u(t)],
Response)
=4
?u(t)+12u(t)
(7.22)
andlet y(0) and?y(0) bethe initial
The Laplace transform
of the differential
equation
values of y(t) and?y(t),
yields the following
alge-braic
equation:
s2Y(s)-sy(0)?y(0)+3
[sY(s)-y(0)]+2Y(s)
=4sU(s)+12U(s)
which can be rearranged
(7.23)
as follows,
5
Y(s) = s2+3s+2U(s)+
s2+3s+2
1 ?y(0)+ s2+3s+2y(0)
s+3
:=
G(s)U(s)+
? G1(s) G0(s) ?
??
(7.24)
y?(0)
y(0)
and therefore the total response of the system is the sum of the zero-state response and the zero-input
response is as follows:
y(t) =L-1 [G(s)U(s)]+L-1
?
? G1(s) G0(s) ?
? ??
y?(0)
y(0)
:= yU(t)+yI(t)
(7.25)
Notethat the denominators ofG(s), G1(s),and G0(s)are all equal to the characteristic polynomial
of the system.
The result of Example 7.8 can be extended to a more general linear time-invariant
by the differential equation,
system described
y(n)(t)+an-1y(n-1)(t)+
+a1
?y(t)+a0y(t)
(7.26)
=bmu(m)(t)+bm-1u(m-1)(t)+
+b1
?u(t)+b0u(t)
and the generalized result
y(t)
of Equation 7.25 will be
=L-1 [Y(s)]
=L-1 [GU(s)U(s)]+L-1
[GI(s)y0] := yU(t)+yI(t)
(7.27
190
7
Fundamentals
where y0 = ?
of Feedback
Control
Systems
?T
y(n-1)(0)
?y(0)
y(0) andthe
polynomial
of the system,
denominators of GU(s) and GI(s) are equal to the char-acteristic
sn+an-1sn-1
+an-2sn-2
+ +a1s+a0
(7.28)
Remark
7.9
(Conditions
fortheZero-Input
Response
tobeZero
ast ?8and
theZero-State
Re-sponse
to be Bounded)
Since the rational function
GI(s) is strictly proper, it can be seenfrom Equation 7.27 that
lim yI(t)
= limsGI(s)y0
t?8
=0
s?0
if all the roots of the characteristic equation are in the strictly left half of the complex plane. That means,
thezero-input
response
ortheresponse
due
totheinitial
conditions
willonly
decay
tozero
ats ?8
if the real
part
of the roots
of the characteristic
equation
are strictly
negative.
On the other hand,
yU(t),
thezero-state
response,
ortheresponse
due
totheinput
only,
ingeneral
may
benonzero
as
t ?8.
However, for the system to be meaningful, yU(t) needsto be bounded if the input u(t) is bounded. Since
the transfer function GU(s) is proper and its poles are assumed all in the strictly left half of the
complexplane,yU(t) =L-1 [GU(s)U(s)] will be boundedif u(t) is bounded.
The remark
can also be recapitulated
system represented by the following
in the state-space
setting.
Consider a linear
time-invariant
state-space model:
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
Let X(s) and U(s) be the Laplace transform of x(t) and u(t), respectively, and x0represent the initial
state vector x(0). Then the Laplace transform of the state equation will lead to the following solution of
the state equation:
x(t)
=L-1
?(sI-A)-1? ?(sI-A)-1BU(s)
?
x0 +L-1
(7.29)
Since
the
inverse
Laplace
transform
of(sI-A)-1
isthestate
transition
matrix
F(t),
L-1
?(sI-A)-1?
= F(t)
Equation 7.29 can be rewritten in terms of the state transition
x(t)
= F(t)x0 +
matrix asfollows:
?tF(t-t)Bu(t)dt:=xI(t)+xU(t)
(7.30
0
Combining this state equation solution
following:
y(t)
=CF(t)x0
+
with the output equation of the state-space model, we havethe
??tF(t-t)Bu(t)dt+Du(t)
?
C
:= yI(t)+yU(t)
(7.31)
of the state-space
model consists of
0
As its counterpart,
shown in
Equation
7.27, the output response
two parts: Oneis yI(t), which is the output response dueto the initial state only, and the other part is
yU(t), which is the output response due to the input u(t) only. It is easy to seethat the response due to
the initial
state,
7.3
limCF(t)x0
Stability
of Linear
Systems
191
=0
t?8
which
means
thatyI(t)willapproach
tozero
ats ?8,if theeigenvalues
oftheAmatrix
(i.e.,
the
poles of the system), are all in the strictly left half of the complex plane. Meanwhile, with the same
condition of system poles being all in the strictly left half of the complex plane, the output response
due to the input only,
yU(t)
=C
?tF(t-t)Bu(t)dt+Du(t
0
will be bounded for every bounded input
following
u(t). For ease of reference, the result is summarized in the
Lemma.
Lemma 7.10 (Sufficient
Condition for
BIBO Stability and Internal
Assume the system poles (i.e., the eigenvalues
plane, then
(1)
ofthe
Stability)
A matrix) are all in the strictly
left ofthe complex
we have the following:
The response
due to the initial
state x0, lim
t?8
t ?8),and
F(t)x0
= 0 (i.e., xI(t)
will approach to zero as
?t
(2)The
response
due
tothe
input
u(t),
yU(t)=C0F(t-t)Bu(t)dt+Du(t),
isbounded
for
every bounded input u(t).
7.3.2 BIBO Stability and Internal
Stability
From the discussion regarding how a system would respond to any initial state x0 or any bounded input
u(t), it is meaningful to define system stability based on the output response of the system due to input
u(t) only or based onthe state response dueto the initial state x0 only.
Definition
7.11 (BIBO
Stability)
A system is BIBO (bounded-input/bounded-output)
in a bounded output.
stable ifand
BIBO stability is defined based on the input-output relationship.
only if every bounded input results
Forthe cart-inverted pendulum sys-tem
consideredin Section4.5,the outputis the angular displacement? of the pendulum(the stick) and
the objectiveis to keepthe pendulum at or aroundthe equilibrium, ? = 0.If the systemis uncompen-sated,
the pendulum will departfrom the equilibrium (i.e., ? will increase without bound) underthe
influence of a tiny disturbance input.
The uncompensated cart-inverted pendulum system is obviously an unstable system. But some sys-tems
mayonly produce unbounded output for a special kind of bounded input. For example, the notorious
collapse of the Tacoma Narrows Bridge was caused by the oscillation resonance that occurs whenthe
input frequency matchesthe natural frequency of the system.
Hence,the key wordin the definition of BIBO stability is every.
For a system to be BIBO stable,
every bounded input hasto result in bounded output. But, how canit possible to examine every bounded
input? A practically doable approach is using the impulse response integral as described in the following
theorem.
Theorem
7.12 (Impulse
Alinear time-invariant
Response Integral
Check for
BIBO
Stability)
system with impulse response g(t) is BIBO stable ifand
only if
192
7
Fundamentals
of Feedback
Control
Systems
?8
|g(t)|dt=M<8
(7.32)
0
Example
7.13 (Check
BIBO
Stability
Using the Impulse
Response Integral)
Consider
thesystem
ofExample
7.6,whose
impulse
response
isg(t)=be-at,
t =0,where
aand
b
?8
be-atdt
=-bae-at
?8?
are constants. Since
ba if
0 =
8
0
a >0
if a < 0
the system is BIBO stable if and only if a > 0.
This impulse response integral approach is doable; however, the computation of impulse response
integral is tedious whenthe system becomes complicated. Now, we will revisit Lemma 7.10 to investi-gate
if the sufficient condition for BIBO stability is also a necessary condition. This sufficient condition
is indeed also a necessary condition since a system with a pole or poles on the imaginary axis of the
complex plane is not BIBO stable.
Example 7.14 (A System That Has Poles on the Imaginary
Axis Is Not BIBO Stable)
Considerasystem G(s) = 1/(s2 +?2
n ) that has poles on the imaginary axis of the complex plane at
s =j?n.The
output
ofthesystem
isbounded
foralmost
allofthepossible
bounded
inputs.
However,
if theinputis a sinusoidalfunction u(t) =cos?t or u(t) =sin?t, withits oscillationfrequency ? equal
to the natural frequency
?n of the system, then the output will be an unbounded function of time t; hence,
the system in not BIBO stable.
Thesinusoidalfunction, u(t) = cos?t and u(t) = sin?t arethe real partandthe imaginary part,re-spectively,
of the complex function ej?t. Forcomputationalsimplicity, u(t) = ej?t = cos?t +jcos?t is
chosen asthe input, then the output y(t) =yRe(t)+ jyIm(t) is also a complex function of time consisting
of yRe(t) and yIm(t) asits real and imaginary parts, respectively.
Withthe input
U(s)
U(s) and the system G(s) given as:
ej?t?
s-j?
1
=L ?
and
=
G(s) =
1
1
s2 +?2 =
(s+j?)(s-j?)
Then the output Y(s) can be written as
Y(s) = G(s)U(s)
=
1
c1
(s+j?)(s-j?)2=s+ j?
c2
+
c3
+
s-j? (s-j?)2
wherethe partial fractional expansion coefficients are
c1 =
Recall that the parameter ain the following
L-1
1
1
-1
4?2
, c2=-c1=4?2 ,
??
1
s+a
= e-at,
c3 = 2?
Laplace transform
L-1
e-jp/2
pairs can be a complex number:
?(s+a)2=
?te-at
1
Hence, we have the output of the system y(t) dueto the input u(t)
= ej?t in the following
7.3
y(t)
=c2
ej?t
-e-j?t?
?sin?t-?tcos?t?
+c3tej?t
?
The equation is rearranged
y(t)
j2sin?t
=
t
+
4?2
2?
ej(?t-p/2)
Stability
of Linear
sin?t
=j 2?2
+
Systems
193
t (sin?t-j cos?t)
2?
as
=
t sin?t
+j
2?
= yRe(t)+ j yIm(t)
2?2
where
yRe(t)
=(1/2?)t
sin?tand
yIm(t)
=(1/2?2)
(sin?t-?tcos?t)
are
theoutput
responses
due
to the inputs u(t) = cos?t and u(t) = sin?t, respectively. Note that both are unbounded, growing
with time. Therefore, the system is not BIBO stable if it has poles on the imaginary axis.
Since the sufficient
condition
for
BIBO stability
described in Lemma 7.10 is also a necessary condi-tion,
the lemma now can be modified asthe following
Theorem 7.15 (Necessary and Sufficient
A linear time-invariant
theorem.
Condition for BIBO Stability)
system is BIBO stable ifand
only if all its poles are in the strictly left
halfof
the complex plane.
This theorem provides an easily verifiable necessary and sufficient condition for BIBO stability. It
also makesit clear that all the closed-loop system poles are required to be placed in the strictly left
half of the complex plane to achieve stability.
In addition to BIBO stability, which is defined based on the external input-output relationship, there
is another meaningful stability definition called internal stability.
Definition 7.16 (Internal
Stability)
Thelinear time-invariant
system
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
is internally stable ifthe solution x(t) of
x?(t)
= Ax(t)
withinitial state x(0)
=x0
tends
toward
zero
ast ?8forarbitrary
x0.
As described in
Lemma 7.10, the condition
that all the eigenvalues
of the A matrix are in the strictly
left half of the complex plane is sufficient for the system to be internally stable. Similar to the case of
BIBO stability,
with eigenvalues
Example
this condition
is also a necessary condition
on the imaginary
7.17 (A System
axis
would be internally
with Eigenvalues
for internal
stability.
If it is not, then a system
stable.
on the Imaginary
Axis Is
Not Internally
Stable)
Consider the system
x?(t)
= Ax(t)
where x10 and x20 are arbitrary.
=
?0 -? ?
?
0
x(t)
with initial state x(0)
= x0 =
Notethat the eigenvalues of the system,
??
x10
x20
which are the roots of the
characteristic
equation
det(?I-A)
=?2+?2=0,areonthe
imaginary
axis
atj?. Sinc
194
7
Fundamentals
of Feedback
Control
Systems
? ?-1
?? ?s -???? ?x10s-x20?
s
?
1
x10
x20
(sI-A)-1x0
= -? s
=
x10
x20
? s
s2 +?2
1
=
s2 +?2
x10?+x20s
we havethe solution of the state equation as
x(t)
?(sI-A)-1x0
?=?x10cos?t-x20sin?t?
=L-1
x10sin?t +x20cos?t
which
obviously
willcontinue
tooscillate
instead
ofapproaching
tozero
ast ?8.Therefore,
the
system is not internally
Since the sufficient
stable if it has eigenvalues on the imaginary
condition
for internal
stability
condition, the lemma now is modified asthe following
Theorem 7.18 (Necessary and Sufficient
Thelinear time-invariant
is internally
complex
described in
axis.
Lemma
7.10 is also a necessary
theorem.
Condition for Internal
Stability)
system
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
stable if and only if the eigenvalues
of the
A matrix are all in the strictly
left
half of the
plane.
Remark 7.19 (Internal
Stability Is a Stronger
Condition
Than BIBO Stability)
Internal stability is a stronger stability condition than BIBO stability since BIBO stability only re-flects
the attributes of the system that are observable from the output and controllable from the input.
There mayexist an unstable hidden-mode statethat leads to internal instability while the BIBO stability
measureis unable to detect it. However, if all the states of the system are controllable and observable,
then BIBO stability is equivalent to internal stability. Detailed discussion regarding controllability and
observability will be given in Chapters 10 and 11.
Theorem 7.18 provides an easy wayto check the internal stability of an existing system. It also can be
employed to design a stabilizing controller to guarantee the internal stability of the closed-loop system.
In the following section we will discuss how to employ this theorem and a simple state-feedback
control to stabilize an originally unstable system.
7.4 Similarity
Transformation
As discussed in the previous
of a system is not unique
chapters,
in State Space
particularly
since the state variables
in
Section 6.6, the state-space
theoretically
can be chosen
model representation
arbitrarily.
However,
in practice the state variables are not chosen randomly. They usually are selected based on two
considerations.
Oneis to associate the state variables with the physical variables of interest, and the
other is to choose state variables so that the state-space modelis in a special form for the purpose
of analysis or control system design. Onsome occasions, we may need to transform a given state-space
modelto a special form. Thistransformation is called similarity transformation.
Under similarity
transformations,
the characteristic
the A, B, and C matrices in the state-space
equation,
and the poles and the zeros
model may change, but the transfer
will remain invariant.
function,
7.4
Assume a linear time-invariant
system is described
Similarity
Transformation
by the following
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
state-space
in
State Space
195
model:
(7.33)
Letx(t)
bea newstatevector,definedasx(t)
compatible
=T-1x(t), whereTis a nonsingular matrix withdimension
with the state vector. Then wehavethe newtransformed state-space modelin the following:
x?(t) = T-1?x(t) = T-1Ax(t)+T-1Bu(t)
= T-1ATx(t)+T-1Bu(t)
(7.34)
y(t)
=Cx(t)+Du(t)
=CTx(t)+Du(t)
Thatis, the similarity transformationx(t) = T-1x(t) hastransformed the original state-space modelin
Equation7.33to the new state-space model withthe newstatevectorx(t),
x?(t)
=Ax(t)+
Bu(t)
(7.35)
y(t) =Cx(t)+Du(t)
where
A = T-1AT, B = T-1B, C =CT
Theorem 7.20 (Invariance
Under Similarity
(7.36)
Transformation)
Thetransfer function, the characteristic equation, and poles and zeros are invariant
under the simi-larity
transformation that transform the state-space model(A,B,C) to (T-1AT,T-1B,CT).
Proof:
G(s)=
C(sI-A)-1
B =CT(sI-T-1AT)-1T-1B
=CT(TsI-TT-1AT)-1B
=C(TsIT-1
-TT-1ATT-1)-1B
=C(sI-A)-1B
=G(s)
The ability
to use the similarity
transformation
to transform
a state-space
representation
from
one form to another is a great advantage in the analysis and design of control systems. The special
forms of state-space representations include diagonal form, companion form, controller form, observ-ability
form,
noncontrollable
10 and 11, the similarity
canonical
form,
transformation
and nonobservable
canonical
will be employed to achieve the
form.
Later on in
Chapters
Kalman decomposition,
which
breaks down the state spaceinto four parts according to controllability and observability. Moreover,the
similarity transformation also can be utilized to identify the stable subspace in the state space. In the
following,
we will introduce
the diagonal form
and the companion
form
and their advantages in control
system analysis and design.
7.4.1 Diagonalization
of A Matrix Using Similarity
Transformation
Considerthe state-space modelgiven by Equation7.33. Find a similarity transformationx(t)
=T-1x(t)
that transforms the state-space model of Equation 7.33to the new state-space modelshown in Equation
7.35 so that the
A = T-1AT is diagonal.
?
?
?
?1
T-1AT = ? =
?
?
?
?2
??AT
=T?
?
...
?
?
?n
(7.37
196
7
Let ei, i
Fundamentals
= 1,2, ...,n,
of Feedback
be column
Control
Systems
vectors of T. Then,
?
Ae1e2 en = e1e2 en
?
?
?1
?
?
?
?
?2
?
?
??Aei
=?iei
?
...
?
i
?
= 1,2, ...,n
(7.38)
?
?n
where?i must be an eigenvalue of A and ei a corresponding eigenvector. A nonsingular T can be
found if and only if A has n linearly independent eigenvectors.
Example 7.21 (Diagonalization
of a State-Space
Model)
Considerthe system withthe following state-space model,
x?(t)
= Ax(t)+Bu(t)
y(t)
To find a similarity
=Cx(t)
transformation
=
= ? 12
?-2 -2? ?-1?
2
3
x(t)+
u(t)
1
(7.39
16 ? x(t)
that transforms
a state-space
model to a diagonal form, the first step
is to obtainthe eigenvalues?1,?2andtheir associatedeigenvectors
e1,e2.
? ?
? ? ? ???=0 ? e1=? ?
?-2 2-3? ?-2 -1??
? ??
-2 =0 ?e2= -2
? +2
2
det(?I-A)
=det -2 ? -3 =?2-?-2=0 ? ?1=-1,
?2=2
and
-1+2 2
-2 -1-3
2+2
2
e2 =
Let the similarity transformation
1
e1 =
2
4
2
-2
-2
1
1
-2 -4
1
1
matrix be T =[e1 e2]. Then the diagonalized state-space modelis
obtained as follows:
x?(t) = T-1ATx(t)+T-1Bu(t)
=
?? ? ?
1/3
-10 x(t)+
0 2
-1/3
u(t)
(7.40)
y(t) =CT
x(t) =? -8 -20?x(t)
Wehave learned that a given state-space
model can be transformed
to a diagonal form if and
only
ifthenn Amatrix
ofthemodel
has
nlinearly
independent
eigenvectors.
The
diagonal
form
has several advantages.
The poles of the system appear conspicuously
A, and all the off-diagonal
on the diagonal line of the
elements of the A matrix are zero. The diagonal form
matrix
not only saves data stor-age
size, it also tremendously reduces computational complexities, especially for high-order systems.
The corresponding diagonal state diagram on Figure 7.9 shows even more clearly that the system
has
been
decomposed
intotwosubsystems:
onewith
itspole
ats =-1and
theother
with
apole
at
s = 2.
Another important
form
of state-space
model is the companion
form,
implement a pole-placement control system design later in this chapter.
which
will be employed
to
7.4
Fig. 7.9: Similarity
7.4.2
Obtaining a State-Space
To reduce the notational
to
Consider the state-space
y(t)
without loss
demonstrate
=Ax(t)+
State Space
197
Transformation
state-space
model is em-ployed
model to a companion
form.
model given by
where
det(?I-A)
=?3+a2?2
+a1?
+a0
=Cx(t)
(7.41)
= T-1x(t) that transforms the state-space
companion form:
Bu(t)
whereA = T-1AT =
y(t) =Cx(t)
Let the similarity
follows
a third-order
a state-space
The objective is to find a similarity transformation x(t)
x?(t)
in
state diagram.
Using Similarity
of generality,
how to transform
= Ax(t)+Bu(t)
modelto the following
Transformation
to diagonal form
Modelin Companion Form
complexity,
in the following
x?(t)
transformation
Similarity
?
?
0
0
1
0
0
1
?
?,
??
??
0
B = T-1B = 0
-a0-a1-a2
transformation
(7.42)
1
matrix be T =[t3 t2 t1]. Then the equations can be rearranged as
A[t3 t2 t1] =[t3 t2 t1]
?
?
0
1
0
0
0
1
?
?
?
?,B=[t3
t2t1]
??
0
0
-a0-a1-a2
1
which lead to
t1 = B,
t2 = At1 +a2t1,
t3 = At2 +a1t1
or
T = ? A2B+a2AB+a1B
Notethat the similarity
matrix
C(A,B)
matrix
(7.43)
The definition and physical
will be introduced
later
in
Chapter
meaning of controllabil-ity
10. A state-space
model can be
to a companion form if and only if the system is controllable.
The companion
form
similarity
models, and by induction
formula
B?
matrix T is nonsingular if and only if the controllability
= ? B AB A2B? is nonsingular.
and controllability
transformed
transformation
AB+a2B
is obtained
transformation
procedure
can be repeated
for
higher-order
a more general n-th order companion form similarity
and summarized
in the following
theorem.
state-space
transformation
198
7
Theorem
Fundamentals
of Feedback
7.22 (Companion
Form
For the n-th order state-space
x?(t)
y(t)
Control
Similarity
Transformation)
model given by
= Ax(t)+Bu(t)
where
det(?I-A)
=?n+an-1?n-1
+ +a1?+a0
=Cx(t)
The similarity transformation x(t)
companion
Systems
= T-1x(t) that transforms the state-space model to the following
form,
?
0
x?(t)
(7.44)
=Ax(t)+
where
A =
y(t) =Cx(t) =CT
x(t)
...
0
?
Bu(t)
? ??
? ??
1
0
0
?
?
?
?
?
...
?
?
?
1
, B =
?
?
?
?
...
?
(7.45)
?
?tn
???????
=An-1
+an-1An-2
+++a2A+a1I
BB
tn-1=
An-2
+an-1An-3
+a3A+a2I
?
?
?
0
?
-a0 -a1
?
1
?
?
0
?
1
-an-2-an-1
can be obtained using the following formula:
?
?
?
?
T =[tntn-1 t2t1], where...
t2=(A+an-1I)B
(7.46)
???????
t1 = B
Example
7.23 (Companion
Form
Transformation
of a State-Space
Model)
Considerthe system withthe following state-space model:
x?(t)
y(t)
The characteristic
polynomial
= Ax(t)+Bu(t)
=Cx(t)
=
= [12
? ? ??
-2 -2
2
3
-1
x(t)+
1
u(t)
(7.47)
16] x(t)
of the system is
?
? +2
2
?
det(?I-A)
=det -2 ? -3 =?2-? -2=?2+a1?
+a0
Thesimilarity transformationx(t)
=T-1x(t) that transformsthe state-spacemodelto the companion
form:
x?(t) =Ax(t)+ Bu(t)
y(t) =Cx(t)
can be obtained
= T-1ATx(t)+T-1Bu(t)
=CT
x(t)
using Equation
7.46 from
T =[t2 t1]
where
A =
? ? ??
0
1
, B =
-a0-a1
0
1
Theorem 7.22 as follows:
=[(A+a1I)B
B] =
?1 -1?
0
1
(7.48
7.4
Then the companion
form state-space
x?(t)
Example
Equation
=CTx(t)
7.21, a state-space
7.40 using a diagonal
state-space
=[12
model of Equation
form
form,
similarity
respectively,
these state-space representations
x?d =
in
State Space
199
? ? ??
0 1
2 1
0
x(t)+
u(t)
1
(7.49
14]x(t)
7.39
was transformed
transformation.
was transformed
companion form similarity transformation.
and the companion
=
model of Equation
7.39
Transformation
as follows:
= T-1ATx(t)+T-1Bu(t)
y(t)
In
model is found
Similarity
Similarly,
to the companion
to the
in
diagonal
Example
form in
form
in
7.23, the same
Equation
7.49 using a
Thetwo state diagrams associated with the diagonal form
are shown in Figure
7.9 and Figure 7.10. For ease of reference,
are also shown in the following:
?? ??
-1 0
0
2
1/3
xd(t)+
-1/3
u(t)
= Adxd(t)+Bdu(t)
(7.50)
y(t)=[-8 -20]
xd(t)=Cdxd(t)
and
x?c =
y(t)
These two state-space
There
must be some
? ? ??
0 1
xc(t)+
2 1
=[12
0
1
14] xc(t)
u(t)
relationship
(7.51)
=Ccxc(t)
models appear quite different,
mathematical
= Acxc(t)+Bcu(t)
and yet they actually represent the same system.
between them. Indeed, the following
theorem
provides
a wayto find the similarity transformation connecting them. In the following, a minimal state-space
model meansthat the number of state variables or the dimension of the A matrix is minimal. A
minimal state-space modelis also controllable and observable. The definition and physical mean-ing
of controllability,
observability, and minimal realization will be discussed in Chapters 10 and
11.
Theorem 7.24 (Minimal
Assume (A1,B1,C1)
State-Space
Models Related by a Unique Similarity
and (A2,B2,C2)
are two
minimal state-space
Transformation)
models of a system. Then there
exists a uniquesimilarity transformation T sothat A2 = T-1A1T, B2 = T-1B1, andC2 =C1T, andthis
similarity
transformation
matrix is
T =C(A1,B1)
C-1(A2,B2)
where
C(A,B)
=? B AB
where C(A,B) is the controllability
In the following
employed
transfer
function
Example
are connected
and characteristic
7.25 (Similarity
(7.52)
matrix ofthe state-space model(A,B,C).
example, the diagonal-form and the companion-form
to verify that they
?
An-1B
by a similarity
transformation
state-space models will be
and they share the same
equation.
Transform
Between the
Diagonal
and the
Companion
Forms)
The controllability
matrices C(Ad,Bd) and C(Ac,Bc) of the diagonal andthe companion state-space
models,respectively, are computed in the following:
C(Ad,Bd)
=[Bd
AdBd] =
?1/3 -1/3?
-1/3 -2/3
,
C(Ac,Bc)
=[Bc
AcBc] =
??
0 1
1 1
(7.53)
200
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.10: Similarity transformation to companion form state diagram.
Hence, the similarity
transformation
matrix is
?
???-1
?
1/3 -1/3
T =C(Ad,Bd)
C-1(Ac,Bc)
= -1/3
-2/3
0
1
1
1
=
?
-2/3 1/3
-1/3 -1/3
which transforms the diagonal state-space model (Ad,Bd,Cd) to the companion
(7.54)
model (Ac,Bc,Cc)
as
verified in the following:
? ?-1
? ?? ?? ?
-2/3 1/3?-1
?-1/3
? ???
-1/3 -1/3
?
CdT=[-8 -20]?
T-1AdT =
-2/3 1/3
-1/3 -1/3
-1 0
0
=
-2/3 1/3
-1/3 -1/3
The transfer
functions
of both state-space
-1/3 -1/3
0
1/3
T-1Bd =
-2/3 1/3
2
=[12
1
14]
=
0
2
1
1
= Ac
= Bc
=Cc
models can be obtained
using
matrix computation,
4s+12
Cd(sI-Ad)-1Bd
=Cc(sI-Ac)-1Bc
=s2-s-2
or by applying
Masons gain formula to the state diagrams shown in Figure 7.9 and Figure 7.10,
respectively. The characteristic equations of both models are alsoinvariant under similarity trans-formation.
They are
det(?I-Ad)
=det(?I-Ac)
=?2-? -2=0
7.5 Pole Placement Control in State Space
The basic principle of pole placement control in state space is rather simple. Suppose we are given an
n-th order system with a state-space model
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)
(7.55
7.5
and
we have learned that the
behavior
of the system is
Pole Placement
Control in
mainly determined
by the
State Space
201
poles of the system,
which arethe roots of the characteristic equation,
det(sI-A)
=sn+an-1sn-1
+ +a1s+a0
(7.56)
=(s-s1)(s-s2)(s-sn)=0
If all the state variables in the state vector x(t) are available for feedback, then the control input can
be set as
u(t) = Fx(t)
(7.57)
so that the state equation
of the closed-loop
system
will be
det
[sI-(A+BF)]=sn+an-1sn-1
+ +a1s+a0
(7.58)
=(s-?1)(s-?2)(s-?n)=0
Theorem 7.26 (Necessary and Sufficient
Condition for State-Feedback
Pole Placement)
For the system described by Equation 7.55, there exists a state-feedback control u(t)
=Fx(t) so that
theroots
oftheclosed-loop
system
characteristic
equation
det
[sI-(A+BF)]=0canbearbitrarily
placed in the complex plane ifand
only ifthe
controllability
? B AB
rank
7.5.1
State-Feedback
There are two
straightforward
the order of the
transformation,
Pole Placement:
Direct
matrix ofthe system is offull rank, or
? =n
An-1B
(7.59
Approach
ways to apply the pole placement theorem.
Oneis a direct approach, which is
and conceptually simple, but computationally
can become very complicated when
system is high. The other is a transform approach, which requires a similarity
but the computation
will not become much more complicated, even when the or-der
of the system
becomes
very
high.
Example 7.27 (State-Feedback
Pole Placement Control Using Direct Approach)
Considerthe system
x?(t) = Ax(t)+Bu(t)
=
? ? ??
-2 -2 x(t)+
2
-1
3
1
u(t)
Findastatefeedbacku(t) =Fx(t) sothatthe closed-loopsystemhasdampingratio ? = 0.8andnatural
frequency ?n = 5rad/s. Thatis, the desired closed-loopsystem characteristic equation should be
s2 +2??ns+?2n =s2 +8s+25
=0
(7.60)
or,equivalently,
thedesired
closed-loop
poles
are-4 j3.The
characteristic
equation
oftheuncom-pensate
system is
?
s+2
2
|sI-A| = -2 s-3
?=s2
-s-2
=(s+1)(s-2)
=0
?
?
?
?
?
?
which shows that the system is unstable. It also can be seen that the controllability
matrix
202
7
Fundamentals
of Feedback
Control
Systems
[B
AB] =
??
-1 0
1
1
is nonsingular so there exists a state-feedback controller u(t) = Fx(t) so that the closed-loop system
poles can be placed at any desired location of the complex plane. Let F =[ f1 f2], then
A+BF
=
? ?? ? ?
-2 -2
2
and the closed-loop characteristic
?
s+2+
-1 [ f1
+
3
1
equation
f1
?
-2- f1 -2- f2
f2] =
2+ f1
3+ f2
will be
2+ f2
?=s2
+(
f1-f2-1)s+(-f1
-2)
=0
?
(7.61)
|sI-(A+BF)| = -2- f1 s-3- f2
?
?
?
?
?
By comparing the coefficients of the two characteristic
f1-f2-1 =8
equations, Equations 7.60 and 7.61, we hav
? f1 =-27,f2 =-36
-f1-2 =25
The
state-feedback
control
u(t)=Fx(t),
where
F= -27-36?, has
successfully
placed
theclosed-?
loop
system
poles
atthedesired
location,
s =-4 j3,inthecomplex
plane.
The direct approach works well for low-order systems, but the complexity of the computation
involved will grow exponentially as the order of the system or the dimension of the A matrix in-creases.
Whenthe matrix dimension is morethan 4, the symbolic computation of the determinant of
thesI-(A+BF)matrix
willbecome
very
complicated,
even
with
thehelp
ofasymbolic
computing
software.
Fortunately, the high-order computation issue can beresolved using similarity transformations. Since
the characteristic equation and the system poles are invariant under similarity transformations, a given
state-space model can betransformed to a special form like a diagonal form or a companion form, where
each pole or each coefficient of the characteristic equation can be easily altered independently.
7.5.2 State-Feedback Pole Placement: Transform
Approach
The state-feedback pole placement problem is briefly recited as follows:
the following
Given a system represented by
state equation,
x?(t)
(7.62)
= Ax(t)+Bu(t)
with the characteristic equation
|sI-A|=sn+an-1sn-1
+ +a1s+a0
=0
the objective is to find a state-feedback control u(t)
F =?
so that the closed-loop
system
has the following
= Fx(t),
where
f1f2 fn?
desired characteristic
(7.64)
equation
|sI-(A+BF)|=sn+an-1sn-1
+ +a1s+a0=0
Withthe similaritytransformation,
x(t)
be transformed
to the following.
(7.63)
(7.65)
= T-1x(t), the state-feedback
poleplacementproblemcan
Consider the same system represented
by a transformed
state equation
7.5
x?(t) =Ax(t)+ Bu(t)
Pole Placement
Control in
State Space
= T-1ATx(t)+T-1Bu(t)
203
(7.66)
with the same characteristic equation,
?sIA?=|sI-A|
=sn
+an-1sn-1
++a1s+a0
=0
?
(7.67)
?
Theobjective
istofind astate-feedback
controlu(t) =Fx(t), where
f1f2 fn?
F =?
so that the closed-loop system hasthe following
(7.68)
desired characteristic equation
?sI-(
A+
B
F)
?=sn
+an-1sn-1
++a1s+a0
=0
?
Transform
?
(7.69)
Approach Procedure
Step1: UseTheorem7.22to find asimilarity transformation x(t) = T-1x(t) sothat the newstateequa-tion
is in companion
form
as follows:
x?(t) =Ax(t)+ Bu(t)
?
where
0
1
0
?
A =
= T-1AT
x(t)+T-1Bu(t)
? ??
? ??
0
...
0
?
?
?
?
...
?
?
?
1
0
?
?
?
-a0 -a1
and the similarity transformation
(7.70)
?
?
?
?
?
?
?
1
?
, B = ...
?
?
?
-an-2-an-1
0
(7.71)
?
1
matrix T is
?tn
???????
=An-1
+an-1An-2
+++a2A+a1I
BB
tn-1=
An-2
+an-1An-3
+a3A+a2I
?
?
?
?
T =[tntn-1 t2t1], where...
t2 =(A+an-1I)B
(7.72)
???????
t1
Step 2:
Let the state-feedback
control
= B
be
f1f2 fn?x(t)
u(t) =Fx(t)
=?
(7.73)
Then the closed-loop system state equation becomes
x?(t)
?
where
?
A +BF
=
0
1
...
0
= ?A
+BF ?x(t)
0
...
(7.74
0
0
...
...
?
...
...
?-a0+f1 -a1+f2
...
?
1
0
0
0
?
?
?
?
?
?
?
?
?
?
0
0
1
-an-2+fn-1-an-1+fn
?
(7.75)
204
7
Fundamentals
of Feedback
Note that this companion-form
Control
Systems
state equation reveals that the closed-loop
system characteristic
equa-tion
is
sn+?an-1
-fn? sn-1
+ + ?a1-f2? s+?a0-f1? =0
(7.76)
Step 3: Forthe closed-loop system to have a desired characteristic equation as Equation 7.65, the state-feedback
controller parameters fi, i = 1,2,...,n needto be chosen sothat the two characteristic equa-tions,
Equations 7.76 and 7.65, are equivalent. Hence, we have
f1=a0-a0
a0-f1 =a0
a1-f2 =a1
f2=a1-a1
?
...
(7.77
...
fn=an-1
-an-1
an-1-fn =an-1
? f2 fn? is
Step
4:Note
that
the
state-feedback
gain
matrix
F =f1
designed based on the trans-formed
state equation. Tofind the state-feedback gain matrix F for u(t) = Fx(t), recall that
u(t) =F x(t)
=FT-1x(t)
= Fx(t)
Hence,
? T-1
F=FT-1=? a0-a0a1-a1 an-1
-an-1
wherethe similarity transformation
In the following
example,
Example 7.28 (State-Feedback
matrix Tis given by Equation 7.72.
we will employ the same system considered
how to utilize the transformed
(7.78)
approach to implement
by Example
the state-feedback
Pole Placement Control Using Transform
7.27 to demon-strate
pole placement.
Approach)
Considerthe system
x?(t)
= Ax(t)+Bu(t)
=
? ? ??
-2 -2
2
3
x(t)+
-1
1
u(t)
(7.79)
Use a transform approach to design a state feedback u(t) = Fx(t) so that the closed-loop system has
damping ratio ? = 0.8 and natural frequency ?n = 5rad/s. That is, the desired closed-loop system
characteristic equation should be
s2 +2??ns+?2n =s2 +8s+25
=s2 +a1s+a0
=0
(7.80)
orequivalently,
thedesired
closed-loop
system
poles
are-4 j3.
The characteristic
equation
of the uncompensated
?
s+2
2
system is
?=s2
-s-2
=s2
+a1s+a0
=0
?
|sI-A| = -2 s-3
?
?
?
?
?
(7.81)
which shows that the system is unstable. It also has been shown that the controllability
is nonsingular, so there exists a state-feedback controller u(t)
poles can be placed at any desired location
of the complex
plane.
matrix ? B AB ?
= Fx(t) so that the closed-loop system
7.6
Recallthat in Example 7.23 the state-space
model in companion
x?(t)/dt
Revisit
Cart-Inverted
Pendulum
System
model of Equation 7.47 wastransformed
205
to the state-space
form,
= T-1ATx(t)+T-1Bu(t)
via the similarity transformation x(t)
=Ax(t)+ Bu(t)
=
? ? ??
01
x(t)+
21
0
1
u(t)
= T-1x(t), wherethe similarity transformation matrix wasob-tained
from Equation 7.48 as
T =[t2 t1]
=[(A+a1I)B
B] =
?1 -1?
0
1
LetF = ?f1 f2 ?, then the closed-loop system state equation becomes
x?(t)
where
A +BF =
+BF ?x(t)
? ??? ?
0 1
0
+
2 1
Note that this companion-form
= ?A
1
0
? f1 f2 ? =
1
2+f1 1+f2
state equation reveals that the closed-loop
?
system characteristic
equation
is
s2+? -1-f2? s+? -2-f1? =0
By comparing
the coefficients
of the two
characteristic
-1-f2 =8
-2-f1 =25
equations
(7.82)
Equations
7.80 and 7.82,
we have
? f1 =-27,f2 =-9
Hence,the state-feedback gain matrixis
F =FT-1
=?
? ?-1
= -27-9 ? ?
= -27-36
-1
-27-9? 1
0 1
?
?
1 1
?
01
?, has
The
state-feedback
control
u(t)=Fx(t),
where
F= -27-36
successfully
placed
theclosed-?
loop
system
poles
atthedesired
location,
s =-4 j3,onthecomplex
plane.
As
expected,
theresult
obtained
here using the transform
approach is the same asthat
obtained in Example
7.27 using the direct
approach.
7.6 Revisit Cart-Inverted Pendulum System
In the following, the cart-inverted pendulum system westudied in Section 4.5 will be briefly reviewed
before it is employed for the design and analysis of a stabilizing control system that would convert the
originally unstable equilibrium to a stable one.
Figure 7.11 shows the schematic
and the nonlinear
coupled
equations
of
motion of the cart-inverted
pendulum system. The physical data of the system are also shown in the figure. These data do not show
the units, but they are all in MKS units, like the massin kilograms, length in meters,and so on.
206
7
Fundamentals
of Feedback
Fig. 7.11: Schematic
7.6.1
State-Space
For ease of
Control
Systems
and equations
Model of the
of
Cart-Inverted
managing the nonlinearities
motion of a cart-inverted
Pendulum
pendulum
system.
System
of the system, the nonlinear
coupled
differential
equations
of the
system are converted into a state-space model. Let the state variables be
x1 = ?,
Then the equations
x2 =??,
x3 =s,
shown in Figure 7.11 can be rewritten
?
1.288
x4 =?s
in the following
(7.83)
matrix form:
-2x2?? ?
??31.07sinx1
??
2 -25x4
?T
x?2
x?4
3.17cosx1
3.17cosx1
189.4
Solving Equation 7.84 for ?x?2x?4
=
+
3.17sinx1x2
0
100
u
(7.84)
yields
-378.8x2
-10sinx1
cosx1x2
2 +79.25x4
cosx1
-317cosx1u
?? ?5885sinx1
?
x?2
x?4
1
=
?
(7.85)
4.083sinx1x2
2 -32.2x4
-98.5sinx1
cosx1+6.34x2
cosx1
+128.8u
where
?=244-10cos2x1
(7.86)
Then we have a state-space representation for the nonlinear cart-inverted system:
x?(t) :=
? ?? ?
?x
?3(t)
?=? ?:=f(x,u)
x?1(t)
x?2(t)
?
?
?
?
?
?
x?4(t)
f1(x,u)
f2(x,u)
f3(x,u)
f4(x,u)
(7.87)
?
?
where
f1(x,u)
= x2
2 +79.25x4
cosx1
-317cosx1u)
f2(x,u)
=(1/?)(5885sinx1
-378.8x2
-10sinx1
cosx1x2
(7.88)
f3(x,u)
= x4
f4(x,u)
= (1/?)(4.083sinx1x2
2 -32.2x4
-98.5sinx1
cosx1
+6.34x2
cosx1
+128.8u)
?=244-10cos2x1
This state-space
dynamics
First, we will utilize this
system
model can be employed
to conduct
analysis,
design,
and simulations.
modelto find the feasible operation points or the equilibrium
points of the
7.6
7.6.2
Equilibriums
of the
Cart-Inverted
Pendulum
Revisit
Cart-Inverted
Pendulum
System
207
System
An equilibrium of the systemis a point or a vector (x*,u*) that satisfiesthe equation f(x*,u*)
Note that
= 0.
? ???
=0or
p,x=
20,xis3
arbitrary,
x4
=0 (7.89)
? ?=???x1
?
?
f1(x,0)
f2(x,0)
f3(x,0)
f4(x,0)
?
?
?
?
0
0
0
0
?
?
At equilibriums the angular velocity of the stick, x2, and the velocity of the cart, x4, should be both
zero and the angular displacement of the stick should be either 2kp or (2k+1)p,
where k is any integer.
Hence, physically there are two groups of equilibriums: Oneis the upright equilibriums,
x*up =[0
at which the stick is at the upright
3
0 x*
0]T
(7.90)
position, ? = 0, and the other is the downward
x*down=[p
0 x*
3
equilibriums,
0]T
(7.91)
at whichthe stick is at the downward position, ? = p. Forthe stick, there are only two equilibriums:
Oneis the up position, which is unstable, and the other is the down position,
cart, it can be at anywhere on the rail.
which is stable. For the
A system can only be stabilized at the equilibriums. For the upright stick position equilibrium, al-though
it is unstable the stick can still rest on the upright position if ? can be kept at zero by feed-back
control or other means. Onthe other hand, it is impossible to stabilize the pendulum at any non-equilibrium
position.Forexample,to keepthestickat ? =10?,
the cart wouldhaveto movecontinuously
with a constant acceleration, which is impossible to sustain.
A common practice in control system design is to first identify an operating equilibrium of interest,
and find a linearized dynamics model at the equilibrium. The linearized model can then be employed
in the design of control system to achieve stabilization, regulation, tracking, and so on. For the cart-inverted
pendulum system, the objective is to stabilize the pendulum at the originally unstable
upright equilibrium.
The two equilibriums of interest are
x*U =[0
representing the upright equilibrium,
0 0 0]T
(7.92)
and
x*D =[p 0 0 0]T
representing
the
downward
equilibrium.
At these two
equilibriums,
(7.93)
the cart
position
is assumed
x3 = 0, which usually is considered the center of the rail.
Example 7.29 (Phase Plane Trajectory of the Uncompensated Pendulum
Subsystem)
Considerthe pendulum subsystem of the cart-inverted pendulum system described bythe state-space
model equations in Equations 7.87 and 7.88. Withthe assumption that the cart is at rest (x4 = 0), the
dynamics equation for the pendulum subsystem will be
???(5885sinx1
?
-378.8x2
-10sinx1
cosx1x2
2)/(244-10cos2x1)
x?1
x?2
x2
=
(7.94
208
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.12: Simulation diagram for the uncompensated nonlinear pendulum subsystem.
Based on Equation 7.94, a Simulink model program, CinvPx1x2.mdl,
shown in Figure 7.12, is
constructed to conduct simulations for the uncompensated nonlinear pendulum subsystem. The time
responsesof the two statevariables,x1(t) = ?(t) and x2(t) =??(t), dueto the initial condition:
x1(0)
= 0.001 rad and x2(0)
= 0 rad/s are shown in the left-hand side of Figure 7.13. Theinitial
is very close to the equilibrium
point [0 0]T with the upward pendulum,
state
but since this equilibrium
is unstablethe pendulum beforelong starts to moveaway toward the stable equilibrium [p 0]T.
The simulation
results are obtained by running the
will automatically
The
MATLAB
%
Filename:
%
Run
%%
d2r
call the
Simulink
code is listed
model program
May
CSD_fig7p13.m
program
6,
2020
to
automatically
to conduct
which
the simulation.
response
call
CinvPx1x2.mdl
Simulation
=
pi/180,
x10=0.06*d2r,
x20=0,
sim_options=simset('SrcWorkspace',
open('CinvPx1x2'),
%%
CinvPx1x2.mdl,
CSD fig7p13.m,
asfollows:
CSD_fig7p13.m
this
MATLAB program:
Plot
x1(t),
sim_time=10,
'current',
sim('CinvPx1x2',
x2(t),
x2
vs
subplot(1,2,1),
plot(t,x1,'b-',t,x2,'r-'),
subplot(1,2,2),
plot(x1,x2,'b-'),
[0,
'DstWorkspace',
sim_time],
'current');
sim_options);
x1
grid
on,
grid
on,
grid
minor
grid
minor,
During the transition, both x1(t) and x2(t) exhibit decaying oscillations before they reach a steady
state at the new equilibrium point. This decaying oscillation phenomenon is relevant to the under-damped
case study of the second-order systems in Chapter 3. Thesimulation results can also be seen
from the phase plane trajectory shown on the right-hand side of Figure 7.13. The phase plane trajectory
exhibits the relationship of the angular velocity ?? andthe angulardisplacement?, andclearlyshows
how the trajectory spirals from the initial state into the final state. Theinitial state (0.001,0) is very close
to the origin (0,0), the unstableequilibrium, andthe final state(p,0) is a stableequilibrium.
Example 7.30 (Phase Portrait
The simulation
of the Uncompensated Pendulum Subsystem)
process described can be employed
the system due to a variety
of initial
to repeatedly
states on the phase plane.
generate phase plane trajectories
A collection
of
of the phase plane trajectories
like those shown on Figure 7.14is called a phase portrait. This phase portrait reveals that the origin (0,0)
is a saddle
point,
without any tiny
any infinitesimal
which is an unstable
error, ideally
the perfect
deviation in reality
equilibrium
point. If the initial
upward pendulum
would stay
state is situated exactly
without falling.
at (0,0)
But, of course,
will cause the pendulum to drift away from the unstable equilibrium.
The phase portrait shows two stable equilibrium
points: one on the right at (p,0) and another onthe lef
7.6
Revisit
Cart-Inverted
Pendulum
System
209
Fig. 7.13: Time responses and phase planetrajectory of the uncompensated pendulum subsystem.
at(-p,0).The
pendulum
at(0,0)
can
either
falltotheright,
spiraling
intothestable
equilibrium
(p,0),
ortotheleftinto(-p,0).These
twoequilibrium
points
seem
different
onthegraph,
butactually
they represent the same physical angular position of the pendulum.
Fig. 7.14: Phase portrait of the uncompensated nonlinear pendulum system.
7.6.3
Linearized
State-Space
The Linearized State-Space
Models at the
Equilibriums
Model at the Upright
Equilibrium
Forthe upright equilibrium at (x*U,0),the linearized state-space modelis
x?(t)
where AU and BU are obtained using the following
?
?
?
AU =
?
df1
df1
df1
dx1
dx2
dx3
df1
dx4
d f2
dx1
d f2
dx2
df2
dx3
d f2
dx4
d f3
dx1
d f3
dx2
df3
dx3
d f3
dx4
d f4
dx1
d f4
dx2
df4
dx3
d f4
dx4
?
?
?
(7.95)
= AUx(t)+BUu(t)
Jacobian matrices,(see Appendix C)
? ?
?
25.15-1.6188
00.33868
?
? ?
0
1
0
0
0
0
0
1
?
?
?
=
?
?
?
?
?
?
?
?
-0.42094
0.027094
0 -0.1376
(x*U,0)
(7.96
210
7
Fundamentals
of Feedback
Control
Systems
?? ? ?
?? ? ?
and
df1
0
du
d f2
-1.3547
?
?
BU =
du
?
d f3
du
?
?
?
=
?
?
?
?
?
?
0.5504
d f4
du
(7.97)
?
0
?
(x*U,0)
The poles ofthe linearized state-space modelatthe equilibrium x*Uare the eigenvalues of AU:
?[AU]={0,4.2682,
-5.8926,
-0.13193}
(7.98)
which implies that the system is unstable at the upright equilibrium since there is a pole in the right half
of the complex plane.
The Linearized
State-Space
Model at the
Downward
Equilibrium
Forthe other equilibrium (x* D,0), the linearized state-space modelis
x?(t) = ADx(t)+BDu(t)
where AD and BD are obtained using the following
?
?
?
AD =
?
?
?
?
df1
d f1
d f1
d f1
dx1
dx2
dx3
dx4
df2
df2
df2
df2
Jacobian matrices,
? ?
?
-25.15-1.6188
0 -0.33868
?
? ?
?? ? ?
?? ? ?
0
dx1
dx2
dx3
dx4
df3
df3
df3
df3
dx1
dx2
dx3
dx4
df4
dx1
d f4
dx2
d f4
dx3
d f4
dx4
and
(7.99)
1
0
0
?
?
?
=
?
?
?
?
?
0
?
0
0
1
(7.100
?
?
-0.42094
-0.027094
0 -0.1376
(x*D,0)
df1
?
BD =
?
?
?
du
d f2
du
d f3
du
0
?
?
=
?
?
?
?
0
?
?
(7.101)
?
0.5504
df4
du
1.3547
?
(x*D,0)
The poles of the linearized state-space model at the equilibrium x*Dare the eigenvalues of AD:
?[AD]={0,-0.13194,
-0.81223
j4.9487}
Initial
State
Response
It is clear that the upright
Analysis
(7.102)
and Simulation
equilibrium
is unstable since the linearized
state-space
model at this equilib-rium
has a pole s = 4.2682 in the right half of the complex plane. The pendulum system can stay at this
equilibrium
only if the pendulum
are all perfectly
kept at zero,
angular displacement,
which is practically
pendulum
impossible
to
angular
velocity,
and the cart velocity
maintain.
Onthe other hand,thelinearized state-space modelatthe downward equilibrium x*D =[p
0 0 0]T
has
three
poles:
-0.13194,
-0.81223
j4.9487
inthestrictly
lefthalfofthecomplex
plane,
but
there
is one at the origin s = 0. Strictly speaking, this equilibrium is not stable since there is a pole on
the imaginary axis. This s = 0 pole is caused by the lack of a natural mechanism to bring the cart
7.6
to the x3 = 0 position.
However, if x3 is excluded
in the stability
can be regarded as a stable equilibrium
converge
to the equilibrium
at steady
Revisit
Cart-Inverted
Pendulum
consideration,
the
System
downward
since the other three state variables
211
equi-librium
will naturally
state.
A Simulink computer simulation program built based on the nonlinear state-space model shown in
Equation
7.87 is employed to conduct simulations
simulation
results
of the pendulum
with initial
for the cart-inverted
pendulum.
Figure 7.15 shows the
condition
x(0)=[-0.5240 0.50]T
The
initial
pendulum
displacement
x(0)=-0.524
rad=-30?,
which
isnotanequilibrium,
has
to move
toward a stable equilibrium. If there is no external intervention
a shortest
path to reach a nearby stable
equilibrium.
other than the gravity, it
The pendulum
would choose
moves counterclockwise
towards
1 =-p,overshoots
it,turns
around,
and
thenoscillates
several
times
for
=-p.
the nearbystable equilibrium x*
about6seconds,it settlesatthestableequilibriumx*
1
Fig.7.15:
Response
due
totheinitial
state
x0=[-0.5240 0.50]Tofthecart-inverted
pendulum
system.
Theinitial
the cart
cart position is at x3(0)
only fluctuates
a tiny
= 0.5m. Since the gravity
bit due to a very small
reaction
has no direct influence
force from
the swinging
on the cart,
pendulum.
Therefore,
atsteady
state,
x2=0,
x4=0,and
x1=-p,but
thecartposition
isstillatx3=0.50019m,
whichis very closeto its initial position. The overshoot and oscillation of x1(t) and x2(t) are mainly
caused
bytare
hecomplex
poles
-0.81223
j4.9487
whose
corresponding
damping
ratio
and
natural
? = 0.162 and ?n = 5.015rad/s, respectively.
frequency
7.6.4
Design of a Stabilizing
Controller
Objective of the Feedback Control
The objective
of the cart-inverted
at the upright
position,
Upright
Equilibrium
Design
pendulum
x1 = 0, x2
for the
control
system
= 0, and x4 = 0. It
position, x3 = 0. Therefore, the operating
design is not only to stabilize
will also stabilize
equilibrium
x = x*U = [0 0 0 0]T
the pendulum
the cart at the center of the rail
is
when
u =0
(7.103)
Analysis of the Uncompensated System
Recallthat the linearizedstate-spacemodelof the cart-invertedpendulumsystematthe upright x*
equilibrium
is
212
7
x?(t)
Fundamentals
of Feedback
= Ax(t)+Bu(t)
=
Control
Systems
? ? ?
?25.15-1.6188
00.33868
?x(t)+?
?u(t)
?
0
1
0
0
0
0
0
1
0
?
?
?
?
?
?
-0.42094
0.027094
0 -0.1376
-1.3547
(7.104)
?
0
0.5504
?
The poles ofthe linearized state-space modelatthe equilibrium x*Uare the eigenvalues of A,
?[A]={0,4.2682,
-5.8926,
-0.13193}
(7.105)
The corresponding open-loop characteristic equation is
|sI-A| =(s-0)(s-4.2682)(s+5.8926)(s+0.13193)
=s4+1.7564s3
-24.936s2
-3.3181s
=s4+a3s3
+a2s2
+a1s+a0
=0
Obviously, the uncompensated system is unstable. However,the controllability
(7.106)
matrix
? B AB A2B A3B?
(7.107)
is nonsingular; hence,there exists a state-feedback controller
u(t)
= Fx(t)
(7.108
so that the closed-loop system poles can be placed at any desired location of the complex plane.
Selection of the Desired Closed-Loop System Poles
The desired closed-loop system poles are tentatively
complex conjugate poles as follows:
chosen to include two real poles and a pair of
s1=-2,s2=-8,s3=-4+j3,s4=-4-j3
That is, the desired closed-loop system characteristic
equation is given as:
|sI-(A+BF)| =(s+2)(s+8)(s+4-j3)(s+4+j3)
= s4 +18s3 +121s2 +378s+400
Design of a Pole-Placement State-Feedback
Although the direct approach, introduced
is conceptually
simple, the complexities
the system increases.
For this fourth-order
= s4 +a3s3 +a2s2 +a1s+a0
(7.109)
=0
Controller
in Section 7.5.1, for the state-feedback
involved
in the computation
system, it is still
pole placement
grow exponentially
design
as the order of
manageable using the direct approach,
but
the transform approach, discussed in Section 7.5.2, is much more efficient in computation, espe-cially
if we need to repeat the design process for different sets of desired closed-loop system poles.
In order to take advantage of the transform approach, wefirst needto find a similarity
x(t)
following
= T-1x(t), to transform the state-spacemodelof the plantin Equation(7.104)to the
state-space
modelin companion form:
x?(t) =Ax(t)+ Bu(t)
where
transfor-mation,
= T-1AT
x(t)+T-1Bu(t)
(7.110)
7.6
?
0
0
0
A=?
?
?
1
0
0
0
1
0
Revisit
Cart-Inverted
Pendulum
System
? ??
0
0
1
0
0
0
1
?,
B=??
?
?
?
?
-a0-a1-a2-a3
213
(7.111)
?
?
Since the controllability condition, Equation 7.107 is satisfied, the similarity transformation
T can be constructed using Equation 7.46in Theorem 7.22 asfollows:
?
????t4
=A3
+a3A2
+a2A+a1I
B? T=t4t3t2t1
????
-1.3547
? 0 2.75210-62.75210-6
?
-1.3547
T=?-13.272
0.85428 ?
?
and
matrix
?
t3
= ? A2 +a3A+a2I ? B
t2
= (A+a3I)B
t1
=B
?
?
(7.112)
0
0
0
?
?
0.5504
?
0
0
(7.113)
?
-13.272 0.85428 0.5504
Let
u(t) =Fx(t) =?f1 f2 f3 f4 ?x(t)
(7.114)
Then the closed-loop system state equation becomes
x?(t)
?
0
1
0
0
0
0
1
0
0
0
0
1
A+
B
Fx(t)
=?
=?
?
?
?
?
(7.115)
?
-a0+f1 -a1+f2 -a2+f3 -a3+f4
Notethat this companion-form
is
?
x(t)
?
state equation reveals that the closed-loop system characteristic equation
??
?sIA+
B
F?
=s4
+(a3
-f4)s3
+(a2
-f3)s2
+(a1
-f2)s+(a0
-f1)
=0
1.7564f4 =18
1.7564-18
=f4
?a3a2-f4-f3 =a3
?
?
-24.936f3 =121 ????
-24.936-121
=f3
=a2 ????
????
?
?
a1-f2 =a1
-3.3181f2 =378
-3.3181-378
=f2
a0-f1 =a0 ????
0-f1 =400
0-400=f1
????
????
?
By comparing
(7.116)
?
the coefficients
of the two
characteristic
equations
Equations
7.109 and 7.116,
we have
(7.117
Hence,
F =?f1 f2 f3
f4? =[-400-381.32
-145.94
-16.244]
(7.118)
Since
u(t)=Fx(t)=FT-1x(t)
=Fx(t)? F=FT-1
(7.119)
we havethe state-feedback gain matrix
F =FT-1 =[139.31 24.45230.13830.67]
The state-feedback
control
u(t)
= Fx(t),
with F given, has successfully
(7.120)
placed the closed-loop
system
poles
atthedesired
location,
s =-2,-8,and-4
j3onthecomplex
plane.
214
7
Fundamentals
Nonlinear
of Feedback
Closed-Loop
The nonlinear
dynamics
of state-feedback
simulation
Systems
Simulations
model shown in Equations 7.87 and 7.88 will be employed in the simu-lation
stabilizing
control
of the cart-inverted
diagram shown in Figure 7.16, four integrators
are employed
to implement
Thefunction information
function
System
Control
the state-space
dynamics
model equations
system.
In the Simulink
blocks f2, f4, g2u, and g4u
given in Equations
7.87 and 7.88.
of f2(x), f4(x), g2(x)u, and g4(x)u in Equation 7.88 needto be typed into the
blocks f2, f4, g2u, and g4u in the Simulink
the function
pendulum
and four function
block f2 and f4 should
be shown
program, respectively.
The function
expressions for
as the following:
(5885*
sin(u[1])-10*
sin(u[1])
*cos(u[1])
*u[2]2-378.8*u[2]+79.25*u[3]
*cos(u[1]))
/(244-10*cos(u[1])2)
and
(4.083*
sin(u[1])
*u[2]2-32.2*u[3]-98.5*
sin(u[1])
*cos(u[1])+6.34*u[2]
*cos(u[1]))
/(244-10*cos(u[1])2)
respectively. Notethat u[1], u[2], and u[3] represent x1, x2, and x4respectively. Thereis no x3 or u[4] in
the expressions.
Fig. 7.16:
Closed-loop
simulation
diagram for the cart-inverted
Similarly, the function expressions for the function
lowing:
pendulum
control system.
block g2u and g4u should be shown asthe fol-(-317*cos(u[1])
*u[2])/(244-10*cos(u[1])2
and
7.6
Revisit
Cart-Inverted
Pendulum
System
215
(128.8*u[2])/(244-10*cos(u[1])2)
respectively. Notethat u[1] still represents x1, but u[2] here represents the control input uinstead of x2.
The state-feedback
you can either
controller
is implemented
type the following
by the sf gain
block
and the
Saturation
block;
F matrix
F = ? 139.31 24.452 30.138 30.67 ?
into
the sf gain block
or just type
Fin the block and define the
m-file or in the CommandWindow
before running
matrix values elsewhere like in a MAT-LAB
the Simulink
simulation.
The Saturation
block is
to mimic
thereal
control-input
constraint
assuming
the
input
applied
force
isrestricted
to 80Newtons.
In the simulation
simulation
program
trajectory
diagram, five
ToWorkspace
data in arrays of compatible
can be written
blocks:
x1, x2, x3, x4, and t are there to record
dimension sothat after the simulation
to plot the state
variables
as function
of t (time)
a simple
or draw
the
MATLAB
a phase
plane
of x2 versus x1.
Figure
7.17
shows
that
thependulum
with
initial
state
x0=[-0.524
0 0.50]would
move
towards
?T
andsettle atthe upright equilibrium x*U = ? 0 0 0 0 , now is a stable equilibrium, shortly after one sec-ond
without much overshoot and oscillation.
This desired transient response is the direct result of the
choice
oftheclosed-loop
system
poles
atthedesired
location,
s =-2,-8,and-4
j3onthecomplex
plane. Notethat the cart position x3 is also brought to zero from its initial position x3(0)
Fig. 7.17:
A desired stabilizing
performance
of a cart-inverted
pendulum
system
= 0.5m.
achieved
by a state-feedback
control.
The graph of the control input u(t) is also shown in Figure 7.17. Initially the pendulum position is
atx1=?=-30?.
Apparently,
thependulum
would
falltothe
leftif nocontrol
action
istaken
intime.
The controller hasto apply a negative control-input (i.e., actuator) force to quickly movethe cart to the
left in order to reverse the motion of the pendulum. The pendulum did swing back moving towards th
216
7
Fundamentals
upright
equilibrium.
of Feedback
Control
The controller,
Systems
the actuator, the cart, and the pendulum
and the sensor have to
work
together continuously to readjust the control-input force according to the control law (i.e., the F matrix)
in order to bring the pendulum
system to the desired equilibrium.
The closed-loop system simulation
MATLAB
program:
results shown in Figure 7.17 are obtained by running the follow-ing
CSDfig7p17.m,
CInvP4sNL.slx,
to conduct
results
Part 2B block
%
BC
Program
%
1.
Chang,
Drexel
Part
1:
7.15 can also be obtained
by Part 2A,
=-30,
%%
2A:
Part
[0
%%
Part
r1
=
%%
Desired
0
r2
2.
For
0
initial
%
CSD PlotCInvP.m
The open-loop system
the same program
with
code F=[0 0 0 0].
two
3,
ze
simulation
real
=
wn
=
eq:
a2h
0;25.15
a0h];
-r1,
0
0]
and
one
pair
of
complex
poles
0;
=roots(ChEqh),
system
0
0.5504];
with
0
B
to
0
C=eye(4);
D=[0
1;-0.42094
0
0
0.027094
0]';
ChEq=[1
a3
a2
a1
a0];
Poles_OL=roots(ChEq),
form
t4=(A3+a3*A2+a2*A+a1*eye(4))*B;
T=[t4
feedback
-0.1376];
co=ctrb(A,B);Controllability_sv=svd(co),
companion
t2=(A+a3*eye(4))*B;
0
G=ss(A,B,C,D);
a2=Ev(2)*Ev(3)+Ev(2)*Ev(4)+Ev(3)*Ev(4);
a0=0;
A,
frictions
0.33868;0
controllability');
t3
t2
t1];
t1=B;
Ah=inv(T)*A*T,
Bh=inv(T)*B
Fh
f2h=a1-a1h,
Acl_h=Ah+Bh*Fh,
3
-r2,
Poles_desired
t3=(A2+a3*A+a2*eye(4))*B;
Part
0
s4+a3h*s3+a2h*s2+a1h*s*a0h=0
pendulum
-1.6188
disp('checking
% Using
F=[0
1
a3=-(Ev(2)+Ev(3)+Ev(4));
f1h=a0-a0h,
control
a0h=r1*r2*wn2;
a1h
a1=-Ev(2)*Ev(3)*Ev(4);
% State
no
a2h=wn2+(r1+r2)*2*ze*wn+r1*r2;
-1.3547;
% Convert
versions
u_sat=100
with
poles
0.95,
inverted
Ev=eig(A);
later
CSD_PlotCInvP.m
x30=0.5,
Open-loop
Select
a3h
0
or
3.
0]
=
% Linearized
%%
model pro-gram
conditions
a1h=(r1+r2)*wn2+r1*r2*2*ze*wn;
1
R2015a
CInvP4sNL.slx
characteristic
ChEqh=[1
B=[0;
Simulink
5/07/2020
MATLAB
x10=x10_deg*pi/180,
2B:
2,
%
a3h=2*ze*wn+r1+r2;
A=[0
program,
by running
which is the single line
on
sequence:
Choose
x10_deg
=
The plotting
The MATLAB code is listed asfollows.
University,
running
CSDfig7p17.m
F
Figure
call the
CSDfig7p17.m
%
%
in
codes replaced
% filename:
%%
shown
will automatically
the simulation.
will also be called to plot the figures.
simulation
which
f3h=a2-a2h,
eig(Acl_h),
f4h=a3-a3h,
Fh=[f1h
F=Fh*inv(T),
Acl=A+B*F,
the
and
f2h
f3h
f4h],
eig(Acl)
Simulation
the
following
to
run
simulink
plot
the
simulation
results
sim_time=10,
sim_options=simset('SrcWorkspace',
open('CInvP4sNL'),
'current',
sim('CInvP4sNL',
'DstWorkspace',
[0,sim_time],
'current');
sim_options);
run('CSD_PlotCInvP')
The stabilizing
controllers
for the cart-inverted
pendulum
system
are not unique.
As long
as a con-troller
is able to place all the closed-loop system poles strictly in the left half of the complex plane,
the closed-loop system will be stable. However, the performance may be different from controller to
controller. In common practice, a control system designer will usually consider the following fac-tors:
response time, overshoot and oscillation, steady-state response, robustness, and control-input
constraint. These factors mayconflict with each other; for example, a faster response mayinvite
larger overshoot and oscillations or require
more control-input
power. Hence,the designer may
need to consider
all factors
and find
a best trade-off
design
7.7
Routh-Hurwitz
Stability
Criterion
217
Fig. 7.18: Response of the compensated cart-inverted pendulum system with a moreconservative design
consuming less control resources.
Figure 7.18 shows the response graphs of the cart-inverted pendulum system with a more con-servative
controller design. It can be seen that the response time is slower. It takes morethan seven
seconds to reach the steady state, but it requires muchless control-input force to complete the task.
The conservative controller employed in the Figure 7.18 simulation was designed to have its closed-loop
system
poles
toinclude
tworealpoles
ats =-2and
s =-3,and
apairofcomplex
poles
that
associate with the damping ratio ? = 0.95 and natural frequency
gain matrix of this controller is
F =[31.726
4.4123 0.45207
?n = 1rad/s.
The state-feedback
1.5148]
The control-input constraint or the actuator limit can be a great concern if the control-input signal
issued by the controller is beyond the range of the actuator. Underthis situation, the actuator will saturate.
A saturation of the actuator maycause the control system to become unstable. Hence, a control
system should be designed so that its control inputs are always inside the working range of the
actuators.
7.7 Routh-Hurwitz Stability Criterion
The Routh-Hurwitz criterion provides an easy wayto determine the number of roots of a polynomial
equation in the right
half of the complex
of the linear time-invariant
root in the right
plane without the need of computing
system is determined
half of the complex
plane, the
a system is stable. Although the root-finding
polynomial
control
equations in a split of second, the
systems
by whether the systems
Routh-Hurwitz
the roots.
characteristic
criterion
Since the stability
equation
has any
is a perfect tool to check if
computer programs nowadays can solve for the roots of
Routh-Hurwitz
criterion
is still important
in the design of
218
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.19: Construction of the Routh array for a polynomial equation.
Theorem 7.31 (Necessary
Condition for
Consider a system with the following
Stability)
characteristic equation
ansn
+an-1sn-1
+an-2sn-2
+an-3sn-3
+ +a1s+a0=0
(7.121)
Withoutloss ofgenerality, the coefficient ofthe snterm, an,is assumedto be positive. Then for the system
to be stable, the characteristic equation must have no missingterms and all its coefficients are strictly
positive.
This theorem provides an easy-to-check necessary condition for stability. Just by inspection, wecan
tell if the necessary condition is satisfied. If there is a missingterm or some of the coefficient is negative,
then the system
is unstable.
On the other hand, even if the necessary
conditions
are all satisfied,
the
system can still be unstable becausethese conditions are not sufficient.
The Routh-Hurwitz criterion provides a necessary and sufficient condition for stability, but a Routh
array, as shown in Figure 7.19 needsto be constructed before the criterion can be employed to determine
the stability of the system.
To construct the Routh array, the first step is to build the label column on the left side of the array.
The
label
column
starts
from
snonthetop,and
then
isfollowed
bysn-1,
sn-2,
sn-3, , alltheway
to s0. Step 2 is to populate the first two rows directly from the coefficients of the polynomial equation in
Equation
7.121:
an,an-2,
an-4, , onthesnrow,
andan-1,
an-3,
an-5, , onthesn-1
row,as
shown in the figure.
The elements of the third row (the sn-2 row) are generatedby the elements of the first two rows as
follows.
an-1an-2
-anan-3
-anan-5
-anan-7
,
b2 = an-1an-4
,
b3 = an-1an-6
,
(7.122)
an-1
an-1
an-1
Note
thatthethird-row
elements
share
thesame
denominator,
an-1,
which
isthefirstelement
onthe
b1 =
left
of the second-row.
We circled
this
element,
as shown in the figure,
and
would like to call it as
7.7
cornerstone
element since it not only serves as the denominator
Routh-Hurwitz
Stability
Criterion
219
of the entire row it also plays an impor-tant
role in the numerator computation of the row. The numerator of b1is the cross-product difference
an-1an-2
-anan-3
ofthefourelements
connected
bythetwoblue
lines.
Similarly,
thenumerator
of
b2isthecross-product
difference
an-1an-4
-anan-5
ofthefourelements
connected
bythetwored
lines,
and
thenumerator
ofb3isthecross-product
difference
an-1an-6
-anan-7
ofthefourelements
connected by the two purple lines. It is noted that all the numerator cross product differences begin with
thecornerstone
element,
an-1.
Thefourth row (the sn-3 row) can be constructedin the same mannerusingthe data of the second
and the third
rows and regarding
row.
c1 =
the
b1 element
as the cornerstone
b1an-3
-an-1b2
,
c2
b1
=
element in the construction
b1an-5
-an-1b3
,
(7.123)
b1
The same process is repeated until the last row, which is the s0 row, is completed.
array constructed as above, the Routh-Hurwitz criterion is given in the following.
Theorem 7.32 (Routh-Hurwitz
Criterion:
Necessary and Sufficient
Atypical
feedback
control system
Withthe Routh
Condition for Stability)
For the system associated withthe Routh array shown in Figure 7.19,it is stable ifand
elements onthe first column ofRouth array are strictly positive.
Fig. 7.20:
of this
only ifall the
block diagram.
The block diagram in Figure 7.20 shows a typical feedback control system structure. G(s) repre-sents
the system to be controlled and K(s) is the controller to be designed to achieve some desired
performance.
One of the important
properties
of the closed-loop
system is stability;
since the closed-loop
system is not only useless,it can be harmful if it is unstable. In the following, the Routh-Hurwitz
criterion will be employed to assessthe stability of a control system or to determine the range of the
design parameters for the closed-loop
Example 7.33 (Determine
system to be stable.
the Stability
Range of a Proportional
Control System)
Consider the feedback control system shown in Figure 7.20. Assumethe plant G(s), the system to
be controlled,
is given as
1
G(s)
=
s(s+1)(s+2)
and the controller is a proportional controller,
K(s), where Kis a constant to be determined so that the
closed-loop
By the time the design is finalized,
system has a desired performance.
to know the range of Kin
The closed-loop
which the closed-loop
system characteristic
system is stable.
equation i
the designer
may want
220
7
Fundamentals
of Feedback
Control
Systems
K
1+G(s)K
=0 ? 1+s(s+1)(s+2)
=0 ? s3+3s2
+2s+K
=0
The Routh array for this characteristic equation is constructed as follows:
s3
1
s2
3
s1
b1
s0
K
2
3 2-1K
K
where
b1 =
3
=
6-K
3
According to the Routh-Hurwitz criterion, the system is stable if and only if the first-column
of the Routh array are all strictly positive. Hence,the system is stable if and only if
elements
K>0 and6-K>0
which is
0
The Routh
equation
line.
array
to compute
<K<
not only shows the stability
the pole locations
6
range,
on the imaginary
When K = 6, b1 will be zero and the
auxiliary
0 < K < 6, it
axis
equation
also provides
an auxiliary
when the value of Kis on the border
right
above this
zero element
will be
3s2+6=0,which
has
tworootsj1.414
ontheimaginary
axis.
Example 7.34 (Determine
a Two-Parameter
Stability
Region of a Control System)
Consider the feedback control system shown in Figure 7.20. Assumethe plant G(s), the system to
be controlled,
is given as
s+2
G(s)
and the controller
=
s(2s+1)
is
K(s) =
K1
K2s+1
The objective is to determine the stability region of the closed-loop system on the K2-K1 plane. The
closed-loop system characteristic equation is
1+
s+2
K1
s(2s+1)
K2s+1
=0 ? 2K2s3
+(K2
+2)s2
+(K1
+1)s+2K1
=0
The Routh array for this characteristic
According to the
the
s3
2K2
K1 +1
s2
K2 +2
2K1
s1
b1
s0
2K1
Routh-Hurwitz
Routh array are all strictly
criterion,
positive.
equation is constructed
where
b1 =
as follows:
(K2+2)(K1
+1)-4K1K2
K2 +
the system is stable if and only if the first-column
Hence, the system is stable if
elements of
and only if all the following
inequalities are satisfied.
K1>0, K2>0, and(K2+2)(K1
+1)-4K1K2
>0
three
7.8
Exercise
Problems
221
Fig. 7.21: Stability region on K1-K2 plane for Example 7.34.
As shown in Figure 7.21, the stability region is in the first quadrant (K1 > 0 and K2 > 0) under the red
curve
of(K2+2)(K1
+1)-4K1K2
=0.The
twoblue
dash
lines,
which
are
theasymptotes
ofthered
curve, are inside the stable region. Any point inside the region will guarantee the stability of the closed-loop
system. Forinstance, if K1 = K2 = 1 are chosen then b1 = 2/3 > 0, andthe first column elements
of the
Routh array are all strictly
of the region, is chosen, then
greater than zero. If
b1 = 0 and the auxiliary
K1 = 2, K2 = 1.2,
which is a point on the border
equation
(K2+2)s2
+2K1
=3.2s2
+4=0 ? roots=j1.118
reveals
thattheclosed-loop
system
has
twopoles
j1.118
ontheimaginary
axis.
7.8 Exercise Problems
1
P7.1a: Consider the typical feedback control system shown in Figure 7.22. G(s) = s+1represents
the
system to be controlled, and K(s) = Kp + Ki
is the PI controller to be designed to improve the perfor-mance
s
of the system. Notethat there are two external inputs: the reference input r(t) andthe disturbance
input d(t). The objective of the control system is to design a controller K(s) so that the closed-loop
system is stable and the output y(t) will follow r(t) as closely as possible despite the variation of the
disturbance d(t). Let the transfer function from R(s)to Y(s) andthe transfer function from D(s) to Y(s)
of the closed-loop system be denoted by Gyr(s) and Gyd(s),respectively. Thenthe output response Y(s)
will be
(7.124)
Y(s) = Gyr(s)R(s)+Gyd(s)D(s)
Find the transfer functions
Gyr(s) and Gyd(s)in terms of the parameters Kp and Ki by using
Masons
gain formula.
P7.1b:
Letr(t) =10us(t)
andd(t)=us(t-5),where
us(t)
istheunitstep
function.
Show
that
the
steady-state tracking error is zero or the final value of y(t) is 10,
lim e(t)
t?8
=0
or
lim y(t)
= 10
t?8
by using Equation 7.124 and Theorem 2.27, the final-value theorem
222
7
Fundamentals
of Feedback
Control
Systems
Fig. 7.22: Atypical feedback control system.
P7.1c: Construct a Simulink program according to the closed-loop system block diagram shown in Fig-ure
7.22,
where
r(t)=10us(t),
d(t)=us(t-5),G(s)
=s+1, and K(s) = Kp + sKi with
1
Kp and Ki as
free design parameters.
P7.1d: Find the closed-loop characteristic equationin terms of Kpand Ki, and useit to determinethe
values of Kpand Kisothat the dampingratio ? = 1 andthe naturalfrequency ?n = 1rad/s. Thenrun
the Simulink program constructed in P7.1c and plot the output y(t) andthe control input u(t) in separate
graphs versus time from t = 0 s to t = 10 s.
P7.1e: Varythe value of the dampingratio ? while keepingthe naturalfrequency ?nat a constant. Then
run the Simulink program and plot the output response y(t) andthe control input u(t) in separate graphs
versustime from t = 0 sto t = 10 s. Observehowthe dampingratio ? affectsthe values of Kpand Ki,
the output response y(t), and the control input u(t).
P7.1f: Varythe value of the naturalfrequency ?n while keepingthe dampingratio ? at aconstant. Then
run the Simulink program and plot the output response y(t) andthe control input u(t) in separate graphs
versustime from t = 0 s to t = 10s. Observehowthe naturalfrequency ?naffectsthe values of Kpand
Ki,the output response y(t), and the control input u(t).
P7.1g: Basedonthe experienceyou gainfrom P7.1eand P7.1f,find a bestdesign of Kpand Kisothat
y(t)has
anoptimal
performance
while
satisfying
thecontrol-input
constraint,
u(t)<20,fort =0.Plot
the optimal output response y(t) and the control input u(t) in separate graphs versus time from t
to t
=0s
= 10 s.
P7.2a: Consider the dual-loop feedback control system shown in Figure 7.23. G(s) = s+1is
1
the system
to be controlled,
and the two
parameters
K1 and
K2in the dual-loop
controller
are to be designed to
improve the performance of the system. Notethat there aretwo external inputs: the reference input r(t)
and the disturbance input d(t). The objective of the control system is to design a controller K(s) so that
the closed-loop system is stable and the output y(t) will follow r(t) as closely as possible despite the
variation of the disturbance d(t). Let the transfer function from R(s) to Y(s) and the transfer function
from D(s)to Y(s) of the closed-loopsystem be denoted by Gyr(s)and Gyd(s),respectively. Thenthe
output response Y(s) will be
Y(s) = Gyr(s)R(s)+Gyd(s)D(s)
(7.125)
Findthe transfer functions Gyr(s) and Gyd(s)in terms of the parametersK1and K2by using Masons
gain formula.
P7.2b:
Letr(t) =10us(t)
andd(t)=us(t-5),where
us(t)
istheunitstep
function.
Show
that
the
steady-state tracking error is zero or the final value of y(t) is 10
7.8
Exercise
Problems
223
Fig. 7.23: A dual-loop feedback control system.
lim
t?8
e(t)
=0
or
lim y(t)
t?8
= 10
by using Equation 7.125 and Theorem 2.27, the final-value theorem.
P7.2c: Construct a Simulink program according to the closed-loop system block diagram shown in Fig-ure
7.22,
where
r(t)=10us(t),
d(t)=us(t-5),G(s)=s+1, andthetwo constantsK1andK2ofthe
1
dual-loop controller are free design parameters.
P7.2d: Find the closed-loop characteristic equation in terms of K1 and K2, and useit to determine the
values of K1and K2sothat the dampingratio ? = 1 andthe naturalfrequency ?n = 1rad/s. Thenrun
the Simulink program constructed in P7.2c and plot the output y(t) andthe control input u(t) in separate
graphs versus time from t
= 0 s to t
= 10 s.
P7.2e: Varythe value of the dampingratio ? while keepingthe naturalfrequency ?nat a constant. Then
run the Simulink program and plot the output response y(t) andthe control input u(t) in separate graphs
versus time from t = 0 s to t = 10 s. Observehow the damping ratio ? affects the values of K1and K2,
the output response y(t), and the control input u(t).
P7.2f: Varythe value of the natural frequency
run the Simulink
program
versus time from t
?n while keeping the damping ratio ? at a constant. Then
and plot the output response
= 0 s to t
K2, the output response y(t),
y(t) and the control input
u(t) in separate graphs
= 10 s. Observe how the natural frequency ?n affects the values of K1and
and the control input
u(t).
P7.2g: Based onthe experience you gain from P7.2e and P7.2f, find a best design of K1and K2so that
y(t)has
anoptimal
performance
while
satisfying
thecontrol-input
constraint,
u(t)<20,fort =0.Plot
the optimal output response y(t) and the control input u(t) in separate graphs versus time from t
to t = 10 s.
=0s
P7.3: Problems P7.1 and P7.2 consider exactly the same problem withthe same objective, but have dif-ferent
controller structures; one is the PI controller and the other is the dual-loop controller. Comment
on the pros and cons of these two approaches based on the results of Problems P7.1g and P7.2g.
P7.4:
Consider the following
system:
Y(s) = G(s)U(s)
=
1
s(s+1)
U(s)
It is easy to seethat the system is not BIBO stable since there is a pole atthe origin, whichis onthe imag-inary
axis, according
to
Theorem
7.15. But according
to the definition
of BIBO stability,
there shoul
224
7
Fundamentals
exist a bounded input
of Feedback
Control
Systems
so that the output is unbounded.
response of the system driven by this input
P7.5:
Explain the difference
P7.6a: Consider the following
between
Find this bounded input
will be unbounded.
BIBO stability
and internal
stability.
system:
?4-30? ??
? ?x(t)+
??u(t)
0 1 1
x?(t)
and show that the output
= Ax(t)+Bu(t)
:=
0 0 2
The objective is to find a state-feedback
controller
u(t)
0
0
1
= Fx(t) so that the closed-loop
(7.126)
system poles, or
theeigenvalues
ofA+BF,
are
placed
ats =-1,-2,-3.Here
wewillemploy
thetransform
approach
instead of the direct approach since the transform approach is easier in computation. Thefirst stepis to
transform the state equation into a companion form. Find a similarity transformation matrix T and define
a newstate vectorx(t)
= T-1x(t) sothat the newstate equation
x?(t)
=Ax(t)+
Bu(t)
is in companion form.
P7.6b: Designastate-feedback
controlleru(t) =Fx(t) sothatthe eigenvaluesofA+ BF are placedat
s =-1,-2,-3.
P7.6c: Use
F and Tto determinethe state-feedbackgain matrix Ffor the original stateequationsothat
the eigenvaluesof A+BF arethe same asthose ofA+ BF. Verify yourresults.
P7.7a: Considerthetypical feedback control systemshownin Figure 7.22. Let G(s) =
tive is to design a controller
s2-s
. The
objec-1
K(s) so that the closed-loop system is stable. Can a proportional controller
K(s) = Kpstabilize the closed-loopsystem?If yes, verify yourresults.If not, explain why.
P7.7b: RepeatProblem P7.7a withthe PI controller K(s) =(Kps+Ki)/s.
P7.7c: Repeat Problem P7.7a with the following
first-order controller,
K(s) =
b1s+b0
s+a0
where a0, b1, b0 arereal parameters. Hint: Use Routh-Hurwitz stability criterion theorem
8
Stability, Regulation,and Root Locus Design
I
N Section 7.6.4, we witnessedthat an originally unstable cart-inverted pendulum system wasstabi-lized
by a state-feedback controller that utilizes a pole placement approach to place all the closed-loop
system poles at desired locations in the strictly left half of the complex plane. The controller
not only keeps the pendulum at the upright position, but also movesthe cart to the desired center of
the rail. In Section 7.1.1, a dual-loop feedback control structure with aregulation integrator and two de-sign
parameters were employed to regulate the speed of a DC motorsystem. Theintegrator guarantees
zero steady-state
error in step speed tracking,
and the two
parameters
K1 and K2 are chosen to achieve
a desired transient response by placing the closed-loop system poles at a best possible location in the
complex plane.
This chapter will provide a complete discussion of compensator design via the root locus methodto
achieve stability, regulation, and a best possible transient response implied by pole locations. It begins
with a study of steady-state error and introduces the internal model principle andthe concept of system
type.
The concept
of system type
and internal
model principle
explains
how to select a feedback
con-troller
structure to achieve zero steady-state error response for certain group of references/disturbances.
A cruise control
objectives,
example is used to provide
an overview
of the design process, including
the role of feedback in achieving those goals, as well asthe notion
performance
of performance
robustness
with respect to model uncertainty.
The root locus methodis presented in some detail, including the basic construction rules, whythey
are useful in design, and how to do the computations using MATLAB. Simple examples illustrate key
points throughout
this
discussion.
a DC motor sinusoidal
model principle
along
More expansive
position tracking
with root locus
controller
examples
are given in the last three sections.
is used to illustrate
the application
First,
of the internal
design.
The next section examines the longitudinal flight path control of the F/A18 aircraft. In this ex-ample,
manual control is simulated to illustrate the difficulty level of controlling the extremely low-damping,
long-period (phugoid-mode) oscillations. Asophisticated root locus design with integral reg-ulation
and state-feedback
last section illustrates
pole placement is employed
how altitude regulation
to achieve stability
can be accomplished
and flight
using the flight
path tracking.
The
path angle tracking
controller.
8.1 Type of Feedback Systemsand Internal
Model Principle
In the design of a feedback control system, the designer would usually consider the following: the stabil-ity
of the closed-loop
system, the transient
response, the steady-state
error, and the robust
performanc
226
8
Stability,
Regulation,
against plant uncertainties.
and Root Locus
In the previous
Design
chapter,
welearned
how the pole locations
stability and the transient behavior of the system. In this section,
controller
so that the closed-loop
system
would affect the
we will discuss how to design a
will have a least steady-state
error.
8.1.1 Steady-State Error
Before the discussion
control
problem
of a more general steady-state
considered in Example
response issue,
we will revisit the
DC motor speed
5.26. The DC motor speed control system block
diagram shown
in Figure 8.1 is almost identical to that shown in Figure 5.22. The only difference is the controller; now
the controller is a proportional controller K(s) = KPinstead of anintegral controller KI/s.
Example
8.1 (Speed
Control
of the
DC
Motor System
Using a Proportional
Controller)
Let the Laplace transforms of ?r(t) and ??(t) be Or(s) and O?(s), respectively. Then we havethe
closed-loop
transferfunctionfrom Or(s)to O?(s)in the following:
O?(s)
G(s)K(s)
=
Or(s)
(8.1)
1+G(s)K(s)
Plugging the expressions of K(s) = KPand G(s) into the closed-loop transfer function,
145.5KP
O?(s)
Or(s)
=
145.5KP
(s+43.14)
1+ 145.5KP
=
b
=
s+43.14+145.5KP
we have
xss
=
s+a
(8.2)
ts+1
(s+43.14)
Notethat this closed-loop transfer function belongs to the category of the typical first-order system we
studied in
Chapter 2. Recall that the dynamic
behavior
of the typical
first-order
system is characterized
bythetime constantt andthe steady-state
stepresponsexss.FromEquation8.2, wehavethe following
two equationsthat relatethetime constantt andthe steady-stateresponseconstantxssto the proportional
constant KP:
t =
1
43.14+145.5KP
,
xss =
145.5KP
43.14+145.5KP
Dueto the restriction of the proportional control structure, there is only one design parameter KPto
be determined, we only have one degree of freedom: choosing either the time constant t or the
steady-state step response constant xss.
Fig. 8.1: A DC motor speed control system with a proportional controller.
If wechoose KP = 0.5the time constant will be t = 0.0086s, andthe steady-statestep response
constant will be xss = 0.628. The time constant is faster than
is 37.2% away from the desired steady-state response.
The steady-state
be t
error can be reduced
by increasing
we need, but the steady-state step re-sponse
That meansthe steady-state
the value of KP. If
error is 37.2%.
KP = 1, the time
constant
will
= 0.0053 s, and the steady-state step response constant will be xss = 0.771, which improves the
steady-state
error to 22.9%.
Theoretically,
the steady-state
error could be reduced to zero if
KPincrease
8.1
Type of Feedback
Systems and Internal
Model Principle
to infinity. However, by increasing the proportional control constant KP,the control input
increase accordingly and up to some point it may saturate the actuator.
The simulation results are shown in Figure 8.2. Onthe left-hand-side
227
u(t)
will
graph of the figure, the red
horizontalline onthe top representsthe desired motorspeed ?r(t) = 20us(t) rad/s. WhenKP = 0.5,the
step responsereachesthe steadystate ??(t) = 12.56rad/s, shortly after t = 0.04 s. Therise time and
settling time
are fast, but the steady-state
error is large.
The steady-state
error is 7.44 rad/s away from
thedesired
20rad/s,
which
is(20-7.44)/20
=37.2%.
If KP=1ischosen,
thestep
response
willrise
to the steadystate ??(t) = 15.42rad/s, aroundt = 0.03s. Therise time andsettling time arefaster, but
the steady-state
error is still large.
The steady-state
error is 4.58 rad/s away from the desired
20 rad/s,
which
is(20-15.42)/20
=22.9%.
Fig. 8.2: Simulation results of the proportional feedback speed control of the DC motorsystem.
The control-input u(t) plots are shown in the right-hand-side of Figure 8.2. It can be seen that the
maximum control input occurs almost right after t = 0 s with u(t) = 10 V when KP = 0.5. When KP
increases to KP = 1,the required
inputs
always
the capability
control input increases
have their limitations
of the control
inputs;
in
accordingly
to about 20 V. Note that the control
magnitude or in rate of change.
otherwise, the system
may perform
The system
poorly
needs to
work within
or become unstable.
For
this particular DC motorspeed control problem, the proportional control design structure is not a good
one since the steady-state error is not acceptable, even whenthe control input is allowed to increase to
20 V. For comparison, both the integral control design in Example 5.26 and the dual-loop control
design in Example 6.1 are much better designs since zero steady-state error is guaranteed for step
responses while their transient performances are satisfactory and their control inputs are well within
reasonable range.
8.1.2
Type of Feedback
Systems and Internal
Model Principle
From the DC motor speed control example, we observe, for the same plant G(s), and assume the ref-erence
input
is a step function;
both of the integral
control
design in
Example
5.26 and the dual-loop
control design in Example 6.1 achieve zero steady-state error response, but the proportional control de-sign
in
Example
8.1 is unable to reduce the steady-state
control-input constraint. In the following,
steady-state error.
error to an acceptable level
we will investigate
within a reasonable
what are the deciding factors for the
The block diagram shown in Figure 8.3 is a typical feedback control system. The plant G(s) repre-sents
a system to be controlled, which can be atransfer function or a state-space model. The objective of
the feedback control system is to design a realizable controller
K(s), which can be also represented as
228
8
Stability,
Regulation,
and Root Locus
Design
transfer function or a state-space model,so that the closed-loop system is stable, and the error e(t) is as
small as possible, subject to the control-input u(t) constraints. In this section, we will only focus on the
steady-state
error,
which is
(8.3)
ess =lim e(t)
t?8
If possible,
we would like to design a controller
Fig. 8.3:
Atypical
so that the steady-state
feedback
error
essis zero.
control system structure.
The feedback control system shown in Figure 8.3 hastwo inputs: Oneis the disturbance input d(t)
and the other is the reference input r(t). Since our interest is in e(t) or its Laplace transform E(s), we
would like to know how these two inputs, R(s) and D(s), affect E(s). Thetransfer function E(s)/R(s)
and E(s)/D(s)
can be obtained using Masons gain formula or the algebraic equations approach as
follows:
1
E(s)
E(s)
R(s)
=
and
1+G(s)K(s)
D(s)
=
-1
1+G(s)K(s)
Hence, E(s) can be written asthe sum of the two responses dueto R(s) and D(s), respectively.
E(s) =
1
1
1+G(s)K(s)
R(s)-1+G(s)K(s)
D(s)
(8.4)
According to the Laplace transform final-value theorem, if the real part of all the poles sE(s)
are strictly negative, then the final value of e(t) can be computed in the frequency domain as
follows:
-sD(s):= essR+essD
ess:= lim e(t) =limsE(s) =lim
+lim
t?8
s?0 1+G(s)K(s) s?0 1+G(s)K(s)
s?0
sR(s)
(8.5)
The disturbance input d(t) and the reference input r(t) in general are different, but the ways they
affect the steady-state error are fundamentally
identical.
Without loss of generality, we will just
addressthe steady-state error essRdue to the reference input R(s).
sR(s)
essR = lim
(8.6)
s?0 1+G(s)K(s)
The reference
or disturbance inputs
of interest
are those
with poles at the origin of the complex
plane
and those with conjugate poles on the imaginary axis. In this subsection, we will consider the reference
input described by the following m-degree polynomial time-function and its Laplace transform,
r(t)
= R0 +R1t +
R2
2 t2 + +
where mis the degree of the reference input
Rm
tm
m!
?
polynomial.
R(s) =
R0
s
R1
+
R2
s2 + s3 +
+
Rm
sm+1
(8.7
8.1
In general, the loop transfer function
G(s)K(s)
Type of Feedback
Systems and Internal
Model Principle
229
G(s)K(s) can be written as
=
k(tas+1)(tbs+1)
(tps+1)
si(t1s+1)(t2s+1)
(tns+1)
(8.8)
wherethe superscript i is an integer greater or equal to 0, andthe system is called type i system.
Combining the above three equations,
essR =lim
s?0
we have
s2+ + Rm
R0 + R1
s + R2
sR(s)
=lim
1+G(s)K(s)
sm
(8.9)
s?0 1+ k(tas+1)(tbs+1)(tps+1)
si(t1s+1)(t2s+1)(tns+1)
According to Equation 8.9, the steady-state error essRis mainly determined by two index numbers
m and i, where mis the degree of the reference input polynomial time function and i is the type of
the feedback system. In the following, we will investigate the steady-state error of the typical feedback
control system for each reference input case, starting from m = 0, the step input case, then m = 1, the
ramp
input
case,
and
thenm=2,theparabolic
input
case,, etc.Foreach
case,
the
typeoffeedback
system will be determined to guarantee zero steady-state error.
Case 0: Steady-State
Error Dueto a Step Reference Input
Case 0: Thereference input is r(t) = R0us(t) with its Laplace transform R(s) = R0/s, where R0is an
arbitrary constant. Based on Equation 8.9, the steady-state error for eachtype of feedback system
can be obtained asfollows.
R0
essR = lim
Type 0 system :
1+k/si
=
s?0
Type 1 or highersystem:
R0
= R0
1+k
1+k/s0
(8.10)
R0
essR = lim
1+k/si
s?0
Example
8.2 (Case
The loop transfer
0 Examples
function
=0
with Type 0 and Type 1 Systems)
G(s)K(s)
of the
DC motor proportional
speed control
system in Exam-ple
8.1 can be rewritten in the form of Equation 8.8 as
G(s)K(s)
=
145.5KP
s+43.14
=
k
s0(t1s+1)
whichis a Type 0 system with k = 145.5KP/43.14 = 3.373KP. Hencethe steady-state error will
be
essR =
R0
1+k
R
=
1+3.373KP
If KP = 1 and R0 = 20 rad/s, then the steady-stateerror is essR= 20/(1+
3.373) = 4.57rad/s,
which is 22.9% of error.
Onthe other hand, the loop transfer function
system in
G(s)K(s) of the DC motorintegral speed control
Example 5.26 is
G(s)K(s)
=
145.5KI
s(s+43.14)
k
=
s1(t1s+1)
whichis a Type1 system with k = 145.5KI/43.14 = 3.373KI. Hence,the steady-state error is
essR = 0.
230
8
Stability,
Regulation,
and Root Locus
Fig. 8.4: Case 0: Comparison
Design
of Proportional
and Integral
Controls of DC motor speed.
This example and Equation 8.10 explain why the integral control is able to accomplish zero
steady-state error for step reference input while the proportional control cant. Asrevealed in
Figure 8.4, the proportional KP = 1 controller provides fast response, but it renders a large 22.9%
steady-state error even with a high control input, u(t) = 20 V. Onthe other hand,the integral Ki =
3.95 controller reaches the steady state shortly after 0.2s and only requires about 6V of control input.
The next caseto be considered is Case1, whenthe reference input is a ramp function.
Case 1: Steady-State
Error Dueto a Ramp Reference Input
Case1: Thereferenceinput is r(t) =(R0 +R1t)us(t) withits Laplacetransform R(s) =R0/s+R1/s2,
where R0 and R1 are arbitrary
constants.
Based on Equation
8.9, the steady-state
error for each type
of feedback system can be obtained as follows:
Type 0 system :
essR =lim
Type 1 system :
essR =lim
R0+R1/s
s?0
s?0
R0+R1/s
s?0
Type 2 or higher system :
=lim R1/s
1+k = 8
1+k/s0
= R1
(8.11
k
1+k/s1
R0+R1/s
essR = lim
s?0
=lim
1+k/s2
R1/s
s?0k/s2
=0
Example 8.3 (Case 1 Examples with Type 1 and Type 2 Systems)
The loop transfer function
5.26 is
G(s)K(s) of the DC motor integral speed control system in Example
G(s)K(s)
=
145.5KI
k
=
s(s+43.14)
which is a Type 1 system with k = 145.5KI/43.14
s1(t1s+1)
= 3.373KI.
Hence,the steady-state error
will
be
R1
essR =
k
R1
=
3.373KI
If KI = 3.95 and R1 = 1 rad/s2, then the steady-state error is essR = 1/(3.373)(3.95)
= 0.075 rad/s2,
which is 7.5% of error.
In contrast, a new controller KR(s) = (s+3)/s2
can be designedto achieve Type 2 requirement
for the loop transfer function G(s)KR(s) as follows:
G(s)KR(s) =
k(tas+1)
145.5(s+3)
s2(s+43.14)
=
s2(t1s+1)
8.1
Type of Feedback
which is a Type 2 system with k = (145.5)(3)/43.14
ramp reference input will be essR = 0.
Fig. 8.5:
Case 1:
Comparison
of integral
Systems and Internal
Model Principle
231
= 10.12. Hence, the steady-state error for
and double integral
controls
of
DC
motor speed for ramp
tracking.
The simulation results comparison of the integral control and the double integral control are shown
in Figure 8.5. Theintegral controller K(s) = Ki/s only makesthe loop transfer function G(s)K(s)
a Type 1 system. It can be seen from the graph on the left that the vertical gap between the green
and the red lines is 0.075 rad/s, which exhibits the steady-state error essR = 0.075 rad/sa
7.5%
of error. With the new double integral controller KR(s) = (s+3)/s2,
the loop transfer func-tion
G(s)KR(s) is of Type 2; hence, the steady-state error is zero. It can be verified from the
simulation that the green and the blue lines coincide together shortly after t = 1 s.
Case 2: Steady-State
Case 2:
R1
s2
Error
Due to a Parabolic
Reference Input
= R0 +R1t + R2
2 t2us(t),
The reference input is r(t)
withits Laplacetransform R(s) = R0
s +
+ s3,
R2 whereR0, R1and R2arearbitrary constants. Basedon Equation8.9,the steady-stateerro
for each type
of feedback
system can be obtained as follows:
R0+R1/s+R2/s2
essR =lim
Type 0 system :
s?0
1+k/s1
s?0
Type 2 system:
1+k/s2
s?0
Type3 or highersystem:
essR =lim
s?0
R2/s2
1+k = 8
= 8
(8.12)
R0+R1/s+R2/s2
essR = lim
=lim
s?0
R0+R1/s+R2/s2
essR = lim
Type 1 system :
1+k/s0
= R2
k
R0+R1/s+R2/s2
1+k/s3
=lim
s?0
R2/s2
k/s3
=0
...
Case m: Steady-State
Error
Dueto a Polynomial
Time Function
Case m: Thereference input is
r(t)
with its
Laplace transform
= R0 +R1t +
R2
2 t2 + +
Rm
tm
m!
Reference Input
232
8
Stability,
Regulation,
and Root Locus
Design
R0
R(s) =
R1
+
s
R2
Rm
s2 + s3 +
+
sm
where
R0,
R1,
R2,, and
Rm
are
arbitrary
constants.
Based
onEquation
8.9,
thesteady-state
error
for each type of feedback system can be obtained asfollows:
R0+R1/s++Rm/sm
=lim
essR =lim
Type 0 system :
s?0
essR = lim
Type 1 system :
s?0
1+k/s0
Rm/sm
s?0
1+k
=
R0+R1/s++Rm/sm
= 8
1+k/s1
...
(8.13)
...
Type
m-1system
:
R0+R1/s++Rm/sm
= 8
essR = lim
R0+R1/s++Rm/sm
= Rm
essR = lim
Type msystem :
1+k/sm-1
s?0
1+k/sm
s?0
Type m+1 or highersystem:
k
R0+R1/s++Rm/sm
=0
essR = lim
1+k/sm+1
s?0
Remark
8.4 (Internal
Model Principle)
From these discussions, it can be observed that to achieve zero steady-state error against a step ref-erence
or step disturbance inputs, whosefrequency-domain representation is c/s with c an arbitrary
constant, the loop transfer function needsto be at least of Type 1 (i.e., G(s)K(s) needsto include the
internal model 1/sin it). Similarly, for a ramp reference or ramp disturbance inputs, whosefrequency-domain
representationis c/s2 withc an arbitrary constant,theloop transfer function G(s)K(s) needsto
includetheinternal model1/s2in it in orderto guaranteezerosteady-stateerror. Thisperfectsteady-state
tracking concept is called the internal model principle, which has been widely employed in
the design of feedback control systems. Theinternal
model principle [Francis and Wonham, 1976,
Huang, 2004] also can be applied to regulate sinusoidal disturbances or to track sinusoidal refer-ence
input.
Example 8.5 (Sinusoidal
Tracking
Using the Internal
Model Principle)
Consider the same DC motor speed feedback control block diagram shown in Figure 8.1. The plant
G(s) = 145.5/(s+43.14)
is still the same, but the reference input has changed to the following:
?r(t)
= Asin(?0t
+f)
(8.14)
wherethe frequencyis assumedto be ?0 = 2rad/s, butthe amplitude Aandthe phasef arearbitrary. To
achieve steady-state sinusoidal tracking, the loop transfer function
G(s)K(s) needsto have the internal
model1/(s2 +22)in it to match
the referencesinusoidalfunctionfrequency ?0 = 2rad/s. Therefore,
the controller
K(s) should be of the form
K(s) =
where k is the design parameter to be determined
s+k
s2
so that the closed-loop
response is optimized subject to the control-input
Theloop transfer function
(8.15)
+22
system is stable and the tran-sient
constraints.
G(s)K(s) of the DC motorsinusoidal speed control system now is
G(s)K(s)
=
145.5(s+k)
(s+43.14)(s2
+22)
8.2
which consists of the internal modelfunction 1/(s2+22)
The closed-loop characteristic equation now is
(s+43.14)(s2
An Automobile
Cruise
Control
Example
233
to guarantee zero tracking error at steady state.
+22)+145.5(s+k)
=0
which can be rearranged into
1+k
145.5
s3 +43.14s2 +149.5s+172.56
=0
and an optimal k = 0.616 is obtained using the root locus design approach, which will beintroduced in
the next section.
Fig. 8.6: Comparison of integral and sinusoidal controls of DC motor speed for sinusoidal tracking.
The simulation
reference input
results
?r(t)
are shown in
Figure
8.6. The graph on the left
shows three functions,
the
= sin2t in green,the output response ??(t) in blue associated with the sinusoidal
controller KS(s) = (s + 0.616)/(s2 + 22), and the output response ??(t) in red associated withthe
integral controller KI(s) = 3.95/s. It can beseenthat the sinusoidal controller KSis ableto achieve
perfect steady-state tracking
within 2 seconds while the integral control response has a time-delay
steady-state error. The graph on the right shows that both controllers use about the same amount of
control input.
8.2 An Automobile
Cruise Control Example
Recallthat in Section 7.1 we briefly introduced the features of feedback control that include: (1) it is a
naturally
perfect
characteristics
disturbance
mechanism for automatic
control;
(2) it provides
of the system to achieve better performance;
response
rejection;
(4) it has the ability
an easy wayto
modify the dynamics
(3) it has the ability to achieve steady-state
to achieve steady-state
reference
input tracking
or
regulation; and (5) it achieves robust stability and robust performance against plant uncertainties.
Later,in Section 8.1, westudied the steady-state error issues of feedback control systems andlearned
that steady-state reference input tracking and steady-state disturbance response rejection are mathemati-cally
the same.
types
A feedback
of reference
and/or
controller
disturbance
can be designed to achieve zero steady-state
inputs,
including
step
disturbances
error
for certain
with arbitrary
magni-tude,
ramp disturbances with arbitrary slope, and sinusoidal disturbances with arbitrary
am-plitude
and phase. This perfect steady-state design approach is based on the celebrated interna
234
8
Stability,
model principle
the dynamics
Regulation,
with
and Root Locus
which the loop
Design
transfer
function
of the closed-loop
system is embedded
with
model of the reference and/or disturbance inputs.
In Example 8.2, we compared the performances of the integral controller versus the proportional
controller in a DC motorspeed control system, and witnessedthat the integral control is able to accom-plish
zero steady-state
that the loop transfer
error for step reference input
function
of the closed-loop
G(s)K(s)
=
while the proportional
system
with integral
control
controller
cant.
The reason is
is
145.5KI
s(s+43.14)
=
s(t1s+1)
whichis equipped withthe internal model1/s, andthe referenceinput r(t) is astepfunction witharbi-trary
magnitudeR0, whoseLaplacetransform is R(s) = R0/s.
The basic idea of the DC motor speed control system and the internal
and applied to the automobile
control
design problem
cruise control
model principle can be ex-tended
system. In the following,
based on the brief discussion
in
Section 1.4.
we consider
a basic cruise
The goal is to introduce
the pri-mary
goals and basic methods of feedback control design in a context familiar to most readers. The
objective of basic cruise control is to regulate the vehicle speed to match a commanded speed. In
this
discussion,
bounce,
pitch, sideslip,
roll,
and yaw are ignored.
Our focus is on speed (i.e.,
velocity,
v,in the body x direction). Relativelysmall deviationsof v from the nominal speed,v*, are of concern.
The control input is the engine throttle u, whose manipulation alters the force f ultimately imposed on
the vehicle by the engine through the gear train and tires. Wedo not consider gear shifting in view of
the small deviation of speed assumption. Wind gusts are not considered, although aerodynamic drag is.
Roadslope, ?,is the only disturbanceconsidered. Roadslopeinduces a gravitationalforce component
that contributes to vehicle acceleration.
The block diagram of Figure 1.1 illustrates the typical configuration of a basic cruise control system.
Our goalis to design the control block. The cruise control system responds to a change in speed com-mand
or a disturbance (a change in road slope) by adjusting the throttle position to maintainthe desired
speed. The criteria for design includes the following:
1. Steady-state error: Speed error in steady statein response to command or disturbance
2. Transient response time including rise time, peaktime, and settling time
3. Transient overshoot or undershoot
4.
Robustness: performance
tolerance
to unmodeled
dynamics
or parameter
variation
8.2.1 Assembling a Model
Webegin by establishing
a mathematical
model for each element
described by the following nonlinear first-order
representing the velocity of the vehicle,
m
dv(t)
dt
of the system.
The vehicle
dynamics is
differential equation with one single state variable v(t)
=f(t)-mgsin?
(t)-cv2(t)
(8.16)
where mis the massof the vehicle, f is the force generated by the engine to movethe vehicle, c is the
viscosity coefficient of the vehicle traveling in the air, and ?(t) is the slope angle of the road. Notethat
sin? ? i f |?|=0.262
rad
The
disturbance
termmgsin?
can
bereplaced
bymg?
if|?|=0.262
rad,
orequivalently
theslope
of
the road is less than
15 degrees. Thus, Equation 8.16 can be rewritten
as
8.2
v?(t)
-cv2(t)+
=
1
m
m
An Automobile
Cruise
Control
f(t)-g?(t)
Example
235
(8.17
Assme(v*, f*) is a nominal operating equilibrium of the vehicle sothat
f* =c(v*)2
and let v(t) and f(t), respectively, bethe perturbed velocity and force variables from the equilibrium.
Thus,the relationship between the real physical variables (v, f), the equilibrium (v*, f*), andthe per-turbed
variables (v, f) are
v(t) =v(t)+v*
and
f(t)
=f(t)+
f*
(8.18)
ThenEquation8.17 can belinearized aboutthe equilibrium(v*, f*) usingthe Jacobianapproach
shown in
Equation
C.5 from
v?(t)
=
Appendix
?
?v
?
-cv2
m
C as follows:
+
1
f
m
?? ?
?
+
?
?
?
v*, f*
-cv2
?f
m
+
1
f
m
??
?
?
?
v*, f*
-g?
which is
v?(t)
where
c
=(2c/m)v*.
=
-2cv*v(t)+
m
1
1
f(t)-g?(t):=-cv(t)+mf(t)-g?(t)
m
(8.19)
Takingthe Laplacetransform of Equation8.19 weobtain
1
sV(s)=-cV(s)+F(s)-gT(s)
m
(8.20)
and therefore the velocity of the vehicle is
V (s) =
1/m
g
s+ c F(s)- s+c
T(s)
(8.21)
Turning now to the engine dynamics, assumethe engine thrust response to throttle input is very fast
compared to the vehicle dynamics. Then we mightignore any dynamics and take the ideal
engine
transfer function to be GE(s) = 1. We will design the controller using this assumption andthen evaluate
the impact of engine dynamics by testing the resultant controller with an alternative transfer function
that reflects engine delay. Thus, we will usethree different engine models, all of the form
GE(s) =
? ?2
1
ts+1
(8.22)
with
1. ideal model (used for design), t
2. fast model, t = 0.3
3. slow model, t = 0.6
=0
This engine modelis a typical second-order system with damping ratio ? = 1 and natural frequency
?n = 1/t. It also can be considered as a cascade connection of two identical typical first-order system
with time constant t.
The block diagram for the control system with the component
modelsis shown in Figure 8.7. Note
that Equation
8.21describes
howtheengineoutput
forcef(t) andtheslopeoftheroad?(t) wouldaffect
the vehicle speedv(t), and Equation 8.22is the transfer function of the engine between the throttle u(t)
sothat the closed-loopsystemhasa desired
and the force f(t). Thecontroller K(s)is to be designed
performance.
236
8
Stability,
Regulation,
Fig. 8.7:
and Root Locus
Cruise control
Vehicle Data and a Nominal
block
Design
diagram
with component
transfer
functions.
Operating Equilibrium
The vehicle considered is a typical mid-size SUV (sport utility vehicle) with mass m= 1,929 kg and
a dimensionless drag coefficient Cd = 0.35. The drag force is
Fd = cv2 =
1
2
?CdAv2
(8.23)
where ? = 1.225 kg/m3 is the air density, Cd = 0.35 is the drag coefficient, and Ais the front surface
area, whichis approximately computed as A= 3 m2from the height 1.67 mandthe width 1.8 m. Hence,
the viscosity coefficient of the vehicle is
c =
1
2
?CdA =
1
(1.225)(0.35)(3)
2
= 0.643 kg/m
The nominal operating velocity is chosento be v* = 30 m/s, which is about 108 km/hr or
64.8 mile/hr,andthus the nominal operatingforce for v*is
f* = c(v*)2 = (0.643)(302) = 578.7 N
Notice that the linearized
model of the system shown in
Equation
8.19 or in
Figure
8.7 was obtained
from linearizing the nonlinear system of Equation 8.17 at the nominal operating equilibrium,
(v*, f*)
= (30 m/s,578.7
N)
and the linearized modelis expressed in terms of the perturbed variablesv(t) andf (t),
whoserelation-ship
with the real physical velocity and force is given by Equation 8.18. Thereference input r(t) shown
in Figure 8.7 is also a perturbed velocity variable with respect to the equilibrium velocity v* = 30 m/s.
Thatis,r(t)
=5 m/s means
thatthe desiredperturbedsteady-state
velocityisv(t)
= 5 m/sorthereal
physical velocityis v(t) = 35 m/s.
The parameters g, m, andc in the linearized model described by Equation 8.19 or Equation 8.21
are the gravity g = 9.8 m/s2, the mass of the vehicle m = 1,929 kg, and the characteristic
value is
c- =-2
mv* =-0.02.
It is obvious that the MKS system of units is adopted in the computation of the cruise control prob-lem
in this section. Forthe slope angle of the road, ?(t), although we maydisplayit in degrees,it is
imperative
to adopt the unit radian in computations
is approximately
huge error!
as commented
in
Remark 4.16. Recall that
57.3 degrees. Mistakenly regarding 1 degree as 1 radian in computation
1 radian
will lead to a
8.2
8.2.2 Design of the Controller
An Automobile
Cruise
Control
Example
237
K(s)
The classical control design processis to choose a structure for the controller andthen determine the con-troller
parameters to meetthe performance criteria. In this case wechoose a PI controller (proportional
plus integral
controller):
?t
u(t) =kpr( (t)-v(t))+ki r( (t)-v(t))dt
(8.24)
0
Note
that
there
are
twoterms,
thefirstproportional
tothespeed
error,
e(t)=r (t)-v(t),and
the
second proportional to the time integral of error. The design parameters are the two constants kp and ki.
In terms of e(t), Equation 8.24 is
u (t)
= kpe(t)+ki
?t
e(t)dt
0
Differentiating
with respect to t this becomes
u?(t)
Taking the
Laplace transform
(8.25)
= kp
?e(t)+kie(t)
yields
U (s) =
kps+ki
s
E(s) :=
(8.26
K(s)E(s)
Closed-Loop Transfer Functions
In the closed-loop system shown in Figure 8.7, the output variable of interest isv(t),
by whichthe veloc-ity
of the vehicle, v(t) =v(t)+v*
can beeasily obtained. Meanwhile,there aretwo input variablesr(t)
and ?(t). Theformer is the referenceinput (or called commandinput), andthe latter is the road slope
disturbance input. The objective is to design a controller K(s) so that the output v(t)
will follow the
reference input r(t)
as closely as possible under possible influence of the road slope disturbance
?(t).Since
e(t)=r(t) -v(t),thecontrol
problem
canberephrased
asto make
e(t)as mall
as
possible subject to control-input
constraints.
Hence, we will need to investigate how the controller design will affect the following four closed-loop
transfer functions: GE
R(s), GET(s), G
FR(s), and G
FT(s) that satisfy the following two equations:
E(s) = GE
R(s)R(s)+GET(s)T(s)
F (s) = G
FR(s)R(s)+G FT(s)T(s)
Thesetransfer functions can be obtained by applying Masons gain formula (see Section 6.2.2, or Ap-pendix
D)to the closed-loop system block diagram shown in Figure 8.7. According to Masons gain
formula, we have
GE
R(s)
where
=
1
,
?
GET(s) =
gGp(s)
?
,
FR(s)
G
=
K(s)
?
,
FT(s)
G
=
gGp(s)K(s)
?
238
8
Stability,
Regulation,
and Root Locus
Design
1
? =1-?1=1+m Gp(s)K(s)
and therefore
ms2 +(m
c+kp)s+ki
=
ms(s+c
we have
E(s) = GE
R(s)R(s)+GET(s)T(s)
=
(8.27)
mgs
ms(s+c)
R (s)+ ms2+(m
c+kp)s+kiT(s)
ms2+(m c+kp)s+ki
and
F (s) = G
F R(s)R(s)+G FT(s)T(s)
= m(s+
c)(kps+ki)
R(s)+
ms2+(m c+kp)s+ki
(8.28)
mg(kps+ki)T(s)
ms2+(m c+kp)s+ki
Notice that all of the four transfer functions havethe same denominators (the same poles), but differ-ent
numerators (different zeros). Thestability, damping, and oscillation frequency are mainly determined
by the poles while the steady-state response andthe phase of the transient response are affected by the
zeros.
Analysis
oftheSecond-Order
System
fortheCase
With
ki?=0
It can be seen from
Equations (8.27)
and (8.28) that the closed-loop
ki?=0,and
itscharacteristic
equation
is
ms2 +(mc+kp)
Recall that the roots
s+ki
of the characteristic
=0
or
equation
system is a second-order
??
kp
s2 + c +
m
s+
ki
m
system if
(8.29)
=0
are the poles of the system. In
most of the practical
control system applications, the dominant poles are chosento be a pair of complex numbers as
?
-a j? =-??n j?n 1-?2
The damping factor a and the frequency
? can be chosen to determine the decay rate and
the oscillation frequency of the transient response, respectively. Alternatively, the damping ratio ?
and the natural frequency ?n can be selected to specify the maximum overshoot and the reaction
time ofthe transient response.Therelationshipofthe PIcontrollerparameters
(kp,ki) to (a,?) orto
(?,?n) can beeasily establishedviathe connectionsofthe following characteristicpolynomials:
s2 +2??ns+?2
n
?
? ?
kp
s2 + c+
m
s+
ki
?
s2 +2as+a2
+?2
(8.30)
m
Forexample,if ki =100andkp = 500arechosen,thenthe dampingratio andthe naturalfrequencyfor
the closed-loop system will be
?2
n
=ki/m
2??n =c+kp/m
?
?n =
vki/m
=
v100/1929
? =(c+kp/m)/2?n
= 0.2277rad/s
=(0.02+500/1929)0.4554
= 0.6131
Onthe other hand, if the damping ratio ? = 0.707 and the natural frequency ?n = 1 rad/s are
chosen for the PI cruise control closed-loop system, then the PI controller
should
satisfy the following
ki/m = ?2
n
c+kp/m
= 2??n
parameters
kp an ki
equations:
ki = m?2
n = 1929
? kp=m(c+2??n)=1929(-0.02+1.414)
=2689.03
8.2
An Automobile
Cruise
Control
Example
The closed-loop system poles or their associated damping ratio and natural frequency
the stability,
they
damping,
oscillation,
and transient
alone do not determine the steady-state
reaction time
response
of the closed-loop
of the closed-loop
system.
239
mainly deter-mine
system;
however,
As discussed in
Sec-tions
8.1.1 and 8.1.2, the steady-state tracking/regulation
error is determined by the type of refer-ence/disturbance
inputs, and by the internal
model embedded in the loop transfer function of the
feedback systems.
r(t)
The reference input r(t)
is assumed to be a step function with arbitrary magnitude. That is,
= Rvus(t), where us(t) is a unit step function and Rvis an arbitrary real number representing
a differential
change
of the desired
vehicle
velocity
from the nominal
v* = 30 m/s. For example,if the desired vehicle speedis v(t)
v(t)
operating
equilibrium
velocity,
= 33 m/s, which meansthe desired
is 3 m/s; hence, Rvshould be chosen as 3 m/s. Onthe other hand,if the desired vehicle speed is
v(t)=25m/s,which
means
thedesired
v(t)is -5m/s;
therefore,
Rv
should
bechosen
as-5 m/s.
Similarly,the road slope disturbanceinput ?(t) is assumedto be astepfunction with arbitrary mag-nitude.
Thatis, ?(t) = ?dus(t), whereus(t) is a unit step function and ?dis an arbitrary real number
representing
the roadslopeanglein radian. Forexampleif the vehicleencountersan uphill 5?ramp,
then ?d should be chosen as 0.0873 rad.
Now,assume
thereference
inputis R(s) = Rv/s,andtheroadslopedisturbance
inputis T(s) =
?d/s, and both Rvand ?d are arbitrary real numbers. Thenthe closed-loop error response E(s) can be
expressed as follows:
1
E(s)
=
?
?d
Rv
?(s)
s
+gGp(s)
s
?
where ?(s) = 1+
1
m
Gp(s)K(s)
According to the final-value theorem, if the closed-loop system is stable, we have the steady-state error,
ess =lim e(t)
t?8
1
(Rv +g?dGp(0))
s?0 ?(s)
=limsE(s)
=lim
s?0
Notethat the loop transfer function in ?(s) is embedded withthe internal model1/s, whichis coming
from the integrator in the controller.
lim ?(s) = 1+
s?0
1
m
Hence, we have
Gp(0)limK(s)
= 1+
s?0
Therefore, the steady-state tracking/regulation
ess =
1
lim
?(s)
1
m
Gp(0)lim
kps+ki
s?0
s
=8 if ki?=0
(8.31)
error is
(Rv+g?dGp(0))
=0 if ki?=0
s?0
Analysis of the First-Order
If ki
= 0 in
Equation
System for the Case with ki = 0
8.26, the transfer
function
of the controller
will become
K(s) = kp
which is a proportional controller, and the closed-loop transfer functions in Equations 8.27 and 8.28 will
reduce to the following
240
8
Stability,
Regulation,
and Root Locus
Design
m(s+c)
E(s) = GE
R(s)R(s)+GET(s)T(s) =
R(s)+
ms+(mc+kp)
mg
ms+(mc+kp)
T(s)
(8.32
T(s)
(8.33)
and
F (s) = G
FR(s)R(s)+G FT(s)T(s)
m(s+c)kp
=
R (s)+
ms+(mc+kp)
mgkp
ms+(m
c+kp)
It can be seen from Equations 8.32 and 8.33 that the closed-loop system now is afirst-order system,
and its characteristic equation is
m
ms+(m c+kp)
=0
or
s+1
c+kp
m
= 0
(8.34)
which is characterized by the time constant
t
=
m
c+kp
m
Thetransient response is governed by an exponential function
with time constant specified by the previ-ous
equation. Thetime responseis faster if the time constantis smaller or if kpis larger.
To investigate the steady-state response of the first-order
proportional
controller,
we will follow
the same procedure
closed-loop cruise control system with
we did a while ago for the system
controller. Thereferenceinput andthe road slope disturbanceinput are assumedto ber(t)
with PI
= Rvus(t)
and ?(t) = ?dus(t), respectively, and both Rv and ?d are arbitrary real numbers. Thenthe closed-loop
error response E(s) can be expressed as follows:
E(s) =
?
Rv
1
?(s)
s
+gGp(s)
?d
s
?
where
?(s) = 1+
1
m
Gp(s)kp
According to the final-value theorem, if the closed-loop system is stable, we havethe steady-state error
asfollows:
ess = lim
t?8
e(t)
= limsE(s)
s?0
=
1
lim ?(s)
(Rv +gGp(0)?d) =
gGp(0)?d
Rv
?(0)
+
?(0)
:= essR +ess? (8.35)
s?0
Since the loop transfer function Gp(s)kp/m in ?(s) does not have the internal model 1/s in it,
the steady-state error essis not zero. The steady-state tracking/regulation error is contributed from two
sources: essR due to the reference input and ess? due to the road slope disturbance. For example, if
?d = 0 and Rv = 3 m/s,then we havethe steady-state error
ess = essR =
Rv
?(0)
3
=
3
1+Gp(0)kp/m
=
1+50kp/1929
3
=
1+0.02592kp
The error can be madearbitrarily small if kp can be madearbitrarily large. However, kp cannot be
madearbitrarily large becausethere is a physical constraint onthe engine force f. If kp is chosen to be
kp = 500, then
ess = ess R =
The referenceinput r(t)
3
1+0.02592kp
3
=
1+(0.02592)(500)
= 0.215 m/s
(8.36)
= 3us(t) meansthat the desired vehicle speedis v(t) = 33 m/s. Sincethe
steady-stateerror is ess= 0.215 m/s,the real steady-statevehiclespeedis 32.785 m/s.
8.2
An Automobile
Cruise
Control
Example
241
For another example, assume ?d = 0.0873rad and Rv = 0 m/s, then we havethe steady-state
error:
ess = ess?
=
gGp(0)?d
(9.8)(50)(0.0873)
=
?(0)
1+50kp/1929
The error can be madearbitrarily small if kp can be madearbitrarily large. However,kp cannot be
madearbitrarily large becausethere is a physical constraint onthe engineforce f. If kpis chosento be
kp = 500,then
ess = ess? =
(9.8)(50)(0.0873)
42.777
=
1+50kp/1929
(8.37)
= 3.064 m/s
1+(50)(500)/1929
The reference input r(t)
= 0us(t) meansthat the desired vehicle speed is v(t) = 30 m/s. Since the
steady state error is ess = 3.064 m/s, the real steady-state vehicle speed is 26.936 m/s.
8.2.3 Simulation
Results with Ideal
Engine
In Section 8.2.2, a simple PI (proportional
ki
wasintroduced
Model GE(s) = 1
plus integral) controller structure with two parameters kp and
to achieve cruise speed control
under the influence
of road slope
disturbances.
We
also discussed at great length on how the control design would affect the steady-state andthe transient
responses
of the closed-loop
and by utilizing
In this
the internal
subsection,
system
by selecting
pole locations,
damping ratio,
and natural frequency
model theory.
we will present
the results
of four
simulations
in
which the
engine
model
is assumed ideal with transfer function
GE(s) = 1. Then in the next subsection, Section 8.2.4, two
simulations are employed to show how the unmodelled engine dynamics
1
GE(s) =
(ts+1)2
maydestabilize the system. We will alsolearn that a moreaggressive controller for the ideal
not be robust against unmodelled dynamics.
Furthermore,
in
Section 8.2.5, a feed-forward
compensation
is employed
to reduce the
model may
disturbance
response if the disturbance information is available.
The Simulink program CruisePI.mdl
shown in Figure 8.8 will be employed to conduct sim-ulations
in this and the next subsections. This Simulink program is constructed based on the
closed-loop system block diagram shown in Figure 8.7, wherethe transfer function of the PI con-troller
is K(s) =(kps+ki)
is Gp(s) = 1 ?(s+ c).
same assumption
?s, the engine dynamics is GE(s) = 1
?
(ts+1)2,
andthe vehicle dynamic
Forthe four simulationsin this section,the time constant t is setto zerothe
on which the PI controller
was designed.
The step input block labeled by Rrepresents the reference input r(t)
= Rvus(t). The other step input
blocklabeledby Slope servesastheroadslopedisturbance
input, ?dus(t).Theoutputblocklabeledas
v arecordsthe actualvehicle velocity v(t) =v(t)+v*,
the block errlabeled as er recordsthe tracking
error,e(t), andthe blocklabeled as f a recordstheforce f(t) =f(t)
+f*. Theconstantblocksv n
and f nrepresentthe nominal operatingequilibrium velocity andforce, v* and f*, respectively.
Example 8.6 (Speed Rv = 3 m/s Step Tracking
0,100,500)
Responses with kp = 500 and Three Values of ki =
In this simulation, the vehicle initially is assumed operating at the equilibrium, v(0)
= 0 m/s, or
v(0) = v* = 30 m/s. Letr(t)
= Rvus(t), where Rv = 3 m/s, and ?(t) remains at 0 rad all the time
242
8
Stability,
Regulation,
and Root Locus
Fig. 8.8: Simulink
during the simulation.
from v(t)
diagram
Design
CruisePI.mdl
There is no road slope
= 0 m/sto v(t)
for cruise control
disturbance,
simulations.
and the car is commanded
= 3 m/s,or, equivalently, from v(t)
= 30 m/s to v(t)
to change speed
= 33 m/s.
The three speed tracking responses corresponding to the three PI controllers
with kp = 500
and three values of ki = 0,100,500 are shown in Figure 8.9. Theleft graph is the velocity graph v(t),
the middle oneis the error graph e(t), andthe one onthe right is the control force graph f(t). The ki = 0
responseis shown in pink;it risesfrom v(0) = 30 m/sexponentially withtime constant t computed
based on Equation
8.34,
t
to the steady state v(t)
8.36.
1929
m
=
=
cm+kp
= 3.58 s
(1929)(0.02)+500
= 32.785 m/s with a steady-state error, ess = 0.215 m/s,as calculated in Equation
The ki = 100 and ki = 500 responses are shown in blue and black, respectivelythey
all converge
to the steady state v(t) = 33 m/s with zero steady-state error as expected. The damping ratio and natural
frequency pair (?,?n) associated with the ki = 100 and ki = 500 designs are (0.613,0.228 rad/s) and
(0.274,0.509 rad/s), respectively; hence, the ki = 500 response has larger maximum overshoot and
shorter oscillation period than the ki = 100 response. It also can be seenthat the ki = 500 design requires
alarger swing of control force
-220N < f(t) <3000N
than the ki = 100 design.
The closed-loop system simulation results shown in Figure 8.9 are obtained by running the follow-ing
MATLAB program: CSDfig8p9.m,
which will automatically call the Simulink model pro-gram
CruisePI.mdl,
to conduct the simulation.
The plotting program, Plot3C.m
will also be
called to plot the figures. The MATLAB code is listed asfollows.
% CSDfig8p9.m
% MATLAB
Cruise
R2015a
% simulation
and
% Gravity:
Data:
% Drag
coefficient:
% Front
surface
% Drag
force
cntrl,
BC
Chang,
CSDfig8p9.m
Plot3C.m
to
Drexel
will
plot
call
University,
8/10/2019
CruisePI.mdl
to
Fig8.9
m/s2
(A
m=1929
PI
later.
use
g=9.8
% Vehicle
% Mass:
or
typical
mid-size
SUV
like
Lexus
R350)
Kg
Cd
area:
Fd
=
=
0.35
A=(width)*(height)=(1.8)*(1.67)=3
0.5*rho*Cd*A*v2
:=
c*v2
m2
conduct
8.2
An Automobile
Cruise
Control
Example
243
Fig. 8.9: Speed Rv = 3 m/stracking responses with kp = 500andthree values of ki = 0,100,500.
% where
v
% Thus,
% c
is
c
=
the
=
0.5*rho*Cd*A
% Choose
the
=
30
% f_n
=
c*(v_n)2
% Hence,
=
=
m/s,
c_bar
1929,
c
which
=
viscosity
rho
=
is
=
108
Km/Hr,
0.643*(302)
0.02
=
=
,
air
g
0.643
9.8,
or
64.8
578.7
v_n
=
1.225
Kg/m3
Kg/m
at
miles/hr
N
2*0.643*30/1929
=
coefficient
density
equilibrium
=(2c/m)*v_n
c_bar
the
0.5*1.225*0.35*3
operating
=
is
where
nominal
% v_n
m
velocity,
0.5*rho*Cd*A,
=
=
30,
0.02
f_n
=
594.2,
kp=500
%Initialization
tau=0,
R0=v_n*0.1,
%%
1st
ki
=
Run
0,
Slope_deg=0,
Slope_rad=Slope_deg*pi/180,
Simulation
sim_time=50,
sim_options=simset('SrcWorkspace',
open('CruisePI');
'current','DstWorkspace',
sim('CruisePI',
[0,sim_time],
sim('CruisePI',
[0,sim_time],
sim('CruisePI',
[0,sim_time],
'current');
sim_options);
run('Plot3C')
%%
2nd
Run
ki=100,
Simulation
sim_time=50,
sim_options=simset('SrcWorkspace',
open('CruisePI');
'current',
'DstWorkspace',
'current');
sim_options);
run('Plot3A'),
%%
3rd
ki
=
Run
500,
Simulation
sim_time=50,
sim_options=simset('SrcWorkspace',
open('CruisePI');
'current',
'DstWorkspace',
'current');
sim_options);
run('Plot3B')
Example 8.7 (Road Slope?d = 0.0873rad Disturbance Responseswith kp = 500 and Three Values
of ki = 0,100,500)
The same PI controllers of Example 8.6 will be employed in the following
simulation to
maintain
the vehicle speed atthe equilibrium speed v* = 30 m/sunderthe influence of road slope disturbance
?(t) = ?dus(t), where?d = 0.0873rad.
The three road slope disturbance
responses corresponding
to the three
PI controllers
with
kp = 500 and three values of ki = 0,100,500 are shown in Figure 8.10. Theki = 0 responseis shown
in pink; it drops from v(0) = 30 m/sexponentially, with the sametime constant t = 3.58 s, as computed
in Example 8.6, to the steady state v(t) = 26.936 m/s with a steady-state error, ess = 3.064 m/s,as
calculated in Equation 8.37. It also can be seen from the graph that the steady-state force required
to achieve this steady-state velocity is approximately 2100 N, which can be verified by the followin
244
8
Stability,
Regulation,
and Root Locus
Design
Fig. 8.10: Roadslope ?d = 0.0873rad disturbanceresponse with kp = 500 and three values of ki =
0,100,500.
computations.From Equation8.33 withr(t)
= 0and ?d = 0.0873rad = 5?, wehavethe force at steady
state as
f(t)
= f*
+f(t)
= f* +
mgkp(0.0832)
cm+kp
= 578.7+
(1929)(9.8)(500)(0.0832)
= 2111 N
(1929)(0.02)+500
The ki = 100 and ki = 500 responses are shown in black and blue, respectivelythey
back to steady state v(t)
all converge
=30 m/s with zero steady-state error as expected. The damping ratio and natural
frequencypair(?,?n) associatedwiththe ki =100andki = 500designsarethe sameasthoseshownin
Example
8.6; hence, the ki = 500 response
haslarger
maximum overshoot
and shorter
oscillation
period
than the ki = 100 response. It also can be seenthat the ki = 500 design requires alarger swing of control
force
578.7 N < f(t)
< 3000 N
than the ki = 100 design.
The closed-loop
system
MATLAB
simulation
program:
results
shown
CSDfig8p10.m,
in
Figure
which
8.10 are obtained
will automatically
by running
call the
the fol-lowing
Simulink
model
program CruisePI.mdl,
to conduct the simulation. The plotting program, Plot3C.m
will also
be called to plot the figures. The MATLAB code is listed asfollows. Note that this program is the
same as CSDfig8p9.m,
except the Initialization
line
where
R0 and
Slope
deg are changed
to
0
and 5, respectively.
% CSDfig8p10.m
%
MATLAB
%
simulation
Cruise
R2015a
or
and
PI
cntrl,
later.
use
Plot3C.m
BC
CSDfig8p10.m
to
Chang,
Drexel
will
plot
call
University,
CruisePI.mdl
8/10/2019
to
conduct
Fig8.10
:
:
%Initialization
tau=0,
R0=v_n*0,
Slope_deg=5,
Slope_rad=Slope_deg*pi/180
:
:
Example 8.8 (Speed Rv = 3 m/s Step Tracking Responses with ? = 0.707 and Three Values of
?n = 0.25,0.5,1 rad/s.)
The speed tracking control simulation to be considered here is the same as that conducted in Exam-ple
8.6, except that the PI controller parameters kp and ki are chosen based on the desired damping rati
8.2
An Automobile
Cruise
Control
Example
245
? and naturalfrequency ?nfor the closed-loopsystem.
Recall that from Equation 8.30, we havethe following
relationship
equation between (kp,ki)
and
(?,?n):
?2
n = ki/m
ki = m?2
n
? kp=m(2??n
-c)
2??n =c+kp/m
Basedon Equation8.38 andthe vehicle data, m= 1929 kg andc
pairsassociated withthe three (?,?n) pairsasfollows:
? = 0.707, ?n = 0.25rad/s
?
? = 0.707, ?n = 0.5rad/s
? = 0.707, ?n = 1 rad/s
?
?
(8.38)
= 0.02s-1, weobtainthe three (kp,ki)
kp = 643.3, ki = 120.6
kp = 1325, ki = 482.3
kp = 2689, ki = 1929
Fig. 8.11: Speed Rv = 3 m/s tracking responses with ? = 0.707 and three values of ?n =
0.25,0.5,1 rad/s.
The three
speed tracking
responses
? = 0.707 and three values of?n
speed tracking
all the three
corresponding
to the three
PI controllers
associated
with
= 0.25,0.5,1 are shown in Figure 8.11. All three have perfect steady-state
with zero steady-state error. They also havethe same maximum overshoots since
closed-loop
systems
have the same damping ratio.
However, they
differ in rise time
and
settling time. The system with larger ?n will have faster transient response. The speed track response
associated with ?n = 1 rad/s is shown in pink, which is the mostaggressive one with quickest time
response. However,the pink one requires
designs (4500
much more control force (above 8000 N)than the other two
Nand 2500 N).
The closed-loop system simulation results shown in Figure 8.11 are obtained by running the fol-lowing
MATLAB program: CSDfig8p11.m,
which will automatically call the Simulink
model
program CruisePI.mdl,
to conduct the simulation. The plotting program, Plot3A.m
will also
be called to plot the figures. The MATLAB code is listed asfollows. Note that this program is the
same as CSDfig8p9.m,
except the six lines shown in the following list.
% CSDfig8p11.m
%
MATLAB
%
simulation
Cruise
R2015a
or
and
PI
cntrl,
later.
use
BC
CSDfig8p11.m
PlotA.m
to
Chang,
Drexel
will
plot
Fig8.11
:
%%
1st
ze=0.707,
Run
Simulation
wn=0.25,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
call
University,
CruisePI.mdl
8/10/2019
to
conduct
246
8
Stability,
Regulation,
and Root Locus
Design
run('Plot3A'),
%%
2nd
Run
Simulation
ze=0.707,
wn=0.5,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
:
run('Plot3B'),
%%
3rd
Run
Simulation
ze=0.707,
wn=1,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
:
run('Plot3C')
Example 8.9 (Road Slope?d = 0.0873rad Disturbance Responsewith ? = 0.707 and Three Values
of ?n = 0.25,0.5,1 rad/s)
The same PI controllers
will be employed
of Example
in the following
8.8, designed
simulation
to
based on the damping ratio
maintain the vehicle
and natural frequency,
speed at the equilibrium
speed
v* = 30 m/sunderthe influence of road slope disturbance ?(t) = ?dus(t), where?d = 0.0873rad.
The three
road slope disturbance
responses
corresponding
to the three
PI controllers
designed
based on ? = 0.707 and three values of ?n = 0.25,0.5,1 rad/s are shown in Figure 8.12. All three
responses
dip right
after the disturbance
occurs, then reverse
and recover
to the
nominal
speed
with
no steady-state error. However, the ?n = 1 rad/s response (in pink) recovers faster than the other two
designs.
All three requires the same
maximum
control force
f(t),
2,600
N, but the pink one demands
f(t) to act faster.
Fig. 8.12: Road slope ?d = 0.0873 rad disturbance response with ? = 0.707 and three values of ?n =
0.25,0.5,1 rad/s.
8.2.4
Robustness to
Model Uncertainty
In general, the dynamics
practical
system.
model employed in the design of control system is not exactly the same asthe
The discrepancy
may come from the unmodeled
dynamics,
uncertainties,
and pertur-bations
due to the change of environment. Therefore, a well-designed control system is required to be
robust against the foreseeable uncertainties and perturbations in the plant dynamics.
Recallthat in the cruise control system design subsection, Section 8.2.2, we employed GE(s) = 1
asthe ideal dynamics model of the engine, which is a good approximation of the second-order engine
dynamics
model,
GE(s) =
1
(ts+1)2
(8.39
8.2
An Automobile
Cruise
Control
Example
247
if the time constant t is close to zero. However,the time constant will not be zeroin reality, and so
it is important to know howrobustthe system will be againstthe variation of t. In the following two
examples,
we will evaluate
Example 8.10 (Robustness
1 rad/s.)
the robustness
of two
PI controller
Evaluation of the PI Controller
designs considered
in Section
8.2.3.
Design Based on ? = 0.707 and ?n =
In this example, we will conduct simulations to evaluate the robustness of the more aggressive con-troller
design considered
in
Example
8.8,
which is the PI controller
designed
based on damping
ratio
? = 0.707 and naturalfrequency ?n = 1rad/s.
The simulations are similar to those conducted in Example 8.8, except that the unmodeled engine
dynamics shown in Equation 8.39is employed to evaluate howthe variation of the time constant t
in Equation 8.39 will affect the closed-loop system performance.
in
Figure
8.13. There are three speed tracking
The simulation
responses shown in the figure.
results are shown
The response curve in blue
color with t = 0 is exactly the same asthe pink responsein Figure 8.11 wherethe engine dynamics
model wasassumedideal, GE(s) = 1. Thet = 0.3sresponseshown in black coloris apparently worse
than the blue one. It has larger overshoot and more oscillations. If the time constant is increased to
t = 0.6 s, the response will deteriorate to the response curve in pink whose oscillation amplitude is
growing
with time; that is, the system
becomes unstable!
Fig. 8.13: Robustnessevaluation of the cruise control system with the PI controller designed based on
? = 0.707 and ?n = 1 rad/s.
The closed-loop system simulation results shown in Figure 8.13 are obtained by running the fol-lowing
MATLAB program: CSDfig8p13.m,
which will automatically call the Simulink
model
program CruisePI.mdl,
to conduct the simulation. The plotting program, Plot3A.m
will also
be called to plot the figures. The MATLAB code is listed asfollows. Note that this program is the
same as CSDfig8p11.m,
except the six lines shown in the following list.
% CSDfig8p13.m
%
MATLAB
%
simulation
Cruise
R2015a
or
and
PI
cntrl,
later.
use
BC
CSDfig8p13.m
Plot3A.m
to
Chang,
Drexel
will
plot
call
Fig8.13
:
%%
1st
tau=0,
%
Simulation
sim_time=8,
Run
Simulation
ze=0.707,
wn=1,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
University,
CruisePI.mdl
8/10/2019
to
conduct
248
8
%%
Stability,
2nd
Run
tau=0.3,
%
Regulation,
and Root Locus
Design
Simulation
ze=0.707,
wn=1,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
wn=1,
ki
=
m*wn2,
kp=m*(2*ze*wn-c_bar),
Simulation
sim_time=8,
:
%%
3rd
Run
tau=0.6,
%
Simulation
ze=0.707,
Simulation
sim_time=8
:
Fig. 8.14: Robustnessevaluation of the cruise control system with the PI controller designed based on
? = 0.707 and ?n = 0.25rad/s.
In the
aggressive
next example,
we will evaluate
PI controller.
Usually
may not have an optimal
a less
the robustness
aggressive
of the cruise
controller
control
has a better
system
robustness
with a less
although
it
performance for the nominal system.
Example 8.11(Robustness Evaluation of the PI Controller Design Basedon ? = 0.707 and ?n =
0.25 rad/s.)
In this example, we will conduct simulations to evaluate the robustness of the less aggressive con-troller
design considered
in
Example
8.8,
which is the PI controller
designed
based on damping
? = 0.707 and natural frequency ?n = 0.25 rad/s. Note that the natural frequency
example is only 25% of that chosen by the aggressive controller in Example 8.10.
ratio
chosen in this
The simulations are similar to those conducted in Example 8.10, except that the PI controller is
designed based on damping ratio ? = 0.707 and natural frequency ?n = 0.25 rad/s. The simulation
results
are shown in
Figure
8.14.
response curve in blue color with t
There are three
speed tracking
responses
shown in the figure.
The
= 0 is exactly the same as the blue response in Figure 8.11 where
the engine dynamics model wasassumedideal, GE(s) = 1. The t = 0.3s and t = 0.6sresponsesin
Figure 8.14 are in black and pink, respectively. Both are worsethan the blue one. However,they remain
stable. This less aggressive (or more conservative) controller design has better robustness against
unmodeled dynamics than the aggressive one.
Remark 8.12 (Trade-Off
Between Performance
and Robustness)
The essential lesson learned from Examples 8.10 and 8.11 is that pushing for high performance
often leads to a non-robust
result. If
we choose to go with the robust
design, that is the least aggressiv
8.2
controller,
we have relatively
poor performance,
An Automobile
as can be seen in
Cruise
Control
Example
Figures 8.11 and 8.12.
249
However, the
controller is morerobust. Performance degrades with the perturbed engine dynamics, but still workable.
In general,
context
the trade-off
between
of the controller
If system parameters
performance
and robustness
design. If high performance
will change with environment
8.2.5 Disturbance Feed-Forward
needs to be evaluated
is desired, then
accurate
or age, then robustness becomes essential.
Compensation
One approach to performance enhancement is to add disturbance feedforward
control.
Disturbance
slope, and taking
feed-forward
is based on the notion
action immediately
to provide a control
in the overall
models are required.
rather than
while using the robust
of sensing the disturbance,
in this case road
waiting to respond to error. The sensed signal is used
command that is added directly to the control input.
Such a configuration
is shown
in Figure 8.15. Including the engine dynamics (GE) andthe feed-forward compensator (Gf), the system
error response is given by
E(s) =
1
V (s)+
1+(1/m)K(s)GE(s)Gp(s)
((1/m)Gf(s)GE(s)-g)Gp(s)
T(s)
(8.40)
1+(1/m)K(s)GE(s)Gp(s)
Fig. 8.15: Cruise control block diagram with disturbance feed-forward compensation.
Notethat the feed-forward transfer function Gf can include control enhancement and/or filter ele-ments
as well as sensor dynamics. Gf will alter the zeros of the disturbance to error transfer function
but
will not affect the
poles.
Consequently,
it
has no effect on system
stability.
Figure 8.16 illustrates
the effect of feed-forward.
The error response
due to
a road
slope step disturbance
(? = 0.707,?n = 0.25rad/s) controller withslow (t
is illustrated
using the less aggressive
= 0.6s) engine dynamics withfeed-forward com-pensation
Gf = 0.5mg (the purple response, notethat mand g arethe massand the gravity, respectively)
and without feed-forward Gf = 0 (the black response). The blue color response, which assumes ideal
engine(t = 0)andnofeed-forwardcompensationGf =0,is includedto showthattheresponsewill de-teriorate,
as shown in black for the slow engine t
= 0.6 s. Whereasfeedback allows the control designer
to alter the system poles that determine the damping ratio and natural frequency, feed-forward does not.
However, feed-forward
does alter the zeros of the transfer function
GET(s) from disturbance to error.
The effect of that, in this case, is to reduce the effect on peak overshoot
and undershoot.
If Gfis chosen as Gf = mg,the disturbance response would have perfect cancellation for the
case of ideal (t = 0 s) engine. However, it would cause more oscillation in disturbance response
for the case of a slow engine with t = 0.6 s. Forthis reason, a compromise choice Gf = 0.5mg was
adopted in the simulation
of Figure 8.16
250
8
Stability,
Regulation,
and Root Locus
Fig. 8.16: Comparison
Design
of the three step responses
due to a road slope disturbance.
8.3 Root Locus Preliminaries
The classical approach to feedback controller
of a particular
of commonly
structure,
used compensator
(i.e.,
design involves two basic steps. First, choose a com-pensator
select a compensator
transfer
functions,
four
of
transfer
which
function).
There are a number
we have already seen: the integral
compensator in Example 5.26, the proportional compensator in Example 8.1, the dual-loop feedback
controller in Example 6.1, and the PI controller in Section 8.2. Each compensator or controller has a
number of free design parameters that need to be determined to meetthe demands of the system
performance goals. The root locus method is an approach to selecting those parameters to achieve
desired transient performance of the closed-loop system.
The root locus
approach
usually
in Figure 8.17, where G(s) is a fixed
dynamics,
begins
with a typical
feedback
control
system structure,
as shown
dynamics model consisting of both the plant and the controller
and Kis a design parameter to be determined
to locate
a desired feasible
pole locations
in the
complex plane. The root locus approach shows a portrait of how the closed-loop system poles move
through the complex plane as the parameter K varies from 0 to 8. The closed-loop characteristic
equation of the following form is called atypical root locus equation.
1+KG(s)
= 1+K
N(s)
D(s)
(8.41a)
=0
where
N(s)
D(s)
=
sm+bm-1sm-1
+ +b1s+b0
(s-z1)(s-z2)(s-zm)
=
sn+an-1sn-1
+ +a1s+a0(s- p1)(sp2) (s-pm)
(8.41b)
Fig. 8.17: Atypical feedback control system block diagram for root locus analysis and design.
Note that the design parameter
usually represents the transfer
function
Kin Equation 8.41 is a scalar parameter instead of K(s), which
of a controller.
In some applications,
the control system structur
8.3
may not
match the structure
shown in
Figure 8.17 and the closed-loop
Root Locus Preliminaries
characteristic
equation is not in
the form of typical root locus equation. In this case, if the closed-loop characteristic
linear
into
function
the form
of the
parameter
of typical
K, then the closed-loop
root locus
equation
as shown
characteristic
in
Example
251
equation
polynomial is a
can be converted
8.13.
Fig. 8.18: A controller design candidate suitable for the root locus approach.
Example
Locus
8.13 (Convert
a Closed-Loop
Characteristic
Equation
Into
the Form
of the
Typical
Root
Equation)
Consider the feedback
control
system
shown in
Figure 8.18.
Recall that this
problem
originating
from the ramp input tracking control problem of Example 8.3. In order to achieve steady-state ramp
function tracking, the loop transfer function needs to include the internal model 1/s2. Butjust setting
the controller to K(s) = 1/s2 is not good enough since the closed-loop system will not be stable. Hence,
the controller is chosen to be K(s) = (s +K)/s2, as shown in Figure 8.18, where Kis a free design
parameter to be determined so that closed-loop system is stable and the transient response is optimized,
subject to control-input constraints.
The closed-loop characteristic equation is
1+
145.5
(s+K)
s2
s+43.14
=0
which is not atypical root locus equation. Butthe characteristic polynomial equation
s3 +43.14s2 +145.5s+145.5K
=0
can be rearranged into
1+K
145.5
s3 +43.14s2 +145.5s
= 1+KG(s)
=0
(8.42)
which now is atypical root locus equation.
Notethat the closed-loop system in Figure 8.18 andthe closed-loop system in Figure 8.17 with G(s)
defined
Equation
by Equation
8.42 are not identical,
8.42 can be employed
Definition 8.14 (Loop Transfer
but they
share the same closed-loop
to place the closed-loop
system
system poles for the system
poles.
Hence,
of Figure 8.18.
Function and LTF Poles and Zeros)
In the root locus equation, Equation 8.41, 1+KG(s)
= 0, KG(s) is referred as the loop transfer
function (LTF), and the poles and zeros ofKG(s) as LTF poles and LTF zeros, respectively.
Remark 8.15 (Loop
Transfer Function and LTF Poles and Zeros)
The KG(s) of the root locus equation, Equation 8.41, 1+KG(s)
= 0, is also referred asthe open-loop
transfer function, and consequently, the poles and zeros of KG(s) are regarded as the open-loo
252
8
Stability,
Regulation,
and Root Locus
poles and zeros in the literature.
Design
These terminologies
make sense for simple
cases like the one shown in
Figure 8.17. Butfor more general caseslike the onein Figure 8.18, and another onein Figure 8.38, which
will be introduced
later in this chapter, these terminologies
will become
misleading and confusing
since
the open-loop poles and zeros in general are not the same asthe LTF poles and zeros of KG(s).
A Trivial
Root Locus Analysis and Design Problem
Before we start to learn how to draw root loci diagrams using the root loci construction rules, we will
consider atrivial second-order system by whichits root loci diagram can be easily constructed andthen
be employed to demonstrate the basic concept of root locus analysis and design. Furthermore, this sim-ple
system can also serve as the first example to verify the root loci construction rules, which will be
introduced shortly.
Considerthe feedback control system in Figure 8.17 with G(s) = 1/(s2+4s+3).
equation is
1
1+K
s2 +4s+3
whosetwo roots are
1
s1
=
s2
2
=0
? s2 +4s+3+K
?-4 ?42-4(3+K)?=-2
Table 8.1: The roots of the characteristic
equation s2
+4s+3+K
The characteristic
=0
v
1-K
(8.43)
= 0 as K varies from
0 to
8.
KK=0 K=0.75K=1 K=1.25K=5 K?8
s1
-1
s2 -3
-1.5
-2 -2+j0.5 -2+j2 -2+j8
-3.5 -2 -2-j0.5-2-j2 -2-j8
The
trajectories
oftheroots
for0 =K<8are
called
theroot
lociofthesystem.
Based
onEquation
8.43 wecan construct Table 8.1 that lists how the values of the two roots vary as K changesfrom 0to 8,
which in turn can be usedto draw the root loci of the system, as shown in Figure 8.19.
From
either
therootlociotrhetable,
it can
beseen
thatwhen
K=0thetworoots
are
s =-1and
s =-3,which
are
thetwopoles
ofG(s)
=1/(s+1)(s+3).
AsKincreases,
the
tworoots
move
onthe
real
axis
towards
each
other
until
theymeet
ats =-2when
K=1. When
Kischanging
fromK=1to
K>1,thedouble
roots
split
intotwocomplex
conjugate
roots
-2 j?. The
point
s =-1iscalled
a
break-out point, and after the roots break away from the real axis, the two roots will continue to stay on
thetwovertical
straight
lines,
one
going
upand
theother
down
alltheway
toinfinity
when
K?8.
This simple root loci diagram can be employed to place the closed-loop system poles at desired
feasible locations. For example, if we would like the closed-loop system to have a damping ratio ? =
0.707, wecan draw a ? = cos45? blue dashline, as shownin Figure 8.19. Thenthe intersection point
willgive
theclosed-loop
system
pole
locations
ats =-2 j2,and
reveal
thedamping
ratio,
natural
frequency, and the value of K.
8.3.1 Root Loci Construction
Rules
Forthe previous example, which is a second-order system with a design parameter K,it is easyto solve
for the two roots as functions
of K and then the trajectories
of the roots (i.e., the root loci)
can be draw
8.3
Root Locus Preliminaries
253
Fig. 8.19: Rootloci of the simple system.
accordingly onthe complex plane. Butthis trivial approach is not viable for high-order systems. Theroot
locus approach is one of the main classical control tools in the analysis and design of control systems.
It provides a set of root loci construction rules that makethe sketch of root loci possible for high order
systems without computing the roots. Moreimportantly, it sheds light on how the pole-zero pattern
of G(s) affects the root loci of 1+KG(s)
= 0, which is very helpful in determining the structure
of the controller. In this subsection we will demonstrate how to apply the root loci construction rules
without explaining the detailed theory.
Consider the feedback control system as shown in Figure 8.20. Theloop transfer function
system is
s+3
K
s(s+1)(s+2)(s+4)
which is the product
of the transfer
functions
of the plant, the controller,
design parameter K is a part of the controller.
and the sensor.
of the
Note that the
Hence,the characteristic equation of the closed-loop
system is
s+3
1+K
= 1+KG(s)
s(s+1)(s+2)(s+4)
which is equivalent to the following
characteristic
equation in polynomial
s(s+1)(s+2)(s+4)+K(s+3)
The objective is to find
from
characteristic
function
K = 0 to
out how the closed-loop
Equations
(8.44)
form,
=0
system
K = 8. Note that the closed-loop
equation shown in
= 0
(8.45)
poles vary on the complex
system
poles are the roots
8.44 and 8.45. They are NOT the
plane as K in-creases
of the closed-loop
poles of the loop transfer
KG(s) unless K = 0.
From Equation 8.45, it can be seenthat at K = 0 the closed-loop system poles are the poles of the
loop
transfer
function,
s =0,-1,-2,-4.Notice
that
in Figure
8.21
each
looptransfer
function
pole
is
indicated
byacrossand
thesingle
zero
oftheloop
transfer
function
isrepresented
byacircle
?at
s =-3
254
8
Stability,
Regulation,
and Root Locus
Design
Fig. 8.20: Anintroductory
root loci construction example.
Fig. 8.21: Rootloci sketch of the introductory
Now, we will sketch the root loci
rules.
The detailed
of the system of Equation
explanation
and theorem
example.
8.44 following
related to these rules
the basic root loci
con-struction
will be given later in the
chapter.
Root Loci Construction
Rules
Rule 1: Number of Branches. The number of branches of the root loci is equal to the number of the
loop transfer function poles.
For the introductory
poles.
example, there are four branches since the loop transfer function
Rule 2: Symmetry. Theroot loci are symmetric
The root loci
are symmetrical
or conjugate complex numbers
hasfour
about the real axis.
with respect to the real axis since the roots are either real numbers
8.3
Root Locus Preliminaries
255
Rule
3:Starting
andEnding
Points.
Each
branch
begins
(K=0)atanLTFpole
and
ends
(K?8)
at an LTF zero or at infinity.
Here,the LTF poles stand for the poles of the loop transfer function
KG(s) in the root locus equation 1+KG(s).
The LTF zeros are defined similarly.
The
start
pointK=0points
areatheLTFpoles
s =0,-1,-2,-4.One
oftheK=8end
point
isattheLTF
zero
s = -3,and
theother
threeK =8endpoints
willbeattheinfinity,
defined
by
the asymptotes given by Rule 5.
Rule 4:
Real Axis Segments. For
point
K> 0, if the total number of real poles and zeros to the right of a
on the real axis is odd, this
As seen from
a very right
point lies
on the root loci.
Figure 8.21, assume a man is
end with no pole or zero behind
walking
him.
on the real axis line toward the left from
There would be no root loci
until he encounters
the first pole or zero to begin with the first root loci segment on the real axis. Thefirst segment will
end
when the person
successively
meets the second
until the negative infinity
pole or zero.
Hence, the root loci
will alternate
on and off
end of the real axis. For this example, there are three real axis
segments:
between
s =0and
s =-1,between
s =-2and
s =-3,and
from
s =-4alltheway
to
s =-8.
Rule 5:
Behavior at Infinity.
The root loci branches that tend to infinity
do so along asymptotes
with
angles ?, and these asymptotes all intersect at a common point s, on the real axis as defined
by:
?? =
?
(2?+1)p
n-m , ?=0,1,2,,n-m-1;
s
pj -?z
-an-1=
= bm-1
n-m
n-m
j
i
Forthis introductory example, there will be three asymptotes. The angles of these three asymp-4-1
totes and the intersection of these asymptotes with the real axis are
?0 = p
3 =60?,
4-1=p
s = 0+(-1)+(-2)+(-4)-(-3)
4-1
The three asymptotes
=p =180?,
?2=5p
4-1=5p
3 =-60?
?1 = 3p
= -4
3
=-1.33
are shown in dashed yellow lines in the graph.
Rule 6: Real Axis BreakOut and BreakIn Points. The root locus breaks out from the real axis
where the gain Kis a (local) maximum on the real axis, and breaks in to the real axis where K
is a (local) minimum.
Tolocate candidate break points simply solve
d
ds
??
1
G(s)
=0
or
N(s)
dD(s)
ds
-
dN(s)
ds
D(s) = 0
where G(s) = N(s)/D(s), andthen find the value of K atthe break point by
K =
In Figure 8.21, the root loci
segments
1
|G(s)|
on the second and the third real axis segments
for all values of K. The only break-out
point
will stay on their respec-tive
will occur at the first real axis segment
between
s =0and
s =-1.Since
N(s)=s+3andD(s)
=s4+7s3
+14s2
+8s,
we
have
N(s)
dD(s)
ds
-
dN(s)
ds
D(s)
=0 ? 3s4
+26s3
+77s2
+84s+24
=0
and
itsroots
at-3.311
j0.68124,
-1.6097,
and-0.43492.
Hence,
thebreak-out
point
occurs
at
256
8
Stability,
Regulation,
and Root Locus
Design
s =-0.43492.
The value of K at the break-out point can be computed from the following.
1
?
K =?
?
?
?
?=0.53459
?
G(-0.43492)
?
Rule 7: Imaginary Axis Crossings. Usethe Routh stability test to determine values of Kfor
loci cross the imaginary axis.
The characteristic
equation
of Equation
8.45 can be rewritten
s4 +7s3 +14s2 +(8+K)s+3K
as the following
polynomial
which
equa-tion:
(8.46
=0
The Routh array for this characteristic equation is constructed asfollows:
s4
1
14
s3
7
8+K
s2
b1
3K
s1
c1
s0
3K
3K
7(90-K)
? 90-K
K2+65K-720
b1 = 1
where
c1 = -1 ?
?
According to the Routh-Hurwitz criterion, the system is stable if and only if the first-column ele-ments
of the Routh array are all strictly positive. Hence,the system is stable if and only if all the
following
three inequalities
are satisfied:
K>0, 90-K>0, and K2+65K-720
<0
which is equivalent to
0
Hence, when K = 9.6456, two
<K<
of the root loci
9.6456
branches cross the imaginary
axis.
The intersection
points can be computed from the auxiliary equation,
b1s2
+3K=11.4792s2
+28.9368
=0 ? s =j1.588
Therefore,
when
K=9.6456
theroot
lociintersect
theimaginary
axis
ast = j1.588.
Up
tonow,
withthe information
obtained from Rule 1to Rule 7, weare able to complete the root loci diagram
sketch in Figure 8.21 for the introductory
Rule 8:
root loci
example
without the need of Rule 8.
Angle of Departure from Poles or Angle of Arrival to Zeros. Assumes is at the vicinity
of
anyLTFpole
orzero,
pz,
then
theangle
ofdeparture
fromortheangle
ofarrival
topz,?s(
pz), can be computed from the following
m
equation:
n
G(
s) = ??(
s-zi)-??( s- pj) =(2?+1)p 0 =K<8
i=1
j=1
Example 8.16 (Angle of Departure from a Complex Pole on a Root Loci)
Assume the LTF poles and zero shown
on the root loci
diagram
of Figure 8.22 are p1 = 0, p2 = j1,
p3=-j1,p4=-3,and
z1=-1,and
acomplex
number
s isatthevicinity
ofp2.Then
theangle
ofdeparture
fromp2,?(
s- p2)can
becomputed
from
8.3
Root Locus Preliminaries
257
Fig. 8.22: An example demonstrating angle of departure.
?(
s-z1)-?(s- p1)-?(
s- p2)-?(
s- p3)-?(
s- p4)=(2?+1)p
which can be rewritten as follows:
?(
s- p2)= ?(p2
-z1)-?(p2
- p1)-?(p2
- p3)-?(p2
- p4)-(2?+1)p
Noticethat thes on the right-hand side of the equationhas beenreplaced by p2sinces and p2 are
virtually
the same point observed from far away. Therefore, the angle of departure from
p2 is
?(
s- p2)=45?-90?-90?-18.4?
+180?
=26.6?
With this LTF pole-zero
pattern, there
are four root loci
branches.
One started from
p1 = 0 moving
totheleftand
ended
atthezero
z1=-1.The
second
branch
begins
atp4=-3,heading
tothe
left on the negative real axis all the wayto the 180? asymptote. The other two of the four branches
willdepart
frompoles
p2=j1andp3= -j1with
theangles
ofdeparture,
26.6?
and-26.6?,
respectively,
into the right
half plane.
Since the other two
of the three asymptotes
are heading into
therighthalfofthecomplex
plane
with60?angles,
thetwobranches
leaving
fromp2=j1 and
p3=-j1 willmerge
tothetwoasymptotes,
and
have
nochance
toturnaround
going
back
totheleft
half of the complex plane. Hence, there exists no value of K > 0 so that the closed-loop system
can be stable. We will address this issue later in Section 8.5 regarding a sinusoidal position
tracking problem. One remedy is to add LTF zeros on the negative real axis to change the
departure angle and the asymptotes.
For the introductory
example, all the LTF poles and zero are on the three real axis segments. It
is easyto see onthe first segmentthe angle of departurefrom the pole at s = 0 will be 180?since
the root trajectory
of this branch hasto go to the left to cover this
part of root loci
according
to
Rule
4.For
thesame
reason,
theangle
ofdeparture
forthepole
ats =-1 willbe0?.
Inthefollowing
example, we will verify the departure angles of these two poles.
Example 8.17 (Angle of Departure from the Real Poles on the Introductory
The LTF pole-zero
pattern of the introductory
root loci
example is shown
Root Loci)
again in Figure 8.23,
where
p1=0,p2=-1,p3=-2,p4=-4,and
z1=-3,and
acomplex
number
sisatthevicinity
ofp2.
Then
theangle
ofdeparture
fromp2,?s( p2)
can
becomputed
from
thefollowing
equation
258
8
Stability,
Regulation,
and Root Locus
Design
?(
s-z1)-?(s- p1)-?(
s- p2)-?(
s- p3)-?(
s- p4)=(2?+1)p
which can be rewritten as follows,
?(
s- p2)= ?(p2-z1)-?(p2- p1)-?(p2
- p3)-?(p2-p4)-(2?+1)p
Fig. 8.23: Angle of departure for the introductory
example.
Noticethat thes on the right-hand side of the equation has been replaced by p2 sinces and p2 are
virtually
the same point seen from far away. Therefore the angle of departure from
p2 is
?(
s- p2)=0-p-0-0+p =0
which
means the root loci
departs from the pole p2to the right.
Ifs is chosen to be at the vicinity of z1,then the angle of arrival to z1 will be
?(
s-z1) =?(z1
- p1)+?(z1
-p2)+?(z1
-p3)+?(z1
-p4)-(2?+1)p
and
?(
s-z1)=p+p+p+0-3p=0
Hence,
the angleof arrivalto the zeroz1is 0?, which means
the root loci comesfrom theright to
arrive at z1.
8.3.2 Root Loci Construction
Using MATLAB
In applications, the root locus plot can be generated by computation using established functions like
MATLABs
rlocus.
The two graphs in Figure 8.24 show the same root locus plot generated using the following
code:
% CSD
Fig8.24
Use
rlocus(G,K)
for
the
intro
MAT-LAB
example
s=tf('s');
G=(s+3)/((s)*(s+1)*(s+2)*(s+4));
figure(7)
K=logspace(-3,2,1000);
rlocus(G,K)
The graph on the left shows the root loci
the sgrid command
on the
Command
with the grid on. The grid can be turned
Window or simply
by a right
click of the
on by executing
mouse over the root loci
graph to pop up a small window manual wherethe grid can be turned on or off. The grid shows the
constant damping ratio (?) lines and the constant natural frequency (?n) circles. The black squar
8.3
Fig. 8.24: The root loci of the introductory
Root Locus Preliminaries
259
example plotted using MATLAB rlocus.
cursor that appeared on the root loci was obtained by aleft click on the root loci. The black square
cursor can befurther dragged along the root loci to a position of interest.
Whilethe black cursor
is moving, a small display panel also moves with it, showing the associated pole location, value
of K, damping ratio, and natural frequency.
Onthe left graph, the black cursor is positioned at the
intersection
oftheroot
lociand
theimaginary
axis,
atwhich
thecomplex
conjugate
poles
arej1.59
and the gain K = 9.65, and the damping ratio and the natural frequency are ? = 0 and ?n = 1.59 rad/s,
respectively. If the black cursor is movedto the break-out point, the display panel will show that the
double
poles
ats =-0.435,
thegainK=0.534,
thedamping
ratio
? =1,and
thenatural
frequency
?n = 0.435rad/s.
The graph on the right shows the same root loci but with the grid off. The black cursor is
moved
totheposition
where
thecomplex
conjugate
poles
are
at-0.398j0.398,
and
thegainK=0.923.
At these pole locations, the corresponding damping ratio and the natural frequency are ? = 0.707 and
?n = 0.563 rad/s.
By executing the MATLABs command [K,Poles]=rlofind(G),
a new cursor consisting
of a long horizontal line and a long vertical line appears to hover over the entire Root Locus
window.
Bypositioning
thecursor,
the
intersection
ofthese
twolong
lines,
atthepoint
-0.398+
j0.398
on the loci marked by the previous black square cursor, the value of the gain Kand all associated closed-loop
poles are identified, as shown in the following:
%
>>
CSD
Fig8.24
Use
rlofind
[K,Poles]=rlocfind(G)
point
in
-3.9777e-01
K
the
graphics
+
to
all
a
window
selected_point
3.9790e-01i
=
9.2300e-01
Poles
locate
Select
=
-4.0383e+00
+
0.0000e+00i
-2.1661e+00
+
0.0000e+00i
-3.9777e-01
+
3.9790e-01i
-3.9777e-01
-
3.9790e-01
poles
=
260
8
Stability,
Regulation,
It also can be seen that
and Root Locus
Design
a red cross appears on each branch of the loci
at all of the identified
poles. If
thevalue
ofK=0.923
ischosen
inthecontroller
design,
thetwopoles
-0.398j0.398
willbethe
dominant
ones to
determine
the
ratio and natural frequency
A Simulink
simulation
behavior
of the closed-loop
system,
and the
associated
damping
will be ? = 0.707 and ?n = 0.563 rad/s, respectively.
program is built based on the block
of the feedback
system
shown in Figure
8.20. The step response simulation
When Kis chosen to be K = 0.923, the step response is underdamped with about 5%
system
feedback
which occurs at t
with the same damping ratio
damped. If
step response
function
K = 0.534 are shown
with the step response
and natural frequency,
control system has only about 1% higher in
faster in the peaktime.
critically
= 6.5 s. Comparing
with K = 0.923 and
control
in Figure 8.25.
maximum overshoot,
results
diagram
of the typical
the step response
maximum overshoot
second-order
of this fourth-order
and is only about 1.5 seconds
Onthe other hand, when Kis chosen to be K = 0.534, the step response will be
Kis chosen between these two values, the step response
waveforms. If
Kis chosen to be K = 9.65, the step response
will be between these two
will be undamped
sinusoidal
with oscillation frequency equal to ?n = 1.59 rad/s.
Fig. 8.25: The step response
simulation
results
with K = 0.923 and
K = 0.534.
The two step response graphs in Figure 8.25 are obtained from running the generated using the
following
MATLAB program:
%
CSDfig8p25.m
%
Design
A:
K=0.923,
damping
ratio
0.707
%
Design
B:
K=0.534,
damping
ratio
1,
%%
5/08/2020
Intro
Loci
Example
critically
damped
System
s=tf('s');
%%
G1=(s+3)/s;
Design
K=0.923,
A
step
G2=1/((s+2)*(s+4));
t=linspace(0,20,201);
GR=(K*G1*G2)/(1+K*G1*G2*G3);
R=tf(1,1);
[y,t]=step(GR,t);
%%
Design
K=0.534,
[r,t]=step(R,t);
[u,t]=step(GU,t);
B
step
run('plot2a')
response
t=linspace(0,20,201);
GR=(K*G1*G2)/(1+K*G1*G2*G3);
GU=(K*G1)/(1+K*G1*G2*G3);
[y,t]=step(GR,t);
where plot2a.m
G3=1/(s+1);
response
GU=(K*G1)/(1+K*G1*G2*G3);
%
Root
R=tf(1,1);
[r,t]=step(R,t);
[u,t]=step(GU,t);
run('plot2b')
is given in the following,
and plot2b.m
is the same except replacing
plot2a.m
figure(5),
grid
subplot(1,2,1),
minor,
plot(t,u,'b-'),
title('r
plot(t,r,'r-',
&
grid
on,
y'),
hold
grid
t,y,'b-'),
on,
minor,
grid
subplot(1,2,2),
title('u'),
hold
o
on,
b-by
k--.
8.4
Root Locus
Analysis and
Design
261
8.4 Root Locus Analysis and Design
Notwithstanding the ease of creating a root locus plot using a computer, intelligent use of root locus as
a design tool requires understanding the underlying nature of the root loci behavior. Only with this un-derstanding
can we manipulate the loci by selecting the right compensator and adjusting its parameters.
Recall,in the example of the previous section, weconsidered a system with a PI compensator and exam-ined
the root loci with respect to the gain parameter K. In fact, the PI compensator hastwo parameters,
the gain and the zero location.
The root locus is defined in terms
could choose the zero location
rather than the gain.
of a single parameter. It is true that
we
Or we could develop two separate plots, one for each
parameter. Butthen we would haveto sort outthe coupling between the two. It is generally bestto focus
on the gain K asthe root locus parameter and use our underlying knowledge of root locus behavior to
choose the correct compensator
it
would affect the loci.
and
manipulate its other parameters according
To this end, in this section
Our ultimate goal is to usethe root locus
locations.
So, what are acceptable
we explore the basic theory
to our knowledge
of the root locus
of how
process.
method as a way to achieve acceptable closed-loop pole
pole locations?
With the root locus
method we do not specify
specific
desired pole locations. Instead weidentify an acceptable region in the complex plane within which, with
some exceptions,
region
we wish all poles to reside.
may be specified.
Figure 8.26 shows
Note that any pole residing
ratio ? and damping factor
one simple
within the shaded region
and effective
way such a
will have a better damping
a.
Considerthe feedback loop of Figure 8.27, wherethe loop transfer function is
KG(s)
= K
N(s)
D(s)
sm+bm-1sm-1
+ +b0 = K(s-z1) (s-zm)
sn+an-1sn-1
+ +a0
(s-p1) (s-pn)
= K
(8.47)
Here, weassumethat N(s) and D(s) are completely known but Kis a parameter wecan adjust. Thus, we
know the poles and zeros of the loop transfer function. Ourintent is to determine how the closed-loop
poles vary as we manipulate
Definition 8.18 (The
K. For this
purpose,
we define the root locus
problem
as follows.
Root Locus Problem)
Generate a sketch in the complex plane ofthe closed-loop pole trajectories
parameter K. Thesetrajectories are referred to asthe root loci.
as a function ofthe gain
Fig. 8.26: The goal is to design a compensator such that the closed-loop system poles lie in the shaded
region. Notethat the region is characterized by two parameters:the damping ratio ? (or ?) andthe
dampingfactor a
262
8
Stability,
Regulation,
and Root Locus
Design
Fig. 8.27: Atypical feedback control system block diagram for root locus analysis and design.
8.4.1 The Root Locus
The closed-loop
Method
system poles are the roots (or zeros) of
1+KG(s)
Equation
8.48 can be expressed
(8.48)
=0
as
D(s)+KN(s)
=0
(8.49)
or
KG(s) = K
N(s)
D(s)
(8.50)
=-1
Moreover,
-1=ej(2?+1)p
forallintegers
?.This
establishes
thefollowing
result.
Theorem 8.19 (Magnitude
and Angle Equations)
Due
toEquation
8.50
and-1=ej(2?+1)p,
forallintegers
?,theclosed-loop
poles
satisfy
KG(s) = K
which can be decomposed
N(s)
D(s)
into two equations:
=ej(2?+1)p,
?=0,1,2,
the
magnitude
(8.51)
equation,
m
1
|G(s)|= K
?
?|s-zi|
n
i=1
?
1
= K, 0 =K<8
?spj
?
(8.52)
?
?
j=1
and the angle equation,
m
?G(s)
= (2?+1)p
n
??(s-zi)- ? ?(s-pj) =(2?+1)p,0 =K<8
?
i=1
(8.53)
j=1
According to Theorem 8.19, if a point s is on the root loci it has to satisfy the angle equation,
Equation 8.53 and its associated gain Kshould satisfy the magnitude equation, Equation 8.52.
Example
8.20 (Application
The introductory
root loci
of the
Angle Formula)
example
discussed in Section 8.3.1 is employed
here to illustrate
formula of Theorem 8.19. Recallthat the loop transfer function of the system i
the angle
8.4
Root Locus
Analysis and
Design
263
Fig. 8.28: A graphical view of the application of the angle formula.
s+3
KG(s) = K
s(s+1)(s+2)(s+4)
and its associated root loci diagram wasgiven in Figure 8.24.
Thegraphontheleft of Figure8.28illustratesthetest points =j1.588satisfiesthe angleequation
and therefore is on the root loci.
The magnitude equation
will also be used to compute the associ-ated
gain
K,which
issupposed
tobeK=9.65.
For
thegraph
ontheright,
thetestpoint
iss =-2+j2,
which is not on the root loci,
Fors
should
not satisfy the angle equation.
=j1.588, the algebraic sum of the angles is
?(
s-z1)-(?(s- p1)+?(
s- p2)+?(
s- p3)+?(
s- p4))
tan-1(1.588/0)+tan-1(1.588/1)+tan-1(1.588/2)+tan-1(1.588/4)
?
=tan-1(1.588/3)=27.894?
-(90?
+57.800?
+38.450?
+21.653?)
=27.894?
-207.903?
=180.009?
whichsatisfiesthe angleequation,Equation8.53.Therefore,
the points =j1.588indeedis onthe root
loci. Furthermore, from the magnitude equation, Equation 8.52, we have
K =|s||s+1||s+2||s+4|
|s+3|
=|j1.588||j1.588+1||
j1.588+2||
j1.588+4|
|j1.588+3|
= (1.588)(1.8766)(2.5538)(4.3037)
= 9.649
3.3944
which matchesthe gain value K = 9.65 obtained in Section 8.3.1.
For
s =-2+j2,thealgebraic
sum
oftheangles
is
?(
s-z1)-(?(s- p1)+?(
s- p2)+?(
s- p3)+?(
s- p4))
?tan-1(2/-2)+tan-1(2/-1)+tan-1(2/0)+tan-1(2/2)
?
=tan-1(2/1)=63.435?
-(135?
+116.565?
+90?
+45?)
=-323.13?
=36.87?
?=(2?+1)180?
which
does
notsatisfy
theangle
equation
Equation
8.53,
and
therefore
thispoint
s =-2+j2isnoton
the root loci.
8.4.2 Explaining the Root Loci Construction
In Section 8.3.1,
welisted the root loci construction
Rules
rules and employed them to construct the root loci
a system with four poles and one zero in Figure 8.21. In the following,
behind these rules.
of
we will investigate the reasoning
264
8
Rule 1:
Stability,
Regulation,
Number of
and Root Locus
Branches.
Design
The number of branches
of the root locus is n, which is the number of
LTF poles (the poles of the loop transfer function KG(s)). This follows from the fact that the order
of the polynomial N(s)+KD(s)
is the order of D(s).
Rule 2: Symmetry. The root locus is symmetric
complex poles occur in conjugate pairs.
about the real axis.This follows from the fact that
Rule
3:Starting
andEnding
Points.
Each
locibegins
(K =0)atanLTF
pole
and
ends
(K?8)atan
LTF zero or atinfinity.
This is easily established by noting the following two facts:
D(s)+KN(s)
=0 ? D(s)
=0 as K?0
KD(s)+N(s)
=0 ? N(s)
=0 as K?8 ifsisbounded
1
Rule 4: Real Axis Segments. For K > 0, real axis segments to the left of an odd number of poles and
zeros are part of the root locus.
Proof: According to the angle equation, Equation8.53, assume
s is a point onthe real axis;then
theangles
?(
s-zi) or?(
s-p j)from
theconjugate
poles
orzeros
willcancel
each
other
outand
the
anglesfromthe polesor zerosontheleft ofs will bezero. Hence,onlythe real polesandzeroson
the right-hand side ofs needto becountedin usingthe angle equation.If the number of the poles
and zeros on the right-hand side ofs is odd, then the algebraic sum of the angles is (2?+1)p,
and therefore the points is part of the root loci. Otherwise, it is not on the root loci.
Rule 5:
Behavior
at Infinity.
The root locus
branches that tend to infinity
do so along asymptotes
with
angles ?, andthese asymptotesall intersect at a common point s onthe real axis as definedby
?? =
(2?+1)p
n-m , ? =0,1,2,,n-m-1;
s
zi
-?
-an-1=
= bm-1
i
+? pj
j
n-m
n-m
(8.54
where
bm-1,
an-1,
n,m,
zi,andpj aregiven
in Equation
8.47.
Proof: Recallthe closed-loop characteristic equation:
D(s)+KN(s) = 0
?
D(s)
N(s)
=-K
Weare interested in the situation wheresis a very large complex number. It is easier to consider the
very small complex number e = 1/s, so rewriting in terms of e, we obtain the approximations
1?
1
?
D(s)=sn+an-1sn-1
+ +a0 =en(1+an-1e
+ +a0en)=en 1+an-1e+O(e2)
where
an-1=-(p1 + + pn)
Similarly
N(s) =
1?
?
em 1+bm-1e+O(e2)
where
bm-1=-(z1+ +zm)
Now, transforming
back to s,
D(s)
N(s)
sn-m
an-1
+s
bm-1
+s =-K
8.4
Root Locus
Analysis and
Design
265
-an-1 ???
? bm-1
=(-K)1/(n-m)
(m-n)s
so we can write
s
1+
1
+O
or
s+
s2
bm-1
-an-1=(-K)1/(n-m)
(m-n)
Now,
write
sass ?s+?e
j?and
substitute
s+?e j? +
bm-1
-an-1= K1/(n-m)ej(p+?2p)/(n-m)
(m-n)
which leads to
s =
bm-1
-an-1
, ? =K1/(n-m)
??=(2?+1)p/(n-m)
(n-m)
This meansthat each asymptote is a straight line, which may be viewed as a linear function of the
parameter ?. As ? varies from 0to 8, each asymptote evolves from the point s along a line at one
of the possible angles ??.
Rule 6:
Real Axis Break-Out
the gain Kis a (local)
and
Break-In
Points.
The root locus
breaks out from the real axis
where
maximum on the real axis and breaks in to the real axis where Kis a (local)
minimum.
Tolocate candidate break points simply solve
??
d
1
G(s)
ds
where G(s) = N(s)/D(s),
=0
or
N(s)
cross the imaginary
Rule 8:
ds
-
ds
(8.55)
D(s) = 0
and then find the value of K at the break point by
K =
Rule 7: Imaginary-Axis
dN(s)
dD(s)
1
|G(s)|
Crossings. Usethe Routh stability test to determine values of Kfor whichloci
axis.
Angle of Departure from Poles or Angle of Arrival to Zeros. Assumes is at the vicinity
of
any
LTF
pole
orzero,
pz;
then
theangle
ofdeparture
fromortheangle
ofarrival
topz,?s( pz),
can be computed from the following
m
equation:
n
G(
s) = ??(
s-zi)-??( s- pj) =(2?+1)p 0 =K<8
i=1
Remark 8.21 (Angle
(8.56)
j=1
of Departure and Arrival)
The equation, Equation 8.56, employed to compute the angle of departure or arrival from the poles or
to the zeros, is basically the same as the angle equation
angle of departure (or arrival) in root locus analysis
of Equation
8.53. In fact, the computation
and design is one of the two important
of the
applications
of the angle equation. Thetest points is assumedin the vicinity of pz, whichis a pole or zero of
interest,
but
itsorientation
?(
s- pz)
isunknown.
Since
s and
pizssoclose,
they
can
beconsidere
266
8
virtually
Stability,
Regulation,
the same point,
and Root Locus
as observed
Design
from
the
rest
of the
poles and zeros.
Therefore in
Equation
8.56,
every
s except
theone
in ?(
s- pz)willbereplaced
bypztosimplify
thecomputation.
Acouple
of
angle of departure (arrival)
important
application
computations
were illustrated
of the angle equation
in
Example
was demonstrated
in
8.16 and Example
Example
8.17.
Another
8.20 to determine
whether a
given test points is on the root loci.
8.5 A Sinusoidal Position Tracking Control System
In this section,
steady-state
a feedback
sinusoidal
controller
position
will be designed
tracking
for
and a desired
a DC
transient
motor system
response.
to achieve
a perfect
The dynamics
model of
the plant, whichis a DC motorsystem with geartrain and load, is given by
145.5
T(s)
U(s)
=
(8.57)
s(s+43.14)
from Equation 5.68in Chapter 5. The reference input is
?R(t) = Asin(?0t +f)
(8.58)
wherethe frequency is ?0 = 10 rad/s, but the amplitude A and the phase f are arbitrary. The
objective is to design a controller so that the closed-loopsystemis stable and the plant output ?(t),
the angular displacement, willfollow the sinusoidal referenceinput ?R(t) as closely as possible.
According to the internal
model principle (Remark 8.4), the loop transfer function needsto havethe
internal model1/(s2 +102)in it to matchthe reference sinusoidal function frequency ?0 = 10rad/s.
Thus, the structure
of the controller
is chosen to be of the form,
(s+c1)(s+c2)
K
s2
(8.59)
+102
Notice
thattwozeros,
s = -c1ands = -c2,areadded
tothecontroller
in Figure
8.29,where
c1
and c2 are positive real numbers to be determined in the design process. Thereis a practical reason for
adding these two zeros to the loop transfer function. It can be explained from the root locus diagrams in
Figure 8.30.
Without adding the two zeros, the loop transfer
function
of the system would be
145.5
KG(s)
= K
s(s+43.14)(s2
+102)
which
gives
fourpoles
atp1=0,p2=j10,p3=-j10,
andp4=-43.14,
butnozeros.
The
root
loci
of the closed-loop characteristic
equation 1+KG(s)
= 0 are shown on the left graph of Figure 8.30.
The
fourbranches
ofroot
locibegin
atthefourpole
locations
when
K=0,and
eventually,
asK?8,
they will approach the infinity locations defined by the four asymptotes, respectively. The angles of the
four asymptotes
and their
common intersection
?0 = 4
p,
point are
?1 = 3p
4,
4 , ?2 = 5p
4 = -3p
?3 =7p
4 = -p
4
s = 0+(j10)+(-j10)+(-43.14)
4
=-10.79
There is only one real root loci
segment,
which is between the two
LTF
poles p1 = 0 and p4
=
-43.14.
The
twobranches
starting
fromp1=0andp4=-43.14
willmove
toward
each
other
andmeet
atthebreak-out
point
s =-31.85,
which
isone
otfheroots
oftheequation,
Equation
8.55
8.5
dD(s)/ds
A Sinusoidal
Position
= 4s3 +129.42s2 +200s+4314
Tracking
Control
System
267
=0
After breakout, the two branches will become complex and movetowards the two asymptotes with
angles
135?
and-135?.
The other two root loci
branches
will start at the two
LTF poles on the imaginary
axis,
p2 = j10,
p3= -j10,when
K=0.AsKincreases,
theroot
lociwilldepart
from
these
twopoles.
The
angle
of
departure
computed
based on Rule 8 or Equation
8.56
will reveal
which direction the root loci
move. The angle of departure from the pole p2 can be obtained from the following
would
equation:
-(?(s- p1)+?(
s- p2)+?(
s- p3)+?(
s- p4))=(2?+1)p
? -(?(p2
-p1)+?(
s- p2)+?(p2
-p3)+?(p2
- p4))=(2?+1)p
? =(2?+1)p
s- p2)+(p/2)+tan-1(10/43.14)
? -?(p/2)+?(
? -(?(
s- p2)+13?)
=0? ? ?(
s- p2)=-13?
Fig. 8.29:
Fig. 8.30: Root locus
Hence, the root loci
diagrams
A sinusoidal
explaining
will depart from
position tracking
control system.
why the two zeros are added to the controller
p2into the right
half of the complex
plane
structure.
with the departure
angle
?(
s - p2)= -13?.
Then
thisbranch
ofroot
lociwillcontinue
to move
intherighthalfplane
towardsthe asymptote withangle ?0 = 45?. The branchoriginating from p3 will also moveinto the right
halfplane
with
departure
angle
?(
s-p3)=13?
and
then
approach
theasymptote
with
angle
?3=-45?
268
8
Stability,
These two
value.
branches
Regulation,
and Root Locus
Design
will never come to the left
Therefore, the controller
half of the complex
plane by changing the
K > 0 gain
structure
1
K
s2 +102
does not work since the closed-loop system is not stable for any value of
K> 0.
Next,
wewill
investigate
if adding
azero
s =-10
tothecontroller
would
help.
Byadding
thiszero,
function will become
the loop transfer
145.5(s+10)
KG(s) = K
s(s+43.14)(s2
+102)
which
stillkeeps
thesame
fourpoles
atp1=0,p2=j10,p3=-j10,p4= -43.14.
The
root
loci
of the closed-loop characteristic
equation
1+KG(s)
= 0 are shown on the right graph of Figure
8.30.
The
fourbranches
ofroot
locibegin
atthefourLTF
pole
locations
when
K=0,and
asK?8,
thebranch
originating
fromp1=0 willarrive
attheLTF
zero,
z1=-10,while
theother
three
branches
will movetowards the infinity locations defined by the three asymptotes, respectively. The angles of the
three asymptotes and their common intersection point are
?0 = 3,
p
?1 = p,
?2 = 5p
3 = -p
3
s = -(-10)+0+(j10)+(-j10)+(-43.14)
3
There are two real root loci
segments;
one starts from
=-11.05
p1 = 0 moving toward the left
and ends up at
z1=-10when
K=8,and
theother
starts
fromp4=-43.14
moving
totheleftonthenegative
real
axis, whichis the asymptote withangle ?1 = p.
The other two root loci branches will start at the two LTF poles on the imaginary axis, p2 = j10,
p3= -j10,when
K=0.AsKincreases,
theroot
lociwilldepart
from
these
twopoles.
The
angle
of
departure computed based on Rule 8 or Equation 8.56 will reveal which direction the root loci
move. The angle of departure from the pole p2 can be obtained from the following equation:
would
?(
s-z1)-(?(s- p1)+?(
s- p2)+?(
s- p3)+?(
s- p4))=(2?+1)p
? ?(p2
-z1)-(?(p2
-p1)+?(
s- p2)+?(p2
- p3)+?(p2
-p4))=(2?+1)p
? =(2?+1)p
s- p2)+(p/2)+tan-1(10/43.14)
? (p/4)-?(p/2)+?(
? 45?
-(?(
s- p2)+13?)
=0? ? ?(
s- p2)=32?
Hence,the root loci
will depart from p2into the right half of the complex plane with the departure
angle
?(
s-p2)=32?.
Then
thisbranch
ofroot
lociwillcontinue
to move
intheright
halfplane
towards
the asymptotewithangle?0 = 60?.Thebranchoriginatingfrom p3 will also moveinto the right half
plane
with
departure
angle
?(
s- p3)=-32?
and
then
approach
theasymptote
with
angle
?2=-60?.
These two
value or by
branches
will never come to the left
half of the complex
moving the added zero z1to anywhere
Therefore, the controller
plane by changing the
K > 0 gain
on the real axis.
structure
K
s+c
s2 +102
still
does not
work since the closed-loop
system is
not stable for
any value
of K > 0 and any value
ofc.
Nevertheless,
byadding
thetwozeros,
s =-c1and
s =-c2,
tothecontroller,
asshown
in
Equation 8.59, and with the zero locations carefully selected, the angles of departure from th
8.5
poles on the imaginary
axis can be
the complex plane. Furthermore,
the basic root locus
consideration
managed so that
Position
the root loci
Tracking
Control
will depart into
System
269
the left
half of
the asymptotes can also be confined in the left half plane.
and the sinusoidal
is selected so that the loop transfer
A Sinusoidal
function
tracking
objective in
mind, the controller
With
structure
is
145.5(s+1)(s+2)
KG(s) = K
s(s+43.14)(s2
+102)
(8.60)
which
stillhas
fourLTFpoles
atp1=0,p2=j10,p3= -j10,andp4=-43.14,
butwith
two
added
LTFzeros
atz1=-1andz1= -2.The
rootlocioftheclosed-loop
characteristic
equa-tion
1+KG(s)
= 0 are shown on Figure 8.31. The graph on the left of the figure shows the entire root
loci diagram, but around the origin area two zeros and one pole are crowded together. The graph on the
right provides
an expansion
Fig. 8.31:
view of the area close to the imaginary
Root locus
There are two real root loci
design for sinusoidal
segments
axis and the origin.
position tracking
on the real axis.
One starts from
control system.
p1 = 0, moving to the left,
and
ends
atz1= -1,andanother
starts
fromp4= -43.14,
moving
toward
theright,and
terminates
atz2=-2.Theother
twobranches
ofrootlocibegin
attheimaginary
pole
locations
p2=j10and
p3= -j10when
K=0,andasK ? 8,theywillapproach
theinfinity
locations
defined
bythetwo
asymptotes, respectively. The angles of the two asymptotes andtheir common intersection point are
?0 = p
2,
?1 = -p
2
(8.61)
s = -[(-1)+(-2)]+[0+( j10)+(-j10)+(-43.14)]
4-2
Notice that the two asymptotes completely
reside in the left
=-20.07
half of the complex
plane. It is
mainly
theresult
ofadding
thetwoLTF
zeros,
z =-1and
z =-2,tothecontroller
structure.
The
addition of the two zeros also affects the angle of departure in a big wayfrom the following
of the angle of departure from p2 = j10:
computation
?(
s-z1)+?(
s-z2)-(?(s- p1)+?(
s- p2)+?(
s- p3)+?(
s- p4))=(2?+1)p
? ?(p2
-z1)+?(p2
-z2)-(?(p2
-p1)+?(
s- p2)+?(p2
-p3)+?(p2
-p4))=(2?+1)p
?(p/2)+?(
? =(2?+1)p
s- p2)+(p/2)+tan-1(10/43.14)
? tan-1(10/1)+tan-1(10/2)-
? 84.29?
+78.69?
-(?(
s- p2)+13?)
=0? ? ?(
s- p2)=150?
(8.62
270
8
Stability,
Regulation,
Hence, the root loci
and Root Locus
will depart from
Design
p2 into the left
half of the complex
plane
with the departure
angle
?(
s-p2)=150?.
This
branch
ofroot
lociwillcontinue
to move
inthe
lefthalfplane
toward
theasymp-tote
withangle?0 =90?.Similarly,the branchoriginatingfrom p3 willalso move
intothe left half plane
with
departure
angle
?(
s- p3)=-150?
and
then
approach
totheasymptote
with
angle
?1=-90?.
Notice
thatfromEquation
8.62,
adding
thetwozeros,
s =-1and
s =-2,totheloop
transfer
function
has
changed
theangle
ofdeparture
?(
s- p2)
from-13?
to150?
due
totheirrespective
84.29?and 78.69?angleshift contributions.
The two graphs in Figure 8.31 are obtained from the same root locus plot generated
following
MATLAB code:
% CSD
Fig8.31
Use
s=tf('s');
rlocus(G,K)
for
sinusoidal
Gp=145.5/(s*(s+43.14));
G=Gp*Gr,
figure
(6),
tracking
using the
control
Gr=(s+1)*(s+2)/(s2+100),
K=logspace(-3,1.5,500);
rlocus(G,K)
Onthe root loci diagram, the black square cursor is
movedto the position
where the complex
conjugate
poles
areat -13.5j14.7with
thegainK=5.06.
Atthese
pole
locations,
thecorre-sponding
damping ratio and the natural frequency
are ? = 0.676 and ?n = 19.9 rad/s.
By executing the MATLABs command [K,Poles]=rlofind(G),
a new cursor consisting
ofa long horizontal line and along vertical line appear to hover over the entire Root Locus window.
Bypositioning
thecursor,
theintersection
ofthese
twolong
lines,
atthepoint-13.5+
j14.7ontheloci
marked by the previous
black square cursor, the value of the gain
K and all associated closed-loop
poles
are identified, as shown in the following:
%
>>
CSD
Fig8.31
Use
rlocfind
to
locate
all
poles
[K,Poles]=rlocfind(G)
Select
a
point
selected_point
K
=
in
the
=
-1.3438e+01
graphics
window
+
1.4709e+01i
5.0642e+00
Poles
=
-1.3475e+01
+
1.4671e+01i
-1.3475e+01
-
1.4671e+01i
-1.5958e+01
+
0.0000e+00i
-2.3273e-01
+
0.0000e+00i
It also can be seenthat a red cross sign appears on each branch of the loci at the identified
pole location.
TheMATLAB simulation program, CSDfig8p32sinu.m,
is built based on the block diagram
of the feedback control system shown in Figure 8.29. This simulation demonstrates the effective-ness
of the the sinusoidal
position
tracking
controller
5.06(s+1)(s+2)
s2 +102
as shown in Figure 8.32. The MATLAB program is shown in the following.
%
CSDfig8p32sinu.m
5/08/2020
Sinu
Tracking
Example
Gm=145.5/(s*(s+43.14))
%
Design
A:
Gk=5.06*(s+1)*(s+2)/(s2+100)
%
Design
B:
Gk=3.95*tf(1,1)
%%
System
s=tf('s');
r=sin(10*t)
Gm
=
145.5/(s*(s+43.14));
t=linspace(0,2,201);
8.6
%%
Design
A
step
Controller
Design for
F/A18
Flight
Path Control
271
response
Gk=5.06*(s+1)*(s+2)/(s2+100);
GY=(Gk*Gm)/(1+Gk*Gm);
GU=Gk/(1+Gk*Gm);
[y,t]=lsim(GY,r,t);
[u,t]=lsim(GU,r,t);
run('plot2a')
%%
Design
B
step
response
Gk=3.95*tf(1,1);
GY=(Gk*Gm)/(1+Gk*Gm);
[y,t]=lsim(GY,r,t);
where the plotting
GU=Gk/(1+Gk*Gm);
[u,t]=lsim(GU,r,t);
programs
plot2a.m
run('plot2b')
and plot2b.m
are the same as those given at the end of Sec-tion
8.3.2.
It can be seenthat the blue ?(t) catches up and coincides with the sinusoidal reference input,
the green ?R(t), within 0.25seconds. In contrast,the proportional controller with gain K = 3.95is
unable to track the sinusoidal
reference input,
although it is a Type 1 controller
capable
of perfect step
referencetracking.Thered ?(t) continuesto lag behindatsteadystate.
Fig. 8.32: Simulation
results
8.6 Controller
Design for F/A18
The F/A18
flight
aircraft
dynamics
of the sinusoidal
position tracking
control
system.
Flight Path Control
model [Buttrill
et al., 1992,
Chakraborty
et al., 2011,
Chang et al.,
2016] will be employed as a test bed in the following flight path control system designs. The lateral
subsystem is setto fly straight without any roll or yaw motion; hence, we will only focus onthe control of
the longitudinal subsystem. Thelongitudinal state vector is x= [V a q ?]T, whereV, a, q, and ? arethe
air speed, the angle of attack, the pitch rate, andthe pitch angle, respectively. Thelongitudinal control-input
vectoris u =[de dT]T,wheredeand dTarethe elevatorcontrolandthrustcontrol,respectively.
Assumethe nominal flight trim; here, atrim is a desired equilibrium, chosen to be
x*=[436
ft/s 10?0?/s10?]T,
u*=[-1.26?
5470.5
lbf]T
whichis a level flight
(8.63)
with 10?angle of attack. Atthis trim, alocal linearized model[Chan et al., Dec.
2019] can be obtained as follows
272
8
Stability,
Regulation,
and Root Locus
x?(t)
= x(t)+B
A
u(t),
Design
where
? ?-0.051450 ?
A=? 0 -2.212-0.25320 ?,B=?-2.8791 0 ?
?-0.00033-0.362 1
-0.02389
-28.32 0
0
?
?
?
?
?
?
?
0
Notethat x =x+x*
-3.8114
0.000952
-32.2
?
0
and u =u+u*
1
0
0
(8.64
0
since(x,u) arethe differentialvalues,orthe perturbedvaluesof
(x,u) from the trim (x*,u*). In other words,the origin of the (x, u) coordinate is (x*,u*) while the origin
of the (x,u) coordinate is (0,0). For example, the differential air speedV = 44ft/s meansthat the real
air speedis V =V+436ft/s
For the flight
= 480ft/s.
path control
problem to be considered in this section, the variable to be regulated is the
flight
path
angle
?=? -aand
thecontrol
input
isde,while
thethrust
dTisfixed
atthe
trimmed
value
5,470.5 lbf all the time. Therefore, the local linear state-space model with input de and output?
x?(t)=A
x(t)+Be
de(t)=A
x(t)+[-3.8114
-0.05145
-2.8791
0]T
de(t)
?(t) =C
x(t) =[0 -10 1]
x(t)
will be
(8.65)
wherethe A matrix remains the same asthat in Equation 8.64, Beis the first column vector of the matrix
de
BinEquation
8.64,
andC=[0 -101].Italso
can
beseen
that
trim givenby Equation
8.63,d*
e
=de +1.26?and?=
?sinceatthe
= -1.26?
and
?*iszero.
The
plant
transfer
function
from
theelevator
aircraft at the nominal trim, given by
control input de to the flight path angle output ? of the F/A18
Equation 8.63is
Gp(s) =
G(s)
? e(s)
=
0.051453(s+4.3935)
(s-4.1059)
(s-0.03491)
(s2 +0.6189s+2.286)(s2
+0.02027s+0.01026)
(8.66)
The plant hastwo pairs of complex conjugate poles,
p1
p2
p3
=-0.010133
j0.10077
:=-aL j?L and p4 =-0.30945
j1.4799
:=-aS j?S
(8.67)
Notice that the first pair of poles, p1 and p2, has a small damping factor aL = 0.010133 and a low
frequency ?L =0.10077 rad/s, and the other p3 and p4 pair has arelatively larger damping factor, 0.3095,
and a higher frequency, ?S = 1.48 rad/s. In traditional aircraft parlance, this lower frequency behavior is
referred to asthe long-period
modeor phugoid mode,since the period is TL = 2p/?L
= 2p/0.1008
=
62.4 s. Onthe other hand, the higher frequency behavior is referred to asthe short-period modesince the
short period is TS = 2p/?S = 2p/1.48 = 4.25 s. The subscripts L and Sstand for long period and short
period, respectively. The damping ratio ? andthe natural frequency ?n, associated with the long-period
and short-period
modes, can be obtained
using Equation
3.60, respectively
long-period mode:
? = 0.1,
short-period mode:
? = 0.205,
as follows:
?n = 0.101rad/s
?n = 1.51rad/s
(8.68)
The performance of the uncompensated system is very poor since the damping ratio ? = 0.1 and the
natural frequency ?n = 0.101 rad/s of the dominant Phugoid modeare so small that it would cause un-acceptably
long
and large
up and down oscillations
The plant also has three real zeros, and two
in longitudinal
of them
motion.
are in the right
half of the s-plane,
z1=-4.3935,
z2=0.034907,
andz3=4.1059
so the system is a nonminimum
phase system.
(8.69)
8.6
Controller
Design for
F/A18
Flight
Path Control
273
The computational results relevant to the F/A18 subsystem from the elevator control input deto
the
flight
path
angle
output
?=a-?areobtained
byrunning
thefollowing
MATLAB
program:
% CSD_eq8p66
A
=[
F/A-18
Ele
-2.3893e-02
to
Gamma
-3.2923e-04
-3.6208e-01
4.0491e-11
=[
-3.8114e+00
0;
0;
1.0000e+00
0];
0;
B(:,1);
=
Zeros
-3.8506e-07;
0
G_ge
Poles
-3.2200e+01;
9.5196e-04;
-2.8791e+00
=
tf
0
-2.5319e-01
0
-5.1453e-02
b
model
1.0000e+00
-2.2115e+00
0
B
ss
-2.8317e+01
c
=
0];
[0
-1
ss(A,b,c,d),
0
1];
d
zpk(G_ge),
=
0;
eig(A),
damp(A)
[Z,Kz]=zero(G_ge),
P=pole(G_ge)
In the following, two controller structures will be considered to work together with the root lo-cus
design approach in designing a feedback controller to achieve closed-loop stability, steady-state
tracking,
and desired transient
performance
feedback
controller
are the
controller
structures
for the F/A18
PI (proportional
aircraft
flight
and integral
path control
controller,
problem.
The two
and the state-feedback
withintegrator.
Before getting into the controller design, we will first analyze the open-loop system and observe how
the elevator control input de will affect the longitudinal state variables V, a, ?, andthe flight path angle
?=? -a.
8.6.1 Open-Loop ManualLongitudinal Flight Controlof F/A18 by de
As described in Equations 8.67 and 8.68, the longitudinal
flight
dynamics
have two pair of complex
con-jugate
polesone
associated with the long-period (phugoid) modeandthe other with the short-period
mode. The damping ratios and damping factors of both modes are poor, which implies large over-shoot
and long oscillation periods, especially for the long-period
mode dynamics. Furthermore, the
nonminimum-phase
zerosthe
zeros in the RHP, as shown in Equation
will causethe system response to go to the opposite direction, initially,
direction.
controlling
The analysis and simulation
an uncompensated
An open-loop
aircraft
manual control
flight
simulation
to be conducted
8.69are
can be implemented
since they
before reversing back to the in-tended
in the following
system is quite a challenging
not helpful
will reveal that
manually
job.
based on the block diagram shown in
Figure 8.33. The relationship among the real control input u,the control input trim u*, andthe differential
control input u is described by the following:
?d?e ?? ??
u =u-u*, where
u = dT
,
d*
e
u* =
d*
T
,
de
dT
u =
(8.70)
Similarly, the real state vector x, the state vector trim x*, and the differential state vectorx arerelated
by the following
equation,
x =x+x*,
where x =
?? ?? ??
*??
??,x=??,x=
V
V
a
?
?
q
?
V*
a
?
?
?
?
q
?
a*
?
?
?
?
?
q*
?*
?
(8.71
274
8
Stability,
Regulation,
Fig. 8.33:
and Root Locus
Design
Open-loop longitudinal
In this simulation, the initial
manual control
block diagram for F/A18.
condition is at the trim
x(0)=x*=[436
ft/s 10?0?/s10?]T,
u(0)=u*=[-1.26?
5470.5
lbf]T
and
theelevator
input
de(t)
isassumed
tohave
a2?jump
from-1.26?
to0.74?
att =0(i.e.,
d
e(t)
=
2?us(t),
whilethe
thrust
input
continues
to stay at
de(t)=2?us(t)
-1.26?)
orequivalently
dT(t) = 5470.5lbf throughout the simulation.
The main objective of the 2?jump of the elevator input is to increase the flight path angle
?=? -aviathechanges
ofthepitch
angle
?and
theangle
ofattack
a.However,
theaction
of
the elevator
will affect all the longitudinal
transfer function
state variables including
the air speed
V. In addition
to the
? e(s)to theflight pathangleG(s), whichhasbeengivenin Equation
from the elevator
8.66,the transfer functions from the elevator? e(s) to the pitch angleT (s), the angleof attackA (s), and
the air speedV (s), respectively,canalsobe derivedfrom the state-spacemodelof Equations8.64and
8.65 as follows.
T (s)
? e(s)
=
A (s)
? e(s)
=
-2.8791(s-0.001872)(s+0.34832)
(s2 +0.6189s+2.286)(s2
-0.051453(s+56.185)(s2
+0.02361s+0.01056)
(s2
V (s)
? e(s)
=
+0.6189s+2.286)(s2
+0.02027s+0.01026)
(8.72b)
-3.8114(s+6.6232)(s+0.18029)(s-6.5705)(8.72c)
(s2 +0.6189s+2.286)(s2
The poles of the four transfer
They are the two
(8.72a)
+0.02027s+0.01026)
pairs of complex
functions
in
conjugate
Equations
+0.02027s+0.01026)
8.72a, 8.72b,
poles shown in
Equation
8.72c, and 8.66 are identical.
8.67one
pair associates
with
the long-period dynamics modeand the other with the short-period mode. However, the zeros of these
four transfer functions are quite different. We will seethat the time-domain responses of the longitudinal
flight
dynamics systems
The open-loop
are
mainly determined
manual elevator
control
by the poles and zeros of these transfer
step response
simulation
results
functions.
are shown in
Figure 8.34.
The upper-right graph shows that the real elevator control input applied to the aircraft is de(t) =
de(t) = 2?us(t) to the input of the linearized state-?
-1.26?
+2?us(t),
which
isequivalent
toapplying
space modelshown in Figure 8.33. The output of the linearized
x(t)
=
V (t) a (t) q(t) ? (t)
modelis
?T
Note
thattherealangleofattackresponse,a(t) =a(t)+10?, andtherealpitchangleresponse,
?(t) =
? (t) + 10?,respectively, are recorded on the upper-right and the bottom-left graphs. The upper-lef
8.6
Controller
Design for
F/A18
Flight
Path Control
275
graph
shows
thestep
response
ofthe
flight
path
angle
?(t)=?(t)-a(t)=?(t)-a(t),while
the
real
airspeedV(t) =V(t)+436
Steady-State
Response
ft/s is recordedonthe bottomright graph.
Analysis
The simulation shows that the values ofV, a , ? , and ? at t
= 1000 s are
? (1000)
V(1000)
=480.5
ft/s,a(1000)
=-2.60?,
= 0.16?, ?(1000) = 2.76?
(8.73)
At the end of the 1,000-second simulation, all the state variable values seem to have reached their
steady state.
The steady-state
the final-value
theorem
values or the final
(Theorem
2.27).
values can also be obtained in frequency
The theorem
and the transfer
functions
in
domain
Equations
using
8.72a,
8.72b,8.72c, and 8.66are employedto obtainthe steady-statevaluesofV, a , ? , and? in the following.
2? p
-3.8114(s+6.6232)(s+0.18029)(s-6.5705)
180?= 44.5ft/s
V (8) =lim V(s)
s
=lims
s?0 (s2 +0.619s+2.286)(s2
s?0
a (8)
? (8)
=lim sA(s)
=lim sT(s)
=lim s
s?0
(8.74a)
-0.051453(s+56.19)(s2
+0.02361s+0.01056)
s?0 (s2 +0.619s+2.286)(s2
=lim sG(s)
+0.02027s+0.01026)
s
=-2.60?(8.74b)
2?
-2.8791(s-0.001872)(s+0.34832)
= 0.16?
(8.74c)
2?
0.051453(s+4.3935)
(s-4.1059)
(s-0.03491)
= 2.76?
(8.74d
s?0 (s2 +0.619s+2.286)(s2
=lim s
s
2?
s?0
s?0
?(8)
=lim s
+0.02027s+0.01026)
+0.02027s+0.01026)
s?0 (s2 +0.619s+2.286)(s2
+0.02027s+0.01026)
s
s
Notethat the steady-state real air speed is
V(8) =V(8)+436
Hence, the simulation
values in Equation
Remark
steady-state
ft/s
responses
8.74 obtained by final-value
8.22 (Control
of the
The control of the flight
Flight
Path
= (44.5+436) ft/s
shown
in
Equation
= 480.5ft/s
8.73
match perfectly
with the final
theorem.
Angle)
path angle ? is achieved via the changes of the pitch angle ? and the
angle of attack a. Fromthe simulation resultsin Figure 8.34,it is observedthat a 2?increase of the
elevator control input de has causedthe angle of attack to go downfrom a = 0?to
a =-2.60?
at
steady state, but it has very little effect on the steady-state response of the pitch angle with final
theflight pathanglewill goupto
value, which slightly moves up from ? = 0?to ? = 0.16?.Thus,
?=
?- a =0.16?-(2.60?)
=2.76?.
On
theother
hand,
you
can
conduct
asimilar
simulation
using
the
thrust control dT asthe input and you will observethat the increase of the thrust will causethe pitch
angle
? andtheflight pathangle?to goupatsteadystate withoutmuch
effectonthesteady-state
value of the angle of attack.
Transient
Response Analysis
Recall that longitudinal
dynamics
has two pairs of complex
conjugate
poles as shown in Equation
8.67.
The
pair-aL j?L
isassociated
with
the
long-period
dynamics
mode,
and
theother,
-aSj?S,with
mode. Since the damping factor aSis about 30 times large than aL, the short-period
the short-period
276
8
Stability,
Regulation,
mode will decay
and Root Locus
much faster
Design
than the long-period
mode. Therefore,
under normal
circumstances
the
long-period modewill bethe dominant one. Asshown in Figure 8.34,the step responsesof ?(t), ? (t),
and V(t) oscillate with the samelong period TL = 2p/?L = 62.4 s. Theoscillation period of ?(t) can
bemeasured
from
theduration
between
thefirstand
thesecond
peaks
asT?=105-42=63s,which
agrees with TL. The amplitude of oscillation is large dueto the small damping ratio ? = 0.1. Theselong-period
step responses take morethan 500 seconds to converge to their steady-state values.
Fig. 8.34:
Open-loop longitudinal
manual control
step responses
of F/A18
due to the elevator input
de(t)=-1.26?
+2?us(t).
The open-loop longitudinal
manual control step responses of F/A18
due to the elevator input
de(t)=-1.26?
+2?us(t)
areobtained
byrunning
thefollowing
MATLAB
program:
%%
CSDfig8p34.m
5/09/2020
d2r=pi/180;
A
=[
Intro
-2.3893e-02
-2.8317e+01
-3.2923e-04
-3.6208e-01
4.0491e-11
=[
-3.8114e+00
0
0]
0;
-1
0];
0
1];
G_te=ss(A,b,c_theta,d);
[1
0;
1.0000e+00
-3.8506e-07;
c=[0
display('G_ge'),
0;
9.5196e-04;
0
b=B(:,1);
-3.2200e+01;
-2.5319e-01
0
-2.8791e+00
=
Example
1.0000e+00
-2.2115e+00
-5.1453e-02
c_V
Loci
0
0
B
Root
r2d=180/pi;
d=0;
G_ge=ss(A,b,c,d);
c_alpha=[0
0
0];
G_Ve
=
zpk(G_ge),
1
0
0];
c_theta=[0
G_ae=ss(A,b,c_alpha,d);
ss(A,b,c_V,d);
display('G_te'),
zpk(G_te),
0
0
1];
8.6
display('G_ae'),
%%
Step
zpk(G_ae),
Controller
Design for
display('G_Ve'),
F/A18
Flight
Path Control
277
zpk(G_Ve),
responses
t=linspace(0,500,5001);
[gamma,t]=step(G_ge,t);
[theta,t]=step(G_te,t);
[alpha,t]=step(G_ae,t);
[v,t]=step(G_Ve,t);
[u,t]=step(tf(1,1),t);
Alpha=alpha*2+10;
Gamma=gamma*2;
Ele=u*2-1.26;
V=v*2*d2r+436;
Theta=theta*2+10;
run('plot22A')
where plot22A.m
%filename:
is given in the following:
plot22A.m
ts=t(1:501);
grid
figure(4),
minor,
subplot(2,2,1),
title('Gamma'),
plot(t,Gamma,'b-'),
plot(ts,Ele(1:501),'r-',ts,Alpha(1:501),'b-'),
grid
title('Elevator,AoA'),
subplot(2,2,3),
grid
minor,
title('Pitch')
grid
minor,
title('Speed
grid
on,
subplot(2,2,2),
on,
grid
minor,
plot(t,Theta,'b-'),
subplot(2,2,4),
plot(t,V,'b-'),
grid
on,
grid
on,
V')
One maywonder whythe angleof attack step responsea (t) is not dominatedby the long-period
mode.It is becausethe angleof attacktransfer function A (s)/?e(s) hasone pair of com-plex
conjugate zeros that are almost identical to the long-period poles. The effect of the long-period
mode poles is greatly reduced by these nearby zeros. Hence,the a (t) step responseis dominated
bythe short-period mode,with muchhigher oscillationfrequency. Theoscillationperiodofa (t)
can be roughly
measured by the time
duration
between the second
and the third
which agrees with the oscillation period computed based on ?S, TS = 2p/?S
peaks as Ta
= 4.5 s,
= 4.25 s.
It is also observed that the initial movementsof ?(t) and?(t)
werein the opposite direction
of the intended motion. Both?(t) and? (t) movedownin the first 10 secondsbeforereversing up to
slowly converge toward the steady-state values. These phenomena are caused by the RHP zeros of
their respectivetransferfunctionsG(s)/ ?e(s) andT (s)/ ?e(s).
Theangleof attacktransferfunctionA (s)/?e(s) hasno RHPzeros;hence,
it doesnothavethe
initial
opposite
movement issue.
The air speed transfer
function
However, not all
V (s)/ ?e(s)
RHP zero,
will lead to an initial
opposite
hasa RHPzero ats = 6.5705, butits initial
movement.
movementis
in the intended up direction; apparently it has no initial opposite movement. This RHP zero does not
cause any initial opposite movement issue due to its large distance from the imaginary
axis.
Remark 8.23 (Poor
Transient Performance
The uncompensated
longitudinal
of the Uncompensated
dynamics
is sensitive
Longitudinal
and susceptible
to
Dynamics)
disturbances
and
control actions. Dueto the extremely small damping factor, poor damping ratio, long oscillation period,
and the initial
very
difficult,
opposite
if
movement caused by the
not impossible,
8.6.2 PI Controller
Designfor
to
manually
F/A18
RHP zeros that
control
the aircraft
are close to the imaginary
without
a feedback
axis, it is
compensator.
Flight Path Control
The block diagram of the feedback PI control system is shown in Figure 8.35, wherethe state-space
model
(A,Be,C)was
given
in Equations
8.64and8.65,andthetransfer
function
from
deto ?,
Gp(s)
=G(s)/
?e(s)=C(sI-A)-1Be
was given in
Equation 8.66
278
8
Stability,
Regulation,
The loop transfer
and Root Locus
function
KG(s)
= K
Design
of the PI feedback
control
system shown in Figure 8.35 is
0.051453(s+4.3935)
(s-4.1059)
(s-0.03491)
(s+1)
s(s2 +0.6189s+2.286)(s2
+0.02027s+0.01026)
(8.75)
Note
thatthePIcontroller
has
added
one
pole,
p5=0,and
one
zero,
z4=-1,totheloop
transfer
function
KG(s). Hence,the closed-loop characteristic equation is
1+KG(s)
= 0
and the root loci diagram of the closed-loop system is shown in Figure 8.36. The upper-left graph of the
figure shows a morecomplete root loci diagram, but the part of root loci involving p1, p2, p5, and z2 are
crowded together around the origin area and only can be seen in the upper-right graph with expanded
view.
Fig. 8.35: PI controller
for F/A18
flight
path angle control
problem.
The objective is to find a best gain value of Kto achieve closed-loop system stability, zero or small
steady-state error, and an optimal transient response subject to the control-input constraints. When
K = 0, all the roots (poles)
are at p1, p2, p3, p4, and p5. As K increases
from
K = 0, the root loci
branch
originating
fromp5=0willmove
ontherealaxis
toward
theleftand
reach
atz4=-1when
K ?8.Meanwhile,
thetwobranches
originating
from
theshort-period
poles
p3=-0.3095+
j1.48
and
p4=-0.3095j1.48willmove
totheleftand
bend
toward
therealaxis
andmeet
atthebreak-in
point
s =-8.08
when
K=251,
then
split
intotwobranches,
both
ontherealaxisone
toward
theright
tothezero
z1=-4.3935
and
theother
toward
theinfinity
onthenegative
realaxis,
which
istheonly
one asymptote of this root loci diagram.
The
lasttwobranches
start
from
thelong-period
poles
p1=-0.010133
+j0.10077
andp2=
-0.010133j0.10077.
AsKincreases,
thetwobranches
willmove
totherightand
intersect
theimag-inary
axis
at j0.0828,
when
K=0.00856.
After
crossing
totheright
halfplane,
thetwobranches
will
continue to bend toward the real axis and
meet at s = 0.0811
when K = 0.0758.
Then, as K continues
to increase, the two branches now are both on the real axisone
moving toward the right to the zero
z3 = 4.1059, and another toward the left to the zero z2 = 0.034907.
The four graphs on Figure 8.36 are obtained from the same root locus plot generated using the
following
MATLAB
% CSDfig8p36
Ki=1,
code:
Use
s=tf('s');
rlocus(G,K)
for
F/A-18
PI
Np=0.051453*(s+4.3935)*(s-4.1059)*(s-0.03491);
Dp=(s2+0.6189*s+2.286)*(s2+0.02027*s+0.01026);
Gp=Np/Dp;
figure(10),
Gc=(s+Ki)/s;
K=logspace(-3,3,1000);
G=Gc*Gp;
rlocus(G,K
control
8.6
Controller
Design for
F/A18
Flight
Path Control
279
Fig. 8.36: Rootlocus design of the PI controller for F/A18 flight path control problem.
From the root loci diagram, we can observe how the gain K will affect the root locations onthe five
branches,
and the closed-loop
system
We will see some of the root locations
behavior
are
will be determined
more dominant
system poles that are closer to the imaginary
mainly by these five root locations.
than the others. In general, the closed-loop
axis are the dominant
ones. To choose the right value of the
gain K,the first step is to find the range of Kthat will guarantee the stability of the closed-loop system.
The stability range of Kis 0 < K < 0.00856 since the two branches originating from p1 and p2 will
cross the imaginary
axis entering into the right
half s-plane if
K > 0.00856 and the root before entering
the real branch between p5 and z4 wasin the right half of s-plane when K< 0.
In the following,
we will choose two
K= 0.004 in the lower-right
pole locations
and time-domain
K values,
K = 0.002 as shown in the lower-left
graph, and
graph of Figure 8.36, and compare their corresponding closed-loop system
performances.
Root Locus Design with K = 0.002
If Kis selected as K = 0.002, the five closed-loop system poles are
s1
s3
s5=-0.003
=-a5,s2 =-0.00833
j0.0967
:=-a1 j?1, s4 =-0.31j1.48
:=-a3 j?3
(8.76)
and their associated time
constant and damping ratio,
natural frequency
are obtained as follows
280
8
Stability,
s5:
Regulation,
t5 = 1/a5
and Root Locus
Design
= 333s
s1, s2:
t1 = 1/a1 = 120s, ? = 0.086,
?n = 0.097ad/s,
s3, s4:
t3 = 1/a3 = 3.23s,
?n = 1.51rad/s,
? = 0.205,
o.s. = 76.3%
o.s. = 51.8%
Notethatthe dampingfactor a3is 103times and37times,respectively,
largerthan a5and a1;there-fore,
the effect of the poles at s3 and s4 will decay much earlier and become irrelevant.
Comparably,
since the damping factor a1is only 2.8 times larger than a5,it has slight influence at the early
time to cause some oscillations
on, the
dominant
that
will die out exponentially
with time
pole s5 will be the only one left to dictate the time
constant
response.
120 s. From then
The dominant
pole
s5=-0.003
isreal,with
large
timeconstant
333
s;hence,
it causes
slow
response
butcontributes
no oscillations,
as shown in the graphs of Figure 8.37.
Fig. 8.37: Simulation results of F/A18
A
MATLAB
simulation
program
aircraft using PI controller and root locus design.
is created
based on the
PI feedback
control
system
shown
in Figure 8.35, andthe flight path angle commandis assumedto be a 2? step function. The graphsof
Figure 8.37 show two simulation results: one with K = 0.002 in blue, and the other with K = 0.004
in red. Thegraph onthe upperleft shows the two flight pathangle responses?(t) dueto the reference
input?R(t) =2?us(t)for bothK =0.002and K =0.004,respectively.
Theupper-right
graphreveals
the control-input efforts de(t) dictated by the feedback controller. The angle of attackresponses a(t)
are also shown in this graph. The pitch angle ?(t) andthe air speed V(t) responsesareshown in the
lower-left
and the lower-right
graphs, respectively.
Fromthe bluestepresponses?(t) and ?(t), shown onthe upper-left andlower-left graphsof Figure
8.37, it is observed that the step response
rises extremely
slowly
as an exponential-like
curve
with
time constant close to t5 = 333 s. It also can be seenthat the small oscillations in the first 300 s are
contributed by the pair of complex conjugate poles at s1 and s2. The period of the oscillations can be
approximately
measured
from
thepeak-to-peak
time
otfhe?(t)plot
asTpp
=192s-127s
=65s,
whic
8.6
Controller
Design for
F/A18
Flight
Path Control
281
closely matches
the periodcomputed basedonthe frequency ?1 =0.0967rad/s: T1 =2p/0.0967=65 s.
The pair of complex conjugate
poles at s3 and s4 virtually
The design with the PI feedback control
open-loop
manual control.
have no visible effect on the step re-sponse.
with K = 0.002 has shown great improvement
However, the flight
path angle
rises very slowly,
and
over the
only reaches
95%
of its target value 2? after 1000s oftime. The responsetime can beimproved if the pole s5 on the
lower-left
graph of Figure 8.36 can be movedto the left.
pole s5 to the left, the
should
poles s1 and s2 will
not be moved too
s2 become dominant
The four
graphs
of the simulation
CSDfig8p37Step.m
results
5/09/2020
d2r=pi/180;
A
=[
of the
-2.3893e-02
-2.8317e+01
-3.2923e-04
-3.6208e-01
PI flight
Root
Loci
Example
b=B(:,1);
c=[0
[1
0
K=0.002,
0;
0;
1.0000e+00
0]
0];
-1
0
1];
G_te=ss(A,b,c_theta,d);
Design
-3.2200e+01;
0;
0
d=0;
G_ge=ss(A,b,c,d);
c_alpha=[0
0
0];
A
G_Ve
with
=
1
0
ss(A,b,c_V,d);
c_theta=[0
0];
Gc=tf([1
1],[1
Del=1+K*Gc*(G_ge);
G_gR=K*Gc*G_ge/Del;
G_VR=K*Gc*G_Ve/Del;
G_eR=K*Gc/Del;
zpk(G_gR),
display('G_tR'),
zpk(G_tR),
zpk(G_aR),
display('G_VR'),
zpk(G_VR),
display('G_eR'),
zpk(G_eR),
responses
t=linspace(0,1000,10001);
[gamma,t]=step(G_gR,t);
[theta,t]=step(G_tR,t);
[alpha,t]=step(G_aR,t);
[u,t]=step(G_eR,t);
Gamma=gamma*2;
V=v*2*d2r+436;
Design
Theta=theta*2+10;
B
K=0.004,
with
[v,t]=step(G_VR,t);
Alpha=alpha*2+10;
Ele=u*2-1.26;
run('plot22a')
K=0.004
Del=1+K*Gc*(G_ge);
G_tR=K*Gc*G_te/Del;
G_gR=K*Gc*G_ge/Del;
G_VR=K*Gc*G_Ve/Del;
G_eR=K*Gc/Del;
G_aR=K*Gc*G_ae/Del
display('G_gR'),
zpk(G_gR),
display('G_tR'),
zpk(G_tR),
display('G_aR'),
zpk(G_aR),
display('G_VR'),
zpk(G_VR),
display('G_eR'),
zpk(G_eR),
responses
t=linspace(0,1000,10001);
[gamma,t]=step(G_gR,t);
[theta,t]=step(G_tR,t);
[alpha,t]=step(G_aR,t);
[u,t]=step(G_eR,t);
Gamma=gamma*2;
V=v*2*d2r+436;
Theta=theta*2+10;
where plot22a.m
%filename:
figure(4),
1];
G_aR=K*Gc*G_ae/Del
display('G_aR'),
% Step
0
0]);
display('G_gR'),
%%
0
G_ae=ss(A,b,c_alpha,d);
K=0.002
G_tR=K*Gc*G_te/Del;
% Step
of F/A18
-3.8506e-07;
-2.8791e+00
=
poles. If s1 and
9.5196e-04;
-5.1453e-02
%%
path angle control
-2.5319e-01
0
-3.8114e+00
c_V
dominant
MATLAB program:
0
-2.2115e+00
[v,t]=step(G_VR,t);
Alpha=alpha*2+10;
Ele=u*2-1.26;
run('plot22b')
is given below, and plot22b.m
is a copy of it
with b replaced
by r.
plot22a.m
subplot(2,2,1),
plot(t,Gamma,'b-'),
movethe
will be oscillatory.
1.0000e+00
0
=[
Intro
becoming
K to
Hence, the pole s5
r2d=180/pi;
4.0491e-11
B
at the same time.
much to the left to prevent s1 and s2 from
poles, the associated time response
Figure 8.37 are obtained from using the following
%%
However, as weincrease
move to the right
grid
on,
title('Gamma'),
ahown
on
282
8
Stability,
Regulation,
and Root Locus
Design
hold
on,
subplot(2,2,2),
plot(t,Ele,'b-'),
hold
on,
subplot(2,2,3),
plot(t,Theta,'b-',t,Alpha,'b--'),
title('Pitch,AoA'),
hold
title('Speed
V'),
Root Locus
If
Design
on,
hold
grid
on,
title('Elevator,AoA'),
grid
subplot(2,2,4),
plot(t,V,'b-'),
on,
grid
on,
on
with K = 0.004
Kis selected to be K = 0.004, the five closed-loop
system poles are
s1
s3
s5=-0.00661
=-a5,s2 =-0.00623
j0.0925
:=-a1 j?1, s4 =-0.31j1.48
:=-a3 j?3
(8.77)
and their associated time
s5:
constant and damping ratio,
natural frequency
are obtained as follows:
t5 = 1/a5 = 151s
s1, s2:
t1 = 1/a1
= 160s, ? = 0.0672,
?n = 0.0927rad/s,
s3, s4:
t3 = 1/a3
= 3.23s,
?n = 1.51rad/s,
By increasing
Kfrom
? = 0.205,
0.002 to 0.004, the real branch pole s5 has slightly
o.s. = 80.9%
o.s. = 51.8%
moved to the left to change
the dampingfactor a5from 0.003to 0.00661, whichleadsto the reduction of the time constantt from
333 s to 151 s. Meanwhile, the complex poles s1 and s2 have moved closer to the imaginary
axis to
changethe dampingfactor a1from 0.00833to 0.00623, or equivalently,to reducethe dampingratio ?
from
0.086 to 0.0672.
The real
pole s5 and the complex
Onthe upper-left and lower-left
faster
rise than the
poles s1 and s2 are both dominant.
graphs of Figure 8.37, the red step response waveform shows its
blue one, but at the
expense of
more oscillations
and s2. The period of the oscillations can be approximately
the
due to the complex
poles s1
measuredfrom the peak-to-peak time of
?(t)plot
asTpp
=(200-132)
s=68s,which
closely
matches
theperiod
computed
based
onthe
frequency?1 = 0.0925rad/s:T1 = 2p/0.0925 = 68s. Thepairof complexconjugatepolesats3ands4
do not have perceivable
effect on the step response.
A close-up
view of these two
graphs at the first
seconds, both red and blue step responses all went to the opposite direction before reversing
to the intended
positive
direction,
which is the
nonminimum-phase
effect caused
20
back
by the right-half
plane zeros.
Remark
8.24 (PI
Controller
Design for
From the root locus analysis
are three
groups of root loci
F/A18
of the PI flight
branches.
The two
Flight
Path
Control)
path control for the F/A18
aircraft, it is clear that there
branches originating
p3 and p4 can only
from
move to
the left to increase their already large damping factor a3 = aS = 0.309; hence, the poles s3 and s4 on
these two branches
will not have much noticeable
seconds no matter what value ofK is chosen.
effect on the closed-loop
response
Meanwhile, the real pole s5 is leaving
except in the first
from
3
p5 at the origin
toward
theleftforz4=-1and
thetwocomplex
poles
s1and
s2originating
fromp1andp2areheading
to the right toward the imaginary
very small (e.g.
axis. Thesetwo groups are moving in opposite directions.
K = 0.002) s5 is the dominant
pole that leads to a terribly
On the other hand, s1 and s2 will increase their influence
When Kis
slow response.
or even become dominant,
which leads to
large up and down oscillations when Kincreases. Apparently there is no acceptable compromise; the
response at K = 0.004, which is approximately
a better compromise, is still slow and oscillatory.
This issue still remains even the zero z4is placed elsewhere.
Adding zeros and/or poles to the PI controller, as we did in Section 8.5, or augmenting an inner
feedback loop to the controller
structure
may be helpful in providing
a more friendly
pole-zero
patter
8.6
in the root locus
diagram;
Controller
however, the procedure is tedious
Design for
and it still
F/A18
Flight
employs
Path Control
only the flight
283
path angle
information, instead of utilizing the full available information, in the feedback control system design.
8.6.3 State-Space Pole Placement and Root Locus Design for
F/A18
Flight Path Control
The root locus design approach with the PI controller structure discussed in the previous subsection,
Section 8.6.2, does not provide
an acceptable
solution
for the
F/A18
flight
path tracking
control
prob-lem.
The response is either too slow or too oscillatory. In this subsection, we will employ the state
feedback structure with tracking integrator, shown in Figure 8.38, in the root locus design to im-prove
the flight
path tracking
control
performance.
In the block diagram shown in Figure 8.38, the state-space model (A,Be,C) was given in Equations
8.64 and 8.65. The state-feedback gain matrix F is chosen to place the closed-loop system poles at
some favorable
pole locations
for the loop
s-1 is employed to serve asthe internal
and the integral
closed-loop
parameter
system
transfer
function
in root locus
design.
The integrator
modelto achieve zero-error step response at steady state,
Kis to be determined
poles are at desired locations
using the root locus
to provide
design approach
desired transient
so that
the
response.
The state-space pole placement approach discussed in Section 7.5 can be employed to find the state-F
feedback gain matrix
(8.78)
=[ -0.19318
0.9557
3.2363
6.2859
]
sothat
theeigenvalues
ofA+BeF
are
at-0.17,
-5.9,
and-1.6j1.8.
Fig. 8.38: The state-feedback
controller
with integrator
tracking
for F/A18
flight
path control.
Recallthat the root locus diagram is constructed based on the root locus equation 1+KG(s)
= 0 as
defined in Equation 8.41, in which KG(s) is called the loop transfer function of the closed-loop system
shown in
Figure 8.17.
Hence, the loop transfer
function
of the
closed-loop
system in
Figure 8.38 i
284
8
Stability,
Regulation,
and Root Locus
Design
KG(s),
where
-G(s)
isthetransfer
function
fromuKtoxiassuming
theblock
Kisremoved
from
the
loop. This G(s) can be found as a rational function in s,
(s-4.1059)
(s-0.034907)
(8.79)
G(s)
=s-1C(sI-(A+BeF))-1Be
=0.051453(s+4.3935)
(s+1.6)2 +1.82
s(s+0.17) (s+5.9)
?
?
??? ??
??? ??
?? ??
or in the form of a state-space model,
x?
x?i
G(s):
A+BeF
=
xi =[ 0
x
xi
0
C
0
x
:=CG
xi
1]
Be
+
0
uk:=
AG
x
xi
+BGuk
(8.80)
x
xi
Notethat these two expressions represent exactly the same system: the former in frequency domain
and the latter in time
domain. It also can be seen that the numerator
8.79 is the same as that of the plant transfer function
polynomial
of G(s) in
Equation
from the elevator control input de to the
flight path angle output?, Gp(s) =G(s)/ ?e(s), in Equation 8.66. Butthe polesofG(s) are nowthe
eigenvalues of A+BeF instead of the poles of Gp(s)in Equation 8.66. Thesetwo facts verify that the
state-feedback
control
may alter the pole locations,
but it is unable to
Now we are ready to draw the root locus diagram
move the zeros.
using the following
MATLAB code according to
the state-feedbackmatrix F of Equation8.78andthe G(s)state-spacemodelof Equation8.80.
% CSD
%
Fig8.39
Draw
A_x
the
=[
State-feedback
root
rlocus(G,K)
locus
diagram
-2.3893e-02
(8.64)
-3.2200e+01;
0;
0;
1.0000e+00
0]
9.5196e-04;
-5.1453e-02
-3.8506e-07;
-2.8791e+00
0;
0
0];
Be=B_x(:,1);
% Find
% at
Eq.
-2.5319e-01
0
-3.8114e+00
\gamma-tracking
in
1.0000e+00
-2.2115e+00
0
A=A_x;
F/A-18
G(s)
0
-3.6208e-01
4.0491e-11
=[
to
-2.8317e+01
-3.2923e-04
B_x
for
according
a
state
-0.17,
feedback
-5.9,
P=[-0.17
-5.9
D_G=0;
that
C=[0
the
and
-1.6+1.8i
damp(A+Be*F);
A_G=[A+Be*F
so
-1.6+j1.8,
eigenvalues
-1.6-1.8i];
-1
zeros(4,1);
0
% track
0];
flight
B_G=[Be;
G=ss(A_G,B_G,C_G,D_G);
[zero_G,gain_G]=zero(G);
A+BF
F=place(A,Be,P);
1];
C
of
are
-1.6-j1.8
0];
pole_G
=
figure(12);
F=-F;
path
angle;
C_G=[zeros(1,4)
pole(G);
1];
zpk(G),
K=logspace(-1,6,1000);
rlocus(G,K)
Theroot locus diagram associated withthe closed-loop system in Figure 8.38 with the state-feedback
gain matrix F of Equation 8.78is shown in Figure 8.39. Theleft graph of Figure 8.39 shows a morecom-plete
overview
details.
of the root loci
The enlarged
diagram,
one on the right
but the root loci
around the origin area is too small to show the
graph gives a better view of the root loci
near the imaginary
axis.
The poles of the G(s) in Equation 8.79 or Equation 8.80 include the eigenvalues of A+BeF together
with the integrator
pole at the origin:
p3
p1=-0.17,
p2=0, p4 =-1.6 j1.8,p5 =-5.9
8.6
Controller
Design for
F/A18
Flight
Path Control
285
and the zeros of G(s) are the same asthose of Gp(s) in Equation 8.66:
z1=-4.3935,
z2=0.034907,
z3=4.1059
When
K=0,alltheroots
(closed-loop
system
poles)
areatp1,p2,p3,p4,andp5.AsK?8,the
roots will either moveto the three zeros, z1, z2, and z3, or approach to the two asymptote straight lines
intersecting the real axis at
s =
-(z1+z2+z3)+(p1
+p2+p3+p4+p5)
5-3
with
angles
?0=p/2and
?1=-p/2,respectively.
Fig. 8.39: Rootlocus design of the integral tracking controller
F/A18 flight path control problem.
=-4.5087
with state-feedback compensation for the
When
Kisincreased
fromK=0,therootlocibranch
originating
fromp5= -5.9willmove
on
therealaxis
toward
therightand
arrive
atz1= -4.3935
asK?8.The
tworootlocibranches
started
fromp3andp4,-1.6 j1.8,willmove
totheleftbending
upanddown,
respectively,
and
approach
tothetwoasymptotes
asK?8.The
other
twobranches
originating
fromp1=-0.17
andp2=0
will movetoward each other on the negative real axis.
When K = 2.52, they
meetat the break-out
point
s = -0.0475
andsplit
intotwobranchesone
up,and
theother
downfollowing
thecir-cular
trajectory
with radius approximately
equal to 0.07825.
When K = 6.12 the two branches will
intersect
theimaginary
axis
at j0.718
andmove
intotheRHP.
AsKincreases
to K=16.4,
thetwo
branches will meetthe positive real axis at the break-in point s = 0.109 and split into two on the posi-tive
real axisone
movesto the left toward z2 = 0.034907 andthe other to the right toward z3 = 4.1059.
From the root loci branches shown in Figure 8.39, it is obvious that the roots on the two
branches originating from p1 and p2 are the dominant poles since the other three branches are
much further away from the imaginary axis. We will choose three values ofK:
K= 3, K= 3.5, and
K = 2.5 as shown on the right graph of Figure 8.39, and compare their corresponding closed-loop
system pole locations and time-domain
performances.
Root Locus Design with K = 3
If
Kis selected to be K = 3, the five closed-loop
system
poles ar
286
8
s1
s2
Stability,
Regulation,
and Root Locus
Design
s3
=-0.0407j0.032
:=-a1 j?1, s4 =-1.65j1.86
:=-a3 j?3,s5 =-5.88=-a5
and their associated time constant and damping ratio, natural frequency are obtained asfollows.
s1, s2:
t1 = 1/a1 = 24.6s,
? = 0.79,
?n = 0.052rad/s,
o.s. = 1.76%
s3, s4:
t3 = 1/a3 = 0.61s,
? = 0.665,
?n = 2.49rad/s,
o.s. = 6.09%
s5:
t5 = 1/a5 = 0.17s
Notethat the dampingfactor a1is 41times and 144times, respectively, smallerthan a3 and a5;
therefore,
Fig. 8.40:
the effect
of the poles at s3, s4, and s5 will decay
Simulation
results
of F/A18
aircraft
much earlier and become irrelevant.
using integral
tracking
controller
with state-feedback
compensation and root locus design.
The Simulink simulation program, Stepfig8p40.slx,
based on the state feedback and integral tracking control
shown on Figure 8.38(b), is created
block diagram shown in Figure 8.38(a).
Theflight path anglecommand is assumedto be a 2?step function. The graphsof Figure 8.40show
three simulation results:
K= 3in blue,
K= 3.5in red, and K= 2.5 in black. The graph onthe upper left
showsthethreeflight pathangleresponses
?(t) duetothereference
input?R(t) = 2?us(t)for K =3,
K= 3.5,and K= 2.5,respectively. The upper-right graphrevealsthe control-input efforts de(t) dictated
bythe feedbackcontroller.Theangleof attackresponsesa(t) andthe pitchangleresponse?(t) are
shown in the lower-left
graph. The air speed V(t) responses are shown in the lower-right
graph
8.6
Controller
Design for
F/A18
Flight
Path Control
287
From the blue step response ?(t) shown on the upper-left graph of Figure 8.40,it is observed
thatthestep
response
initially
goes
down
intheopposite
direction
from0?to -0.8?
att =13sto
reverse
the
moving direction
the peak at t
to rise to 90% of the desired
= 113 s with less than
2% of overshoot.
value at t
The initial
= 73 s, and continue
wrong direction
motion
to reach
was caused
by the nonminimum-phase
effect due to the RHP zero at s = 0.034907. However, this half-period
swing is the only visible oscillation and the step response reaches the steady state shortly after
t = 113 s.
It also can be seen that the half period of the
half-period
swing
can be measured from the valley to
peak
times
inthe?(t)plot:
Tvp
=113-13
=100
s,which
matches
thehalf
period
computed
based
on
the frequency ?1 = 0.032rad/s (i.e., Tvp = p/0.032 = 98.2s).
Root Locus Design with K = 3.5
If the integral
parameter
Kis slightly
changed from
K = 3 to
K = 3.5, the s1 and s2 closed-loop
system
poles
willmove
from-0.0407
j0.032
to -0.0339
j0.044,
asshown
intherightgraph
ofFigure
8.39.
This slight change of K hasincreased the oscillation frequency ?1from 0.032 rad/s to 0.044 rad/s,
which would decrease the half-period of the half-period swing from 98.2 s to p/0.044 = 71.4 s, so
that the step response will rise more quickly. However, the damping ratio will reduce from 0.79 to
0.61, and the maximum overshoot will increase from 1.7% to 8.9%.
Root Locus Design with K = 2.5
Onthe other hand,if the integral
parameter
Kis slightly
changed from
K = 3 to K = 2.5, the two
dominant
closed-loop
system
poles
s1and
s2willmove
from-0.0407
j0.032
back
tothenegative
realaxis
sothatthetwopoles
are-0.0423
and-0.0528.
Hence,
theresponse
willbeoverdamped,
which
has no overshoot
but is slower
than the
K = 3.0 response.
The four graphs of the simulation results of the PI flight path angle control
Figure 8.37 are obtained from using the following
MATLAB program:
% CSDfig8p40.m
% Run
%
call
% Run
5/09/2020
CSDfig8p39.m
simulink
file:
A:
Design
K=3,
obtain
A
K=3,
B_x,
D=[0
0
and
0
...
F
0]',
C_ga=[0
-1
Design
C:
K=2.5
[0,
sim_time],
0
1]
%
Design
B:
K=3.5,
%
Simulation
sim_time=500,
'current',
sim('Stepfig8p40',
'DstWorkspace',
'current');
sim_options);
Plot
Ele=-1.26+u*r2d;
V=x(:,1)+436;
Gamma=yout(:,1)*r2d;
Design
K=3.5,
B
R=yout(:,2)*r2d;
'current',
sim('Stepfig8p40',
Ele=-1.26+u*r2d;
V=x(:,1)+436;
Gamma=yout(:,1)*r2d;
Design
Theta=x(:,4)*r2d+10;
run('plot2x2a');
Simulation
open('Stepfig8p40'),
K=2.5,
Alpha=x(:,2)*r2d+10;
sim_time=500,
sim_options=simset('SrcWorkspace',
%%
Tracking
R_gamma=2*d2r
open('Stepfig8p40'),
%%
Intgrl
program
plot_regA.m
sim_options=simset('SrcWorkspace',
%%
rLocus
this
A_x,
C=eye(4);
r2d=1/d2r;
% Design
%%
to
B=B_x(:,1);
d2r=pi/180;
SF
running
Stepfig8p40.slx,
CSDfig8p39.m
A=A_x;
F/A-18
before
of F/A18
C
R=yout(:,2)*r2d;
Simulation
sim_time=500
'DstWorkspace',
[0,
sim_time],
Alpha=x(:,2)*r2d+10;
run('plot2x2b');
'current');
sim_options);
Theta=x(:,4)*r2d+10;
ahown on
288
8
Stability,
Regulation,
and Root Locus
Design
sim_options=simset('SrcWorkspace',
'current',
open('Stepfig8p40'),
sim('Stepfig8p40',
Ele=-1.26+u*r2d;
V=x(:,1)+436;
Gamma=yout(:,1)*r2d;
'DstWorkspace',
[0,
sim_time],
Alpha=x(:,2)*r2d+10;
R=yout(:,2)*r2d;
'current');
sim_options);
Theta=x(:,4)*r2d+10;
run('plot2x2c');
where plot2x2a.m
is given below. The other two copies, plot2x2b.m
print in red and black colors, respectively.
%filename:
will
plot2x2a.m
figure(4),
subplot(2,2,1),
plot(t,Gamma,'b-'),
grid
hold
on,
subplot(2,2,2),
plot(t,Ele,'b-'),
hold
on,
subplot(2,2,3),
plot(t,Theta,'b-',t,Alpha,'b--'),
title('Pitch,AoA'),
hold
title('Speed
8.6.4
and plot2x2c.m,
V'),
Comparison
on,
hold
of the
grid
on,
on,
title('Gamma'),
title('Elevator,AoA'),
grid
subplot(2,2,4),
plot(t,V,'b-'),
on,
grid
on,
on
PI Controller
and the Integral
Controller
with State-Feedback
Compensation
It seems to
variables
PI controller
usually
be an unfair
comparison
of the longitudinal
only the flight
availableeither
since in the state-feedback
compensation
path angle is utilized
by direct
measurement
in the design.
However, these state
or state reconstruction
state
while in the
variables
using the observer
theory, which will be covered in Chapter 11. Moreinformation
design will certainly
makethe control system perform better.
By comparing
all the four
dynamics system are assumed available for feedback,
are
and es-timation
utilized in the controller
the root locus diagrams in Figures 8.36 and 8.39, it can be seen that the state-feedback
has replaced
a more favorable
the LTF
poles and greatly
set of closed-loop
system
changed
the shape
of the root loci
graph
so that
poles can be obtained.
In Figure 8.36, the best possible range of the K values is between K = 0.002 and 0.004.
With
K=0.002,
thedamping
factor
ofthedominant
pole
s5=-0.003
is0.003,
which
implies
a333
stime
constant and morethan 1,000 seconds of settling time in step response. It is an unacceptable slow step
response.
Furthermore,
thenearby
complex
conjugate
poles
s1and
s2at-0.00833
j0.0967
willadd
some long-period
lightly
damped oscillation
the left to decrease the time
sametime
constant
and
to the response.
make the response
By increasing
a little
K, the s5 pole
bit faster.
will
move to
But doing so will at the
movethe complex conjugate poles s1 and s2to the right to makethe response more oscillatory.
AtK=0.004,
thepair
ofcomplex
poles
s1and
s2areat-0.00623
j0.0925,
which
are
asdominant
as
thereal
s5=-0.00661.
Itis much
more
oscillatory
and
itisstillslow.
There
isnogood
choice
available
out of this root locus
diagram.
On the other hand, the root locus
diagram in Figure
8.39 provides
much better options to find
a set
of closed-loop system poles for an acceptable closed-loop system performance. In Figure 8.39, only
theroot
locibranches
originating
fromp1=-0.17
andp2=0are
relevant
since
thedamping
factors
associated
decay
with the other three root loci
much earlier.
With the choices
of
branches
are
much larger
K = 3.0,
K = 3.5, and
and their
associated response
K = 2.5, and their
corresponding
will
root
locations on the right graph of the figure, the damping factors are about 10times larger than those of the
dominant
poles in Figure
8.36. The damping
ratios are also about
10 times larger than those of the PI
control design. Therefore, the performance of the closed-loop system performance designed by the
state feedback, together with the integral tracking, is about 10 times faster in step response with
muchless oscillation.
The time-domain responses foretold bythese observations of the root-locus diagrams are veri-fied
by the simulation results shown in Figures 8.37, 8.40, and 8.41, and in Table 8.2
8.7
Aircraft
Altitude
Regulation
via Flight
Path Angle Tracking
Control
289
Table 8.2: Performance comparison of the four closed-loop systems withtheir respective controllers K=
3.0, K = 3.5, K = 2.5 in the state-feedback with integral tracking structure and K = 0.002, K = 0.004
in PI controller structure.
K
SfI K = 3.0
SfI K = 3.5
SfI K = 2.5
PI K = 0.002
PI K = 0.004
Rise time
74s
58s
100s
790s
380s
Settle time
80s
110s
110s
1000s
460s
Max overshoot
2%
14%
0%
0%
0%
40%
50%
30%
9%
17s
Osc amplitude
0%
0%
0%
18%
35s
Osc period
N.A.
N.A.
N.A.
65s
68s
0
0
0
0
0
Initial
NMP dip
S.S. tracking
error
de deflection
Fig. 8.41:
Comparison
(-1.26?,
0.2?)
(-1.26?,
0.2?)
(-1.26?,
0.2?)
(-1.26?,
0.29?)
(-1.26?,
0.2?)
of F/A18
integral tracking controller
8.7 Aircraft
In the previous
flight
Altitude
section,
flight
path tracking
responses
using the PI controller
Regulation via Flight Path Angle Tracking
Section 8.6,
path angle of the aircraft
angle tracking
aircraft
and the
with state-feedback compensation.
we discussed
can follow
control is often employed
how to design a feedback
the flight
path angle command.
to regulate the flight
altitude
Control
control system so that the
In practice, the flight
path
of aircraft.
Although a six degree-of-freedom aircraft dynamics has 12 state variables, total air speed V, angle
of attack a, side slip , roll rate p, pitch rate q, yaw rate r, roll angle f, pitch angle ?, yaw angle ?,
latitude position pN,longitude position pE, andthe altitude h,only eight ofthem,V, a, , p, q, r, f, and
?, are directly controlled bythe four controlinputs: aileron da,rudder dr, elevator de,andthrust dT. The
other four state variables,including pN, pE, h, and ?, arefunctions ofthe 8trim (directly controlled)
state variables. The relationship between the derivative of the altitude h? andthe trim state variablesi
described
by
h?(t) =Vsin?
cosacos
-Vsinf
cos?
sin -Vcosf
cos?
cossina
(8.81)
290
8
Stability,
Regulation,
and Root Locus
Design
If thelateral systemis controlledto keepthe aircraft fly straight(i.e., f = 0and
= 0),thenthe altitude
equation can be simplified to the following,
Fig. 8.42: Flight path angle tracking
control is employed to achieve F/A18
altitude regulation.
h?(t) =Vsin?
cosa-Vcos?
sin
:=f(x)
(8.82)
The altitude regulation via the flight path angle tracking control can be achieved based on either one
of the two schematic
block
diagrams
shown in Figure
8.42. These two
block diagrams
are similar.
The
only difference is in the flight path angle tracking controller. Figure 8.42(a) employs the state feedback
with integral tracking control structure, asshown in Figure 8.38 while Figure 8.42(b) adopts the
PI controller, asin Figure 8.35.
Note
thatin the altitudeequationof Equation
8.82,the derivative
ofthe altitude
h? is afunctionof
the actual-valuestate variablesx(t) instead of the perturbed-valuestate variablesx(t) in the linearized
state-space model.Sincethe nominal flight trim is chosento bethe straight-levelflight with 10?angle
attack,as shownin Equation 8.63,the relationship betweenx(t) andx(t) is shown as
x(t) =[V(t)
a(t) q(t) ?(t)]T =x(t)+x*
=[V(t) a (t) q(t) ? (t)]T +[43610p/180 0 10p/180]T
(8.83)
where10?and 0?/s arereplaced by 10p/180 rad and 0 rad/s,respectively, becausein computations we
should
only use radians instead
of degrees as the unit of angles.
Theintegration of?his h = ?h+h0, whereh0is the initial value of h. The altitudereferenceinput is
hR = ?hR +h0, where ?hRis the desired increment of altitude. The objective of the outer-loop control
istocontinuously
adjust
thedesired
flight
path
?Raccording
tothedifference
?hR
-?huntil
thedif-ference
reaches zero. Theinner-loop flight path integral gains K1and K2 were obtained in the previous
section
using the root locus
diagram
as K1 = 3.5 and K2 = 0.004, respectively,
for the state-feedback
with
integral
tracking
and
thePIcontrol
design
approaches.
The
outer
looph-?gain
Kh?1
andKh?2
can
also
bechosen
using
root
locus
design
orsimply
byafewiterative
simulations.
These
h-?gains
arechosen
asKh?1
=210-5andKh?2
=610-6inthealtitude
tracking
control
simulations
result
8.8
Exercise
Problems
291
shown in Figure 8.43.
Note that in these flight path and altitude control examples, only the elevator is employed to
control the flight path with the thrust fixed at the trim. If both the elevator and the thrust are
utilized, the performance would have been better.
Fig. 8.43: Simulations
of altitude regulation
of F/A18
using flight
path angle tracking
control.
8.8 Exercise Problems
P8.1a: Considerthe typical feedback control system shown in Figure 8.44. Letthe plant be G(s) = s+1
1 ,
andthe controllerbea proportionalcontroller K(s) = Kp. Notethat therearetwo externalinputs:the
reference input r(t) and the disturbance input d(t). Let the transfer function from R(s) to E(s) and
the transfer function from D(s) to E(s) of the closed-loop system be denoted by Ger(s) and Ged(s),
respectively. Then the error response E(s) will be
E(s) = Ger(s)R(s)+Ged(s)D(s)
Find the transfer functions
(8.84
Ger(s) and Ged(s)in terms of the parameter Kp.
P8.1b:
Let
r(t) =10us(t)
and
d(t)=-2us(t),
where
us(t)
istheunit
step
function.
Find
thesteady-state
tracking error
lim e(t)
t?8
=limsE(s)
s?0
in terms of the parameter Kpby using Equation8.84 and Theorem2.27,the final-value theorem.
292
8
Stability,
Regulation,
and Root Locus
Design
P8.1c: Repeat Problem P8.1a with the controller replaced by the PI controller
transfer functions Ger(s) and Ged(s)in terms of the parameters Kpand Ki.
K(s)
=Kp + s.
Ki Findthe
P8.1d: Repeat Problem P8.1b with the controller replaced bythe PI controller
K(s)
=Kp + s
Ki. Findthe
steady-state tracking
error
lim e(t)
t?8
=limsE(s)
s?0
in terms of the parameters Kpand Ki by using Equation 8.84 and Theorem 2.27, the final-value theorem.
Fig. 8.44:
P8.1e:
Comment
Atypical
on the steady-state
feedback
tracking
control
error results
system structure.
of Problems
P8.1b and P8.1d, and explain
what causes the difference.
P8.2a: Consider a simplified
version of Equation 8.16 as
x?(t) =
-cx2(t)+
M
1
u(t) :=
M
f(x(t),u(t))
(8.85)
where the single state variable x(t) represents the velocity of the vehicle, u(t) is the force generated
by the engine to movethe vehicle, M = 1,929 kg is the mass of the vehicle, c = 0.643 kg/m is the
viscosity coefficient of the vehicle traveling in the air. The nonlinear function f(x(t),u(t))
is included
for notational consistency with the Jacobian linearization process of Appendix C. The equilibriums can
be obtained by setting?x(t) = 0, or by setting the right-hand side of Equation 8.85 to be zero,
-c(x*)2
+u*=0 ? u*=0.643(x*)2
(8.86)
This equilibrium equationspecifiesthe relationship betweenthe velocity x* andtheinput force u*. Un-like
the simple inverted pendulum example in Section 4.4.2, which has only two equilibrium points,
there areinfinite manyequilibrium points on the equilibrium curve line specified by Equation 8.86. Re-call
that in Section8.2.1, weselected x* = 30 m/s, whichleads to u* = 0.643(30)2 = 578.7 N, and
the linearized state-space modelatthis equilibrium (x*,u*) = (30 m/s,578.7 N) wasfound as Equa-tion
8.19. Now, assume we wantto investigate the system behavior whenthe vehicle velocity is around
20 m/s, whatequilibrium point (x*,u*) shall weselect?
P8.2b: Atthe equilibriumpoint(x*,u*) =(20 m/s,u*), whichyouhavejust chosenatthe endof Prob-lem
P8.2a,find the linearized state equation of the vehicle x?(t)
C.5 of Appendix
= x(t)+B
A
u(t)
according to Equation
C.
P8.3a: Considerthe proportional feedback control system on Figure 8.45. The objective is to draw the
root loci
diagram for the system according
to the root loci
construction
rules in Section 8.3.1 and the
8.8
Fig. 8.45: Review the root loci
construction
Exercise
Problems
293
rules.
determine a value ofK based onthe diagram so that the closed-loop system poles are at optimal locations
to deliver a desired performance.
function
(LTF)
The first step is to follow
of the system, the number
Rules 1, 2, and 3to identify
of LTF poles as n, the
the loop transfer
number of LTF zeros as m, and then
mark
thenLTF
poles
andmLTF
zeros
onthecomplex
plane
as and
?r,espectively.
P8.3b:
to
The second step is to draw the root loci
segments on the real axis of the complex
Rule 4. Although this step is quite straightforward,
it is the
most important
plane accord-ing
step in the root loci
construction. Notethat all root loci are originating from LTF polelocations andterminated either at LTF
zero locations
or at infinity
on the asymptotes,
which
will be given by Rule 5.
P8.3c: Follow Rule 5 or Equation 8.54 to compute the angles and the intersection of the asymptotes with
therealaxis.
Then
draw
these
n-masymptotes
onthecomplex
plane
indashed
lines.
P8.3d: Oneach root loci segment on the real axis, if both ends of the segment are poles, then there will
be a break-out
point
on this segment.
If both ends are zeros, then there
will be a break-in
point on the
segment. The break-out or break-in points can be obtained by Rule 6 or Equation 8.55in Section 8.4.2.
P8.3e: Rule 7is about the imaginary axis crossing. The root loci of the system only intersects with the
imaginary at the origin when K = 0. Rule 8 provides a formula to compute the angle of departure or
arrival. Since all the LTF poles and zeros of the system are onthe real axis, the angles of departure or
arrival are trivialeither
0 or 180 degree. Now, based on the information above, a sketch of the root loci
diagram can be roughly drawn. Meanwhile, a more precise root loci diagram can be obtained using the
MATLAB command rlocus
as demonstrated in Section 8.3.2. Compare your root loci diagram sketch
with the graph generated by the MATLAB program, andfind a best choice ofK based onthe closed-loop
pole locations and their corresponding damping ratios, natural frequencies, and time constants.
P8.3f:
Build a Simulink
or
MATLAB
simulation
program
according
to the feedback
control
system
block diagram in Figure 8.45. Then conduct simulations by applying a step input to r(t), and observe
the steady-state and transient step responses at y(t) and u(t). You may need to determine a trade-off
between the performance on y(t) and the control effort constraint on u(t) by iterating the value of K
several times.
Finally, comment
on your design and the simulation
results.
P8.4a: Consider another proportional feedback control system similar to that shown in Figure 8.45 but
with the plant transfer
function
replaced
by the following:
G(s) =
s+6
s(s+2)2
Find the angles of the asymptotes and the intersection of the asymptotes with the real axis
294
8
Stability,
Regulation,
and Root Locus
P8.4b:
Compute the break-out/break-in
Design
point (or points)
on the root loci,
and compute the value of K at
these points using the magnitude equation, Equation 8.52.
P8.4c: Find the intersections
of the root loci
with the imaginary
axis and compute the value ofK
at these
intersections.
P8.4d: Drawthe root loci diagram by hand based on the information
obtained from P8.4ato P8.4c.
P8.4e: Draw the root loci diagram using MATLAB command rlocus
to verify your hand-drawn dia-gram
from P8.4d.
P8.4f: Usethe root loci diagram to locate the dominant roots that would render damping ratio ? = 0.707.
P8.4g: Compute the value of K at the dominant root locations.
P8.4h: Find the closed-loop system transfer functions,
P8.4g.
Y(s)/R(s)
and U(s)/R(s),
with the Kfound in
P8.4i: Plot the step responses, y(t) and u(t), of the closed-loop system using the transfer functions in
P8.4h.
Fig. 8.46: Rootlocus design with phase-lead compensation.
P8.5a: Consider the feedback control system with phase-lead compensation on Figure 8.46. Before us-ing
the root locus design with phase-lead compensation, we would like to confirm that the proportional
controller Kandthe PI controller (Kps+Ki)/s
are unableto stabilizethe system. Verifythe confirma-tion
using the root locus design theory.
P8.5b: Root locus design is not just about choosing the right value of the parameter K. The controller
structure that determines the LTF (loop transfer function) pole-zero pattern is at least asimportant.
Adding
thephase-lead
compensator
(s+b)/(s+a)
with
b<aisequivalent
toadding
azero
ats =-b
and
apole
s =-atotheLTF
pole-zero
pattern
thatcanchange
thelocation
oftheasymptotes,
the
break-out/break-in
(s+b)/(s+a)
not
points, the moving direction of the roots trajectories, etc. If b > a then the compen-sator
will become a phase-lag compensator. Showthat the phase-lag compensation will
work based on the root locus
design theory.
P8.5c: Vary the values of b and a, one at atime, to observe how they will change the shape of the root
loci
diagram.
Then find a best choice of K based on the closed-loop
damping ratios,
natural frequencies,
and time constants
pole locations
and their
correspond-ing
8.8
P8.5d:
Build a Simulink
or
MATLAB
simulation
program
according
Exercise
Problems
to the feedback
control
295
system
block diagram in Figure 8.46. Then conduct simulations by applying a step input to r(t), and observe
the steady-state and transient step responses at y(t) and u(t). You may need to determine a trade-off
between the performance of y(t) and the control effort constraint on u(t) by modifying the value of K a
few times.
Finally,
comment
on your design and the simulation
results.
P8.6a: For the feedback position control system shown in Figure 8.47,the plant, which is the system to
be controlled, is an extremely lightly damped system with damping ratio ? = 0.0125. In order to achieve
zero steady-state tracking
error, the controller
needs to include
an integrator.
If there is no compensation,
therewould
benoLTF
zeros,
only
three
LTF
poles
attheorigin,
and-0.05j4.The
twobranches
of
the root loci
originating
from the complex
away and have no way of coming
pair
will jump into the right
half of the complex
back. In order to bend the angle of departure of the two
poles toward the left half of the complex plane and movethe asymptotes to intersect
plane right
complex
plant
with the negative
real
axis,
twozeros,
s =-band
s =-c,and
one
pole,
s =-awith
b <aand
c <a,need
tobeadded
to the controller. Show that the controller K(s+b)/(s(s+a)),
compensator, is still not ableto stabilize the system.
anintegral
controller
Fig. 8.47: Rootlocus design with double zeros compensation for an extremely lightly
control system.
P8.6b:
they
Vary the values
of the three compensator
will change the shape of the root loci
parameters
diagram.
with phase-lead
damped tracking
b, c, and a, one at a time, to observe how
Then find a best choice
ofK
based on the closed-loop
pole locations and their corresponding damping ratios, natural frequencies, and time constants.
P8.6c: Build a Simulink or
block
diagram in Figure
MATLAB simulation
8.47. Then conduct
program according to the feedback control system
simulations
by applying
the steady-state and transient step responses at y(t) and u(t).
between the performance
few times.
comment
Or you
of y(t)
and the control effort
constraint
a step input
and observe
You may need to determine a trade-off
on u(t)
may even need to go back to readjust the compensator
on your design and the simulation
to r(t),
by modifying the value of K a
parameters
b, c, and a. Finally,
results.
P8.7a: The plantto be controlled in the feedback position control systemshown in Figure 8.48(a) is
exactly the same as that considered in Problem P8.6, where an integral controller structure with two
zeros/one pole compensation was employed in the root locus design. This P8.6 solution resolves sta-bility
and the lightly
second since the
damped issue;
however, the response is slow
dominant real pole is too
close to the imaginary
with time
axis.
constant larger
than one
This issue can be resolved if the
plant
complex
poles
-0.05j4canberelocated.
The
new
approach
willemploy
astate
feedback
to
movethis complex pair of poles to the new locations with damping ratio ? = 0.9 and natural frequency
?n = 10 rad/s. Find the state-feedback gain matrix F
296
8
Stability,
Regulation,
and Root Locus
Design
P8.7b: Notethat the two block diagrams in Figure 8.48(a) and (b) are equivalent. Show that the loop
transfer
function
(LTF)
ofthesystem
iss-1C[sI-(A+BF)]-1B.
P8.7c: Drawthe root loci diagram associated withthis LTF. Vary the damping ratio andthe natural fre-quency
to obtain a few corresponding F matrices,and then observe how F will affect the shape of the
root loci diagram.
P8.7d: Build a Simulink or MATLAB simulation program according to the feedback control system
block diagram shown in Figure 8.48(a). Then conduct simulations by applying a step input to r(t),
and observe the steady-state and transient step responses at y(t) and u(t). You may need to determine
atrade-off between the performance of y(t) and the control effort constraint on u(t) by modifying the
value of
K a few times.
Finally, comment
Fig. 8.48: Root locus
control system
Or you
may even need to go back to readjust
on your design and the simulation
design
with state-feedback
the state feedback
gain
matrix
results.
compensation
for an extremely
lightly
damped tracking
F.
9
Time Delay,Plant Uncertainty, and RobustStability
I
N Chapter 8, welearned how to select a feedback control system structure to achieve steady-state
disturbance response rejection and steady-state reference input tracking using the concept of the
type of a system and the internal modelprinciple. Wehave also studied, to a great extent, using root
locus design technique to fine-tune the transient response of the closed-loop system. In this chapter, we
will consider the stability issues caused by time delay and plant uncertainties. The Nyquist stabil-ity
criterion,
and
plant
developed
by Nyquist in 1932,
uncertainties
affect the closed-loop
will be employed
system
to characterize
stability.
how the time
The Nyquist approach
delay
also provides
two robust stability measures(i.e., gain and phase margins)to specify the allowable perturbations of the
gain and the phase of the loop transfer function at two specific frequencies for the closed-loop system
to remain stable.
frequencies
These two stability
instead
of the
margin
whole frequency
measures only reveal robust stability
spectrum.
Furthermore,
these two
information
at the two
measures only
work for
SISO (single-input/single-output)
systems. The attempt of extending the concept of gain/phase margins
to MIMO (multi-input/multi-output)
systems was not successful since there is no meaningful definition
for the phase of a matrix loop transfer
The success
all frequencies
of finding
arrived
function.
a more general robust stability
shortly
after the
small
measure that
gain theorema
works for
generalization
MIMO systems
of the
Nyquist
at
crite-rion
to MIMO systemswas
proved by George Zames in 1966 [Desoer and Vidyasagar, 1975]. The
generalized stability margin is defined in terms of the maximum singular value (or the magnitudefor
SISOsystems)of complementary
sensitivityfunction matrix T(j?)
= L(j?)[I+L(
j?)]-1,
L(j?) is the loop transfer function matrixofthe closed-loop system. Thegeneralized stability
works for
frequency
where
margin
both SISO and MIMO systems and provides robust stability information for the entire
spectrum. The frequency-dependent generalized stability margin function gives the maxi-mum
allowable variation of the maximum singular value (or the magnitude for SISO systems) of the
plant for every frequency
so that the closed-loop
system can still remain
stable.
9.1 Stability Issues Caused by Time Delay and Plant Uncertainty
The first
and the
most important
requirement
in the design of feedback
control system is stability,
since
an unstable system is not only useless, it can be harmful or even potentially can cause a disaster. As
discussed in Section 7.3, a linear time-invariant system is stable if and only if all the roots of its charac-teristic
equation
are in the strictly left
half of the complex
plane.
Thus, in the design of feedback
control
systems as described in Chapter 8, the characteristic equation of the closed-loop system 1+L(s)
= 0,
where L(s) is the loop transfer function of the closed-loop system, should have all its roots in the strictly
left
half of the complex
plane
298
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
The stability of feedback control systems can be affected bytime delay and plant uncertainty; hence,
it is imperative
to incorporate
the considerations
the design process to ensure the system
criterion
and the stability
of all possible time
will remain
margins concept
stable under all
derived from the
delays and plant uncertainties
worst-case scenarios.
Nyquist approach
in
Nyquist sta-bility
were the first tools
developed to address the two important stability issues.
9.1.1 Time Delay and Stability
In feedback
control systems, time
delay occurs in almost every sensing and actuation
process.
Hence, its
effect on stability is an important practical issue. However,this issue was not resolved until 1932 when
Nyquist published
one of the
known
as Nyquist stability
is that
a feedback
infinite
mostimportant
criterion.
control
system
control
systems theories in [Nyquist,
The obstacle that prevented the problem from
with time
delay is an infinite
dimensional
1932],
which is now
being resolved
system,
which
earlier
has an
number of closed-loop system poles.
If the system is finite dimensional, or, equivalently, the number of the roots of the characteristic equa-tion
is finite, then the stability ofthe closed-loop system can be determined based onif the characteristic
equation
has a root in the
RHP, which stands for the right
half of the complex
plane.
One wayto know if there exist closed-loop system poles in the RHP is to compute all these poles
by solving for the roots of the associated characteristic equation, or to compute the eigenvalues of the
corresponding state-space model. Alternatively, the Routh-Hurwitz stability criterion approach of Sec-tion
7.7 can be employed to check the number of poles in the RHP without the need of solving for the
roots or the eigenvalues.
ways of choosing
Furthermore,
controller
the root locus
parameters
design approach
so that the closed-loop
described in Section 8.4 provides
system
will not have poles in the
RHP.
However, these approaches do not work for the infinite-dimensional
systems; therefore, they will
not work for the systems with time delay either since the feedback control systems with time delays
are infinite-dimensional
systems.
Fig. 9.1: Afeedback control system with aloop transfer function
L(s) that includes atime delay.
Example 9.1 (A Closed-Loop System with Time Delay Is an Infinite-Dimensional
System)
Considerthe feedback control system shown in Figure 9.1. If the loop transfer function L(s) is given
a
L(s) =
2
s+1
then the closed-loop characteristic equation will be
1+L(s)
= 1+
2
s+1
=0 ? s+3=0
and
therootoftheequation
can
befound
ats =-3toassert
that
theclosed-loop
system
isstable.
9.2
Recall that the loop transfer function
controlled,
the sensor, the actuator,
or actuating
process.
Contour
Mapping and Cauchys
Principle
of the
Argument
299
L(s) usually consists of the plant, which is the system to be
and the controller.
Now, assume the delay time is
L(s) =
In practice,
a time
delay
T. Then the loop transfer
2
may occur in the sensing
function
will become
e-sT
s+1
(9.1)
wherethe term e-sT is the transfer function of atime delay element with delaytime T as describedin
Theorem 2.22. Thusthe closed-loop characteristic equation will turn out to be
F(s) = 1+L(s)
all the infinite
number
1
3
s+1
which apparently is a polynomial
compute
?3+(1-2T)s+T2s2
- T3s3
+ ?
=0
1
=
equation
with infinite
of roots, the stability
numbers of roots. Sinceit is impossible to
analysis
time delay was an unresolved issue before Nyquist stability
9.1.2 Plant Uncertainty
A feedback
control
(9.2)
of feedback
criterion
control
system involving
was discovered.
and Stability
system
usually is
designed
based on a mathematical
model of the system to
be
controlled,
which is called the plant. In practice, the real system to be controlled is not identical to the
ideal
model due to unmodelled
plant
plant dynamics,
specification
tolerance
of components,
and plant
parameter perturbations influenced by the environment conditions. Hence, a feedback control system
not only needs to be stable for the nominal system with the ideal plant model,it should be designed
to achieve robust stability against all possible plant uncertainties.
The Nyquist stability criterion
complex integral
1997].
theorem
was developed by Nyquist in 1932 [Nyquist,
presented by a French
The Nyquist approach
mathematician
not only resolves the stability
control systems, it also provides important
1932] based on Cauchy
Augustin-Louis
Cauchy in 1831 [Smithies,
analysis issue of infinite-dimensional
feed-back
concepts and tools for achieving robust stability
of
feedback control systems, which will be elaborated in later sections of this chapter.
9.2 Contour Mappingand Cauchys Principle of the Argument
Nyquist stability analysis theory was developed based on Cauchys principle of the argument, and the
complex contour mapping wasthe underlying framework of Cauchys complex variable theory, includ-ing
the principle of the argument. Hence, we will briefly review the complex function and complex
contour mapping in the following subsection before discussing Cauchys principle of the argument.
9.2.1 Complex Contour
Complex contour
complex
function
Mapping
mapping is a special complex function, such as F(s), in
is a simple
directional
closed
path
on the complex
which the domain of the
plane.
Here, simple
means
the closed path does not crossitself. The complex variable s is supposed to travel along the closed
contour for exactly one revolution. For each value of s (e.g., s1) on the contour, a corresponding
complex function value F(s1) can be computed and plotted on the complex
image
of the contour
on the s-plane is
mapped onto the F-plane.
F-plane. That meansthe
As s traverses the simple closed contour
once, the image orthe trajectory of F(s) onthe F-plane will also be a closed path, butit maycross itself
and
may encircle
a particular
point
N times,
where N can be any integer, in the same (positive
N) or i
300
9
Time
Delay, Plant
the opposite (negative
Uncertainty,
and Robust Stability
N) encirclement
direction
with the contour
on the s-plane.
To avoid confusion, the closed contour travel direction onthe s-plane can be fixed to be either clock-wise
or counterclockwise. Counterclockwise contour direction has been considered the positive direction
in almost all mathematics books. Onthe contrary, for some practical reasons, the majority of the control
and systems
community
as the positive
the right
Nyquist
has decided to go the other
direction.
half complex
path.
the contour
mapping
compelling
plane is consistent
Therefore,
chosen for the contour
One of the
we will follow
encirclement
reasons
the systems
and control
Note that
counting.
the real function
learning
experience
Fig. 9.2:
contour
rotation
encirclement
Counterclockwise
similarities
of
direction in the
convention
direction
angle
direc-tion
regarding
convention
measurements
in
is only
degrees
direction is positive unless
two examples, we will briefly re-view
and differences,
which
may enhance our
mapping.
A simple real function
Example 9.2 (A Simple Real Function
communitys
mapping,in the following
mapping, and study their
of the complex
the clockwise
this clockwise
For other
or radians, westill follow the universal convention:
otherwise specified.
To better understand complex contour
is that
encirclement
with the positive frequency increasing
direction.
encirclement
way to adopt the clockwise
to demonstrate
the
mapping of a real variable.
Mapping)
Considerthe simple real function
y =f(x) =-x+2
which specifies how the variable x will affect the value of y. Asthe independent variable x varies,
the dependent variable y will change accordingly. The mapping relationship between x and y can be
represented by the tabulated chart,
x -1 0
y
3
2
1
2
3
1
0 -1
(9.3)
or by the graphs shown in Figure 9.2. A graphical representation of a function like the one shown in
Figure 9.2(a) is an effective
real variable
y. But this two-dimensional
complex function
function
way to illustrate
the
mapping relationship
from
graph or even a three-dimensional
mapping since a generalization
mapping would need a four-dimensional
of Figure 9.2(a)s
a real variable
x to another
graph cannot be applied to
graphical representation
to complex
space.
The graph in Figure 9.2(b) looks awkward for real function
mapping. However, the concept can
be easily generalized to complex function
mapping. Thetwo real number lines for the independen
9.2
real variable
x and the
dependent real
Contour
variables
Mapping and Cauchys
y can be generalized
Principle
of the
Argument
301
to the complex
s-plane
and the
complex F(s)-plane.
Fig. 9.3:
A simple
complex function
Example 9.3 (An Illustration
to demonstrate the
of Complex Function
Contour
mapping of a complex
contour.
Mapping)
Considerthe simple complex function
s-0.5
F(s) =
s+0.5
which specifies how the complex variable s will affect the value of the complex function
independent
complex
variable
s varies on the complex
s-plane, the dependent
complex
F(s). Asthe
function
variable
F(s) will change accordingly on the complex F-plane. The domain of the complex function in general
can be any set of complex variables onthe complex s-plane. The domain of the complex function weare
interested,
is a simple
closed contour, similar
to the one shown in Figure 9.3(a). For ease of explanation,
this simple closed contour (or path) Gsis decomposed into two segments. Segment (1), in blue, is on
thej? axis
(the
imaginary
axis)
with
s =j?, where
? =-1 ?0 ?1.Segment
(2),inred,
isa
semicircle
s =ejf, where
f =p/2?0?-p/2.
The F(s)
F(j?)
mappings of segments (1) and (2), respectively,
=
j?-0.5 = ej(p-?)
j?+0.5
= ej(p-2?),
will be
where ? =tan-1 ?
ej?
, ? =-1 ?0 ?1
0.5
and
F(ejf)
=
ejf -0.5=
ejf
The mapping relationship
s
+0.5
0.75+ j sinf
(cosf +0.5)2 +sin2f
,
where f =
p
2
?0 ?
between s and F(s) can be represented bythe following
-j
F(s) e-j53?
-p
2
tabulated chart:
? -j0.5? 0 ? j0.5 ? j ? ej0
?e-jp/2? -1 ?ejp/2 ?ej53?? 1/3
(9.4
302
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
Now we have completed the F(s) mapping GF of the closed contour Gsshown in Figure 9.3(b).
After obtaining the complex contour mapping GF, we are particularly interested in the number and
direction of the encirclements of the origin by GF on the F-plane. It can be seenthat GF encircles
the origin once clockwise, whichis in the same direction of the Gscontour in the s-plane. Therefore, the
number
of encirclement
is
N= 1. In the case that the encirclement
direction
were counterclockwise,
the
number
would
have
been
N=-1.
9.2.2 Cauchys
Principle of the Argument
One version of the Cauchy complex integral theorem is called Cauchys principle of the argument or
Cauchys complex contour mappingtheorem, which is given next.
Theorem 9.4 (Cauchys
Principle of the Argument)
Let Gsbe a simple closed curve in the (complex) s-plane, as shown in the left graph ofFigure 9.4.
F(s) is a rational function having no poles or zeros on Gs. Let GF be the image of Gs under the map
F(s). Then,
N=Z-P
where
Nis the number ofclockwise encirclements ofthe origin by GF ass traverses Gsoncein the clockwise
direction;
Zis the number ofzeros ofF(s)
enclosed by Gs,counting multiplicities;
and
Pis the numberofpoles ofF(s) enclosedby Gs,counting multiplicities.
Proof:
Notethat the rational function
F (s)
= K
(s-z1)(s-z2) (s-zm)
(s-p1)
(s-p2) (s-pn)
(9.5)
can be expressed as:
F(s) = K
?z1ej?z1?z2ej?z2
?p1e
j?p1?p2e
j?p2
K?z1?z2
?zm
?zme
j?zm
ej[(?z1+?z2++?zm) - (?p1+?p2++?pn)]
=
?pn
?pne
j?pn ?p1?p2
(9.6
where
j =1, ,n
?sp
j
are
themagnitudes
ofthephasors
s-ziands- pj,respectively,
and
?zi=|s-zi|, i =1, ,m and
?pj =
?
?,
?
?zi=?(s-zi),i =1, ,m and ?p
j =?(s-pj), j =1, ,n
are
theangles
(arguments)
ofthephasors
s-ziand
s- pj,respectively.
In view of Equation9.6, asstraverses Gsonce,its image GFencirclesthe origin only if atleast oneof
the angles ?j undergoesa changeof 2p radians. Any pole or zero outside of Gsdoesnot produce any
angle change through a circuit of Gs. Onthe other hand, a pole or zeroinside of Gsdoes produce
a 2p angle change. This is easily visualized in Figure 9.4. Equation 9.6 implies that for a complete
clockwise transverse of Gs,each zero inside of Gs produces a clockwise 2p angle change and each
pole inside of Gs produces a counterclockwise 2p angle change. Hence,the net number of clockwise
encirclements
oftheorigin
byGF
is Z-P.
9.2
Example
9.5 (Illustrate
The complex
contour
the
Principle
Contour
of the
mapping example
Mapping and Cauchys
Principle
of the
Argument
303
Argument)
shown in Figure 9.4 is employed
to illustrate
the theorem
on
principle of the argument. Assumethe complex rational function is
F(s) =
s-z1
s-1
s-1
=
=
(s-p1)(sp2)(sp3) (s+1-j)(s+1-j)(s-3) s3-s2-4s-6
(9.7)
and the simple closed path Gsis a circle centered at the origin of the s-plane with radius equals to 2. For
clarity, Gsis partitioned into two segments: Segment (1) is in red, which starts from s = 2, clockwise
and
ends
ats =-2.Immediately,
segment
(2),which
isinblue,
begins
from
s =-2,moving
along
theupper
semicircle
togoback
tos =2tocomplete
one
revolution
alongthe semicircle to s = 2e-j135?
of the circle.
Fig. 9.4: Complex contour
mapping and the principle
of the argument.
Theimage of segment(1) of Gsis shown in the right graph of Figure 9.4 as segment (1) of GF,
to
also
inred.
It canbeseen
thatthestarting
point
s =2is mapped
to F(2)= -0.1,
s =2e-j135?
F(2e-j135?)
= -0.16212j0.38819,
ands = -2findsitsimage
atF(-2) =0.3onGF.Similarly,
thethree
points
onsegment
(2)ofGs:
s =-2,s =2e
j135?
, and
s =2aremapped
to F(-2)=0.3,
F(2e j135?
) = -0.16212+
j0.38819,
andF(2)= -0.1,respectively,
onGF.Note
thatGF
issymmet-rical
with respect to the real axis, and it encircles the origin of the F-plane once counterclockwise.
Hence,
thenumber
ofclockwise
encirclements
oftheorigin
byGF
is N=-1.Recall
that
thenumber
of poles of F(s) enclosed by Gsis P = 2, and the number of zeros of F(s) enclosed by Gsis Z = 1.
Therefore,
N =-1=Z-P=1-2,which
isconsistent
with
theresult
ofTheorem
9.4.
Now
wewillemploy
thisexample
toillustrate
whyN=Z-P.Following
thenotations
ofEquation
9.6, Equation 9.7 can be rewritten asthe following:
F(s)
=
?z1
s-z1
ej[?z1-(?p1+?p2+?p3)]
=|F(s)|?F(s)=
?p1?p2?p3
(s-p1)(sp2)(sp3)
(9.8)
Asthe complex variable s traverses around Gs,both the magnitude and the phase (also called angle or
argument) of F(s) will change accordingly. However, the number Nof clockwise encirclements of th
304
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
origin by GFin Theorem 9.4 is independent of the magnitudes ?z1, ?p1, ?p2, and ?p3. Furthermore, the
poles and zeros of F(s) that are not enclosed by Gsalso do not contribute to the number Nof clockwise
encirclements of the origin by GF. Asshown in the left graph of Figure 9.4, the pole p3is not enclosed
byGs.
Assstarts
froms =2,moving
around
Gs
once
through
thepoints
s = -j2,s =-2,s =j2,
and
back
tos=2,thephase
of?(s-p3)
will
change
frompoint
topoint,
but
theoverall
net
change
of the phase is zero, asit can be clearly seen from
2
s
? -j2
Equation 9.9.
? -2 ? j2
? 2
(9.9)
?p3
=?(s-p3)-180?? -146.3?
? -180?? -213.7?
? -180?
Onthe other hand, each zero or pole of F(s) enclosed by Gs will contribute to increasing or decreas-ing
the number of encirclements of the origin by GF. The zero z1 = 1is enclosed by Gs. After s traverses
around
Gs
once,
thenet
change
ofthephase
?z1
=?(s-z1)
willbe-360?,
asshown
inEquation
9.10.
2 ? -j2 ? -2 ? j2
s
? 2
?z1
=?(s-z1)0? ? -90?? -180?? -270?? -360?
(9.10)
Similarly,
thenet
changes
ofthephases
?p1
= ?(s- p1)and?p2
=?(s-p2)willbealso-360?,
respectively, since these two poles are also enclosed by Gs. Hence,the net change of the phase of F(s)
and the number of encirclements Nof the origin by GF can be summarized asfollows:
?z1
- (?p1+?p2
+?p3)
=(-360?)-[(-360?)+(-360?)+0?]
? N=Z-P=1-2 =-1
Note
thata -360?
phase
change
isequivalent
toone
clockwise
encirclement
oftheorigin
byGF.
In this example, wehave witnessedthat the polesand zeros of F(s) that are not enclosedby Gs will
not affectthe numberof encirclements of the origin by GF. However,they do affect the magnitudeof
F(s), and consequently will change the shape of GF. In the following example, we will see how the
removal of the pole p3from the F(s) of the previous example will affect the contour mapping.
Example 9.6 (A Pole or Zero Outside Gs Does Not Affect the Encirclement
Change the Shape of Contour Mapping GF)
The complex rational function
Number
N, but
Will
F(s) under consideration hereis almost the same asthat considered
in the previous example, except that the pole p3 outside Gsis removed. Thatis,
F(s) =
s-z1
=
s-1
(s-p1)(sp2) (s+1-j)(s+1-j)
=
s-1
s2 +2s+2
(9.11)
The simple closed path Gsis still the same, whichis a circle centeredatthe origin of the s-plane with
radius equal to 2. For the sake of clarity, Gsis partitioned into two segments: Segment (1) is in red, which
isthelower
halfcircle,
starting
from
s =2and
ending
ats =-2,and
segment
(2)isinblue,
which
is
theupper
half
circle,
starting
from
s =-2and
ending
ats =2.
Theimage of segment(1) of Gsis shownin the right graphof Figure 9.5 assegment(1) of GF,alsoin
) = 0.5,
red.
It can
beseen
thatthestarting
point
s =2is mapped
to F(2)=0.1,
s = -j2to F(2e-j90?
) =j1.978,
and
s =-2finds
itsimage
atF(-2)= -1.5onGF.
, s = j2, and s = 2 are mapped
Similarly,
thefourpoints
onsegment
(2)ofGs:
s =-2,s =2e
j136.4?
and s = 2e-j136.4?to F(2e-j136.4?
) = 0.5 and
to F(-2)=-1.5,F(2e
j136.4?
) = -j1.978,
F(j2) =F(2e-j90?
F(2) = 0.1, respectively,
on GF. Notethat GFis symmetrical with respect to the real axis, and it encircles the origin of the F-plane
once, counterclockwise.
Hence,the number of clockwise encirclements of the origin by G
9.2
Contour
Fig. 9.5: Complex contour
Mapping and Cauchys
Principle
mapping for Example
of the
Argument
305
9.6 .
is N=-1.Recall
that
thenumber
ofpoles
ofF(s)
enclosed
byGs
isP=2,and
thenumber
ofzeros
ofF(s)
enclosed
byGs,
isZ=1.Therefore,
N=-1 =Z-P=1-2,which
isconsistent
with
the
result of Theorem 9.4.
Notethat the N, Z, and P numbers are the same asthose in Example 9.5, but the shape of contour
mappingGFis very different from that in Example9.5.
In the next example, we will observe how the contour mapping GF and its clockwise encirclement
number around the origin will be affected if the zero, z1,inside Gsis removed.
Fig. 9.6: Complex
contour
mapping for
Example
Example 9.7 (A Change of the Number of Poles or Zeros Inside
Number N of GF Around the Origin)
The complex rational
function
F(s) under consideration
9.7.
Gs Will Affect the Encirclement
here is almost the same as that
considered
in the previous example, except that the zero, z1,inside Gsis removed. Thatis,
F(s) =
1
1
=
(s-p1)(sp2) (s+1-j)(s+1-j)
1
=
s2 +2s+2
(9.12
306
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
The simple closed path Gsis still the same, which is a circle centered at the origin of the s-plane with
radius equal to 2. Forthe sake of clarity, Gsis partitioned into two segments: segment (1) is in red, which
isthelower
halfcircle,
starting
from
s =2and
ending
ats =-2,and
segment
(2)isinblue,
which
is
theupper
half
circle,
starting
from
s =-2and
ending
ats =2.
Theimage of segment (1) of Gsis shown in the right graph of Figure 9.6 assegment (1) of GF, also
) =
inred.
It can
beseen
that
thestarting
point
s =2is mapped
to F(2)=0.1,
s = -j2to F(2e-j90?
0.1+ j0.2, s =2e-j120?
to F(2e-j120?
to
F(2e-j144?
) =-0.5,
and
s =2e-j144? ) =-j0.688,
and
s =-2
finds
itsimage
atF(-2)=0.5onGF.
Similarly,
theimage
ofsegment
(2)ofGs
canbefound
asthe
conjugate
of the image
of segment (1), as shown in the right
Notethat GFis symmetrical
graph of Figure 9.6.
with respect to the real axis, and it encircles the origin of the F-plane
twice,
counterclockwise.
Hence,
thenumber
ofclockwise
encirclements
oftheorigin
byGF
is N=-2.
Recall that the number of poles of F(s) enclosed by Gsis P = 2, and the number of zeros of F(s)
enclosed
byGs
is Z=0.Therefore,
N=-2=Z-P=0-2,which
isconsistent
with
theresult
of
Theorem
9.4.
9.3 Nyquist Path, Nyquist Plot, and Nyquist Stability Criterion
As mentionedin the beginning of the chapter, for the typical feedback control system shown in Figure
9.7, the closed-loop
system stability
is determined
by the closed-loop
1+L(s)
where L(s)
=G(s)K(s)
is the loop transfer
function,
system characteristic
equation,
= 0
(9.13)
and G(s) and K(s) are the system to be controlled
(called the plant) and the controller, respectively. The closed-loop system is stable if and only if all the
roots of the characteristic equation are in the strictly left half of the complex plane. One wayto check the
stability of the closed-loop system is to solve the equation for all the roots. This approach may become
tedious or evenimpossible if the equation is of high order or hasinfinite number of roots. As described in
Example 9.1, a system with time delay is an infinite-dimensional system whose characteristic equation
has an infinite number of roots.
Fig. 9.7: Atypical feedback control system block diagram for
Nyquist stability analysis.
9.3.1 Nyquist Path
The Nyquist stability
stability
analysis
approach
without the need of finding
provides
an alternative
the roots of the characteristic
way to check the closed-loop
equation.
The tool
system
Nyquist employed
to solve the very important engineering problem wasthe Cauchy complex integral theorem or Cauchys
principle of the argument, which was briefly reviewed in Section 9.2. In order to find if there is any
closed-loop system pole (the characteristic equation roots) in the right half of the complex plane
9.3
Nyquist
Path,
Nyquist
Plot, and Nyquist
Stability
Criterion
307
Nyquist took two brilliant steps: (1) Create a special simple closed path Gs,now called Nyquist path
or Nyquist contour, to enclose the entire right half of the complex plane; (2) Define the mapping
function
(9.14)
F(s) = 1+L(s)
so that the zeros of F(s) are the closed-loop system poles.
The Nyquist pathis a simple closed path Gsthat encloses the entire right half of the complex s-plane,
as shown in Figure 9.8. The mappingfunction F(s), which is the closed-loop characteristic function, is
assumedto have no poles or zeros on Gs.If F(s) has a pole onthe imaginary axis of the s-plane, atiny
semicircle contour withradius e will be employedto go aroundit, asshownin Figure 9.8b.
If F(s) has no poles onthe imaginary axis, the Nyquist path Gs will look like the one shown in Figure
9.8a. For clarity, the Nyquist pathis divided into the following three segments:
1:
Segment (1) starts from the origin of the s-plane,
moving up on the imaginary
axis with s = j?,
?=0?8.
2: Segment
(2)isahuge
semicircle
with
s =limRejf
and
f =90??0??-90?.
R?8
3: Segment
(3)isthecomplex
conjugate
ofSegment
(1),which
starts
from
s = -j8,moving
upon
the
imaginary
axis
back
totheorigin
ofthes-plane
(i.e.,
s=j?, ? =-8?0).
In the casethat F(s) has a pole at the origin, asshown in Figure 9.8b, the Nyquist path Gs will have
four segments
with a tiny semicircle
segment (4) inserted
between segments (3) and (1):
1: Segment
(1)starts
from
s =je,moving
upontheimaginary
axiswith
s =j?, ?=e ?8.
2:
Segment (2) is a huge semicircle
with s = lim
R?8
Rejf
and
f =90??0??-90?.
3: Segment
(3)isthecomplex
conjugate
ofsegment
(1),which
starts
from
s =-j8,moving
uponthe
imaginary
axiswith
s =j?, ? =-8?-e.
4: Segment
(4)isatinysemicircle
with
s =limee
jf and
f =-90??0??90?.
e?0
Fig. 9.8: The Nyquist path is a simple closed paththat encloses the right half of the s-plane
308
9
9.3.2
Time
Nyquist
Delay, Plant
Uncertainty,
and Robust Stability
Plot
Similar to the complex contour mapping we did in Section 9.2, an F(s) mapping of the Nyquist path
Gscan be produced as GF on another complex plane, the F(s)-plane. This GF plot is called the
Nyquistplot of F(s). Then,accordingto Cauchysprincipleofthe argumentin Theorem9.4, wehave
N= Z-P,where
Nisthenumber
ofclockwise
encirclements
oftheorigin
byGF,
andZandPare
the numbersof zerosand poles of F(s) enclosed by Gsin the s-plane. Since F(s) andthe loop transfer
function
L(s) share the same poles, P is the number
of the poles of L(s) in the right
half of the s-plane,
whichis usually given or can beeasily computed. The encirclementnumber Ncan becountedfrom GF,
the Nyquist plot of F(s). Therefore, we will have Z = N+P, which is the number of zeros of F(s)
enclosed by Gs,or, equivalently, the number of the closed-loop system poles in the right half of s-plane.
The closed-loop system is stable if and only if Z = 0.
In the following
example,
we will construct the
and use it to determine the stability
Example 9.8 (Construction
Consider the feedback
the characteristic
Nyquist plot for a simple
feedback
control system,
of the system.
of Nyquist Plot of F(s) for a Simple Feedback Control System)
control system shown in Figure 9.7, where the loop transfer function
function
F(s) of the closed-loop
L(s) = 2
s-1
s+1
L(s) and
system are given as
? F(s)
=1+L(s)
=3s-1
s+1
(9.15)
respectively. It is easy to seethat F(s) has one zero, s = 1/3, in the right half of s-plane and,therefore
the closed-loop
system is unstable.
However,
we will employ this simple
example to demonstrate
how
to usethe Nyquistapproachto determinethe number of zeros of F(s) enclosedbythe Nyquistpath Gs.
Since F(s) = 1+L(s),
we will construct GL,the Nyquist plot of L(s) first, and then shift the graph
of GLto the right by one unit to obtain the Nyquist plot of F(s). The procedure for constructing GL,the
Nyquist plot of L(s), is given asfollows:
1: On
segment
(1)oftheNyquist
path
Gs,
we
have
s =j?,and? =0?1?8.Then
thetrajectory
of
L(j?)
=2
j?-1
j?+1
=2
v ?2 +1ej(p-?)
v?2 +1ej?
= 2ej(p-2?)
will be a red upper semicircle, shown in Figure 9.9(b),
L(j0)
= 2ejp to L(j1)
= 2ejp/2 and ending at L(j8)
where ? =tan-1?
with radius equal to 2, starting
(9.16)
from
= 2.
2: Segment
(2)isahuge
semicircle
with
s =limRejf
andf =90??0? ?-90?.
With
s =Rejf,
R?8
wehave
lim
L(Rejf)
R?8
hence, the
entire
segment
(2) semicircle
= lim 2
R?8
is
Rejf-1 = 2
Rejf
(9.17)
+1
mapped to the single
point
(2, j0)
on L-plane
indi-cated
by a green dot in Figure 9.9(b).
3:
The image
of segment
(3) is the
blue lower
semicircle
with radius
equal to 2, starting
from
L(-j8)=2toL(-j1)=2e-jp/2
andending
atL(j0) =-2,which
isthecomplex
conjugate
of
the image
of segment (1).
Now we havethe Nyquist plot of L(s), GL, on the L-plane, as shown in Figure 9.9b. Since F(s) =
1+L(s), the Nyquist plot of F(s), GF can be obtained by simply moving GLto the right by one unit, as
shown in Figure 9.9c. It can be observed that the Nyquist plot of F(s) on the F-plane, GF, encircle
9.3
the origin
(0, j0) clockwise
one time,
Nyquist
Path,
which indicates
Nyquist
Plot, and Nyquist
that the encirclement
Stability
Criterion
number is
309
N= 1. Recall
thatF(s)
andL(s)
share
thesame
poles,
and
theonlypole
ofL(s)
iss =-1,which
isnotenclosed
by Gs;hence, the number of poles ofF(s) enclosed by Gsis P= 0. Therefore, according to Theorem
9.4, principle of the argument, we have the number of zeros of F(s) enclosed by Gsis Z = N+P =
1+0 = 1, which implies that the closed-loop system is unstable because it has one pole in the right
half of the complex s-plane.
Fig. 9.9: Construction of Nyquist plots of L(s) and F(s) for the simple feedback control system in
Example 9.8.
The Nyquist plot graph on Figure 9.9(b) is obtained using the following
% CSD
Fig9.9b
Nyquist
plot
based
on
MATLAB code:
mapping
figure(5)
theta=linspace(0,pi/2,500);
re_L=2*cos(pi-2.*theta);
im_L=2*sin(pi-2.*theta);
plot(re_L,im_L,'r-'),
axis([-2.5,2.5,-2.5,2.5]),
hold
re_L=2*cos(pi-2.*theta);
on,
theta=linspace(pi/2,pi,500);
im_L=2*sin(pi-2.*theta);
plot(re_L,im_L,'b-');
axis([-2.5,2.5,-2.5,2.5]);
grid,
The same Nyquist plot can also be obtained using the following
% CSD
s=tf('s');
Fig9.9b
Nyquist
plot
L=2*(s-1)/(s+1);
using
MATLAB commands:
nyquist(L)
figure(21),
nyquist(L)
Remark 9.9(The Number of Encirclements of the Origin(0, j0) by GF Equalsto That of the Crit-ical
Point
(-1,j0)byGL)
From Figures 9.9c and 9.9b, it can be easily observed that the number of clockwise encirclements
of the origin (0, j0) by GFis exactly the same asthe number of clockwise encirclements of the critical
point
(-1,j0)byGL.
For
thisreason,
itisnotnecessary
toconstruct
theNyquist
plotofF(s).
From
now
on,wewilljustemploy
theNyquist
plotofL(s),
GL,
together
with
thecritical
point
(-1,j0)to
determine the stability
of the closed-loop system
310
9
9.3.3
Time
Nyquist
Delay, Plant
Stability
Uncertainty,
and Robust Stability
Criterion
From these discussions, the Nyquist stability analysis can be summarized in the following
Theorem 9.10 (Nyquist
Stability
Criterion
theorem.
Theorem)
Supposethe s-plane Nyquist contour Gshas an image in the L-plane that encircles the critical point
-1+j0clockwise
Ntimes.Moreover,
suppose
therearePpoles
ofL(s)
intherighthalfs-plane.
Then
thenumber
ofunstable
closed-loop
system
poles
isZ= N+Psince
N=Z-P,asshown
inTheorem
9.4, Cauchys principle ofthe argument.
Remark 9.11 (Comment
Nyquists theorem
on Proof of the Theorem)
simply restates
Cauchys
principle
of the argument in Theorem 9.4 as the number
ofclosed-loop
system
poles
intheRHP
equals
thenumber
ofclockwise
encirclements
ofthe-1+j0
point in the L-plane plus the number of the loop transfer function polesin the RHP.
Example 9.12(Nyquist Plot of L(s) with a Pole at the Origin and a Parameter K)
Considerthe feedback control system shown in Figure 9.7, wherethe loop transfer function
the characteristic function F(s) of the closed-loop system are given as
L(s) =
K(s-1) and
s(s+1)
F(s) = 1+L(s)
= 1+
K(s-1)= s2+(K+1)s-K
s(s+1)
s(s+1)
L(s) and
(9.18)
respectively. It is easy to seethat the closed-loop characteristic equation is
s2+(K+1)s-K
=0
and according to Routh-Hurwitz stability criterion, the closed-loop system is stable if and only if the
parameter Ksatisfies the following inequality:
K+1>0 and -K>0,
Fig. 9.10:
Construction
of Nyquist plot of L(s)
control system in Example 9.12
orequivalently,
-1 <K<0
with a pole at origin and a parameter
Kfor the feedback
9.3
However, this example is here
transfer function
will then
Nyquist
Path,
mainly to demonstrate
Nyquist
Plot, and Nyquist
Stability
how to construct the
Criterion
Nyquist
311
plot of the loop
L(s) that has a pole at the origin and a design parameter K. The Nyquist plot of L(s)
be used together
with Theorem
9.10, the
Nyquist stability
criterion,
to determine the number
of closed-loop system poles in the RHP. Notethat the Nyquist path Gsshown in Figure 9.10(a) consists
of four segments. Segment (4), atiny semicircle with radius e around the pole at the origin, is added to
fulfill the assumption that Gsshould not have poles on it.
The procedure to constructing
1:
GL,the Nyquist plot of L(s) with K> 0, is given asfollows:
Onsegment(1) of the Nyquist path Gs,the variable s = j?, is assumedto be movingalong the
imaginary
axis
following
? =e ?1?8.The
corresponding
image
onL-plane,
GL,
willbe
K
j?-1) = K vv?2 +1 ej(p-?)
ej(p/2-2?)
where ? =tan-1?
(9.19)
L(j?) = K(
?
j?( j?+1)
which is in red
?2 +1 ejp/2e j?
with label (1) passing through
K
ejp/2
L(je) =lim
= 8ejp/2
e?0e
Notethat L(j1)
= ?
the following
three
points:
K
)e-jp/2
?L(j1)=K ?L(jR)=lim
R?8( R
= 0e-jp/2
(9.20)
= K > 0 meansthat when ? = 1 rad/s, segment (1) of the Nyquistimage GL
intersects the real axis of the L-plane at (K, j0), as shown in Figure 9.10(b).
2: Segment
(2)isahuge
semicircle,
with
s =limRejf
and
f =p/2?0?-p/2. With
s =Rejf,
we
R?8
have
lim
L(Rejf)
K(Rejf
-1)
= lim
R?8Rejf(Rejf+1)
R?8
p
-p
e-jf
=0e-jf,
where
-f=
?0?
2
2
R?8 R
= lim
K
(9.21)
hence, the
entire
segment
(2) semicircle
is
mapped to the single
point
(0, j0)
on the
L-plane,
indicated by a green dot in Figure 9.10(b).
3:
Theimage of segment (3) is in blue with label (3) passing through the following three points:
K
ejp/2
L(-jR)=lim
R?8 R
=0ejp/2
K
?L(-j1)=K?L(-je)=e?0
lime e-jp/2
=8e-jp/2 (9.22)
4: Segment
(4)isatinysemicircle
with
s =e?0
limee
j?and?=-p/2?0 ?p/2.With
s =ee
j?
and?=-p/2?0?p/2,we
have
lim
K
j?-1)
L(eej?) =lim eejK(ee
?(ee
j?+1) =lim eej(p-?) = 8ej(p-?)
e?0
e?0
e?0
(9.23
where
p-?=3p
2 ?p ?p
hence,
thetinysemicircle,
segment
(4),is mapped
tothehuge
semicircle
with
radius
R ?8
passing through the following three points on the L-plane:
2
L(ee-jp/2)
= 8ej3p/2
?L(ee
j0)=8e
jp ?L(ee
jp/2)=8e
jp/2
(9.24)
Now we havethe Nyquist plot of L(s), GL, on the L-plane, as shown in Figure 9.10(b) for the case
withK>0.The
number
ofclockwise
encirclements
ofthecritical
point
(-1,j0)bytheNyquist
plot GLis N= 1. Meanwhile,the numberof the poles of L(s) enclosedby Gsis P= 0. Hence,according
312
to
9
Time
Delay, Plant
Nyquist stability
Uncertainty,
criterion,
and Robust Stability
Z =N+P=
1+0
= 1, which implies
that the closed-loop
system has one
pole in the RHP;therefore, the closed-loop system is unstable.
For the case with K < 0, this procedure can be usedto obtain the Nyquist plot of L(s), GL, which
isshown
inFigure
9.10(c).
Notice
that
thenumber
Nofclockwise
encirclements
of(-1,j0)bythe
Nyquist plot GLis dependent onthe intersection of GL with the real axis. If the intersection
point is be-tween
(-1,j0)and
(0,j0),or,equivalently,
-1 <K<0,theencirclement
number
is N=0.Hence,
according to
Nyquist stability
criterion,
Z = N+P
= 0+0
= 0, which implies that the closed-loop
system
has
nopole
intheRHP;
therefore,
theclosed-loop
system
isstable
when
-1 <K<0.
If K<-1,theintersection
point
oftheNyquist
plotGLwith
therealaxiswillbeontheleft
side
ofthecritical
point
(-1,j0).Under
thiscondition,
thenumber
ofclockwise
encirclements
of
(-1,j0) bythe Nyquist
plotGLwillbeN =2,which
leads
to Z = N+P=2+0 =2.Therefore,
the
closed-loop system is unstable since it hastwo poles in the RHP.
Segments (1) and (3) of the Nyquist plot graph on Figures 9.10(b) are obtained using the following
MATLAB
% CSD
code:
Fig9.10b
Nyquist
plot
Ex9.12
hold
on,
figure(6),
theta=linspace(pi/8,(pi+0.01)./2,500);
re_L=(1./tan(theta)).*cos(pi/2-2.*theta);
im_L=(1./tan(theta)).*sin(pi/2-2.*theta);
plot(re_L,im_L,'r-'),
theta=linspace(-(pi-0.01)/2,-pi./8,500);
re_L=(1./tan(theta)).*cos(pi/2-2.*theta);
im_L=(1./tan(theta)).*sin(pi/2-2.*theta);
plot(re_L,im_L,'b-'),
grid,
The same Nyquist plot graph also can be obtained using the following
% CSD
Fig9.10b
K=1,
s=tf('s');
Nyquist
plot
Ex9.12
use
L=K*(s-1)/(s*(s+1));
MATLAB commands:
nyquist(L)
figure(22),
nyquist(L)
Example 9.13(Construction of Nyquist Plot of L(s) with a Polein RHP and a Parameter K)
Forthe feedback control system shown in Figure 9.7, assumethe loop transfer function L(s) is given
as
L(s)
Then we havethe characteristic function
=
0.6K
F(s) of the closed-loop system as follows,
0.6K
F(s) = 1+L(s)
= 1+
(9.25
(s-0.5)(s2
+s+1)
=
(s-0.5)(s2
+s+1)
s3+0.5s2
+0.5s+0.6K-0.5
(s-0.5)(s2
+s+1)
Note that the loop transfer function L(s) has a pole in the RHP and a design parameter K, which is
to be determined so that the closed-loop system is stable. The Nyquist stability analysis approach
will be employed to achieve the objective.
As wedid in the previous examples, the first step is to construct the Nyquist plot of L(s) andthen use
it together
with Theorem
9.10, the
Nyquist stability
criterion,
to determine
the number
of closed-loop
system poles in the RHP. Sincethe loop transfer function L(s) has no poles on the j?-axis, the Nyquist
path Gsconsists of onlythree segments,as shownin Figure 9.11(a).
The procedureto constructing GL,the Nyquist plot of L(s) with K = 1.3,is given asfollows:
9.3
1:
Nyquist
Path,
Nyquist
Plot, and Nyquist
Stability
Onsegment (1) of the Nyquist path Gs,the variable s = j?, is assumed to
Criterion
313
movealong the
imaginary
axis
following
?=0?0.707
?8.Then,
the
trajectory
of
L(j?)
=
0.6K
(j?-0.5)(-?2
+j?+1)
= 0.6K
-0.5(?2+1)+j?(?2-0.5)
(9.26)
(0.52
+?2)((?2
-1)2+?2)
willbeasolid
green
curve
labeled
as(1),shown
inFigure
9.11(b),
starting
fromL(j0) =-1.2K=
-1.56
to L(j0.707)
= -0.8K= -1.04
andending
atL(j8) =0.Note
thatsegment
(1)ofthe
Nyquist plot GL with K = 1.3, the solid green curve, intersects the negative real axis of the
L-plane at these three points.
2: Segment
(2)isahuge
semicircle
ofGs,
with
s =limRejf
and
f =p/2?0?-p/2.With
s =Rejf,
R?8
wehave
lim L(Rejf)
R?8
0.6K
= lim
R?8
(Rejf-0.5)(R2e
j2f +Rejf+1)
= lim
R?8
K
R3e-j3f =0e-j3f
2 ; hence,the entire segment (2) semicircle is
where
-3f =p
2 ?0? -p
point (0, j0) on the L-plane, as shown in Figure 9.11(b).
3:
The image
three
of segment (3)
with
K= 1.3 is in dashed green
(9.27)
mappedto the single
with label (3) passing through
the follow-ing
points:
K
Re-jp/2
L(-jR)=lim
R?8
= 0e-jp/2
?L(-j0.707)
=-0.8K
=-1.04
?L(-j0)=-1.2K
=-1.56
(9.28)
Fig. 9.11: Construction of Nyquist plot of L(s) with an RHP pole and a parameter Kfor the feedback
control system in Example 9.13.
Withthree K values, we havethree corresponding double-crossing Nyquist plots of L(s), GL, on the
L-plane, as shown in Figure 9.11(b). Notethat each Nyquist plot intersects with the negative axis twice.
The Nyquist plot in green is for the case
with
K= 1.3. The number
of clockwise
encirclements
of the
critical
point
(-1,j0),marked
astheblack
dot,bytheNyquist
image
GL,
is N=1.Meanwhile,
the
number of the poles of L(s) enclosed by Gsis P = 1. Hence,according to Nyquist stability crite-rion,
Z = N+P
= 1+1
= 2, which implies that the closed-loop system has two poles in the RHP
314
9
therefore,
Time
Delay, Plant
the closed-loop
Uncertainty,
and Robust Stability
system is unstable
when K = 1.3.
Forthe case with K = 1, this procedure can be repeated to obtain the Nyquist image of L(s), GL,as
shown in Figure 9.11(b) in red labeled with K = 1. The red mapping contour intersects the real axis
at the following three points:
(9.29)
L(j0) =-1.2K=-1.2
, L(j0.707)=-0.8K=-0.8
, L(j8) =0
It also
encircles
thecritical
point
(-1,j0)once,
counterclockwise;
hence,
thenumber
ofclockwise
encirclement
of(-1,j0)bytheredGL
is N= -1.With
P=1,wehave
thenumber
ofunstable
closed-loop
system
poles
Z=
N+P
=(-1)
+1=0.Therefore,
the
closed-loop
system
isstable
when K = 1.
The Nyquist plot of L(s), GL,associated with K = 0.8 is in blue, labeled with K = 0.8, asshown in
Figure
9.11(b).
The blue
mapping
contour
intersects
the real
axis at the following
three
points:
L(j0) =-1.2K=-0.96,L(j0.707)=-0.8K=-0.64,L(j8) =0 (9.30)
It
does
notencircle
thecritical
point
(-1,j0),which
implies
N=0.Thus,
thenumber
ofunstable
closed-loop system poles is one since Z = N+P = 0+1 = 1. Therefore the closed-loop system is
unstable when K = 0.8.
Next,
we would like to find the range
of the design
parameter
K so that
the closed-loop
system
is stable. It is observed that the Nyquist plot GLconsists of two loops attaching together on the negative
real
axis
atheL =-0.8K
intersection
point.
The
rightend
oftheright
loop
isalways
connected
tothe
origin
oftheL-plane,
but
theleftend
oftheleftloop
isL =-1.2K.
The
encirclement
direction
ofthe
left loop is counterclockwise,
but the right loop is circling
in the opposite
direction.
For the closed-loop
system
tobestable,
thecritical
point
(-1,j0)has
tobeinside
theleftloop.
Therefore,
theclosed-loop
system is stable if and only if the following two inequalities are satisfied:
-1.2K<-1 and -1 <-0.8K
(9.31
which is equivalent to
5
6
The intersection
imaginary
<K<
5
4
points of the Nyquist plot GL with the real axis can be obtained from setting the
part of Equation 9.26 to zero. That is,
v
?(?2
-0.5)
=0 ? ?=0 or ? = 0.5=0.707rad/s
Then we havethe two intersection
L( j0)
points onthe real axis at
= 0.6K
-0.5(0+1)
(0.52+0)((0-1)2+0)
=-1.2K
-0.5(0.5+1)
L(j0.707)
=0.6K
(0.52+0.5)((0.5-1)2+0.5)
=-0.8K
Furthermore,
as??8,theNyquist
plotGLwillalso
intersect
thereal
axis
atheorigin
oftheL-plane,
since
L(j8) =0.
The Nyquist plot graph on Figure 9.11(b) is obtained using the following
MATLAB code:
9.3
% CSD
Fig9.11b
Double-crossing
Nyquist
Path,
Nyquist
Nyquist
plot
Plot, and Nyquist
Stability
Criterion
315
Ex9.13
figure(8)
K=1,
num=0.6*K;
den=[1
0.5
0.5
[re,im]=nyquist(num,den,w);
axis([-1.7,0.1,-0.4,0.4]),
hold
[re,im]=nyquist(num,den,w);
Fig9.11b
K=1,
s=tf('s');
on,
w=-w;
grid
Nyquist
% CSD
w=logspace(-3,2,500);
plot(re,im,'r--'),
axis([-1.7,0.1,-0.4,0.4]),
The same
-0.5];
plot(re,im,'r-'),
plot graph
can also be obtained
Double-crossing
Nyquist
using the following
plot
Ex9.13
L=0.6*K/((s-0.5)*(s2+s+1));
use
MATLAB
commands:
nyquist(L)
figure(22),
nyquist(L)
or
s=tf('s');
L=0.6/((s-0.5)*(s2+s+1));
figure(23),
nyquist(L,1.3*L,0.8*L)
9.3.4 Stability Issue Arising from
As briefly
described in
Feedback Control System with Time Delay
Section 9.1.1, time
the delay time is over some limit.
delay in a feedback
Since feedback
control
control
heavily
system
may lead to instability
depends on the
if
measured or estimated
information of reality to perform continuous control corrections to achieve stability and desired perfor-mances,
the control system maynot be able to perform correct control action in time if the sensor is too
slow to provide
accurate timely
delay can transform
information.
an originally
Since an infinite-dimensional
From the analysis in
simple first-order
system
Example
system to a complicated
has infinite
9.1,
infinite
welearned
that a time
dimensional
poles, there is no way to compute
system.
all of them
and
check if they are all in the left half of the complex plane. As we have studied in the previous sections,
the Nyquist stability criterion is capable of effectively counting the number of infinite-dimensional
closed-loop system poles in the RHP without the need to compute all the poles. In the following,
the Nyquist stability analysis approach will be employed to investigate how the delay time T will
affect the stability of the closed-loop system.
Example 9.14 (Construction
of Nyquist Plot of L(s) with a Time Delay Parameter T)
Considerthe feedback control system shown in Figure 9.7, wherethe loop transfer function
the characteristic function F(s) of the closed-loop system are given as
L(s)
=
2e-sT
s+1
1
,
F(s)
= 1+L(s)
=
s+1
L(s) and
?3+(1-2T)s+T2s2
- T3s3
+ ?
1
3
(9.32)
respectively. Theloop transfer function L(s) has a pole, butit is not in the RHP. Thus,the integer number
Pis zero in the Nyquist stability criterion equation Z =N+P. Therefore, the closed-loop system is sta-ble
if andonly
ifthenumber
ofclockwise
encirclements
Nofthecritical
point
(-1,j0)byGL,
the
Nyquist plot of L(s), is zero. Notethat if the delay time T is nonzero, the characteristic function
will haveinfinite
numbers of zeros, which meansthat the closed-loop system has aninfinite
F(s)
number of
poles.
The construction
of the
Nyquist plot for an infinite-dimensional
system is basically the same as that
forfinite-dimensionalsystems.SinceL(s) hasno polesonthe j?-axis ofthe s-plane,the Nyquistpath
will include
the same three segments, as shown in Figure 9.12(a).
The procedure to constructing GL,the Nyquist plot of L(s) with delay time T = 1.0 s, is given
as follows. We will see how the variation of T affects the performance of the closed-loop system after
the
Nyquist plot is completed
316
1:
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
Onsegment(1) ofthe Nyquistpath Gs,the variables = j? is assumedto movealongthe imaginary
axis
following
?=0 ?2.029
?8.Then
the
trajectory
ofthemapped
contour
willbedetermined
by the following
L(j?)
=
equation:
2e-j?T
1+ j?
2
2
where
? =tan-1?(9.33)
= v1+?2ej?e-j?T= v1+?2e-j(?T+?)
This polar form complex function can also be written in the following
L(j?)
=
2
2
v1+?2 cos(?T+?)- jv 1+?2
sin(?T+?)
rectangular form:
where ? =tan-1?
When?increasesfrom 0to 4.913,the valuesofthe complexfunction L(j?)
to Equation 9.33 to generatethe following
?
(9.34)
will changeaccording
tabulated chart:
0 ? 0.861?2.029? 3.426? 4.913
?L(
j?) 0 ? -p/2 ? -p ? -3p/2? -2p
(9.35)
|L(j?)| 2 ? 1.516?0.884? 0.56 ? 0.399
It starts from L( j0)
= 2 and will spiral around the origin clockwise,
with decreasing radius via
thepoints
L(j0.861)
=1.516e-jp/2,
L(j2.029)
=-0.884,
L(j3.426)
=0.56e-j3p/2,
toarrive
at
L(j4.913) = 0.399e-j2p to complete the first spiraling cycle. As ? continues to increase, the
Nyquist
plotGL
will
spiral
around
theorigin
infinite
times
and
approach
theorigin
as??8.
The trajectory
Fig. 9.12:
is shown in red, labeled as(1), shown in Figure 9.12(b).
Construction
of Nyquist
plot of L(s) for a feedback
control
system
with a time
delay
T in
Example 9.14.
2:Segment
(2)isahuge
semicircle
ofGs,
with
s=limRejf
and
f =p/2?0?-p/2.With
s =Rejf,
R?8
wehave
lim
R?8
L(Rejf)
=
2
lim ? eRT(cosf+jsinf)(Re jf
R?8
+1)
?=0
(9.36
9.3
Nyquist
hence, the entire segment (2) semicircle is
shown in Figure 9.12(b).
3:
Path,
Nyquist
Plot, and Nyquist
Stability
Criterion
317
mapped to the single point (0, j0) on the L-plane, as
Theimage of segment (3) is in blue with label (3),
whichis the conjugate of the image of segment
(1). Thered andthe blue GLtrajectories aresymmetrical withrespectto the real axis.
Now we havethe complete Nyquist plot of L(s), GL,for the system with delay time T = 1s.
Sincethe leftmost intersection point of GL with the negative real axisis on the right-hand side of
thecritical
point
(-1+j0),thenumber
ofclockwise
encirclements
of(-1,j0),marked
asthered
dot, by the Nyquistimage GL,is N = 0. Meanwhile,the number ofthe poles of L(s) in the RHPis
P = 0. Hence, according to Nyquist stability criterion, Z = N+P = 0+0 = 0, whichimplies that the
closed-loop system has no poles in the RHP; therefore, the closed-loop system is stable when the
delay time is T = 1s or less.
The Nyquist plot graph on Figure 9.12(b) is obtained using the following
% CSD
Fig9.12b
K=1,
T=1,
Nyquist
plot
Ex9.14
Effect
of
time
MATLAB code:
delay
w=linspace(0,50,1000);
re=K*2./sqrt(1+w.2).*cos(w.*T+atan(w));
im=-K*2./sqrt(1+w.2).*sin(w.*T+atan(w));
figure(30),
plot(re,im,'r-'),title('Nyquist
plot'),
w=linspace(-50,0,1000);
hold
on,
re=K*2./sqrt(1+w.2).*cos(w.*T+atan(w));
im=-K*2./sqrt(1+w.2).*sin(w.*T+atan(w));
figure(30),
plot(re,im,'b-'),title('Nyquist
plot'),
grid,
The same Nyquist plot graph also can be obtained using the following
% CSD
Fig9.12b
T=1,
num=2;
Nyquist
plot
den=[1
Ex9.14
Time
delay
Use
MATLAB commands:
Nyquist(L)
1];
L=tf(num,den,'InputDelay',T),
figure(25)
nyquist(L)
Critical
Delay Time
From the Nyquist plot graph GLin Figure 9.12(b), it is observed that the leftmost intersection point of
GL will moveto the left if the delay time Tincreases. Based on Equations 9.33 and 9.34, we will have
L(j?) =-1,which
means
GL
intersects
thereal
axis
at(-1,j0),ifthere
exists
adelay
time
Tcand
a
frequency ?cso that the following two equationsaresatisfied:
2
?
1+?2
The solution
of the equations
=1
and
provides the smallest
?c =
?cTc +tan-1?c = p
(9.37)
c
delay time that
v
would destabilize
the system:
2p
3 =1.732
rad/sandTc=3 v3 =1.209s
(9.38)
As shown in Figure 9.13(a), if the delay time is T = Tc = 1.209s, the Nyquist plot GLtrajectory
will
intersect
thenegative
realaxis
at(-1,j0)when
? =?c=1.732
rad/s
(i.e.,L(j1.732)
=-1).
If the
delay time
of the system is greater than the critical
delay time,
T > Tc = 1.209s, the leftmost
intersection
point
ofGL
ontherealaxiswillbeontheleft-hand
side
of(-1,j0),asshown
in Figure
9.13(b),
and
then
thenumber
ofclockwise
encirclements
of(-1,j0)byGLwillbecome
N=1.Hence,
the number of closed-loop
system poles in the
RHP will be one since
Z
=N+P
= 1+0
= 1 according
to the Nyquist stability criterion. Therefore, the closed-loop system is unstable if the delay time Tis
greater than the critical delay time Tc = 1.209s
318
9
Time
Delay, Plant
Uncertainty,
Fig. 9.13:
and Robust Stability
Critical delay time that
will destabilize
the system.
Bode Plot Perspective
In Example
9.14, the
Nyquist plot of the loop transfer
function
L(s) and the
Nyquist stability
criterion
were employed to investigate the stability issue caused by time delay in a feedback control system. In
the following example we will consider the same system regarding the same stability issue, but the main
tool is the Bode plot instead of the Nyquist plot.
Recall that the Bode plot
and the second-order
were introduced
in Section 2.5.4 and Section 3.7, respectively,
systems in the study of steady-state
sinusoidal
response.
for the first-order
For a quick review,
consider a typical closed-loop system, as shown in the block diagram of Figure 9.7, where the loop
transfer function is L(s), and the overall closed-loop transfer function between the input R(s) and
the output Y(s) is M(s) as described in the following:
Y(s) =
L(s)
1+L(s)
R(s) :=M(s)R(s)
(9.39)
Forthe output response y(t) of a stable closed-loop system M(s), dueto aninput r(t), we usually are
interested in both of its steady-state response yss(t) and the transient response ytr(t). However,for many
applications
we may only be interested
function (e.g., r(t)
= cos?(t)).
in the steady-state
response if the input
r(t)
is
a sinusoidal
Then the steady-state output response can be easily obtained as
yss(t)
=Acos(?t
+?) where
A=|M(j?)| and?=?M(
j?)
(9.40)
Note
thatboth
themagnitude
|M(j?)|and
thephase
?M(
j?) are
functions
ofthefrequency
?.
Whenthe frequency of the sinusoidal input signal varies, the amplitude andthe phase of the steady-state
sinusoidal response will change accordingly. The Bode plot consist of the magnitude and phase plots to
explicitly exhibit their values at eachfrequency of interest.
Example 9.15 (Bode Plot Perspective of the Stability Issues Caused by Time Delay)
In addition to being an effective
are important
tools for feedback
graphical
display
control systems
of the frequency
design.
response
However, in almost
of a system,
all feedback
Bode plot
control
sys-tems
design and analysis, weemploy the loop transfer function L(s) instead of the closed-loop
transfer function M(s)in the design/analysis process. For the same reason, the Bode plot of th
9.3
Nyquist
Path,
Nyquist
Plot, and Nyquist
loop transfer function L(s) (not the closed-loop transfer function
the stability
margins of the closed-loop system.
Stability
Criterion
319
M(s)) will be constructed
to de-termine
Forthe loop transfer function of a system with time delay T,
L(s) =
2e-sT
(9.41
s+1
L(j?) can beexpressedin polarform as:
L( j?)
=
2e-j?T
2
2
e-j(?T+?),
where
? =tan-1? (9.42)
= v1+?2ej?e-j?T= v1+?2
1+ j?
Then we have the magnitude and the phase of L( j?) in dB and degree, respectively, asfunctions of
frequency in rad/s in the following:
|L(j?)|dB=20log10
|L(j?)| =20log102-10log10(1+?2)
?L(
j?) =-180(?T+tan-1?)/p
(9.43)
The Bode plot of the loop transfer function L(s) for three delay times, T = 1s, T = 1.209s, and
T = 2s, are shown in Figures 9.14(a), 9.14(b), and 9.14(c), respectively. Thesethree Bode plot are
associated with the three Nyquist plots shown in Figure 9.12(b), Figure 9.13(a), and Figure 9.13(b), re-spectively.
Note
thatthemagnitude
plots
|L(j?)|dB
versus
?areallidentical,
since
thetimedelay
only affects the phase. Larger delay time causes more phase shift.
The Bode plot for the case
with delay time
T = 1.209s is shown
in
Figures
9.14(b).
It can be
seen
thatatthefrequency
? =1.732
rad/s,
thephase
is ?L(
j1.732)
=-180?
and
themagnitude
is
|L(j1.732)|dB
=0dB,
which
means
thattheNyquist
plotoftheL(s)withdelay
timeT =1.209s
in-tersects
, since 0dB
therealaxis
oftheL-plane
atthecritical
point
L(j1.732)
=1 e-j180?
delay time is the critical
than Tc = 1.209s.
delay time Tc,for the system to be stable, the delay time
The Bode plot for the case with delay time T = 1s are shown in Figures 9.14(a).
= 1. This
must be smaller
Whenthe phaseis
-180?,
thefrequency
is ?=2.029
rad/s,and
themagnitude
is|L(j2.029)|dB
=-1.07dB,
which
means
thatL(j2.029)
=0.884
e-j180?
, since-1.07dB
=0.884.
This
1.07dB
=1.131
gainmargin
means
that
the magnitude is allowed to increase by 1.131 times before the system become unstable. The defini-tion
of the stability gain margin will be officially defined in the next section.
The Bode plot for the case with delay time T = 2s are shown in Figure 9.14(c).
Whenthe phaseis
-180?,
thefrequency
is ? =1.145
rad/s,and
themagnitude
is|L(j1.145)|dB
=2.385dB,
which
means
,since
2.385dB
=1.316.
This
-2.385dB
=0.76
gainmargin
means
that
thatL(j1.145)
=1.316e-j180?
the magnitude needs to decrease to 76% for the system to become stable.
The Bode plot graphs in Figure 9.14(a)(b)(c) are obtained using the following
% CSD
K=1,
Fig9.14
T=1,
Bode
plot
Ex9.15
Time
w=logspace(-1,0.36,500);
phase=-(w.*T+atan(w))*180/pi;
subplot(2,1,1),
title('Magnitude
semilogx(w,phase,'r-'),title('Phase
delay
and
stability
mag=K*2./sqrt(1+w.2);
magb=20*log10(mag);
figure(31),
semilogx(w,magb,'r-'),
response
in
dB'),
grid,
response
subplot(2,1,2),
in
deg'),
grid,
MATLAB code:
320
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
Fig. 9.14: Bodeplot of the loop transfer function L(s) = 2e-sT/(s+1)
1.209s,
The same Bode plot can be obtained using the following
% CSD
K=1,
with delaytime T = 1s, T =
and T = 2s, respectively.
Fig9.14
T=1,
Bode
num=2;
plot
Ex9.15
den=[1
1];
Time
9.16 (Bode
figure(25),
Plot and
The Bode plot and the
systems
other very
Use
Nyquist
bode(L,w),
grid
Plot)
Nyquist plot are both important
design and analysis.
bode(L,w)
w=logspace(-1,0.36,500);
L=tf(num,den,'InputDelay',T),
Remark
delay
MATLAB commands:
frequency-domain
graphical tools for con-trol
Either one alone has its pros and cons, but they compensate
well. The Nyquist plot, together
with the
Nyquist stability
criterion,
successfully
for each
address the
stability issue of the infinite-dimensional systems and provide a meaningful measureof robust stability
for control systems. However,the Nyquist plot does not explicitly display the magnitude, the phase,
and the frequency as clearly as the Bode plot. Onthe other hand, the information
provided by the
Bode plot alone is not enough for stability analysis because it only considers s = j?, the positive
imaginary axis of the s-plane, not the whole Nyquist contour.
In some cases,the Bode plot can be a more precise and effective tool in control systems design. For
a feedback
control
negligible.
Then it is clear that the increase
system
with time
delay, usually the system is assumed stable
of the delay time
can only
when the delay time is
make the system less stable. In
this case, the Bode plot can be very effective in the design of a compensator to offset the influence of the
time delay on the closed-loop system.
A Simple Compensation to Improve the Stability of a System with Time Delay
Example 9.17 (A Simple
Stability
Compensator to Offset the Time Delay Influence on the Closed-loop Sys-tem
9.3
From the stability
analysis
Nyquist
of the system
Path,
Nyquist
considered
in
Plot, and Nyquist
Examples
Stability
9.14 and 9.15,
Criterion
321
we have learned
that the closed-loop system will become unstable if the delay time is greater than the critical delay time
Tc = 1.209s.
For the case with delay time
shown in the
Nyquist plot in Figure 9.13(b) and the
By comparing
the
Bode plots in
T = 2s, the closed-loop
Figures
system is apparently
unstable,
as
Bode plot in Figure 9.14(c).
9.14(a)
and 9.14(c),
we can see that the
difference
is in
thesign
o|fL(j?)|dB
athephase-crossover
frequency.
When
thesign
isnegative,
asinFigure
9.14(a),
|L(j?)|dB= -1.07dB,
theclosed-loop
system
isstable.
On
theother
hand,
when
thesign
ispositive,
asinFigure
9.14(c),
|L(j?)|dB=2.39
dB,
theclosed-loop
system
isunstable.
Thus,
itis possible
to
findacompensator
toreduce
the|L(j?)|dB
toanegative
value
sothattheclosed-loop
system
will
become stable.
Fig. 9.15: Bodeplot of the loop transfer function L(s) = 2Ke-sT/(s+1)
the proportional control
with delaytime T = 2s and
K = 1 and K = 0.631, respectively.
Later in Section 9.6, the concept of frequency loop-shaping
will beintroduced in control system
design, which can be employed to target some specific frequency range. For now, we will just insert
a simple constant proportional controller Kto the loop transfer function so that L(s) will be
L(s) =
2Ke-sT
(9.44)
s+1
Thenthe magnitudein dB andthe phasein degreeof L(j?)
will become
|L(j?)|dB=20log10
|L(j?)| =20log102+20log10K-10log10(1+?2)
(9.45)
?L(
j?) =-180(?T+tan-1?)/p
Notethat adding the proportional controller
K only changesthe magnitude while the phaseremains
unchanged.
If 20log10K
ischosen
tobe-4dB,which
isequivalent
to K=0.631,
then
the|L(j?)|d
322
9
Time
magnitude
Delay, Plant
plot curve
Uncertainty,
will
and Robust Stability
move down
by 4dB,
as shown
in
Figure
9.15(b).
Since the
phase
plot
?L( j?) remains the same, and the phase-crossoverfrequency is still at ? = 1.145rad/s, the value
of|L(j?)|dB
at hephase-crossover
frequency
willbe|L(j1.145)|dB
=-1.61dB.
Therefore,
thein-sertion
oftheproportional
control
compensation
with
20log10K
= -4dB
has
transformed
anunstable
system to a stable system
The
Nyquist
with a 1.61dB stability
plot associated
with the
gain
Bode plot
margin.
of the
compensated
system is shown in
Figure
9.15(c).
The
intersection
ofGLwith
thenegative
realaxis
isnow
atL(j1.145)
=-0.831,
which
is
on the
right-hand
side of the critical
point
(marked
as a red
dot).
Thus, the
number
of the
clockwise
encirclements
of(-1,j0)byGLhas
changed
fromN=1to N=0,and
therefore
the
closed-loop
system
has become stable since the
Z = N+P
= 0+0
= 0.
number
of unstable
closed-loop
system
poles is now
9.4 Robust Stability
Now let us consider the essential value of Nyquists
stability
robustness
with respect to
not so much with intentional
variations
variations,
theorem;
it allows
in the loop transfer
such as compensator
us a direct
function
method of evaluating
variation.
Our concern is
gain or pole and zero locations,
but with
unintended and uncertain deviations of the plant parameters from the design model. Webegin with the
assumption
that the system is stable (i.e., it satisfies the
Nyquist criterion).
The Nyquist criterion
ustwo direct and useful measuresof robustness. Thefirst, as noted, is the admissible variation
gain without the loss of closed-loop stability. Thisis called the gain margin.
Recall that the gain
margin concept derives from the fact that
multiplication
of the loop transfer
gives
of loop
func-tion
L(s) by the positive number K expands or contracts the Nyquist image to a point whereit crosses
over
the(-1,j0)point
intheL-plane.
Alternatively,
we
can
consider
rotating
theimage
until
theim-age
crosses
the(-1,j0)point,
thereby
changing
thenumber
ofencirclements
and
violating
thestability
criterion.
Note that because of symmetry
of the
Nyquist image
with respect to the real axis, the rotation
can be clockwise or counterclockwise. The admissible rotation angle is called the phase margin. The
phase margin indicates the additional phase lag of the loop transfer function that can betolerated
without destabilizing the closed-loop system.
The gain and
phase
margins can be determined
from
either
the
Nyquist
or Bode plot, as indi-cated
in Figure 9.16(a) and (b).
9.4.1
Gain and Phase
Margins
The gain and phase
margins can be obtained
Nyquist
clear physical
plot exhibits
either from the
Nyquist
meaning of the gain and phase
plot, or from the
Bode plot.
margins, but its visual
The
presentation
does not provide precise readings of the margins and the frequencies at which these margins are mea-sured.
Onthe other hand, the Bode plot alone do not provide enough information to determine if the
system is stable. However, if the system has been known stable, the Bode plot can be employed to
provide precise readings of the gain and phase margins and the frequencies at which these margins
are measured.
Reading Gain Margin from
Nyquist Plot
In the partial Nyquist plot of L(s) shown in Figure 9.16(a), the gain marginis determined by the position
of the intersection (the blue dot) of the Nyquist image GL(the blue curve) andthe negativereal axis. The
intersection
position is represented
by the complex
number
on the
L-plan
9.4
Robust Stability
?L(
j?p)
??180?
?
323
(9.46)
?
where ?pis called the phase-crossoverfrequency at which the phase of L(j?p) is 180?. The gain
margin is defined
by the following
equation:
j?p)
?L(
j?p) ?L(
GM=?
?
?
dB =-20log10
?
(9.47)
?
?
Fig. 9.16: Positive gain and phase margins defined on Nyquist plot in (a), and the corresponding gain
and phase margins shown on Bode plot in (b).
Ifthisintersection
point
isbetween
theorigin
(0,j0)and
thecritical
point
(-1,j0)(the
reddot),
which
means
theintersection
is ontheright-hand
side
of(-1,j0),then
themagnitude
?L(
j?p)
?will
?
be less than 1 and the gain margin (GM)
margin
does not
mean it is stable.
?
will be positive. A closed-loop system with a positive gain
However, if the system is stable
when the intersection
point is
between
(-1,j0)and
(0,j0),then
thesame
system
willcontinue
tobestable
as
longastheinter-section
point remains
inside
the interval.
The gain marginis a good indication
on how much gain variation is allowed for the system to stay
stable.
For
example,
iftheintersection
isat(-0.1,j0)or0.1?180?,
thegain
margin
willbe
GM=-20log100.1
=20dB
and the system will remain stable if the system gain will notincrease to ten times of its original gain. On
theother
hand,
if theintersection
point
isat(-0.8,j0)or0.8?180?,
thegainmargin
willbe
GM=-20log100.8
=1.94
dB
and the system gain
will only need to increase
by 25% (or become
1.25 times
of its original
value) to
move
the
intersection
point
totheother
side
otfhecritical
point
(-1,j0)todestabilize
thesystem.
Ifthe
intersection
point
moves
totheother
side
otfhecritical
point
(-1,j0),thenumber
ofencirclements
o
324
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
(-1,j0)bytheNyquist
image
contour
willchange;
hence,
thestability
status
oftheclosed-loop
system
will change according to the Nyquist stability criterion.
Therefore, the system will be more stable if the absolute value of gain margin is larger, or if the
intersection point
?L(
j?p)
??180?
isfurther
away
from
the
critical
point
(-1,
j0).
?
?
Reading Gain Margin From Bode Plot
Although the Nyquist plot clearly shows the intersection point of the Nyquist image GL with the negative
real axis, it does not reveal the phase-crossover
frequency.
Since the
Bode plot explicitly
show both the
magnitudeandthe phaseof L(j?) asfunctions ofthe frequency,they provide moredetailedinformation.
As defined in
Equation
9.47, the gain
margin is determined
by the
magnitude of
?L(
j?p)
??180?.
The
?
?
phase-crossover
frequency,
?p,
isthefrequency
atwhich
thephase
ofL(j?)is180?
or-180?.
From
the Bode plot shownin Figure 9.16(b),it is easyto seethat the ?L(j?)
phasecurve intersects the
-180?horizontal
line when
thefrequency
?is thephase-crossover
frequency
?p.Draw
avertical
straightline atthe phase-crossoverfrequency, ?p = 15.8rad/s, and extendthis line upto intersectthe
|L(j?)|dBmagnitude
curve.
Then,
thevalue
of|L(j?)|dBat?p=15.8
rad/scanberead
from
thegraph
as
dB=-8.8dB.Therefore,
thegainmargin
is
?L(
j?p)
?
?
?
?L(
j?p)
GM=?
Reading
Phase
Margin
From
Nyquist
?
dB = 8.8 d
?
Plot
In the partial Nyquist plot of L(s), shown in Figure 9.16(a), the phase margin is determined by the
intersection (the purple dot) of the Nyquist image GLand the unit circle centered at the origin. The
intersection position is represented by the complex number onthe L-plane
1 ?L( j?g)
(9.48)
where ?g is called the gain-crossoverfrequency at which the
?L(
j?g)
?
?
?
magnitude of L(j?g) is one, or
dB = 0 dB. Thephasemargin
is definedbythe following equation:
P.M.
=?L(
j?g)-180?
(9.49)
Just like the perturbation of the gain, the variation of the phase of the loop transfer function L(s) can
also destabilize a system. Recallthat in Section 9.3.4, a time delay will cause a phaselag in the Nyquist
plot of L(s). A system with larger absolute value of phase margin will be more robust against the
phase variations of the system.
Reading Phase Margin From Bode Plot
Asdefinedin Equation9.49,the phase marginis determinedbythe phaseof L(j?) atthe gain-crossover
frequency ?g,the frequency at whichthe magnitude
dB = 0 dB. Thefirst step
is to find the gain-crossoverfrequency ?g. Fromthe Bode plot shown in Figure 9.16(b), it is easyto
?L(
j?g)
?
is
1,
or
?L(
j?g)
see
that
theintersection
ofthe|L(j?)|dB
curve
with
the0dBhorizontal
lineonthemagnitude
plot
?
?
?
?
?
occurs when the frequency
? is the gain-crossover frequency
?g. Draw a vertical straight line at the
gain-crossoverfrequency, ?g = 9.2rad/s, and extendthis line downto the phaseplot to intersect the
?L( j?) curve. The value of ?L(j?g) at ?g = 9.2rad/s can bereadfrom the phaseplot that ?L(j?g)
is -162?.
Therefore,
thephase
margin
isP.M.=-162?
-(180?)
=18?.
9.4
9.4.2
Effect of the
In this section,
Gain of Loop Transfer
we will consider
Function
two examples
on Gain and
that exhibit
Phase
Robust Stability
325
Margins
how the gain of the loop transfer
function
L(s) will affect the stability of the closed-loop system.
Example
9.18 (Stabilize
an
Consider a closed-loop
Originally
system
Unstable
with the following
L(s) =
From the partial
Nyquist
found the gain and phase
System to a Desired
plot and the
Gain
Margin)
loop transfer function:
20000
(9.50)
s3 +55s2 +250s
Bode plot in
Figures
9.17(a)
and 9.17(b), respectively,
we have
margins of the system as follows:
dB=-3.4dB where?p=15.8
rad/s
?L(
j?p)
P.M.
=?L(
j?g)-180?
=-6?where
?g=19rad/s
GM=?
?
?
(9.51)
Notethat the system has negative gain and phase margins,since the Nyquistimage GLintersects the
negative
real
axis
atheleft-hand
side
otfhecritical
point
(-1,j0).Ingeneral,
asystem
with
negative
gain or phase
the partial
margin
Nyquist
does not
plot and the
mean the closed-loop
system is unstable.
Bode plot is not enough to
determine
The information
if the closed-loop
given
by
system is
stable. However,it is quite straightforward to check the closed-loop stability by solving the three roots of
the characteristic equation or using the Routh-Hurwitz criterion, or the Nyquist stability criterion based
on the complete
Nyquist
plot.
The system is indeed
unstable,
but both the gain and phase
margins are
small.
Againmargin
of-3.4dBmeans
thatit onlyrequires
asmall
gain
reduction
ofL(s)
to67%
of its original
gain to stabilize the closed-loop system.
Fig. 9.17: The gain and phase margins of the unstable
closed-loop
Assumethe loop transfer function L(s) structure is slightly
system in
modified t
Example 9.18
326
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
L(s) =
where Kis a constant
gain to be designed.
20000K
(9.52)
s3 +55s2 +250s
The system is the same as the original
one if
K = 1. In gen-eral,
this Kcan be afunction of s, or K(j?) is afunction of frequency ?, sothat the systemcan
be designed to satisfy multiple design objectives like robust stability, disturbance reduction, and
control-input constraints. Here, we will assume Kis just a constant parameterto be determined so that
a robust stability
objective
can be achieved.
Now
theclosed-loop
system
isunstable
with
gain
margin
-3.4
dBand
phase
margin
-6?.Ww
e ould
like to design a simple
improved
constant
controller
Kso that
the closed-loop
system is stable
with gain
mar-gin
to 14.6 dB.
Since the change of K will not affect the phase, the phase plot
will remain the same. From the Bode
plot
inFigure
9.18(b),
itcan
beseen
that
thegain
and
phase
margins
will
improve
ifthe
|L(j?)|dB
curve
is moved down on the magnitude plot while the phase plot remains unchanged.
Fig.
9.18:
Illustration
ofreducing
Kto move
down
the|L(j?)|dB
curve
toimprove
thegain
and
phase
margins of an originally
unstable closed-loop system in Example 9.18.
The graphs in Figures 9.18(a) and (b) show the differences
between the original
system (with
K= 1)
and the modified system (with K = 1/8) in the Nyquist plot and in the Bode plot. The Nyquist plot
shows that the Nyquist image GL has movedits intersection point with the negative real axis from the
black
dotposition,
crossing
over
thereddotcritical
point
(-1,j0),tobecome
stable
toward
theblue
dot
position.
The Nyquist plot shows clearly that an obvious robust stability
improvement
has been achieved,
but it does not reveal the details quantitatively.
The Bode plot shows clearly that if the objective is to achieve a gain
margin of 14.6 dB, then
K needs
tobereduced
toalevel
sothat
the
|L(j?)|dB
curve
canmove
down
by18dBonthemagnitude
plot.
The 18 dB reduction of gain is approximately equivalent to reducing
Kfrom 1to 1/8
9.4
Robust Stability
327
With K = 1/8, the Nyquist plot and the Bode plot have been modified to those shown in
Figures 9.19(a) and (b), respectively. It can be seen that the phase-crossover frequency is still at
?L(
j?p)
-3.4dBto14.6
dB.Although
thephase
plotremains
unchanged,
thegain-crossover
frequency
has
changed
from?g=19rad/s
to?g=6.2
rad/s,
and
thephase
margin
has
improved
from-6?to
?p = 15.8 rad/s, but dueto the 18 dB reduction of
?
?
?
dB,the gain margin hasimproved from
31?.
Fig. 9.19: Illustration of reducing
margin by 37? in Example 9.18.
Kfrom 1 to 1/8 to improve the gain margin by 18 dB and the phase
The Nyquist plot graph in Figure 9.19(a) is produced using the following
% CSD
Fig9.19a
K=1/8;
Nyquist
num=20000*K;
plot
den=[1
Ex9.18
Gain
55
0];
[re,im]=nyquist(num,den,w);
plot'),
[x,y]=meshgrid(t);
Fig9.19b
phase
hold
on,
Bode
num=20000*K;
Figure
9.19(b)
plot
Ex9.18
den=[1
[mag,phase]=bode(num,den,w);
55
margins
plot(re,im,'-r'),
t=2.9:pi/20:4.8;
plot(cos(t),sin(t),'k--');
The Bode plot graphs in
K=1/8;
&
w=linspace(5,100);
figure(103),
title('Nyquist
% CSD
250
MATLAB code:
axis
are obtained
Gain
250
&
0];
equal;
using the following
phase
grid
MATLAB
code:
margins
w=linspace(4,20);
magb=20*log10(mag);
figure(104),
subplot(2,1,1),
plot(w,magb,'-r'),
title('Magnitude
subplot(2,1,2),
plot(w,phase,'-r'),
title('Phase
in
in
deg'),
dB'),
grid,
grid
In the following example, we will revisit the system considered in Example 9.13. The system has
a double-crossing
Nyquist image that intersects the negative real axis twice. Recallthat, according
to the
Nyquist stability
criterion,
the closed-loop
system
will be stable
when these two intersections
are
attheopposite
sides
ofthecritical
point
(-1,j0).Furthermore,
thegain
margins
aredetermined
bythe
The gain margin for the intersection on the right is positive while the
two intersection positions.
other has a negative gain
margin
328
9
Example
Time
Delay, Plant
9.19 (A
System
Uncertainty,
with
and Robust Stability
Double-crossing
Nyquist
Image
Has Positive
and
Negative
Gain
Margins at the Same Time)
Fig. 9.20: Revisitthe double-crossing Nyquist image of Example 9.13that has positive and negative gain
margins at the same time in Example
9.19.
The system considered in Example 9.13 has a double-crossing
9.11. Its loop transfer function is
L(s)
=
Nyquist image, as shown in Figure
0.6K
(s-0.5)(s2
+s+1)
A partial Nyquist plot and the Bode plot of L(s) with K = 1 are shown in Figures 9.20(a) and (b),
respectively. Note that the Nyquist image GLintersects the negative real axis of the L-plane at the
following
two
points:
v
L(j0) =-1.2andL(j 0.5)=-0.8
It has been shown in Example 9.13that the system is stable. According to the definition of gain and
phase margins,the system has phase margin PM = 13? but hastwo gain margins
GM1
=-20log100.8
=1.9382
dB andGM2
=-20log101.2
=-1.5836
dB
The positive gain margin on the right is GM1 = 1.9382 dB. It meansthat the gain of the loop
transfer function
Kis allowed to increase to K= 1.25 without destabilizing the closed-loop system.
On
theother
hand,
thenegative
gainmargin
GM2
=-1.5836
dBindicates
thattheclosed-loop
system can stay stable aslong as Kis not reduced to below K= 5/6. In other words,the closed-loop
system is stable if and only if
5/6
<K<
1.25
Notethat the stability range ofK derived in terms of the two gain marginsis consistent withthe inequality
in
Equation
9.31,
which wasfound
based on the
Nyquist stability
criterion
9.5
The Nyquist
% CSD
plot graph in
Fig9.20a
Figure
Nyquist
9.20(a) is produced
plot
Ex9.19
Generalized
Stability
using the following
Double-crossing
Two
MATLAB
gain
Margins
329
code:
margins
figure(105),
K=1,
num=0.6*K;
den=[1
0.5
0.5
-0.5];
w=linspace(0,10,500);
[re,im]=nyquist(num,den,w);
plot(re,im,'r-'),
axis([-1.7,0.1,-0.4,0.4]),
grid
The Bode plot graphs on Figure 9.20(b) are obtained using the following
% CSD
K=1,
Fig9.20b
Bode
num=0.6*K;
plot
den=[1
Ex9.19
0.5
[mag,phase]=bode(num,den,w);
subplot(2,1,1),
-0.5];
Two
gain
margins
w=linspace(0,0.8,100);
magb=20*log10(mag);
figure(106),
plot(w,magb,'-r'),
title('Magnitude
grid,
Double-crossing
0.5
MATLAB code:
response
subplot(2,1,2),
in
dB'),
plot(w,phase,'-r'),
9.5 Generalized Stability
grid
Margins
As described in the previous section, the gain and phase
perturbation
is allowed
at the phase-crossover
frequency
margins are good indications
and how
of how
much gain
much phase variation is permitted
at
the gain-crossover frequency, respectively. However, these two stability
margin measuresonly reveal
the robust stability information
at two frequencies, not the whole frequency spectrum. Further-more,
these two measures only work for SISO (single-input/single-output)
systems. The attempt of
extending the concept of robust stability marginsto MIMO (multi-input,
multi-output) systems was not
successful until 1966 whenthe small gain theorem was discovered.
Fig. 9.21:
Atypical
feedback
control
system
with unstructured
norm-bounded
plant uncertainties.
9.5.1 Small Gain Theorem and Robust Stability
Consider the feedback control system with plant uncertainties shown in Figure 9.21. The nominal plant
modelis represented by G,and the set of uncertain plants is described by
G =
?
G
~: G
~=(I+?)G,where
?(s)
?RH8
and
s[?(j?)]=?(?)
?
Here, RH8 represents the set of rational function
(9.53)
matrices with real coefficients that are analytic (have no
poles)in the closedright half complex plane,ands [X] standsfor the maximumsingular value of X, or,
equivalently,the squareroot ofthe maximumeigenvalue of X*X, where X*is the conjugatetransposeof
the complex matrix X. The positive real scalar function ?(?) prescribes the maximum magnitude
variation
of the uncertain
plant dynamics at all frequencies
according to the practical system an
330
its
9
Time
Delay, Plant
working environment.
Uncertainty,
In reality,
and Robust Stability
the discrepancy
between the nominal
model and the real system is
larger at higher frequencies, and, thus, extra care needsto be given for high-frequency
dealing
with robust stability
uncertainties in
issues.
The unstructured plant uncertainties include unmodeled dynamics and any variation caused by envi-ronmental
parameters, whichare unknownexceptthe boundednorminformation in ?(?). To determine
a necessary and sufficient
condition
so that the uncertain
closed-loop
system is stable seems to be a very
difficult problem. Surprisingly, this challenging robust stability design problem wassolved in an effec-tive
and elegant fashion,
as shown below in the Small
Gain Theorem.
First, let usintroduce the concept of sensitivity function and complementary sensitivity function.
Remark
9.20 (The
Sensitivity
and
Complementary
Sensitivity
Functions)
Consider the feedback control system shown in Figure 9.21, where Gand
G
~ arethe nominaland
the perturbed modelsof the systemto becontrolled, respectively,and ? representsa set of plant uncer-tainties.
In general, the controller Kis designed sothat the closed-loop system has desired performance.
There aretwo closed-loop transfer functions that are central to understanding the performance of feed-back
systems. These are the sensitivity function, S(s), and the complementary sensitivity function,
T(s). Let the loop transfer function (LTF) be denoted by
L(s) = G(s)K(s)
(9.54
and define
S(s) =[I+L(s)]-1
and
T(s) = L(s)[I+L(s)]-1
For the feedback control system of Figure 9.21, if
input, then the relationship
between the reference input
(9.55)
? = 0 and d = 0, where d is the disturbance
R(s)
=L[r(t)]
and the tracking
error
E(s)
=
L[e(t)] is
E(s) =[I+L(s)]-1R(s)
= S(s)R(s)
Thus, a smaller S(s) will lead to a smaller tracking/regulation
0, then the relationship between the disturbance input D(s) =
error. Similarly, if ? = 0 and r =
L[d(t)] and the output disturbance
response Y(s) =L[y(t)] is
Y(s) =[I+L(s)]-1D(s)
Hence, a smaller S(s) will also imply
= S(s)D(s)
a smaller disturbance response.
Onthe other hand, as will beseenin the Small Gain Theorem below that a smaller complementary
sensitivity function
T(s) will produce better robust stability. That is, to achieve smaller tracking/reg-ulation
error, less disturbance
so that the sensitivity
response,
function
However, it is impossible
and better robust stability,
and the complementary
to reduce
sensitivity
both at the same time
we would like to design a controller
function
are both small, if possible.
and at the same frequency
since the sum
of S(s) and T(s) is a constant:
S(s)+T(s)
Nevertheless,
plant
while the
=[I+L(s)]-1
uncertainties
disturbances
+L(s)[I+L(s)]-1
are
more significant
and the reference
inputs
=[I+L(s)]
in
[I+L(s)]-1
high frequency
usually
range
=I
than
occur in low frequency
low
fre-quencies
range.
Therefore,
thecontroller
can
bedesigned
tominimize
|S(j?)|inlow
frequency
range
while
reduc-ing
|T(j?)|athigh
frequencies.
9.5
Theorem
9.21 (Small
Gain Theorem for
Consider the feedback
control system
Generalized
Stability
Margins
331
Robust Stability)
with plant uncertainty
described
by the block diagram
shown
in Figure 9.21 and by the set of uncertain plants satisfying the bounded norm condition described in
Equation 9.53. Theloop transfer function matrix L(s) andthe complementary sensitivity function matrix
T(s) ofthe nominal closed-loop system are defined in Equations 9.54 and 9.55, respectively.
If the nominal closed-loop system
C(G,K),
which is the closed-loop system
with ? = 0 or
G
~ =G,is stable,thentheuncertainclosed-loop
systemCG
? ~,K? isstableif andonlyif thefollowing
inequality
is satisfied:
s [T( j?)]
Note that the small gain theorem
<
1
for all
?(?)
for robust stability
?
(9.56)
works for both
MIMO and SISO systems,
and
the plant G(s) is not restricted to be a square matrix. The theorem also covers the whole spectrum
of frequencies.
Therefore, it provides
margins. The physical sense of the
may not be very clear to
much more general applications
maximum singular
most of the undergraduate
value,
than the classical
gain and phase
which is the norm employed in the theorem,
students at this
moment. However, for the application
of the theorem to SISO systems, the maximum singular valuess [T( j?)]
ands [?( j?)]
are simply the
magnitudes
|T(j?)|and
|?(j?)|,respectively,
asshown
inthefollowing
corollary.
Corollary 9.22 (Special
Case of the Small Gain Theorem for SISO Systems)
Consider the same feedback control system with plant uncertainties described by Figure 9.21 and
Equation
9.53 except that
all
matrix functions
are replaced
by their
scalar function
counterparts
and
s [?(j?)] =?(?)isregarded
as
|?(j?)| =?(?)
Thenthe uncertainclosed-loopsystemC?G~,K? is stableifand onlyifthe followinginequalityis satis-fied:
1
|T(j?)| < ?(?)
9.5.2 Interpretation
of the
Generalized
Stability
for all ?
(9.57
Margins
In the previous subsection, Section 9.5.1, welearned that a generalized frequency-dependent stabil-ity
margin function can be obtained from the small gain theorem. The uncertain closed-loop system
C
?G~,K ? is stable if and only if
1
where
|?(j?)| =?(?) forall ?
|T(j?)| < ?(?)
For
example,
if|T(j?1)|=0.5,
then
themaximum
allowable
variation
of|?(j?1)|has
tobeless
than
2.
Otherwise, the system
will become unstable.
To guarantee the closed-loop
system to be robustly
stable,
themaximum
allowable
variation
of|?(j?)|has
tobeless
than
1?|T(j?)|forall?. Hence,
naturally,
a generalized stability
margin function
M(?)
M(?) can be defined asfollows:
?
s[T(j?)] =1|T(
j?)|
= 1?
(9.58)
332
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
Fig.
9.22:
The
generalized
stability
margin
function
M(?)
isthe
inverse
of|T(j?)|.
Example
9.23
(Generalized
Stability
Margin
FunctionM(?)
and
|T(j?)|)
Considerthe feedback control system with plant uncertainties shown in Figure 9.21. The loop trans-fer
function of the system is given as
L(s)
= G(s)K(s)
=
5000
s(s2 +55s+250)
(9.59
The complementary sensitivity function of the system is
T(s) = L(s)[I+L(s)]-1
and
themagnitude
plotofT(j?),|T(j?)|dB,
can
beobtained,
asshown
inFigure
9.22.
ForMIMO
systems,this plot wouldhavebeenthe maximum
singularvalueplot ofs [T( j?)].
Onthis complementarysensitivity plot, it can beseenthat at ? = 9.36rad/s, wehave
|T(j9.36)|dB
=10.28
dB ? |T(j9.36)|
=3.27? M(9.36)
=1/3.27
=0.306
which
means
that
themaximum
allowable
magnitude
variation
of|?(j9.36)|,
toavoid
destabilizing
the
system,
is0.306.
Ifthevariation
of|?(j9.36)|
is more
than
0.306,
thesystem
willbecome
unstable.
Similarly,
thevalues
of|T(j?)|at?=13.4
rad/s,?=15.8
rad/s,
and?=19.3
rad/s
are
computed
respectively
as follows:
|T(j13.4)|dB
=0dB ? |T(j13.4)|
=1 ? M(13.4)
=1/1=1
|T(j15.8|dB
=-4.84
dB ? |T(j15.8)|
=0.573? M(15.8)
=1/0.573
=1.745
|T(j19.3)|dB
=-10
dB ? |T(j19.3)|
=0.316? M(19.3)
=1/0.316
=3.165
Therefore,
themaximum
allowable
magnitude
variations
of|?(j?)|atthese
three
frequencies
so
that
the
system
willremain
stable
are
|?(j13.4)|
=1,|?(j15.8)|
=1.745,
and
|?(j19.3)|
=3.165,
respectively.
Note
thatasmaller
|T(j?)| allows
more
magnitude
perturbation
atthatfrequency,
and
the
generalized
stability
margin function is the reciprocal
of the
magnitude of the complementary
sensitivity
function.
Itisalso
observed
that
|T(j?)|isdecreasing
atthehigh
frequency
region,
which
meansthe high frequency components of the system are allowed larger uncertainties.
9.5
Generalized
Stability
Margins
333
The
|T(j?)| plotgraph
onFigure
9.22
isobtained
using
thefollowing
MATLAB
code:
% CSD
Fig9.22
|T(jw)|
num=5000;
den=[1
% Complementary
T
=
plot
55
function
feedback(L,1);
Ex9.23
0];
Generalized
stability
margin
L=tf(num,den);
T=L*inv(1+L);
w1=linspace(1,35);
svTb=20*log10(svT);
grid
250
figure(100),
plot(w1,svTb),
svT=sigma(T,w1);
title('sigma
of
T'),
grid
on,
minor
From these discussions,
welearn that the
maximum singular
value of the complementary
sensitivity
function,s [T( j?)] for MIMOsystems,orthe magnitudeofthe complementarysensitivityfunction,
|T(j?)|for SISO
systems,
provides
important
stabilitymargin
information
forthewhole
frequency
spectrum. This information is critical in the loop-shaping control system design to ensure robust stabil-ity.
In the next subsection, we will discuss the differences between the complementary sensitivity
function T( j?) and the loop transfer function L( j?) in terms of their ability to provide robust
stability information.
We will show that the gain and phase margins can also be obtained from the
magnitude of the complementary sensitivity function.
9.5.3 Relationship
Between Gain/Phase Margins and the Generalized Stability
Margins
In the following we will show that the gain and phase margins can be computed based onthe magnitude
of the complementary sensitivity function.
Theorem
9.24 (Use
?T(
j?p)
?and
?T(
j?g)
?toCompute
the
Gain
and
Phase
Margins)
?
Consider the feedback
?
?
?
control system in Figure 9.21 with the loop transfer
function
L(s)
=G(s)K(s)
andthe complementary
sensitivityfunction T(s) = L(s)[I+L(s)]-1. Assume?pand ?garethe phase-crossover
frequency
and the gain-crossover
frequency,
respectively.
(a) If
Then the gain
?
?
?
?
?
and the
?
?
?
?
(b) If
margin (GM)
?T(
j?p)
?and
?T(
j?g)
?L(
j?p)
?<1,
the
gain
margin
is
??
GM=
20log10
1+1
?T(
j?p)
?L(
j?p)
?>1,
the
gain
margin
is
??
GM=
20log10
1-1
?T(
j?p)
PM
=2sin-1
?0.5
??
?T(
j?g)
PM
=-2sin-1
?0.5
??
?T(
j?g)
phase margin (PM) ofthe system can be computed using
?
?
?
(9.60)
?
(9.61)
?
?
?
?
(c) If?L( j?g) > p,the phasemarginis
?
?
?
(9.62)
(d) If?L( j?g) < p,the phasemarginis
?
?
Proof:
?
(9.63)
334
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
?L(
j?p)
?e
jpand
GM
=
?L(
j?p)
?.
Since
?L(
j?p)
?<1,
the
equation
relating
the
complementa
sensitivity
func-?
1?L(
j?p)
?-1??L(
?=1+
?T(
j?p)
?= ?=?L(
j?p)
j?p)?L(
j?p)
?T(
j?p)
?
?
GM=-20log10
?L(
j?p)
?=20log10
?T(
j?p)
(a) According to the definition of the gain margin, we have L(j?p)
-20log10
?
?
?
=
?
?
tion to the loop transfer function at ? = ?p will become
1
?
1+L( j?p)
?
?
L(j?p)
?
?
?
1
1
?
?
?
?
?
?
?
1
=
?
?
?
?
?
?
?
?
Therefore, the gain
margin is
?
1
1+
?
?
?
?
(b) The proof is similar to part (a) exceptthat nowthe intersection ofthe Nyquistplot of L(j?)
the real
axis is at the left
of(-1, j0).
with
Thus,
?L(
j?p)
?-1
?
=
???L(
?=1?=
?T(
j?p)
j?p)
?T(
j?p)
?L(
j?p) ?L(
j?p)
?
?
GM=-20log10
1?L(
j?p)
?=20log10
?T(
j?p)
?L(
j?g)
?=1,?L(
j?g)
=p+f,
and
the
phase
margin
isPM
=f.
?1+L(
j?g)
j?g)
?T(
j?g)
?= ?=?L(
j?g)?1+L(
?1+L(
j?g)
?.Points
A,
O,
and
Bare
on
the
same
straight
line
whose
?1+L(
j?g)
?T(
j?g)
? ?
?T(
j?g)
1
?
1+L( j?p)
?
?
?
?
?
?
L(j?p)
?
?
?
?
1
-1
= 1+
?
?
?
1
?
?
?
?
Therefore, the gain
?
?
?
margin is
1
?
?
?
?
?
(c) Since ?L( j?g) > p, we have
Theequationrelatingthe complementarysensitivityfunction to the loop transfer function at ?=?g
?
?
becomes
1
1+L( j?g)
?
?
?
?
L(j?g)
?
From the
Nyquist plot graph in
points
A and
Cis
?
?
?
?
?
?
?
?
?
=
?
?
?
?
Figure
?
9.23, it can be seen that the length
of the straight
line con-necting
?
length is the diameter of the circle, whichis 2. Thetriangle ?ABC is a right triangle
?ABC is 2f, wherethe angle f is the phase margin. Hence,
sin
f
2
?
=
andthe angle
0.5
?
?
=
2
?
?
?
and therefore the phase margin is
0.5
PM = f = 2sin-1
?
?
?
(d)
The proofis similar to Part (c) and is left as an exercise.
Example 9.25 (The Magnitude Plot of the Complementary
with the Gain and Phase Margins)
Sensitivity
Function and Its
Relation-ship
The system considered in Example 9.23 with the loop transfer function:
L(s)
= G(s)K(s)
=
5000
s(s2 +55s+250)
(9.64)
9.5
will be employed
of T(j?),
gain
in the following
to compare
the
Nyquist
plot,
Generalized
Stability
Margins
Bode plot, and the
335
magnitude
plot
whichis the complementarysensitivity function, andidentify the relationship among the
margin
GM, the
phase
margin
PM, the
magnitude
of the complementary
sensitivity
function
|T(j?)|,and
thegeneralized
stability
margin
functionM(?).
The Nyquist plot and Bode plot of the feedback control system with loop transfer function
given by Equation 9.64, are shown in Figure 9.23. These plots
L(s),
were shown in Figure 9.16 earlier in Sec-tion
9.4.1 to introduce how to read gain and phase marginsin the Nyquist plot and Bode plot. The plots
in
Figure
9.23 show
more detailed
geometric
information
that
are essential for the proof in
Theorem
9.24.
Now,from Figure 9.23, wehavethe gain-crossoverfrequency ?g = 9.36rad/s at which
L(j?g)
and the phase marginis PM = f
which
= 1ej(p+f)
= 17.5?. The phase-crossoverfrequency is ?p = 15.8rad/s at
we have
?L(
j?p)
?e
jp
?L(
j?p)
?=8.77
dB.
The
|T(j?)|plot,
themagnitude
ofthecomplementary
sensitivity
function,
isshown
inFigure
9.24.
L(j?p)
=
?
?
and
thegainmargin
canbefound
asGM=-20log10
?
?
Asdiscussed
in Example
9.23,
thismagnitude
function
|T(j?)|isthereciprocal
ofthegeneralized
sta-bility
marginfunction
M(?), whichspecifiesthe maximumallowable variations of ?(j?) for each ?
so that the system can remain
In
stable.
Figure 9.24, when the frequency
?T(
j?g)
?
is at ? = ?g, the gain-crossover frequency,
we have
?T(
j?g)
?=3.27
and
its
reciprocal,
the
generalize
stability
?T(
j?g)
??T(
=3.27
value
can
j?g)
?=3.27
? ? ??
?T(
j?g)
?T(
j?p)
?T(
j?p)
?=0.573.
The
reciprocal
of
this,M(?p)
=1/0.573
=
1.745,
isthe
generalize
stabil-it
?T(
j?p)
?=0.573
value
can
be
employed
tocompute
the
gai
??
1+1
?T(
j?p)
=20log10
(1+1.745)
=8.77
dB
?
dB = 10.28 dB, which gives
?
?
?
margin function
M(?g) = 1/3.27 = 0.306. This 0.306 stability margin at the frequency ?g means
that the variation of ?( j?g) < 0.306 will not causethe system to become unstable. In addition to this
?
robust stability information regarding the maximal variation of ?(j?g), the ?
be employed
into
to compute the phase
margin, as promised
by Theorem
9.24. Plugging
?
?
Equation 9.62, we have the phase margin
0.5
PM = f = 2sin-1
= 2sin-1
?
?
0.5
3.27
= 17.6?
?
which is consistent withthe result obtained from the Bode plot.
Atthe frequency ?p,the phase-crossoverfrequency, we have
?
gives
?
?
dB= -4.84dB,which
?
margin. This 1.745 stability margin at the frequency
will not destabilize the system. The
?
margin
?
using
Equation
?p meansthat the variation of ?(j?p)
?
9.60,
GM= 20log10 ?
?
?
?
which is consistent withthe result obtained from the Bode plot.
The Nyquist plot graph on Figure 9.23(a) is produced using the following
% CSD
Fig9.23a
num=5000;
w2=logspace(0.85,2);
Nyquist
den=[1
55
plot
250
Ex9.25
0];
|T(jw)|
&
Gain/phase
margins
L=tf(num,den);
[re,im]=nyquist(num,den,w2);
MATLAB code:
figure(103),
< 1.745
336
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
Fig. 9.23: Gain and phase margins on Nyquist and Bode plots.
Fig.
9.24:
Gain
and
phase
margins
onthe|T(j?)|plot.
plot(re,im,'r-'),
[x,y]
=
grid
on,
title('Nyquist
meshgrid(t);
grid
Fig9.23b
num=5000;
on Figure
Bode
den=[1
plot
55
w3=linspace(7,20);
figure(104),
hold
on,
t
=
axis
0:pi/80:6.28;
equal;
minor
The Bode plot graphs
% CSD
plot'),
plot(cos(t),sin(t),'m--');
9.23(b)
Ex9.25
250
are obtained
|T(jw)|
using the following
&
Gain/phase
MATLAB
margins
0];
[mag,phase]=bode(num,den,w3);
magb=20*log10(mag);
subplot(2,1,1),
title('Magnitude
response
in
dB'),
plot(w3,magb),
title('Magnitude
response
in
dB'),
grid
on,
grid
minor,
subplot(2,1,2),
title('Phase
response
in
deg'),
plot(w3,phase),
title('Phase
response
in
deg'),
grid
on,
grid
mino
code:
9.6
Essential
Closed-Loop
Transfer
Functions
and Loop Shaping
337
9.6 Essential Closed-Loop Transfer Functions and Loop Shaping
In the previous section, Section 9.5, welearned that the robust stability of a closed-loop system is closely
related with the complementary sensitivity function
T(s) = L(s)[1+L(s)]-1
where L(s)
= G(s)K(s)
is the loop transfer
function
(LTF)
(9.65)
of the closed-loop
system, as shown in
Fig-ure
9.25.
Onthe other hand, as discussed in Section 8.1.2, the tracking error response E(s), dueto the refer-ence
input R(s) and the disturbance input D(s) of the same closed-loop system, shown in Figure 9.25,
is given by Equation
8.4:
E(s)
=[1+L(s)]-1R(s)-[1+L(s)]-1D(s)
Note that the tracking
performance
is
mainly determined
by the sensitivity
function,
S(s) =[1+L(s)]-1
(9.66)
where L(s) = G(s)K(s), again, is the loop transfer function (LTF) of the closed-loop system.
It can be seen that the tracking
error
and the
disturbance
response
will be better if the sensitivity
function S(s) can be madesmaller. Wealso learned that the robust stability margin will be larger if the
complementary sensitivity function T(s) can be reduced. However,the two functions S and T can not
be reduced at the same time since S+T = 1. Fortunately, the tracking reference input signal r(t) and
the disturbance input d(t) usually occur in the low frequency range while the plant uncertainties only
become issues at high frequencies. Forthis reason, it has been a common practice to design a controller
K(s)sothat the sensitivityfunction S(j?) is smallin the low frequency range, whilekeepingthe
complementary sensitivity function T(j?) small atthe high frequency end. This control system de-sign
is called aloop shaping approach.
Fig. 9.25: Atypical feedback control system structure.
Atlowfrequencies,
themagnitude
oftheloop
transfer
function,
|L|,usually
is much
greater
than
1
so that the sensitivity function S = (1+L)-1
1
|L(j?)| |S(j?)|
is approximately equal to L-1; hence, wehave
atlowfrequencies
when
|L| ?1
(9.67)
On
theother
hand,
atthehigh
frequency
region,
themagnitude
oftheloop
transfer
function,
|L|,is much
lessthan 1 sothatthe complementary
sensitivityfunction T = L(1+L)-1 is approximatelyequalto L;
hence,
we hav
338
9
Time
Delay, Plant
Uncertainty,
and Robust Stability
athigh
frequencies
when
|L| ?1
|L(j?)| |T(j?)|
(9.68)
Therefore, the loop shaping design can be accomplished by choosing a controller K(s) so that the mag-nitude
of the loop transfer function L(s) = G(s)K(s) is large in the low frequency range, but small at
high frequencies.
Recall that in Examples 9.23 and 9.25, weselected a controller
Kto
manipulate the loop trans-fer
function L=KG, andthus the complementary sensitivity function T =L(1+L)-1, to achieve an
acceptable
robust
stability
requirement.
In the following,
we will evaluate
the tracking
error
and
disturbance response performance of the systemin terms of the sensitivity function S = (1+L)-1.
Example 9.26 (Loop
Shaping for Tracking
Performance and Robust Stability)
The system considered in Example 9.23 with loop transfer function
5000
L(s)
is revisited in the following
same loop shaping
= G(s)K(s)
to reveal the tracking
=
(9.69)
s(s2 +55s+250)
performance and the robust stability
margin on the
diagram.
Fig.
9.26:
Aloop
shaping
diagram
showing
|T(j?)|dB,
|1/S(
j?)|dB,
and
|L(j?)|dB
.
Three
graphs|1/S(
j?)|dB
in blue,
|T(j?)|dB
ingreen,
and
|L(j?)|dB
inredareshown
on
theloopshaping
diagram
in Figure
9.26.
These
graphs
verify
that|1/S(
j?)|dB|L(j?)|dBwhen
? <4rad/s
and
|T(j?)|dB|L(j?)|dB
when
? >20rad/s.
The
controller
designer
would
liketo
make
|L(j?)|dB
large
(i.e.,
todecrease
|S(j?)|)inthelowfrequency
range
andathesame
time
make
|L(j?)|dB
small
(i.e.,
todecrease
|T(j?)|)inthehigh
frequency
range.
The
high
frequency
portion
(? >6rad/s)
ofthe
|T(j?)|dB
graph
was
employed
inExamples
9.23
and 9.25to respectively obtainthe generalizedstability marginfunctionM(?)
andthe gain/phase mar-gins.
Inthelowfrequency
portion
ofthe|L(j?)|dB
or|1/S(
j?)|dB
graph,
forinstance,
when
?=1rad/s,
we hav
9.7
Exercise
Problems
339
|L(j1)|dB
=|1/S(
j1)|dB
=26dB ? |S(j1)|=0.05
which meansthe tracking error of the reference input signal at the frequency
5%. Similarly, when ? = 0.1 rad/s, we have
? = 1 rad/s will be only
|L(j0.1)|dB
=|1/S(
j0.1)|dB
=46dB ? |S(j0.1)|=0.005
which meansthe tracking error ofthe reference input signal at the frequency ? = 0.1 rad/s will
be only 0.5%.
Theloop shaping graphs on Figure 9.26 are obtained using the following
%
Fig9.26
MATLAB code:
CSD_LoopShaping_Ex9.26
num=5000;
den=[1
%
Complementary
T
=
%
Sensitivity
55
250
0];
function
L=tf(num,den);
T
feedback(L,1);
function
S
S=1-T;
figure(34),
sigma(inv(S),'b',T,'g',L,'r',{.1,100}),
grid
on
Fig. 9.27: The loop transfer function (LTF) L(s) of a typical feedback control system.
9.7 Exercise Problems
In the typical feedback control system shown in Figure 9.27, G(s) is the plantthe
system to be
controlledand
K(s) is the controller to be designed to improve the performance androbust stability of
the closed-loop
system subject to control-input
constraints.
particularly focus on the robust stability requirement.
function
(LTF),
which has been playing
In the following
exercise
problems
we will
Noticethat L(s) = G(s)K(s) is the loop transfer
a key role in the root locus
design and Nyquist stability
analysis.
P9.1a: In this exercise problem, we will revisit the dual-loop motor speed tracking control system con-sidered
in
Example 6.1. The block diagram
plant is a DC motor with transfer
of the feedback
control system is shown in Figure 9.28. The
function
G(s) =
b
s+a
where the output y(t) and the control input u(t) represent the motor speed and the electric voltage,
respectively. The controller is composed of one integrator s-1 and two constant parameters K1 and K2
340
9
Time
The integrator
Delay, Plant
is included
Uncertainty,
and Robust Stability
to guarantee zero steady-state
error due to step tracking
input,
and K1 and K2
are to be determined based on the desired transient performance and robust stability subject to control-input
constraints.
First of all, show that the loop transfer
L(s) =
function
is
bK2
s(s+a-bK1)
Fig. 9.28: A dual-loop motor speed tracking control system.
P9.1b:
Let b = 145.5 and a = 43.14.
Find the characteristic
equation
of the closed-loop
determine the values of K1 and K2 so that the damping ratio and the natural frequency
system,
and
of the closed-loop
system are ? = 0.9 and ?n = 50rad/s,respectively.
P9.1c:
With the K1 and K2 designed in
P9.1b,
we are ready to evaluate the performance
of the closed-loop
system. Assumethe control input u(t) is required to be less than 15V to avoid actuator saturation,
and the reference input is expected to be below 40 rad/s. Letthe reference input be r(t) = 40us(t) rad/s.
Plot the closed-loop responses y(t) and u(t). Inspect the performance of y(t) in terms of steady-state
error, rise time,
maximum
overshoot,
and settling time.
Also check if the control-input
constraint
is sat-isfied
for u(t).
P9.1d: Drawthe Nyquist plot and the Bode plot of the loop transfer function
L(s), andfind the phase-crossover
frequency ?p,the gain margin GM,the gain-crossoverfrequency ?g, andthe phase margin
PM of the system.
P9.1e: Find the complementary sensitivity function
T(s) = L(s)[1+L(s)]-1
Plot
themaximum
singular
value
ofT(j?),which
isequal
tothemagnitude
|T(j?)|inthisSISO
case,
s[T(j?)] =|T(j?)|
and comment onthe physical meaning of this plot.
P9.1f: Usethe gain-crossover frequency ?g from the solution of P9.1d, and the value of
?T(
j?g)
?from
?
?
the|T(j?)|plotobtained
inP9.1e
tocompute
thephase
margin
ofthesystem
based
onEquation
9.62
or Equation 9.63. Verify this result
P9.1g: In almost all practical
control
with the PM value obtained from P9.1d.
systems, time
delay
may occur in the process of
measurement, de-cision
,
making,andactuatoractuation,andthelooptransferfunction needsto be modifiedas L(s)e-sTd
where Tdsis the delay time. Let Td = 0.01s, draw the Nyquist plot andthe Bode plot of L(s)e-sTd, an
9.7
Exercise
Problems
341
find the phase-crossoverfrequency ?p,the gain margin GM,the gain-crossoverfrequency ?g, andthe
phase margin PM of the delayed system. Is the system stable?
P9.1h:
Find the critical
delay time
Tc at which the system
will become unstable.
Draw the
and the Bode plot of L(s)e-sTc, and use these plots to verify that Tc is indeed the critical
Nyquist plot
delay time.
P9.1i: Comment onthe controller designed in P9.1b especially based on its ability to address the robust
stability
issue caused by time
delay.
P9.2a: In this exercise problem, we will still consider the same dual-loop integral control system struc-ture
shown in Figure 9.28, the same plant
G(s) =
but the controller
parameters
a better robust stability
b
145.
s+a
=
s+43.14
K1 and K2 will be designed differently
against time
delay. The loop transfer function
so that the closed-loop
will have the same form,
system has
as shown
in P9.1a. Find the characteristic equation of the closed-loop system in terms of the parameters K1 and
K2, and then determine the values ofK1
and K2 so that the damping ratio and the natural frequency
closed-loop system are ? = 0.9 and ?n = 10 rad/s, respectively.
ratio, but reduce the natural frequency
P9.2b:
by a factor
of the
Notice that we keepthe same damping
of 5 in order to slow down the tracking
response.
Withthe new K1and K2 designed in P9.2a, weareready to evaluate the performance of the new
closed-loop
system.
Let the reference input
and u(t). Inspect the performance
be r(t)
of y(t) in terms
= 40us(t)
rad/s.
of steady-state
Plot the closed-loop
error, rise time,
responses y(t)
maximum
overshoot,
and settling time. Also check if the control-input constraint is satisfied for u(t). Comparethe simulation
results with those obtained in P9.1c.
P9.2c: Drawthe Nyquist plot and the Bode plot of the loop transfer function L(s) and find the phase-crossover
frequency?p,the gain marginGM,the gain-crossover
frequency?g,andthe phasemargin
PM of the system.
P9.2d: Find the complementary sensitivity function
T(s) = L(s)[1+L(s)]-1
Plotthe maximumsingular value of T(j?),
s[T(j?)] =|T(j?)|
and comment onthe physical meaning of this plot. Comparethis plot with that in P9.1e.
P9.2e: Usethe gain-crossover frequency ?gfrom the solution of P9.2c, andthe value of
?T(
j?g)
?from
?
?
the|T(j?)|plotobtained
inP9.2d
tocompute
thephase
margin
ofthesystem
based
onEquation
9.62
or Equation
9.63. Verify this result
with the PM value obtained from
P9.2c.
P9.2f: Assumethere exists atime delay in the system and the loop transfer function needsto be modified
, where Tds is the delay time. Let Td = 0.01s, draw the Nyquist plot and the Bode plot of
to L(s)e-sTd
, andfind the phase-crossoverfrequency ?p,the gain margin GM,the gain-crossoverfrequency
L(s)e-sTd
?g,andthe phase marginPM ofthe delayedsystem.Is the systemstable? Compare
the results withthose
obtained in P9.1g.
342
9
Time
Delay, Plant
P9.2g:
Find the critical
Uncertainty,
delay time
and Robust Stability
Tc at which the system
will become unstable.
Draw the
Nyquist plot
, and usethese plots to verify that Tcis indeed the critical delay time.
andthe Bode plot of L(s)e-sTc
Compare the critical
delay time
with that
obtained in P9.1h, and give your comment.
P9.2h: Comment onthe advantages andthe deficiencies of the gain and phase margins as measuresof
robust stability.
Fig. 9.29:
P9.3a: In this exercise problem,
with the same plant
A PI motor speed tracking
we will consider the same DC motor speed tracking
G(s) =
as in
P9.1, but the controller
control system.
structure
b
145.
=
s+a
is different.
s+43.14
It is the
controller
Kp +
well-known
PI (proportional
plus integral)
Ki
s
as shown in Figure 9.29. First of all, show that the loop transfer
L(s) =
control problem
function
is
bKps+bKi
s(s+a)
Note
that
thisloop
transfer
function
has
azero
at-Ki/Kpwhile
theloop
transfer
functions
in P9.1
and
P9.2 have no zeros.
P9.3b: Let b = 145.5 and a = 43.14. Find the characteristic equation of the closed-loop system, and
determinethe valuesof Kpand Kisothat the dampingratio andthe naturalfrequency ofthe closed-loop
system are ? = 0.9 and ?n = 50 rad/s, respectively. Noticethat the poles of this PI control system are
placed at the same locations asthe poles of the system in P9.1.
P9.3c: Now weare ready to evaluate the performance of the closed-loop system with the Kpand Ki ob-tained
in P9.3b. Assumethe control input u(t) is required to beless than 15V to avoid actuator saturation,
and the reference input is expected to be below 40 rad/s. Letthe reference input be r(t) = 40us(t) rad/s.
Plot the closed-loop responses y(t) and u(t). Inspect the performance of y(t) in terms of steady-state
error, rise time,
maximum overshoot,
and settling time,
and check if the control-input
constraint
is satis-fied
for u(t). Comparethe simulation results with those obtained in P9.1c, and give your comments.
P9.3d: Drawthe Nyquist plot and the Bode plot of the loop transfer function
L(s), andfind the phase-crossover
frequency ?p,the gain margin GM,the gain-crossoverfrequency ?g, andthe phase margin
PM of the system.
9.7
P9.3e:
Find the complementary
sensitivity
Exercise
Problems
343
function
T(s) = L(s)[1+L(s)]-1
Plotthe maximumsingular value of T(j?),
s[T(j?)] =|T(j?)|
and comment
on the physical
meaning of this plot.
P9.3f: Usethe gain-crossoverfrequency ?gfrom the solution of P9.1d,andthe value of
?T(
j?g)
?from
?
?
the|T(j?)|plotobtained
in P9.3e
tocompute
thephase
margin
ofthesystem
based
onEquation
9.62
or Equation
9.63. Verify this result
with the PM value obtained from
P9.3d.
P9.3g: Assumethere exists atime delay in the system and the loop transfer function needsto be modi-fied
, where Tdsis the delay time. Let Td = 0.01s, draw the Nyquist plot andthe Bode plot of
to L(s)e-sTd
, andfind the phase-crossoverfrequency ?p,the gain margin GM,the gain-crossoverfrequency
L(s)e-sTd
?g,andthe phasemarginPMofthe delayedsystem.Is the systemstable?
P9.3h: Find the critical delay time Tc at whichthe system will become unstable. Drawthe Nyquist plot
andthe Bodeplot of L(s)e-sTcand usethese plots to verify that Tcis indeed the critical delay time.
P9.3i:
Comment
on the
PI controller
designed in P9.3b, especially
on its differences
from the dual-loop
controller designed in P9.1b.
P9.4a: In this exercise
delay in
problem,
we will revisit the first-order
Example 9.14. The loop transfer
function
L(s) =
where T is the delay time.
closed-loop
control
system
with a time
T = 0). Draw the
Nyquist plot
of the system is
2
s+1
e-sT
First of all, assume there is no time
delay (i.e.,
andthe Bodeplot of L(s) with T = 0, anddeterminethe gain-crossover
frequency]?g rad/s,andthe
phase
margin PM in degrees.
P9.4b: Recallthat we computed the critical delay time Tc based on Equations 9.37 and 9.38. The delay
time
also can be obtained
using the information
of the phase
margin and the gain-crossover
frequency.
Usethe phase marginPM andthe gain-crossoverfrequency ?gin P9.4ato verify that the critical delay
time is Tc = 1.209s.
P9.5a: Consider the feedback control system shown in Figure 9.27 whoseloop transfer function L(s) is
described by the following first-order linear system with time delay,
L(s) =
where b and a are real constant
can be computed
b
s+a
e-s
parameters and T is the delay time.
using the following
formula
Show that the critical
if b > a:
b2-a2??
?
?v
p-tan-1
b2-a2
Tc= v
1
a
delay time
Tc
344
P9.5b:
9
Time
Delay, Plant
What does it
Uncertainty,
and Robust Stability
mean when the condition
b > ais not satisfied?
P9.5c: Vary the parameters b and a, one at a time, and observe how they would affect the value of the
critical delay time Tc
10
State Feedbackand Linear Quadratic Optimization
I
Nthis chapter and the next we will focus on feedback control analysis and design based on the
state-space models. The state-space approach became popular in the early 1960s beginning with
the publications of Rudolf Kalman, [Kalman, 1960a,b, Kalman and Bertram, 1960, Kalman and
Bucy, 1961]. Instead of frequency-domain
methods(i.e., Laplace transform approaches), attention re-turned
to the earlier methods of analysis and design using differential equations (e.g., [Maxwell, 1868]).
The main reasons of this revolutionary
paradigm shift are the following: (1) The state-space
approach resolved basic theoretical problems that hadimpeded the extension of frequency-domain tools
to MIMO systems; (2) The nonlinear system state-space representation is elegant and versatile, allowing
systematic waysto identify the equilibriums of the system, to find alocal (linear or nonlinear) model at
each equilibrium of interest that can be employed in analysis and controller design; (3) The state-space
framework makesit easier to formulate the control problems as constrained optimization problems like
the LQR (linear quadratic regulation), the LQG (linear quadratic Gaussian),the H2optimization, andthe
H8 optimization
control
problems;
(4)
The computing
capability
and the
miniaturization
of the digital
computer hasfacilitated the applications of the state-space control approaches in almost every product
and every
manufacturing
process to achieve automation,
However, the rise of the state-space
approach.
frequency-domain
Instead,
The frequency-domain
any system.
properties
They are inseparable.
by the pole locations
approach
the state-space
performance
precision, reliability,
does not
model framework
requirement
into
properties
In fact, the time-domain
response
enhance-ment.
mean the end of the frequency-domain
has
made it
design of large-scale
and the time-domain
and the frequency
and performance
possible to incorporate
MIMO control systems.
are still the two important
responses, stability,
aspects of
and robustness
of the system, as we have
are dic-tated
witnessed in Sections
2.5, 3.3, 3.4, 8.4, 8.6, 9.3, and 9.4.
In the following section, we will have a brief review of what wehavelearned in the previous chapters
regarding the control systems fundamentals related to the state-space approach. We will reinforce these
key concepts
performance
formulation
theory
quadratic
and employ them in new control system
optimization
with control-input
design applications.
constraints
and discuss the solution to the optimization
will be introduced
optimization
and employed
approach
to optimally
can be extended
in the
we will incorporate
control
problem
problem. In the next chapter, the observer
estimate
to the
Then
state-feedback
the state variables
more general
output
so that the linear
feedback
case
346
10
State Feedback
and Linear
Quadratic
Optimization
10.1 Brief Review of the State-Space Approach
The nonlinear
state-space
model was employed
system in Section 4.4 because it
provides
to describe the dynamics
the
of the simple inverted
best possible framework
to compute
pendu-lum
the equilib-riums
and to obtain the linearized state-space model at the equilibriums
of interest via Jacobian
matrices approach, as shown in Equations 4.41 and 4.42. Thelinear state-space model at each equilib-rium
is described in the form:
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
where
x(t)isthe21 state
vector
consisting
theangular
displacement
and
theangular
velocity
ofthe
pendulum, u(t) is the control input representing the applied force, y(t) is the output representing the
angular
displacement
of the pendulum,
and the dimensions
of the
matrices A, B, C, and
D are assumed
comparable.
The poles of the inverted
pendulum
system at the equilibrium
associated
with the upright
pendulum
position wereshown to be the eigenvalues of the A matrix, which verifies the equilibrium is unstable
since one of the two poles is in the right half of the complex plane, as shown in Equation 4.42. Thelin-ear
state-space modelcan befurther employed to design a state-feedback controller u(t) = Fx(t)
to place the closed-loop system poles, which are the eigenvalues of the matrix A+BF, at desired
locations in the left half of the complex plane to stabilize the closed-loop system and achieve de-sired
performances.
Although the state variables assignments are not unique, usually for
variables like
electrical
displacements
and velocities
systems, the currents through
mechanical systems the physi-cal
are chosen to be state variables, if possible.
the inductors
Similarly,
and the voltages across the capacitors
for
are practi-cal
candidates to serve as state variables. In the control system design or analysis, for computational
reasons, a state-space similarity transformation
may be employed to change the state variables so
that the state-space modelis in some special form, like the company form or the diagonal form.
On most occasions,
state variables
are chosen after a differential
equation
model is obtained.
However,
it is possible to assign state variables before constructing the dynamics model.In Section 5.5, a direct
state-space modeling approach was presented to obtain a state-space model for electrical circuits, as
demonstrated in Examples 5.12 and 5.13.
Chapter 6 covers systems representations
flow
graphs, transfer
functions,
state-space
of interconnected
systems including
models, state diagrams,
block diagrams, signal
and the relationships
among them.
Masons gain formula, the state transition matrix,the Cayely-Hamilton theorem, andthe solution of the
state equation are also the maintopics of the chapter. Astate diagram is a graphical representation of a
state-space
model in the form
of signal flow
graph or block diagram.
Hence, it is trivial
model given a state diagram and vice versa. Furthermore, the transfer function
to find the state-space
can be computed
directly
fromthestate-space
model
using
theformula
G(s)=C(sI-A)-1B+D
in Equation
6.24
or obtained indirectly from the state diagram using Masons gain formula, as demonstrated in Ex-ample
6.23. Conversely, given atransfer function there areinfinitely manycorresponding state diagrams
or state-space models. Depending on the need or preference, the corresponding state diagram can be in
the companion form via direction realization, or in the diagonal form via parallel connection, or in any
other form associated with cascade connection or feedback connection, as demonstrated in Examples
6.24, 6.25, 6.26, and 6.27.
In Chapter 7, morefundamental concepts and tools of feedback control systems are provided to pre-pare
for the more advanced control systems design and analysis, especially in the state space. These
include advantages and limitations of feedback control, characteristic equations, poles and zeros
10.1
BIBO stability,
internal
and their
defined
stability,
similarity
Brief
transforms
State-Space
Approach
in state space, state-feedback
347
pole place-ment
applications to control systems design and analysis. Notethat the internal stability is
based on the status of every state in the state-space
only in terms
Review of the
of the input-output
stronger than the BIBO stability.
after controllability
relationship
model while the
of the system;
We will discuss
more about the difference
and observability are officially
BIBO stability
hence, the internal
stability
was defined
in general is
of these two stability
defini-tions
defined later in this chapter.
Since the performance and behavior of the system are mainly determined by its pole locations and
the pole locations
to improve
can be altered via feedback
the characteristics
of the proportional
controller
control,
of the closed-loop
K will
it is a common
system.
move the closed-loop
practice to utilize
As demonstrated
in Section
system pole locations,
pole placement
3.6, the variation
as shown in the right
graph of Figure 3.16. Dueto the movement of the poles, the damping ratio ?, the natural frequency
and the corresponding
time responses
will change accordingly,
as shown in left
graph of Figure 3.16.
?n,
A
similar idea wasapplied to the state-space modelsin Section 7.5 using state feedback. The state-space
pole placement can be achieved by the direct approach or the transform approach. The direct ap-proach
is conceptually simple, but computationally
can become very complicated when the order
of the system is higher than three. The transform approach requires a similarity transformation
to transform the state-space model to a special companion form initially, but after that the com-putation
becomes extremely simple, even the order of the system is as high as 100. The transform
approach was employed in Section 7.6.4 to place the poles of the closed-loop cart-inverted pendulum
system
inthe
lefthalfofthecomplex
plane
ats =-2,-8,and-4 j3toconvert
anoriginally
unstable
system to become stable.
In
Section 8.6, an aircraft
attempt
longitudinal
wasto use the classical control
flight
path tracking
approach
control
with a PI controller
problem
was considered.
together
with root locus
The ini-tial
design
(Figure 8.35) to achieve the closed-loop system stability, zero steady-state tracking error, and acceptable
transient response. Dueto the RHP zeros and the phugoid (long-period) mode poles,the associated root
locus diagram (Figure 8.36) does not allow much option to choose an acceptable set of closed-loop sys-tem
poles to deliver an acceptable performance. The response was either extremely slow or too much
oscillation, as shown in Figure 8.37.
To address the issue, a state-feedback integral tracking control structure (Figure 8.38) together
with root locus design was employed to achieve the objective. Since all the four state variables are
assumed available
for feedback to provide
moreinformation
than the
PI controller
case where only the
flight path angle is assumed available for feedback, the state-feedback controller is expected to do better.
Like any other viable feedback control scheme, the state feedback could not change the RHP zero po-sitions;
however,
it didreplace
thepoles
oftheloop
transfer
function
tothenew
locations,
-1.7,-5.9,
-1.6 j1.8,ands =0,which
provides
a much
more
favorable
rootlocusdiagram,
asshown
in Fig-ure
8.39. Consequently, an optimal tracking integrator constant K wasselected to lead to a much better
tracking performance, as shown in Figures 8.40 and 8.41.
Although westill have moreto learn about the state-space approaches later in this chapter andthe
following one, wehave already experienced the benefit of employing state feedback, internal
model
principle, and the root locus technique to design a stabilizing, tracking/regulating
closed-loop con-trol
system. In the following section, we will employ a rather simple second-order lightly damped
position control system to demonstrate the state space/root locus pole placement approach and
compare it with the traditional
PID (proportional integral and derivative) control approach
348
10
State Feedback
and Linear
10.2 Control of a Lightly
The poor performance
and oscillate
Quadratic
Optimization
Damped Positioning System
of a lightly
with large amplitude,
damped system,
which can easily drift from its operating
is common in practice; for example, the aircraft flight
equilibrium
dynamics like
the phugoid (long-period)
modeof F/A-18 has a very small damping ratio, ? = 0.1, and a quite long
(62.3 seconds) oscillation period, as described in Equation 8.68. If the system is not adequately com-pensated,
moderate turbulence would cause unacceptably long and large up and down oscillations
in longitudinal
motion and makethe aircraft very difficult, if not impossible, to maneuver.
10.2.1
A Simple Pendulum
Positioning System
In the following, a rather simple nonlinear lightly damped pendulum positioning system, as shown
in Figure 10.1, will be employed to demonstrate how to stabilize and improve the performance of
the system
to a pivot
using the state space/root
locus
pole placement
with bearing so that the rod can perform
approach.
friction. The other end of the rod is attached with a motor/propeller,
produce the torque, T, required in the control system.
Fig. 10.1: A nonlinear, lightly
One end of the rod is connected
one degree-of-freedom
swing
motion
without
much
which serves as an actuator to
damped pendulum positioning system.
Equations of Motion
The procedure
of deriving the
discussed in Section
dynamics
4.4 and Section 4.5.
model of the pendulum
By either the
Newtonian
positioning
or the
system is similar
Lagrange equation
to those
approach,
the equation of motion for the pendulum can be found as follows:
J?(t)+b ??(t)+0.5mg?sin?(t)+Mg?sin?(t) = T(t)
(10.1)
Thetotal moment of inertia is J = J? +JM = m?2/3+M?2,
where ? = 0.5 mis length of the rod and
m = 0.1 kg and M= 0.1 kg are the massof the rod and the motor,respectively. Thefriction coefficient
ofthepivot
isassumed
b=4.510-3Nm/rad/s
and
thegravity
isg=9.81m/s2.
The
output
?(rad)
is the angular displacementof the rod, the control input is the appliedtorque T = ?Fp(?) (Nm), and
the propeller force Fp(?) is determined by the motorspeed ? (rad/s). The DC motor dynamics is
described by the following
equation
10.2
Control
0.0232??(t)+?(t)
where E(V) is the DC motor input
of a Lightly
Damped Positioning
System
= 47.22E(t)
349
(10.2
voltage.
Nonlinear State Equations and Equilibriums
Withthe physical data given, the pendulum equation, Equation 10.1 becomes
? (t)+a1
??(t)+a0
sin?(t)
= b0T(t)
(10.3)
where a1 = 0.135, a0 = 22.073, and b0 = 30. Letthe state variables be x1(t)
then the nonlinear
state equation
x? =
Assume the
associated
with Equation
x2
operating
= f(x,T)
=
equilibrium
is chosen to
=??(t),
10.3 can be written as
???-a0sinx1-a1x2+b0T?
x?1
x?2
= ?(t) and x2(t)
keep the
=
angular
?
f1(x1,x2,T)
f2(x1,x2,T)
displacement
?
(10.4)
of the pendulum
at
?(t) = ?* = 15? = p/12 rad. Thenthe equilibrium ofthe systemcanbefound by solvingthe state
equations
with the derivative
of the state variables set to zero.
Now, we have
??=0 ? x2
=sin15?
0 -a1x2
-a0
+b0T
=0 ? ?
x*?? ?
x*
x?1
x?2
1
15?
=
0
2
Linearized State-Space
, T* = 0.19043Nm
Model
Next, we will find a linearized state-space modelfor the ?* = 15? equilibrium. Atthis equilibrium
(x*1,x*2,T*) = (15?,0?/s,0.19043 Nm), wehavethe linearized state-space model
x?(t)
= x(t)+B
A
T(t)
(10.5)
wherethe matrices A and B are computed via Jacobian matricesJx and JT, respectively, asfollows:
A = Jx =
?? ?
?f
?x
and
=
x*,T*
?f1/?x1
?f2/?x1
x?(t)
=
??
p/12
x*=
0
?
=
,T*
?-a0cos(p/12)-a1?
0
?? ? ? ??
? ?? ?????
B = JT =
That is,
?f1/?x2
?f2/?x2
x?1(t)
x?2(t)
0
=
?f
?T
=
x*,T*
1
-21.32-0.135
?f1/?T
?f2/?T
x1(t)
x2(t)
1
0
=
x*,T*
0
+
30
T (t)
b0
= x(t)+B
A
T(t)
(10.6)
Notethat the relationship amongthe real valuesx(t), T(t), the equilibrium valuesx*, T*, andthe differ-ential
(perturbed) valuesx(t),
T areshownin thefollowing:
x(t) =x(t)+x*,
T(t) =T(t)+T*
Forinstance,xT = ? 20? 0?/s ? meansthat the real state vector is xT = ? 35? 0?/s ?, and aT
revealsthat the real torque is T = T* = 0.19043 Nm.
(10.7)
= 0 Nm
350
10
Analysis
State Feedback
of the
and Linear
Open-Loop
Quadratic
Optimization
System
Recall that the eigenvalues of the A matrix, or the poles of the system are the roots of the following
characteristic equation:
?
|sI-A| =
?
?
s
-1
21.32 s+0.135
?=s2
+0.135s+21.32
:=s+2??ns+?2
2
?
The roots of this characteristic
n =0
?
(10.8
?
?
equation are
-a j? =-0.0675j4.6169
and the corresponding
damping
ratio
and the natural
? = 0.0146
respectively.
and
frequency
are
?n = 4.6174 rad/s
Wehavelearned from the previous chapters, especially Section 3.4.3, that the time-domain
behavior of the systemis closelyrelatedto the dampingfactor a,the frequency ?,the dampingratio ?,
andthe naturalfrequency ?n, which are derivedfrom the roots of the characteristic equation. Hence,it
is possible to get a general idea of how the system will behave based on the information
poles, or, equivalently, the roots of the characteristic equation.
of the system
Before conducting a computer simulation using the mathematical model, we can perform a
simple virtual experiment. Imagine that a pendulum positioning system, as shown in Figure 10.1 is
operating
atthe ? = ?* =15?equilibrium.Thatis, the motorpropeller
is providingaconstant
torque
T* =0.19043 Nmto keep? atthe 15?equilibrium. Now,lets perturbthe systemalittle bit by moving
the pendulum upto the 35? position whilethe appliedtorque remainsthe same at T* = 0.19043 Nm.
Thenreleasethe pendulumatt = 0 s and observethe pendulum angular displacement ?(t) as a func-tion
of time.
What will you see? The system is stable, so eventually the pendulum
will go back to the
equilibrium. Butsincethe dampingfactor a andthe dampingratio ? are very small, the pendulum will
go down toward the equilibrium,
equilibrium
manytimes
pass it
with a large overshoot,
until it is stabilized
and then swing
back and forth
about the
at the equilibrium.
The oscillationfrequencyis ?=4.6169 rad/s, andthusthe oscillation periodis P=2p/?=1.361
s.
The dampingfactor a = 0.0675revealsthat the oscillation amplitude will decreaseexponentially with
time constantt = 1/a = 14.81s. Thatis, the oscillation amplitude will reduceto 36.8% ofits original
value after t = t = 14.81 s, or, equivalently,reduce to 1.83% ofits original value after 4t = 59.2 s
of time.
The oscillation
amplitude
decreasing
? = 0.0146.The maximumovershoot
is
rate
can also be computed
v
OS= e-?p/ 1-?2
using the
damping
ratio
= 0.95516, which meansthat the oscil-lation
amplitude reduces to 95.516% of its value after a half period P/2 = 0.68 s of time. Hence, after
87 half periods, which is 59.16s, the oscillation amplitude will reduce to 1.84% of its original value.
A Simulink simulation
of the pendulum
positioning system of Equation 10.3 or Equation 10.4
withinitial angular displacementat ?(0) = 35?, whichis 20?awayfrom the equilibrium position
?* =15?,canbeconducted
to obtain
theoscillatory
timeresponse,
asshownontheleft graphof Figure
10.2, whilethe control-inputtorque is kept atthe equilibrium torque T* = 0.19043 Nm,asshown onthe
right graph. The oscillation period P can be measuredfrom the peak-to-peak time onthe left graph. The
time at the seventh peak is approximately
overshoot
OS can be computed
9.55s, which implies
based on the
10.5?, or, equivalently 10.5/20 = 52.5% of the initial
have OS14 = 0.525, whichimplies that OS = 0.955.
The simulation
results are fairly
P = 9.55/7
measure of the amplitude
consistent
= 1.364 s. The maximum
at the seventh
peak,
which is
amplitude after 14 half periods. Therefore, we
with the virtual
experiment
and the prediction
based
onthepole
location
-a j?,thedamping
ratio
?,and
thenatural
frequency
?n.
10.2
Fig. 10.2:
An oscillatory
initial
time response
Control
of the nonlinear
of a Lightly
lightly
Damped Positioning
damped pendulum
System
system
351
due to a per-turbed
condition.
10.2.2 State-Feedback
Stabilization
of the Pendulum Positioning System
From this analysis and the simulation results of the pendulum positioning system, the open-loop sys-tem
can drift out of its equilibrium easily dueto a small disturbance, and it also requires along period
of time to recover
difficult
back to the equilibrium.
or even impossible
miscommunication
to
makeit
for
commands.
the lightly
due to the
damping
oscillations
and time delay. One maysuggest to fundamentally
more heavily
damped.
Well, it is like
the system less sensitive to disturbances
control
Furthermore,
manual control
One effective
a double-edged
will at the same time
way to resolve this
Recallthat atthe equilibrium (x*1,x*2,T*)
and
feature
may
man-machine
make it
interface
change the mechanical design
sword. Increasing
the friction
to
make
cause the system to be less responsive
dilemma
is to employ
feedback
to
control.
=(15?,0?/s,0.19043 Nm), wehavethe linearized state-x
space model
?(t)
= x(t)+BT(t)
A
wherethe matrices A and B are given by Equation 10.6:
A =
Now, we will close the loop
?
0
? ??
1
,
B =
-21.32-0.135
0
30
using the state feedback
T (t) = F
x(t)
(10.9)
sothat
theclosed-loop
system
poles
are
ats =-10,-11.
The characteristic equation of the closed-loop system will be
(s+10)(s+11)
=s2 +21s+110 =s2 +2??ns+?2
n ? ?=1.0011,
?n=10.49
rad/s
which implies the closed-loop system response will be slightly overdamped with no overshoot and no
oscillations. The closed-loop control system thus designed will yield the state-feedback gain matrix as
?
F=? -2.956
-0.6955
and the state equation
of the closed-loop
x?(t)
system
(10.10)
will become
=(A+BF) x(t)
=
? ?
0
1
x(t)
-110-21
(10.11
352
10
State Feedback
Assume the initial
and Linear
Quadratic
condition
x(0)T
The solution
Optimization
of the closed-loop
= ? 20? 0?/s ? =? 0.349 0 ?
system state equation,
Equation 10.11 can be found
as
11e-10t
-10e-11t e-10t
-e-11t???
? ? ? ??-110e-10t
+110e-11t
-10e-10t
+10e-11t
x1(t)
x2(t)
= e(A+BF)t
0.349
0
where e(A+BF)t is the state transition
inverse Laplace transform,
0.349
=
0
matrix of the closed-loop system that can be computed using the
?(sI-(A+BF))-1?
e(A+BF)t =L-1
or utilizing the Cayley-Hamilton approachin Section6.4.2. Hence,the pendulumangular position ?(t)
and the applied torque T(t) will be
?(t) =x*1+
x1(t)=15?+220?e-10t
-200?e-11t
and
T(t)=T*+
T(t) =T*+F
x =0.19043+15.3522e-10t
-16.3838e-11t
Nm
respectively. Notethat
to deg and
wealways use rad, rad/s as units in the computation,
deg/s in display
for ease of human
and convert them back
reading.
Fig. 10.3: The nonlinear, linearized, and simulation
models of the pendulum positioning system.
The Simulink simulation diagram in Figure 10.3(c) with the nonlinear plant is employed to
conduct the closed-loop initial state response simulation. In the Simulink simulation diagram, you
will see a function block named Fcn1. Double click the function block to open the pop-up window,
called Function
Block
Parameters:
Fcn1, andtype
-a0*sin(u(1))-a1*u(2)+b0*u(3)
into the
Parameters
function
of the three inputs:
Expression
box.
Note that this
function
u(1), u(2), and u(3) that represent
f(u)=-a0*sin(u(1))-a1*u(2)+b0*u(3)
is a
x1, x2, and T, respectively.
The value of the function then serves asthe input to the Intg2 integrator; that is,?x2=f(u). The initial
statex1(0) = 35?=0.6109
rad, whichis equivalent
tox1(0) =20? =0.349rad,is chosenby double
clicking
the Intg1 integrator
to open the pop-up
window, called
Function
Block
Parameters
10.2
Intg1,
and type the radian
number
Control
of a Lightly
0.6109 onto to Initial
Damped Positioning
Condition
System
box. The input
353
constant
boxes
named xe and Te should beassignedas 0.2618,the radian numberof the equilibrium position x*
1 = 15?,
and 0.19043, the equilibrium
Equation
torque,
respectively.
Of course,
F is the state-feedback
gain, as shown in
10.10.
The simulation
the analysis
results
are shown in Figure 10.4. It can be seen that the results
and computations.
The state-feedback
control
has brought
the
are consistent
pendulum
with
from
the
perturbed 35?positionbackto the 15?equilibrium within 2%vicinityin just 0.6seconds,whichis
100 times faster
than
the slowly
decayed
oscillatory
response
without feedback,
as shown in
Figure
10.2. Therequired torque T(t) to complete the control action shown on the right graph of Figure 10.4 is
between
-0.8Nm
and
0.5Nm,
which
is within
apractical
range.
Fig. 10.4:
10.2.3
Animpressive
Stabilization
of the
disturbance
response
Motor/Propeller-Driven
reduction
accomplished
Pendulum
by state feedback.
Positioning
System
In Section 10.2.2, a state-feedback control wasemployed to greatly improve the stability of the originally
lightly
damped
system.
For simplicity,
the torque can be provided
positioning
the torque
was considered
as the control input.
In
by a motor control subsystem. In the motor/propeller-driven
system shown in Figure 10.1, the torque
T(t) = ?Fp(t)
applied to the pendulum
practice,
pendulum
is
and Fp(t) is proportionalto ?2(t)
(10.12)
where ? (m) is the length of the pendulum and ?(t) (rad/s) is the rotational speed of the motor/propeller.
The motor dynamics
and the torque
applied to the pendulum
are described
by the equation,
??(t) =-am?(t)+bmE(t)
:= f(?,E)
T(t) = (1/k2)?2(t)
(10.13)
:= h(?,E)
where am = 43.1 (1/s), bm = 2034.5 (1/Vs2), and k2 = 26345 (1/Nms2).
Notethat the state equation
of the motoris linear, but the relationship between the torque T and the motor speed ? is nonlinear.
When
the systemis operating atthe 15? pendulumequilibrium,
[x* T*]T =[x*1 x*2 T*]T =[0.2618 rad 0rad/s 0.19043Nm]T
the corresponding motorspeedequilibrium ?* andthe motorinput voltageequilibrium E* can be com-puted
based on Equation
10.13 as follows
354
10
State Feedback
and Linear
Quadratic
Optimization
? ?*= v0.19043k2
=70.83rad/s
T* = 0.19043 =(1/k2)(?*)2
0 =-am?*
+bmE*
? E*=(am/bm)?*
=1.5V
Thelinearized
model of the nonlinear motor/propeller dynamics shown in Equation 10.13 can
be obtained using the Jacobian matrices approach in Section 4.4.2 as follows:
??(t) =-am
?(t)+bm
E(t)
(10.14)
T (t) = (1/k2) ??2
?? ?*?(t) =(2/k2)?*?(t)
?
?
?
Fig. 10.5: The nonlinear,
system
linearized,
with motor/propeller
and simulation
models of the
closed-loop
pendulum
positioning
as the actuator.
Withthe motor/propeller as the actuator, the model diagrams of the closed-loop pendulum position-ing
system shown in Figure 10.3 are now redrawn asthose shown in Figure 10.5. Figure 10.5(a) shows
the closed-loop block diagram of the nonlinear pendulum system with the linear controller including F,
K,andthetransferfunctionofthe motor/propeller
(2?*bm/k2)/(s+am). Notethatthe outputofthe
nonlinear
plant is the real state vector
x while the input
to the linear
controller
is the
differential
(orperturbed)
state
vector
x =x-x*,where
x*istheequilibrium
state
vector.
Similarly,
theoutput
of the linear controller is the differential (or perturbed) torque T whilethe input to the nonlinear
plant is the real torque T =T+T*,
Figure
Figure
10.5(c) is
a Simulink
simulation
10.5(a). In fact, this simulation
for the insertion
The insertion
torque.
of the
of the
But the
feedback
motor/propeller
motor/propeller
action intended
diagram
diagram
motor/propeller
constructed
dynamics
based
on the
is almost the same as that in
transfer function
transfer
by the original
is inserted for compensation
locus
where T*is the equilibrium torque.
function
and the additional
is for the practical
is a low-pass
filter,
F state-feedback
which
design.
block
Figure 10.3(c)
will slow
of generating
down the
reason, the
purposes. An optimal value ofK can be determined
of
except
design parameter
reason
For this
diagram
K.
the
desired
K parame-ter
using the root
design approach.
Figure 10.5(b) is the closed-loop block diagram of the linearized
controller
closed-loop
including
F,
K, and the transfer
function
block diagram is almost the same as that in
motor/propeller transfer function and the K parameter
of the
pendulum system with the lin-ear
motor/propeller.
Figure 10.3(b)
Note that this linear
except for the insertion
of the
10.2
Recall that the root locus
Control
of a Lightly
Damped Positioning
design approach starts from the pole-zero
diagram
System
of the loop transfer
of the closed-loop system shown in Figure 8.17 and Definition 8.14. The loop transfer
(LTF) of the closed-loop system in
355
func-tion
function
Figure 10.5(b) is KG(s), where G(s)is
2?*bm/k2
G(s)=-F(sI-A)-1Bs+am
The
root
locus
diagram
based
onthethree
poles,
p1,p2=-0.0675
j4.62,
p3=-43.1
and
one
zero
z1=-4.25
ofG(s),
can
beconstructed
asshown
inFigure
10.6.
Fig. 10.6: The closed-loop pole locations for K = 2 and K = 3 on the root loci diagrams of the system
shown in Figure 10.5(b).
As shown in Figure 10.6, if the controller parameter is chosen to be K = 2, the closed-loop system
poles will be
-6.38j7.28 and -30.4
wherethe two complexconjugatepolesarethe dominantpoleswith dampingratio ? =0.66, which
implies the system response will be slightly underdamped
with about 6% overshoot. Onthe other
hand, if the controller parameter is chosen to be K = 3, the three closed-loop system poles will be
-10.2and -16.5j10.1
where the real pole with damping factor
response will be slightly overdamped.
a = 10.2 is the dominant
pole, which implies the system
The disturbance recovery simulation results based on the Simulink diagram of Figure 10.5(c) are
shown in
Figure 10.7.
The K = 2 design
The simulation
results
will lead to a slightly
are fairly
underdamped
consistent
with the root locus
response that
brings the
design analysis.
pendulum
down from
? = 35?toward the desired equilibrium position ? = 15? morequickly than the K = 3 response,but
it overshotto ? = 13? before returning to the ? = 15?later, aroundt = 0.7 s. Onthe other hand,the
K= 3responseis slightly overdampedthat comesdownfrom ? = 35?rather slowly to reachthe desired
equilibriumposition? = 15?aroundt =0.65s withoutovershootor oscillations.It turns outthatthe
K= 2.5 designis anapproximately optimal choice.It reaches ? = 15?att = 0.4s withan unnoticeable
overshoot.
When K = 2.5, the three closed-loop system poles locations are at
-9.49j6.92and -24.
356
10
State Feedback
and Linear
Quadratic
Optimization
wherethe complex conjugatepolesarethe dominant poles with dampingratio ?= 0.808, whichimplies
the overshoot is just 1.35%.
Fig. 10.7: The disturbance recovery responses accomplished by the motor driven state-feedback control
system shown in Figure 10.5(c).
Remark 10.1 (The Insertion
Compensation)
of
Motor/Propeller
Dynamics Into the Feedback
Loop and the
K
It is noted that the K = 3 disturbance recovery response in Figure 10.7(a) is about the same asthat
shown in Figure 10.4(a). The K = 2.5 response in Figure 10.7(a) is better than that shown in Figure
10.4(a). Butif the value of the controller parameter reduces to K= 1, the disturbance recovery response
will be oscillatory
performance
with 26% overshoot
actually
is foretold
and
wont reach the steady state until t
by the root locus
= 2 s. This K = 1 poor
analysis that the three closed-loop
system poles are at
-2.72 j6.53 and -37.8
andthe complex conjugate poles arethe dominant poles with dampingratio ?= 0.385that implies the
overshoot is about 27%.
Theinsertion of the motor/propeller dynamics into the feedback loop without the K compen-sation
will slow down the feedback correction and degrade the performance due to the inherent
low-pass filter property of the motor. However, the performance can be recovered or even made
better with an adequate K compensation at K = 2.5.
The two
graphs
MATLAB
% CSD
Fig10.5b
%
Construction
%
shown
am=43.1,
%
10.6
of
in
Fig.
State
1;
Motor
-21.32
Root
on Figure
State-feedback
locus
10.6 are generated
Pendulum
diagram
for
the
xe=[0.2618
at
-0.135],
this
0]',
Te=0.19043,
equilibrium,
B=[0;
30],
xe,
damp(A),
feedback
-0.6955],
speed
k2=26345,
%
plot shown
we=sqrt(k2*Te),
s=tf('s');
Ee=(am/bm)*we,
Design
sysG_F=ss(A,B,F,0);
G_M=(2*we*bm/k2)/(s+am),
K=logspace(-3,2,1000);
damp(A+B*F),
equilibrium
Locus
systems
10.5(b)
model
F=[-2.956
%
rlocus(GK)
root
bm=2035,
Linearized
A=[0
%
of the same root locus
using the following
code:
G_F=tf(sysG_F),
G=-G_M*G_F,
rlocus(G,K
figure(12)
Te
positioning
system
10.2
10.2.4
Tracking
Control
of the Pendulum
Control
Positioning
of a Lightly
System
Damped Positioning
Using State-Space
System
357
Pole Placement
In the previous two subsections, Sections 10.2.2 and 10.2.3, a state feedback/root locus design approach
was employed
practical
to improve
applications,
which is usually
the stability
the operating
unknown
and performance
equilibrium
a priori,
at a specified
operating
equilibrium.
may depend on an external command
and the controller
In
many
or reference input,
is required to be designed to achieve closed-loop
system stability, zero steady-state tracking error, and an optimal transient performance subject to control-input
constraints. The regulation/tracking
issues arise in almost every engineering problem involved
with precision, automation, guidance, navigation, and control. In fact, we have addressed these is-sues
in Sections 8.1, 8.5, 8.6, and 8.7 regarding the internal
model principle, type of systems, and
their applications in DC motor speed control, sinusoidal position tracking control, F/A-18 flight
path control, and aircraft altitude regulation.
In the following,
we will consider the tracking control of the pendulum positioning system
using the internal
model principle, the state-feedback pole placement, and the root locus design
approaches.
The system to be controlled
is still the pendulum
positioning
system
with the same dynamics
scribed by Equation10.4andthe same operatingequilibrium at (x*1,x*
The difference
is in the structure
of the tracking/regulation
block
and capability
controller
diagram in Figure 10.8(b),
of the tracking/regulation
is shown in
control
Figure 10.8. The controller
where the controller
as de-2,T*)
= (15?,0?/s,0.19043Nm).
system.
The structure
is designed
is connected to the linearized
based on the
model
x?(t) = x(t)+B
A
T(t)
wherethe matrices A and Bare given in Equation 10.6. After the design of the controller is completed,
the controller will be connected with the nonlinear model or the real system, as shown in Figure
10.8(a), for simulation or real-time testing.
Fig. 10.8: The state feedback
The integrator
steady
state;
internal
hence, the
and integral
model structure
design
of F and
tracking
control for the pendulum
will guarantee
K will focus
the step tracking
on the stability
positioning
error
system.
to
and the transient
be zero at
response.
There are several ways of designing F and Kto enhance the performance of the closed-loop system.
Onesimple, effective way is to first find a state-feedback gain matrix F so that the eigenvalue
358
of
10
State Feedback
A+BF
are placed
and Linear
at favorable
Quadratic
Optimization
locations
on the complex
design approach to find a value of the integrator
are at desired locations
to achieve
The root locus
can provide.
constant
a best possible
LTF Pole-Zero Pattern and Construction
plane and then
employ
the root locus
Kso that the closed-loop system poles
performance.
of Root Locus Diagram
design of K can only do as good as the
We will see that the eigenvalues
of
LTF (loop transfer
A+BF
function)
pole-zero
pattern
will be main part of the LTF pole-zero
pattern;
therefore, the choice of Fis critical to the success of the overall design. Recallthat we placed the eigen-values
ofA+BF
ast 1=-10
and
s2=-11
toobtain
F= -2.956
-0.6955
. We
willcontinue
touse
?
?
it and call this design as Design b. For comparison, we will have another design, called Design a, which
?.
places
theeigenvalues
ofA+BF
ats1,
s2=-10j5.Design
agives
F =? -3.456-0.6622
To apply
the
root locus
design to
a simple
closed-loop
system
structure,
as shown
in
Figure
10.9(b), the LTF pole-zero pattern required is just the pole-zero pattern of the open-loop system
G(s). However, for a more complicated
pattern
required
open-loop
for
root locus
closed-loop system like
design in general
Figure 10.9(a), the LTF pole-zero
is not the same as the
pole-zero
pattern
of the
system.
Recall that the loop transfer function of the closed-loop system in Figure 10.9(b), before the Kis
cut off the loop, is KG(s), where G(s) can be seen as the transfer function from u to e multiplied by
a negative sign. Similarly, for the closed-loop system in Figure 10.9(a), the loop transfer function
should be KG(s), where G(s) is the transfer
function
from uK to xi multiplied
by a negative sign.
Hence,
G(s)
can
berepresented
bythefollowing
state-space
model
with
uKand-xiastheinput
and
the
output, respectively.
??? ??
??? ??
?? ??
x?
x?i
G(s):
A+BF
=
x
0
-C 0
-xi =? 0 -1?
x
xi
xi
:=CG
+
B
0
uK:= AG
x
xi
+BGuK
(10.15)
x
xi
Fig. 10.9: Derivation of the loop transfer function for the closed-loop state feedback andintegral tracking
control
system shown in Figure 10.8.
Based on Equation
as shown in
10.15, the root locus
Figure 10.10(a)
and Figure
diagrams for
10.10(b),
Design a and
respectively.
Design b can be constructed,
From Equation
10.15, the state-spac
10.2
Control
of a Lightly
Damped Positioning
System
359
Fig. 10.10: Root locus diagrams associated with the integral tracking control system shown in Figure
10.8.
representation of the loop transfer function, it can be seenthat the LTF poles are the eigenvalues of AG,
which are the eigenvalues of A+BF together with the integrator pole at s = 0. Thus,for Design a, the
loop transfer function will have three poles and no zeros. The three poles are
p1
p2
=-10 j5, p3=0
wherethe pair of complex poles are the eigenvalues of A+BF,
the integrator.
andthe pole at the origin is the pole of
When K = 0, the closed-loop system poles (the roots of the closed-loop characteristic equation)
willbeatthethree
LTF
pole
locations,
asK?8thethree
roots
willbeonthethree
asymptotes:
the
three
straight
lines
with
angles
p/3,pa
, nd-p/3intersecting
the
real
axis
ats=(p1+p2+p3)/3=
-6.667.
The
tworootlocibranches
originated
fromthecomplex
conjugates,
p1andp2,willmove
toward
therealaxiswith
departure
angles
-63.6?
and63.6?,
respectively.
These
twobranches
will
break
intotherealaxis
ats =-8.34
when
K=7.71
and
split
intotwobranches
ontherealaxis,
one toward
the left
all the
way to the negative infinity
of the real
axis, and the other
moving to the
right
to meet
thethirdbranch
coming
fromp3attheorigin.
These
twobranches
meet
ats = -5
when
K=8.33
and
break
away
immediately
intothecomplex
plane
toward
thep/3and
the-p/3
asymptotes,
respectively.
These
twobranches
willcross
the
imaginary
axis
atj11.2
intotheright
half
of the complex
On the
plane
when K = 83.3.
other hand, for
Design b, the loop
transfer
function
will have three
poles and no zeros.
The three poles are
p1=-11,p2=-10,p3=0,
allontherealaxis.
There
are
three
asymptotes:
thethree
straight
lineswith
angles
p/3,p,and-p/3
intersecting
the
real
axis
ats =(p1+p2+p3)/3
=-7.The
root
locibranch
originated
from
p1=-11
willmove
lefttoward
thenegative
infinity
oftherealaxis.
The
twobranches
started
fromp2= -10
andp3=0willmove
toward
each
other
ontherealaxis
until
theymeet
ats =-3.48
when
K=5.67
andbreak
away
immediately
intothecomplex
plane
toward
thep/3and
the-p/3asymptotes,
respectively.
These
twobranches
willcross
the
imaginary
axis
atj10.5
intotherighthalf
ofthe
complex plane when K = 77.1
360
10
State Feedback
and Linear
Quadratic
Optimization
Fig. 10.11: Simulation diagram associated with the state feedback andintegral tracking control system
shown in Figure 10.8.
Determination
of Kin Root Locus Design
In Figure 10.10(a), the root locus diagram for
see how the
K value affects the closed-loop
and natural frequencies.
Design a, by moving the little
system poles trajectories
When K = 10, the three closed-loop
s1
s2
and their
black square cursor
we can
associated damping ratios
system poles are at
=-4.36 j2.77, s3 =-11.3
wherethe pair of complex poles are the dominant poles andtheir corresponding damping ratio and nat-ural
frequency are ? = 0.843 and ?n = 5.16rad/s, respectively. There will be alittle bit overshoot at the
peaktimetap = p/2.77 = 1.13s, butthis less-than-1%overshoot
is a worthwhiletradeofffor afaster
response, as will be seenin the simulation results graphs of Figure 10.12.
Fig. 10.12: Simulation
results
of the state feedback
and integral
tracking
control
system.
In the root locus diagram of Figure 10.10(b) for Design b, when K= 6,the three closed-loop system
poles are at
s1
s2
=-3.45j0.938,s3=-14.1
wherethe pair of complex poles are the dominant poles andtheir corresponding damping ratio and nat-ural
frequency are ? = 0.965 and ?n = 3.57rad/s, respectively. There is an unnoticeable overshoot a
10.2
Control
of a Lightly
Damped Positioning
System
361
the peaktime tap = p/2.77 = 1.13s andtbp = p/0.938 = 3.35 s, respectively, whichis slower than
Design a, as will be verified in the simulation results graphs of Figure 10.12.
The Simulink simulation
shown in
diagram of Figure 10.11 is constructed
based on the block diagram
Figure 10.8. The only difference is that in the Simulink program the real pendulum angle
?,instead of? , is fed backto compare withthe desiredreal angle position ?Rinstead of?R. This dis-The
crepancy
does not affect the simulation
results
since the required
correction
error sent to the tracking
integrator is essentially the same
simulation results of both Design a and Design b are shown in Figure 10.12. Figure 10.12(a)
shows that Design a response is faster than Design b since the peak times of Design a and Design b
are
tap = p/2.77 = 1.13s and tbp = p/0.938 = 3.35s
respectively.
Recallthat the peaktime is computed based on Equation 3.69, tp
= p/?,
where ? is the
imaginary part ofthe dominantcomplex poles.Increasing ? maynot always help sinceit maycausethe
damping ratio to decreaseto enlarge the overshoot and oscillations.
The control-input torques required to complete the tracking/regulation for Design a and Design b
are shown in Figure 10.12(b), and the required real torques for both cases are between 0.19 Nm and
0.42
Nm, which are in a reasonable
applicable
range.
The Design a root locus diagram on Figure 10.10(a) is generated using the following
MATLAB
code. The Design b root locus diagram on Figure 10.10(b) can be generated by simply replacing
F=[-3.456
-0.6622]
with F=[-2.956
-0.6966]:
% CSD
Fig10.10
% Root
loci
A=[0
Root
for
1;-21.32
F=[-3.456
loci
the
C=[1
a
in
-0.135];
-0.6622];
s=tf('s');
Design
system
Design
b
with
F=[-3.456
-0.6622]
B=[0;30];
%F=[-2.956
0];
A_G=[A+B*F
C_G=[zeros(1,2)
-1],
D_G=0,
pole_G=pole(G),
[zero_G,gain_G]=zero(G),
K=logspace(-1,2,1000);
&
Fig10.9(b)
-0.6966];
zeros(2,1);-C
0],
B_G=[B;0],
sysG=ss(A_G,B_G,C_G,D_G);
G=tf(sysG),
figure(13),
rlocus(G,K)
In the root locus diagram construction, the main MATLAB command is rlocus(G,K)
in which
Kis the root locus gain parameter to be determined, and Gis the state-space representation of the loop
transfer
function
given by Equation
designer is still required
of the LTF pole-zero
10.15. This tool is versatile
to be proficient
pattern
in the
and helpful.
materials covered in
will affect the trajectories
However, a control system
Chapter 8. Knowing
of the root locus
diagram is even
how a change
more important
than constructing the root loci diagram. The root locus design actually starts from finding
LTF pole-zero pattern rather than starting from a given LTF pole-zero pattern.
If a SISO (single-input/single-output)
system is described by a rational function
G(s)
afavorable
=N(s)/D(s),
where
bothN(s)
andD(s)
are
polynomial
functions
with
real
coefficients
and
deg(D(s))
=deg(N(s)),
then the poles and zeros of the system are the roots ofD(s) = 0 andthe roots ofN(s) = 0, respectively. If
a system is given by a state-space representation G(s) = (A,B,C,D), the poles and zeros can be directly
computed in the state space
without the need of converting
eigenvalues of the A matrix, and the zeros can be computed
eigenvalue problem as described in Remark 10.2.
Remark 10.2 (Zeros of State-Space
to polynomial
functions.
The poles are the
based on the solution of the generalized
Model (A, B, C, D))
The zeros ofthe state-space model(A,B,C,D) arethe generalizedeigenvalues?i so that
362
10
State Feedback
and Linear
Quadratic
Optimization
??iI-A B? ?? ??AC -D-B??
vi =[?iE-Z]vi
-C
D
vi = ?i
I
0
0
0
(10.16)
-
where vi are the corresponding generalized eigenvectors. The MATLAB command
>>[Vz,Dz]=eig(Z,E)
can be employed to obtain the generalized eigenvalues in Dz and the generalized eigenvectors in
Vz.
Example 10.3 (Find the Zeros of a System in State-Space Representation)
s+1
, which
obviously
has
azero
ats =-2.A
state-space
model
ofthesystem
can
befound,
G(s)=C(sI-A)-1B+D
with
A=-1,B=1,C=1,
Consider the system with transfer function,
G(s) = s+2
D = 1. Now, we would like to employ this state-space
of a system in state space using Equation
model to demonstrate
how to compute the zeros
10.16.
? ?
? ?vi=0, ? ?i =-2 ? ? ?vi=0 ? vi=??
B drops
The
zeros
ofC(sI-A)-1B+D
are
thecomplex
numbers
?isothat
therank
of ?iI-A
-C D
below its normal rank. It can be solved as a generalized eigenvalue problem:
?i +1 1
1
1
-2+11
-1 1
-1 1
This generalized eigenvalue problem can also be solved using the following
>>
A=-1;
B=1;
C=1;
D=1;
E=[1
0;
0
0];
Z=[A
-B;
C
(10.17)
MATLAB code:
-D];
[Vz,Dz]=eig(Z,E)
Vz
=
1
1
Dz
0
-1
=
-2
0
0
Inf
This
generalized
eigenvalue/eigenvector
problem
has
aneigenvalue
at?i=-2and
itscorresponding
eigenvector vi =[1
10.2.5
Tracking
In the previous
1]T.
Control
subsection,
of the Pendulum
Section
10.2.4,
Positioning
we employed
System
Using PID
the state-feedback
Control
pole placement
together
with integral tracking compensation to address the tracking/regulation control problem for the pendulum
positioning system. The state-feedback pole placement is employed to replace the poles of the plant
so that the pole-zero pattern of the loop transfer function will become more favorable for root
locus design. In this subsection, we will try to design the best possible PID controller for the lightly
damped pendulum position tracking control problem. Before getting into the PID control system design,
we will explore if the PI controller
Can PI Controller
can do the job.
Work for the Pendulum Positioning
System?
The PI controller shown in Figure 10.13(a) and 10.13(b) is expressed as
s+c0
s
K = KP +KI
1
s
? KP
=K, KI=c0
10.2
Fig. 10.13:
Block diagram
and root locus
where KP and KI are the proportional
is, inserting
a PI controller
Control
of a Lightly
diagram for the PI control
and the integral
to the feedback
coefficients,
loop is equivalent
Damped Positioning
of the pendulum
respectively,
to adding
System
positioning
363
system.
of the PI controller.
a pole at the origin
That
and a
zero
ats =-c0tothepole-zero
pattern
oftheroot
locus
diagram,
and
hoping
achoice
ofKvalue
in the root locus design
will lead to a desirable closed-loop
system performance.
From Figure 10.13(b),
it can
beseen
that
thetransfer
function
ofthelinearized
plant
is C(sI-A)-1B.
Thus,
theloop
transfer
function
for the root locus
design
will be
s+c0
G(s) =
KC(sI-A)-1
s
which hasthree poles and one zero. They are
p1
p2
=-0.0675
j4.6, p3=0, z1=-c0
where the pair of complex poles p1, p2 are the poles of the lightly
damped pendulum system. The
pole
atheorigin,
p3=0,isthepole
oftheintegrator
and
thezero,
z1=-c0,
isassociated
with
the
integrator gain. If c0 > 0.135 is chosen, the root locus diagram will look like the one shown in Figure
10.13(c), in whichthe two asymptotes will bein the RHP andthe real part of the complex roots can only
bebetween
-0.0675
and
0fortheclosed-loop
system
tostay
stable.
But
thedamping
ratioforany
of these complex
roots is extremely
small,
so these
complex
roots
cannot
serve as dominant
poles.
On
theother
hand,
fortherealrootsitting
between
0and-c0tobethedominant
pole,
it needs
to
moveto a position extremely close to the imaginary
slow to
pendulum
be practical.
position
PID Controller
Therefore,
control
there
exists
axis that
PI control
will makethe control
solution
for the lightly
action too
damped
system.
Structure
The PID controller KDs+KP
+KI/s, together
KDs2 +KPs+KI
s
and, therefore,
no viable
the
nonlinear
and the linearized
with the low-pass filter
(s+c0)(s+c1)
w
s+w
=
closed-loop
s(s+w)
pendulum
w/(s+w),
can be expressed as
K
position
(10.18)
systems
with PID con-troller
are depicted in Figure 10.14(a) and 10.14(b), respectively. Notethat the insertion of the PID
controller and the low-pass filter to the feedback loop is equivalent to adding two poles, s = 0 and
s =-w,and
twozeros,
s =-c0and
s =-c1tothepole-zero
pattern
oftheroot
locus
diagram.
364
10
State Feedback
Fig. 10.14:
Construction
and Linear
Quadratic
Optimization
Block diagram for the PID control
of the pendulum
of Root Locus Diagram for PID Controller
positioning
system.
Design
From
Figure
10.14(b),
it can
beseen
that
thetransfer
function
ofthelinearized
plant
is C(sI-A)-1B,
and thus the loop transfer
function
for the root locus
G(s) =
(s+c0)(s+c1)
s(s+w)
design
will be
KC(sI-A)-1B
(10.19
which hasfour poles and two zeros,
p1
p2
=-0.0675j4.6, p3=0, p4 =-w, z1 =-c0, z2 =-c1
wherethe pair of complex poles p1, p2 are the poles of the lightly
(10.20)
damped pendulum system. The pole
attheorigin,
p3=0,isthepole
otfheintegrator.
The
pole
p4=-wisdetermined
bythebandwidth
of the low-path filter,
which is usually chosen to be larger than the actuator
bandwidth.
The zeros
z1=-c0and
z2=-c1are
freedesign
parameters
tobeselected
toobtain
afavorable
root
locus
diagram ready for the design of the K parameter.
The
parameters
w,c0,
and
c1are
chosen
sothatp4=-50,
z1=-1.5,
and
z2=-2lead
totheroot
locus diagram shown in Figure 10.15. Since the loop transfer function has four poles and two zeros,
there are four root loci branches and two of them will approach to the two asymptotes with angles p/2
and-p/2that
intersect
therealaxis
ats =-23.32
when
K?8.The
twobranches
originating
from
p1andp2willleave
thetwopoles
with140?
departure
angles
moving
totheleftand
then
bending
toward
therealaxis.
These
twobranches
willmeet
ats =-10.3
when
K=23.5
and
split
immediately
ontherealaxis,
onemoving
totheright
toward
z2=-2,and
theother
totheleft,which
willmeet
the
branch
thatcame
fromp4ats =-21.2
when
K=25.2,
and
then
break
away
intothecomplex
plane,
one going uptoward the asymptote with angle p/2, andthe other going downtoward the asymptote
with
angle
-p/2.The
branch
outofp3=0is moving
tothelefttoward
z1=-1.5.
Determination
of Kin
Root
Locus
Design
Now we havethe four complete root trajectories onthe root locus diagram, asshown in Figure 10.15. If
K = 30 is selected, the four closed-loop system poles will be at
10.2
Control
of a Lightly
Damped Positioning
System
365
Fig. 10.15: Root Locus diagram for the PID control of the pendulum positioning system.
s4
s3=-0.754,
s1=-5.52,s2 =-21.9j13
(10.21)
where
s3=-0.754
isthedominant
pole.
If K = 23is selected, the four closed-loop system poles will be at
s1
s3=-0.687,s2 =-9.74j2.38,s4 =-30
(10.22)
where
s3=-0.687
isthedominant
pole.
Fig. 10.16:
Simulation
diagram for the PID control
of the pendulum
positioning
system.
The Simulink simulation diagram shown in Figure 10.16 is similar to that in Figure 10.11 ex-cept
the controller part is replaced by the PID controller. Two simulations wereconducted: Oneis
for the
PID controller
The step tracking
associated
responses
with
K = 30, and the other is for the PID controller
for both cases K = 30 and
with K = 23.
K = 23 are shown in Figure 10.17(a).
responses exhibit resemblance in waveform, although the K = 23 is alittle
Both
bit slower. For the K = 30
case,
therealpole
ats3=-0.754
iscalled
thedominant
pole
because
theresponses
due
totheothe
366
10
State Feedback
three poles
will decay
and Linear
Quadratic
much more quickly.
Optimization
But it does not
mean the
In fact, the response associated with a pole with larger
of the real
they
part)
or smaller
will also decay faster.
time
constant
will respond
As shown in Figure
non-dominant
responses
damping factor (the absolute value
to command/disturbance
10.17(a), the
are neg-ligible.
faster
K = 30 step tracking
although
response (in blue)
shoots up from ? = 15?to ? = 35? within 0.1 s. Thisfast responseto the step commandin the first
0.15
sis mainly
contributed
bythepair
ofcomplex
poles
at-21.9j13.But
it also
decays
fast
todip
the response down to 31? before the dominant response catches up at around 0.6 s. Then the response
thereafter
isdue
tothedominant
pole
s3=-0.754
with
timeconstant
t3 =1/0.754
=1.33
s,which
will take another
3 s to reach the steady state.
Fig. 10.17: Simulation results of the PID control of the pendulum positioning system.
For
theK=23case,
therealpole
ats3= -0.687
isthedominant
pole.
Itisalittlebitcloser
tothe
imaginary
axis than the
K = 30 case; hence, it
the K = 30 case, the K = 23 step tracking
will take longer time to reach the steady state. Similar to
response (in red, asshown in Figure 10.17(a)) shoots up
from ? = 15?to ? = 35? within 0.12 s. Thisfast responseto the step command in the first 0.2 s is
mainly
contributed
bythepairofcomplex
poles
at-9.74j2.38.
But
it also
decays
fasttodipthe
response down to 30? before the dominant response catches up at around 0.7 s. Thenthe response
thereafter
isdue
tothedominant
pole
s3=-0.687
with
time
constant
t3 =1/0.687
=1.46
s,which
will take another 3.4 s to reach the steady state.
Control-Input
Constraint
The limitation
of control input
system requires
always
a control-input
needs to be considered in all control
magnitude
beyond the capability
and the input to the plant (the system to be controlled)
the saturation
value.
That
value of the actuator for all controller
means the controller
loses its ability
systems
designs. If a control
of the actuator, the controller
outputs
to control
with
the
magnitude greater than the saturation
plant,
and the system
unstable. For this reason, a good control system design should avoid control-input
The required control-input torque
output
will be different. Theinput to the plant will be
T(t) to provide the tracking
may become
saturation.
performance in Figure 10.17(a) is
shown
inFigure
10.17(b).
The
range
ofthecontrol-input
torque
fortheK=30case
isbetween
-1.5Nm
and
10.5
Nm.
The
range
fortheK=23case
isalittlebitbetter,
but
stillbetween
-1Nm
and
8.2Nm.
For comparison, the required control-input torque T(t) to provide the tracking performance of the state
feedback/integral
tracking
control system in Figure 10.12(a) is shown in Figure 10.12(b).
range of the control-input torque is between 0.19 Nm and 0.42 Nm, which is
than that of the PID control system
Note that the
morethan 25 times less
10.3
The surge of torque at t
Controllability
367
= 0 from T = 0.19 Nmto T = 10.5 Nmin Figure 10.17(b) is to provide the
shoot-upfrom ? =15?to ? = 35? within0.1s in Figure10.17(a).This quick responseat the early
time is unnecessary
Remark 10.4 (An
Control)
since the dominant
employs
up.
Unfair Comparison of the PID Control with the State Feedback/Inetgral
The state feedback/integral
former
pole response is too slow to catch
tracking
more information
control is supposed to do better than the
(all the states information)
than the latter,
PID control
Track-ing
since the
which only utilizes the
output feedback. By using the state feedback, the original plant poles can be replaced by favorable
ones. Onthe other hand, by using the PID controller, the original plant poles remain part of the
pole-zero pattern for root locus design. The only modification the PID control can do is to choose
the locations of the two added zeros andthe low-pass filter pole. The presence of the original lightly
damped plant poles prevents the PID control from obtaining a root locus pole-zero pattern as
favorable asthat of the state feedback/integral
tracking control approach.
10.3 Controllability
Recall that in Section 7.6 welearned that the cart-inverted
can be stabilized.
A sufficient
condition
is given in the
pendulum
paragraph
system is unstable, but the system
right
after
Equation
7.108: If
the
system is controllable, then the closed-loop system poles can be placed anywhere onthe complex plane.
The definition
of controllability
and relevant issues
will be discussed in later chapters.
The concept of controllability is discussed in the following.
Consider a system with state equation
x?(t)=Ax(t)+Bu(t),
x?Rn,
u?Rm
(10.23)
Definition 10.5 (Controllability)
The system is (completely)
controllable if there exists a control input u(t) defined on a finite time
interval [0,t f] for somet f > 0that steersthe systemfrom anyinitial state x0to anyfinal state xf.
Equivalently, there exists t f
> 0 and u(t) for any x0 and xf, such that
xf = eAtf x0 +
?t
f
eA(t
f-t)Bu(t)dt
(10.24)
0
Equation 10.24 will be employed to prove the controllability
10.3.1
Controllability
rank theorem.
Rank Test
Theorem 10.6 (Controllability
Rank Condition)
Consider
thefollowing
nnmcontrollability
matrix:
C=[BABA2B An-1B]
The system(10.23), or the matrix pair (A,B), is (completely) controllable ifand
rank
C = n
(10.25
only if
368
10
Proof:
State Feedback
First,
and Linear
Quadratic
Optimization
we will show that
(A,B)
iscontrollable
? rank
C=n.
In
many cases, it is easier to
statement. If(A,B)
prove by contradiction
for
a false statement
than
directly
prove a true
is controllable but rank C < n, then there exists a nonzero vector v such that
vTC=vT[B
ABA2B An-1B]
=0
or
vTB=0,vTAB
=0,vTA2B
=0, , vTAn-1B
=0
Recall that
by Theorem 6.15 (Cayley-Hamilton
eAt =
theorem)
and Equation
6.14,
n-1
?ai(t)Ai=a0(t)I+a1(t)A+a2(t)A2
+ +an-1(t)An-1
i=0
+an-1(t)An-1B
=0
vTeAtB
=vT
?a0(t)B+a1(t)AB+a2(t)A2B+
Hence,
?
Now,from Equation 10.24, ifx0
= 0, the final state xf will be
xf =
?t
f
?t
f
eA(tf-t)Bu(t)d
0
and
vTxf
= vT
eA(tf-t)Bu(t)dt
=0
0
forallu(t),t =0.Thismeans
thatthecontrollable
subspace
isorthogonal
tothevector
v,which
contra-dicts
the assumption that (A,B) is controllable.
Therefore, the rank ofthe controllability
matrix C has
to be n.
Next, conversely
we will show that
(A,B)
iscontrollable
? rank
C=n
Again, we will prove it based on the contradiction
controllable;
approach. Assume rank
C = n but (A,B) is not
then there exists a nonzero vector v such that
vT
?t
f
eA(t
f-t)Bu(t)dt
=0
0
whichimplies that
vTeA(t
f-t)B=0, 0=t =tf
Differentiate
this equation to obtain the following
derivatives,
vTeA(t
f-t)AB=0, vTeA(t
f-t)A2B
=0, , vTeA(t
f-t)An-1B
=0, 0=t =tf
Let t =t f, and wehave
vTB=0,vTAB
=0,vTA2B
=0, , vTAn-1B
=0
This contradicts the assumption that rank C = n. Therefore, (A,B)
hasto be controllable.
10.3
Example
10.7 (Controllability
Consider a three-state
of a
two-input
MIMO
linear
Controllability
369
System)
system described
by the following
state equation:
??? ???? ???
??=? ???+
??
?
???1 0 -1??
??=Cx
-9 3 7
-5 1 5
-5 3 3
x1
d
x2
dt
x3
=
y2
1
1
1
x3
1
1
u1
u2
= Ax+B
x2
-11 1
matrix can be found
C = ?B
Weeasily recognize
2
x2
x1
y1
The controllability
x1
AB
that the rank of
x3
as
A2B? =
?21 11
?1 1
?
?
-8 1 32 1
-4 1 16 1
-4 1 16 1
Cis 2. Thus, the system is not controllable.
If a system fails to be controllable it is imperative to identify the uncontrollable physical modes.For
example, an uncontrollable mode may be stable and does not interfere with achieving the performance
we desire for this system. Onthe other hand, if the uncontrollable modeis unstable or has a harmful
effect on the system performance, identifying it may provide clues to modifying the system actuation
structure to resolve the problem. One wayto easily makesuch anidentification is to transform the system
to diagonal form (or, more generally into Jordan form). In the following, the system considered in the
previous example will be employed to illustrate which part of the system is uncontrollable.
Example 10.8 (Controllable
Modes)
Consider the system of Example 10.7. We will compute the eigenvalues ?1,?2,?3 and correspond-ing
eigenvectors v1,v2,v3. From the latter define the similarity transformation
matrix T = [v1,v2,v3]
and implement the transformation from the original state x to the new statex, x = T-1x to obtain the
transformed
system
x? = ? T-1AT ?x+ ? T-1B ? u
(10.26)
The eigenvalue-eigenvector pairs of A are:
??
??
??
?1=-4,v1=
??,?=
2-2,
v2
=??,?=31,
v3
=??
2
1
1
1
0
1
1
1
1
Notethat there are three real eigenvalues and two of the modes are stable. The third is unstable. Using
the three eigenvectors weset upthe transformation matrix T and obtain the transformed state equations
x? =
?0 -2 0? ? ?
? ?
x+? ?u
-4 0 0
0
Webegin withthe knowledge (from
0
1
1
0
0
0
0
1
Example 10.7) that there is one uncontrollable
states,x1,x2, x3 denotethe motion associated with modes1, 2, and 3, respectively.
equations
are decoupled.
They can be solved
one at a time,
each independent
mode. The new
Observethat the state
of the others. Inspection
of the transformed B matrix shows that neither of the control inputs affectsx2, indicating that x2
is the uncontrollable
mode.
370
10
State Feedback
10.3.2
The Controllability
and Linear
Quadratic
Decomposition
The ability to identify the controllable
interpretive
mode implications.
10.9 (Controllability
Given a linear
Form
part of a system has control design implications
Here, we discuss the
1963] that separates the controllable
Theorem
Optimization
decomposition
and the uncontrollable
Decomposition
introduced
beyond the
by Kalman [Kalman,
parts of a linear system.
Form)
system in the form
x?(t) = Ax(t)+Bu(t)
y(t)
=Cx(t)
with
x?Rn,u
?Rm,y
?Rp,
ifthesystem
isnotcontrollable
and
rank
C =r
<n
then there exists a nonsingular similarity transformation
the newstatex, x
matrix Tthat transforms the original state x to
= T-1x suchthat the transformed system
x? = ? T-1AT ?x+ ? T-1B u =Ax+ Bu
y =[CT]x
has the form
A =
=Cx
? ? ??
A11A12
0 A22
B =
,
B1
0
C =? C1C2?
,
?A11,
where
A11
?Rrr,
B1?Rrm,
C1?Rprand
thesubsystem
B1,
C1?is
Proof:
Hint: Choose T =[T1
(10.27)
controllable.
T2] with
T1 = ? f1 f2 ... fr ?
where f1, f2, ... fr are linearly independent and span the r-dimensional
columns
ofT2 are chosen so that together
Example 10.10 (Controllability
with the columns
Decomposition
ofT1 span the
range space ofC(A,B).
whole n-dimensional
Form)
Recallthat the system considered in Example 10.7 is
??? ???? ???
??=? ???+
??
?
???1 0 -1??
??=Cx
-9 3 7
-5 1 5
x1
d
x2
dt
x3
-5 3 3
The controllability
2
1
x2
1
1
x3
1
1
u1
u2
x1
y1
y2
x1
=
x2
-11 1
x3
matrix of the system is
C =
-81321?
?21
-41161
?11
11 -41161?
= Ax+Bu
The
space.
10.3
Since rank
C= 2 < 3, any two linearly
range space of C. Let
independent
column
vectors ofC
Controllability
371
will span the two-dimensional
??
??
? ?
? ?
-4 0 3? ? ?
? ? ?? ?
?0 0 -2?
x+
??
? ?
T1 = [ f1
The choice of T2is arbitrary aslong asit
f2]
2
1
1
=
makes T =[T1
T =[T1
T2] =
1
1
1
T2] nonsingular. Let T2 be chosen so that
2
1
0
1
1
0
1
1
1
0
1
Then the transformed state-space model will be
x? =
A11A12
0 A22
x+
y =?C1C2?x =
B1
u
0
1
0
0
1
=
3
1
0
0
1
0
0
-1 x
1
Note
thatneither
u1oru2has
effect
on
A22=-2.Hence,
thedynamics
associated
with
A22
are
notcontrollable, whilethe dynamicsassociatedwithA11 can be modifiedviastatefeedback.
Remark
10.11 (Controllable
and
Uncontrollable
Subspaces)
The controllable subspace is the range space of the controllability
matrix C, while the uncon-trollable
subspace is the null space of CT. The null space of CTis the orthogonal complement of
the range space of C.
For the system in Example 10.10, the controllable subspace is a two-dimensional
by any two linearly
independent
vectors that are the linear
combinations
controllability
matrix C. Hence, the two vectors, [1 0 0]T and [0
vectors of the range space of C, which is the controllable subspace:
of the column
space spanned
vectors
of the
1 1]T can serve asthe basis
??
Range(C)
~
??
The
uncontrollable
subspace
istheone-dimensional
null
space
ofCT
spanned
bythevector
[0 -11]T:
?-1?
CT?
??
1
0
0
1
0
1
0
Null ?
~
1
which is orthogonal to the range space of C.
Notethat the spacespannedbythe vectorT2 =[0 0 1]T of Example10.10is notthe uncontrol-lable
subspace although its associated similarity
If the similarity
range space of
transformation
transform
T still achieves controllability
decom-position.
matrix T =[T1 T2] is chosen according to the basis vectors of the
C and the null space of
CT as
372
10
State Feedback
and Linear
Quadratic
Optimization
?0 1 -1?
? ?
?? ?
? ? ?? ?
?0 0 -2 ?
x+? ?u
?1 -1 ?
T = [T1
then the transformed
state-space
x? =
0 A22
B1
x+
the
numbers
are slightly
u
0
y = ?C1C2 ?x =
Although
0
0
0
1
1
model will be
A11A12
remains the same and the
1
T2] =
=
different,
of
2
1
1
1
0
0
-1 x
-1 2
eigenvalues
-9 10 4
-5 6 2
0
the structure
of the
controllability
decomposition
form
A11 andA22, respectively, areidentical to those obtainedin
Example 10.10.
Example 10.12 (Controllable
Subspace and Stabilizability)
Consider a system described by the following
??? ?????
d
x1
x2
dt
The controllability
=
0
1
x1
1
0
x2
1
+
u = Ax+Bu
1
matrix of the system is
C =[B
Since rank
state equation:
C = 1 < 2, the range space of
AB] =
C is the
??
1
1
1
1
one-dimensional
subspace
spanned
by the vector
[1 1]T,
and
thenull
space
ofCT
istheother
one-dimensional
subspace
spanned
bythevector
[-1 1]T:
??and
??
Null
CT?
1
Range(C)
~1
These vectors can be employed
xto new coordinatesx, x
to form
a similarity
Then the transformed
state-space
x? =
transformation
= T-1x.
T =[T1
equations
~ -1
1
?
T2] =
matrix to transform
old coordi-nates
?1 -1?
1
1
are
? ? ?? ?0 -1???
A11A12
x+
0 A22
B1
0
u
=
1
1
x+
1
0
u
Note
that
thecontrol
input
uhas
noeffect
on
A22=-1.Hence,
thedynamics
associated
with
A22
arenotcontrollable,whilethe dynamics
associated
withA11 =1 canbe modified
viastatefeedback
as
follows.
Withthe state feedback,
u =F x
the closed-loop system state equation will be
=?f1
f2 ?
?
x1
x2
10.4
???0
x?1
x?2
Poles and Zeros of
MIMO
Systems
373
???
-1
1+f1 1+f2
=
x1
x2
Note
that
theuncontrollable
yet
stable
eigenvalue
A22=-1isunchanged,
while
thecontrollable
unsta-ble
eigenvalueA11 = 1 has beenchangedtoA11 = 1+f1. Thesystem will becomestableif f1 is chosen
tobeless
than-1.Therefore,
asystem
canbestabilized
if theuncontrollable
partofthesystem
is
stable.
Fig. 10.18: Controllable and uncontrollable subspaces.
The controllable
Assume the initial
and
uncontrollable
subspaces
state isxinitial at [2
of the system
are illustrated
in
Figure
10.18.
1]T position. Since the eigenvalue associated with the un-controllable
subspace spanned byx2 is stable, the position of the state on thex2 axis will move
from 1 to 0, although it is not controllable.
Onthe other hand, the eigenvalue associated with the
controllable subspace spanned byx1 is unstable; hence,the position ofthe state onthex1 axis will
grow without bound if no control action is taken.
However, a control action can be taken to convert
the unstablestatex1 to becomestable.
10.4 Poles and Zeros of MIMO Systems
We have learned that the behavior
of a system is
mainly determined
by its pole locations
plane, and if the system is controllable and observable (the definition of observability
the next chapter) afeedback
improve
in
the stability
controller
and performance.
Chapter 8 on how they
can be designed to
move the poles to
The zeros also have effect on system
would influence
the root locus
on the complex
will be given in
more favorable
performance,
locations
to
as revealed
diagram to shape up the closed-loop
system
performance. It is notedthat the zerosin the right half of the complex plane are notorious in limiting the
performance
of the closed-loop
system,
as described in
Remark 8.23.
For SISO systems, the poles and zeros can be easily found if their transfer functions are available.
For MIMO systems with state-space model (A,B,C,D), the poles of the system can be evaluated as
the eignevalues of the matrix Aif (A,B,C,D) is controllable and observable. However,the computa-tion
of the zeros of MIMO systems is more complicated. In the following, we will explain the physical
meaning of system zeros, give a precise definition of MIMO system zeros accordingly, and conver
374
10
State Feedback
the computation
and Linear
Quadratic
of zeros to a generalized
The poles and zeros of
Optimization
eigen
problem
that
can be solved
efficiently.
MIMO systems can also be computed based on their transfer function
ma-trices
intheformofG(s)
=N(s)/d(s),
where
G(s)
isapmrational
function
matrix
with
poutputs
and minputs.
The denominator d(s) is the least common denominator of all entries of G(s) and N(s) is
apm polynomial
matrix.
However,
thepolynomial
approach
isinvolved
and
beyond
thescope
ofthe
book.
The book by Kailath [Kailath,
1980] provides
more detailed
discussion
on
MIMO system
poles
and zeros.
The general linear state-space modelis described by
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
(10.28)
where
x?Rn,u
?Rm,y
?Rp.Recall
that
thetransfer
function
ofthestate-space
model
was
derived
in
Section 6.6.1 a
G(s)=C(sI-A)-1B+D
For SISO cases, the transfer function is a rational function G(s) = N(s)/D(s),
where both N(s) and
D(s) are polynomial function of s. The poles of the system are the roots of D(s) = 0, andthe zeros of
the system are the roots of N(s) = 0.
10.4.1
Revisit Poles and Zeros of SISO Systems
Consider
thestate-space
model
(A,B,C,D)
with
one
input
and
one
output
sothatx?Rn,u
?R1,y
?R1.
Then Bis a column
vector
of length
n, denoted b; Cis a row
vector
of length
n, denoted c; and
Dis a
scalar, denoted d. Thetransfer function can be written
[sI-A]b+d
|sI-A| = k n(s)
G(s)
=c[sI-A]-1b+d
=cAdj
d(s)
(10.29)
|sI-A|
where n(s) is a polynomial of degreeless than or equal to n and d(s) is a polynomial of degree n. The
gain parameter k is a normalizing parameter such that the highest order coefficients of both n(s) and d(s)
are both equal to one. The poles of the transfer function are the roots of d(s) = 0 and its zeros are
the roots of n(s) = 0. Thisis a simple and conventional definition of single-input/single-output(SISO)
poles and zeros.
Example 10.13 (F/A-18
Revisited, Poles and Zeros of a SISO System)
Recallthe F/A-18 longitudinal flight dynamics
model of Equation 8.64 from
which if only the el-evator
control
deisconsidered
astheinput
and
theflight
path
angle
?=?-aastheoutput
wewill
have
?
? ? ?
A=? 0 -2.212-0.25320 ?,b=? ?
-0.02389
-28.32 0
?
-0.00033
-0.362 1
-32.2
?
0
0
1
-3.8114
0
?
?
?
?
0
-0.05145
-2.8791
?
?
0
?
c =? 0 -101,
d =0
With
thisdata
wecan
evaluate
thetransfer
function
G(s)
=c[sI-A]-1b+d
using
Mathematica
or use the MATLAB command:
equation of the system is
ss2tf
(state space to transfer
function).
Now the characteristic
10.4
Poles and Zeros of
MIMO
Systems
375
|sI-A|=s4+0.63916s3
+2.3086s2
+0.052669s+0.023445
=0
whoseroots are the poles of the system. Equation 10.29 yields the transfer function, shown in factored
form:
G(s) = 0.051453
(s+4.393)
(s-4.106)
(s-0.03491)
(s2 +0.02027s+0.01026)
(s2 +0.6189s+2.286)
We
clearly
see
thefourth-order
denominator
and
itsroots,
thesystem
poles
-0.0101
j0.101
and
-0.309j1.48,aswellasthethird-order
numerator
anditsroots,
thesystem
zeros-4.393,
4.106,
and 0.03491.
The computational
% CSD
A
results in Example 10.13 are generated using the following
Ex10.13
=[
Eq8.64
-2.3893e-02
State
model
-3.2923e-04
c
=
[0
-1
[NUM,DEN]
0]
0;
0
B(:,1);
0;
1.0000e+00
-3.8506e-07;
-2.8791e+00
=
0;
9.5196e-04;
-5.1453e-02
b
Zeros
-3.2200e+01;
-2.5319e-01
0
-3.8114e+00
Poles
1.0000e+00
-2.2115e+00
0
=[
F/A-18
0
-3.6208e-01
4.0491e-11
B
of
-2.8317e+01
MATLAB code:
0
=
0];
1];
d
=
0;
ss2tf(A,b,c,d)
eig(A)
damp(A)
sys
=
ss(A,b,c,d)
zpk(sys)
10.4.2 Physical
Meaning of System Zeros
For MIMO case,the transfer function
G(s)=C[sI-A]-1B+D
can be rewritten as
G(s) =C
Adj
[sI-A] B+D
|sI-A|
=
CAdj
[sI-A]B+|sI-A|D=
|sI-A|
N(s)
(10.30)
d(s)
Notice
that
thatN(s)
isapm matrix.
The
elements
ofN(s)
arepolynomials
insofdegree
atmost
n.
G(s)
isapm matrix
whose
elements
are
rational
functions
with
least
common
denominator
d(s).
Thus, weseethat the poles of the system are easily identified asthe roots of the polynomial d(s), which
is also the characteristic
polynomial of the matrix A.Identifying the zeros of the transfer matrix G(s)
is more complicated.
During the 1970s an extension of the concept of zeros to MIMO systems wasa substantive topic
of discussion (e.g., [Rosenbrock, 1973, 1974, Davison and Wang,1974, Kouvaritakis and MacFarlane,
1976a,b,
Kalnitsky
definition
and Kwatny, 1977]).
of SISO system
Wecan establish
zeros that does two things.
an alternative,
First, it provides
but equivalent
a new physical interpretation
of a system zero, and second it can be extended to multiple-input/multiple-output
basic idea is the following
theorem.
view to our ear-lier
(MIMO)
systems. The
376
10
Theorem
State Feedback
10.14 (Physical
and Linear
Quadratic
Optimization
Meaning of SISO System
Zeros)
Consider a linear SISO state-space system defined by Equation 10.28. Ifthe complex number ? is a
system
zero
then
there
exists
aninput
u(t)=e?tand
aninitial
state
x(t0)=x0such
thaty(t)=0forall
t =t0.
Proof:
Recall that in
Equation 6.23 of Section 6.6.1,
we have
?
?
Y(s)=C[sI-A]-1x0
+ C[sI-A]-1B+D
U(s)
In the SISO case, Cis a row vector, Bis a column vector and Dis a scalar. Consequently,
G(s) :=
?C[sI-A]-1B+D
?=k
n(s)
d(s
where
d(s)=|sI-A|
Suppose u(t)
= e?t, equivalently,
1
U(s) =
Thus,
Y(s) =
s-?
CAdj
[sI-A]x0+k n(s)
d(s)
d(s)
1
.
s-?
Because G(s) is realizable, the second term is a proper rational function.
perform an (incomplete) partial fraction expansion:
k
n(s)
1
Take the second term and
b
a(s)
=
+
s-? d (s) s-?
From
thebasic
results
ofthepartial
fraction
expansion,
a(s)
isapolynomial
ofdegree
n-1and
bis
d(s)
a constant. Our goal is to determine the constant bin the usual way:
lim
s??
?(s-?)
kn(s)
d(s)
?
s-?
1
(s-?)a(s)+lim (s-?)b
=lim
d(s)
s??
s??
s-?
From this we determine b as
n(?)
b = k
Hence,
Y(s) =
?CAdj[sI-A]x0 ?s-?
s-?
?CAdj[sI-A]x0 ?
CAdj
[sI-A]x0+k n(s)
d(s)
= G(?)
d(?)
=
d(s)
G(?)
a(s)
1
d(s)
+
d(s)
+
It is always possible to determine x0 so that
a(s)
d(s)
+
d(s)
=0
This
follows
from
thefact
thatCAdj
[sI-A]x0isapolynomial
oforder
n-1and
thencomponents
of
x0 can be used to matchits n coefficients to those of a(s). As a result, we have
G(?)
Y(s) =
If ? is a zero of G(s), then Y(s) = 0 and y(t)
= 0.
s-?
10.4
Remark
10.15 (Another
Interpretation
of SISO System
Poles and Zeros of
MIMO
Systems
377
Zeros)
Theorem 10.14 assertsthat if a SISO system has a zero at s = ? (or a pair complex conjugate zeros)
there exists a control input and aninitial state such that the resultant statetrajectory maintains the output
at zero. Another interpretation is that regardless of the initial state, the system output contains only the
homogeneous response (i.e., the specific input is blocked from having any affect on the output).
The importance
of this theorem is that it can be extended to
to the question, What
are the zeros of a MIMO system?
MIMO systems and gives us the answer
Consider once again Equation 10.28, but this
timex ?Rn,u
?Rm,y
?Rp.
This
timewe
ask
ifthere
isa?,ag?Rm
foracontrol
u(t)=ge?t
andan
initialstate
x0?Rn
such
thatx(t)=x0e?t
and
y(t)=0,t =t0.Such
asolution
must
satisfy
?x0e?t = Ax0e?t +Bge?t
and
0 =Cx0e?t +Dge?t
Putting these together
??I-AB???
we get
-C D
Definition 10.16 (MIMO
-x0
= 0.
g
(10.31
System Zeros)
The complex number ? is a zero ofthe system
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
with
x?Rn,u
?Rm,y
?Rp
ifEquation
10.31
has
anontrivial
solution
forx0and
g.
The matrix equation,
Equation
10.31 represents
elements of the vectors x0 and g. A nontrivial
value ? so that
n+ p linear
equations
in the
n+m
unknowns,
the
solution exists if and only if there exists a complex
??I-A B?
-C D
drops belowits normal rank, andthis ?is a zero of the system (A,B,C,D).
The matrix
S(?) =
??I-AB?
(10.32)
-C D
is called the system matrix or Rosenbrock system matrix [Rosenbrock, 1969]. It is the degeneracy of the
system matrixthat leads to system zeros. If Rank(B) =m and Rank(C) = pthen the generic rank of S(?)
(i.e., for generic values of ?)is n+min(m, p). Rank(B) =m meansthat the mcontrols are independent.
Rank(C) = p meansthat the p outputs are independent. If all controls and outputs are independent, then
all zeros are associated with specific values of ? that reduce the rank of S from n+min(m, p). In the
following discussion we assumeindependent controls and outputs.
Wecan identify
different
leads to categorization
conditions
of system
under
zeros into
which nontrivial
several types.
solutions
of Equation
This is accomplished
10.31 exist,
by examining
which
various
conditions that lead to degeneracy of the system matrix S(?).
Thoseparticular values of ? for whichthe rank ofthe Rosenbrocksystem matrix S(?) drops
below its normal rank are called invariant
zeros. Invariant zeros can be of three different types:
378
10
State Feedback
and Linear
Quadratic
Optimization
1. Input decoupling zeros: The eigenvaluesassociated with uncontrollable modes.The values of ?
that satisfy
? <n
? ?I-AB
rank
2.
Output decoupling zeros:
The eigenvalues associated with unobservable
that satisfy
rank
3.
Transmission zeros:
??I-A?
<n
C
All other invariant zeros.
The problem of finding the values ? that
rank is a special
modes. The values of ?
case of the generalized
model (A,B,C,D)
theorem.
makethe system matrix of Equation 10.32 lose its normal
eigenvalue
problem.
The computation
of the zeros of a state-space
based on the Rosenbrock system matrix S(?) is summarized in the following
Theorem 10.17 (Computation
of the Zeros of State-Space
Thezeros ofthe state-space model(A,B,C,D)
Model (A, B, C, D))
are the generalized eigenvalues ?i so that
??iI-A B? ?? ??AC -D-B??
vi=[?iE-Z]
vi
vi = ?i
-C
D
I
0
0
0
-
(10.33
where vi are the corresponding generalized eigenvectors.
The MATLAB command
Dz=eig(Z,E)
or
[Vz,Dz]=eig(Z,E)
can be employed to obtain the generalized eigenvalues and the generalized eigenvectors in
Vz, respectively.
system in
Example
This state-space approach wasemployed to find the zero of a simple first-order
10.3. In
Example
10.13,
we obtained the poles and zeros of a fourth-order
Dz and
SISO
F/A18
SISO system by converting the state-space modelto a transfer function. We will verify the zeros of the
fourth-order F/A18 SISO system in the next example bythe Theorem 10.17 approach.
Example 10.18 (F/A18
State Space)
SISO Longitudinal
Revisited, Computation of Zeros of a SISO System in
Recallthat the F/A18 longitudinal flight dynamics
A =
?
?0
?
? ? ?
-2.212
-0.2532
0 ?,
b=? ?
-0.02389
-28.32 0
-0.00033
-0.362 1
?
0
model considered in Example 10.13 is given by
0
1
-32.2
0
0
?
?
?
?
-3.8114
-0.05145
-2.8791
?
?
0
?
c =? 0 -101,
d =0
With the data (A,b,c,d) obtained from running the Example 10.13 program,
zeros of the system using the following
MATLAB command:
% CSD
Ex10.18
Z=[A
-b;
Eq8.64
c
E=blkdiag(eye(4),zeros(1))
Dz=eig(Z,E)
-d];
Compute
F/A-18
SISO
zeros
we can compute the
10.4
And the results
E
Poles and Zeros of
MIMO
Systems
379
will be:
=
Dz
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
=
3.4907e-02
-4.3935e+00
4.1059e+00
-Inf
Inf
Asexpected,
wehave
obtained
thesame
three
zeros,
0.034907,
-4.3935,
and4.1059,
fortheSISO
system by either the direct computation
eigenvalues of the system matrix.
Example 10.19 (F/A18
in State Space)
Recall the F/A18
of the transfer
MIMO Longitudinal
longitudinal
flight
function
or computing
the generalized
Revisited, Computation of Zeros of a MIMO System
dynamics
model of Equation 8.64, from
which both of the
elevator andthrust controls, deand dT, are consideredasthe inputs and boththe flight path angle ?=
?-aand
theairspeed
Vare
theoutputs,
and
wewill
have
-3.8114
9.5196*10-4
-28.320 -32.2
?-0.02389
? ?-0.05145
?
-0.00033
-0.362 1
-3.8506*10-7
A=? 0 -2.212-0.25320 ?,B=?
?0 -101? ? ?
0
?
?
?
?
?
?
0
C =
1
0
,
0 0 0
1
0
?
-2.8791
0
0
?
0
00
D =
00
Run the Example 10.13 program to obtain the data A, B, then we can find the 2-input/2-output
transfer function
matrix using the following
MATLAB command:
% CSD
C=[0
Ex10.19
-1
0
Eq8.64
1;
1
0
F/A-18
0
0];
Then we have the transfer
From
input
1
to
zpk(sys)
functions from the two inputs to the two outputs:
(s+4.393)
0.02027s
-3.8114
(s-4.106)
(s-0.03491)
+
0.01026)
(s-6.57)
(s2
+
(s+6.623)
0.6189s
+
2.286)
+
2.286)
(s+0.1803)
--------------------------------------------------(s2
+
From
sys=ss(A,B,C,D),
--------------------------------------------------(s2
+
2:
Zeros
output...
0.051453
1:
Poles
D=zeros(2);
input
0.02027s
2
3.8506e-07
to
+
0.01026)
(s2
+
0.6189s
output...
(s+0.8378)
(s2
+
0.2532s
+
2.212)
380
10
1:
State Feedback
and Linear
Quadratic
Optimization
--------------------------------------------------(s2
+
0.02027s
0.00095196
2:
+
0.01026)
(s2
(s-0.01245)
(s2
+
+
0.6189s
0.6392s
+
+
2.286)
2.314)
--------------------------------------------------(s2
+
0.02027s
+
0.01026)
(s2
+
0.6189s
+
2.286)
Notice
that
thefourelements
thatmake
upthe22 transfer
function
matrix
are
rational
functions
with a common denominator that provides a clear identification of the poles. However, the zeros of
each individual
numerators are not the zeros of the system. Although the zeros can be obtained as
the complex
values
of s that
makethe numerator
matrix
N(s) lose its normal rank, the computation
complicated and numerically unreliable. Weobtain the zeros by determining
of the system matrix. Withthe MATLAB command:
Z=[A
-B;
C
is
the generalized eigen-values
-D];
E=blkdiag(eye(4),zeros(2))
Dz=eig(Z,E)
we will havethe following results:
E
=
Dz
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
=
-Inf
-4.4631e+00
4.2099e+00
Inf
Inf
Inf
We
have
thetwozeros:
-4.4631
and
4.2099.
We
can
gofurther
and
confirm
thatthesystem
isboth
controllable
and observable.
Thus, the two system zeros are transmission
zeros.
10.5 State-Feedback Control via Linear Quadratic Regulator Design
Consider a system described by the following
state equation
x?(t)
= Ax(t)+Bu(t)
(10.34)
where
x?Rn,u
?Rm,
and
allthestate
variables
inx(t)are
available
forfeedback.
Recall
that
thebe-havior
and performance of the system are mainly determined by the poles of the system, which are the
eigenvalues
of the A matrix.
Using the state-feedback
u(t)
the state equation
of the closed-loop
system
x?(t)
control to close the loop,
= Fx(t)
(10.35)
will become
= (A+BF)x(t)
(10.36
10.5
State-Feedback
and then the behavior and performance
Control
of the closed-loop
via Linear
system
Quadratic
Regulator
will be dictated
Design
381
by the eigenvalues
of
A+BF.
If the system is controllable, orthe rank of the controllability
matrix
C=[B ABA2B An-1B]
is n, then the eigenvalues
state feedback
wasfirst
of
of the complex
plane.
The concept
of the
unstable simple in-verted
More detailed discussions of the state-feedback pole placement approach were
Sections 7.5.1 and 7.5.2, respectively,
The direct approach
simple
can be placed anywhere
applied in Section 4.4.3 of the book to stabilize the originally
pendulum system.
given later in
A+BF
is implemented
but computationally
according
for the direct approach
to
Equations
can become very complicated
and the transform
10.35 and 10.36,
for high-order
approach.
which are conceptually
systems.
Onthe other hand, the
transform approach requires the state equation be transformed into the companion form, but the effort
of the transformation
certainly
is
worthwhile
as easy as that for the low-order
In
addition
to stabilizing
a system
Section 7.6.4, the state-space
since it
has madethe computation
for high-order
compu-tations
systems.
at an originally
pole placement
approach
unstable
can
equilibrium,
work together
as demonstrated
in
with tracking/regulation
theory and root locus design, as shown in Section 8.6.3 and Section 8.7, respectively, to achieve
aircraft flight path angle tracking control and aircraft altitude regulation.
The reasoning behind the pole placement approach is that the performance of the system is quite
related to the pole locations
on the complex
plane. For example, the typical
second-order
system under-damped
step
response
isdetermined
bythetwocomplex
poles
-a? ortheir
associated
damping
ratio
? and natural frequency ?n, as shown in Section 3.4, particularly demonstrated in Figures 3.10 and 3.11.
However, in general, especially
desired time-domain
for high order systems, the pole locations
performance.
in the pole placement
Furthermore,
design process.
control-input
Hence, substantial
constraints
simulations
may not precisely
are not explicitly
usually
are required
reflect the
considered
to verify the
design.
Thelinear quadratic regulator (LQR) design is a time-domain performance index optimization ap-proach
in which the performance index consists of two parts: one accounting for the performance and
the other representing
the control-input
solution
by solving
can be found
effort.
With a chosen
the optimization
weighting
function,
an optimal
trade-off
problem.
10.5.1 Performance Index and LQR State Feedback
Consider the system described by the state equation Equation 10.34. Assumethe system is stabilizable;
then the LQR control
the closed-loop
design problem is to determine
system is stable and the performance
PI
=
a state-feedback
control law
u(t)
= Fx(t)
so that
index,
?8
?xT(t)Qx(t)+uT(t)Ru(t)
? dt
(10.37)
0
is minimized,
where
Q?Rnn
ispositive
semi-definite
andR?Rmm
ispositive
definite.
The
definitions
of positive definite and positive semi-definite
Theorem 10.20 (Linear
Assume the system
Quadratic
matrices are given in
Appendix E.6.
Regulator State Feedback)
with the state equation
Equation
10.34 is stabilizable,
then the optimal
state-feedback
control that stabilizes the closed-loop system and minimizesthe performance index PI ofEqua-tion
10.37 for any initial state x(0)
= x0i
382
10
State Feedback
and Linear
Quadratic
Optimization
(10.38)
u=Fx withF =-R-1BTX
where
X =XT=0istheunique
positive
semi-definite
stabilizing
solution
ofthe
algebraic
Riccati
equa-tion
[Zhou et al., 1995, Kailath, 1980],
ATX+XA-XBR-1BTX+Q
=0
and the optimal
performance
10.5.2 Stabilizing
(10.39
0 Xx0.
index is PIopt = xT
Solution of the Algebraic
Riccati Equation
One wayto solving for the stabilizing solution of the algebraic Riccati equation, Equation 10.39, is to
use its corresponding
Hamiltonian
matrix [Zhou
H =
et al., 1995,
Kailath,
1980],
?-QA -BR-1BT
?
-AT
(10.40)
The eigenvalue structure of the Hamiltonian matrix hasthe following interesting property.
Theorem 10.21 (Eigenvalues
of the Hamiltonian
Matrix)
The set ofthe 2n eigenvalues ofthe Hamiltonian
matrix H are symmetric with respect to the imagi-nary
axis.
That
is,?isaneigenvalue
ofHifandonly
if-? is.
Assumethe Hamiltonian matrix H has no eigenvalues on the imaginary axis; then H has n stable
eigenvalues in the left half of the complex plane and another n unstable eigenvalues in the right half of
the complex plane. The n eigenvectors associated with the stable eigenvalues can be stacked to form a
2nncomplex
matrix
T?C2nn
with
partition
asfollows:
T =
??
T1
(10.41)
T2
where
T1and
T2areboth
nn complex
matrices.
IfT1
isinvertible,
then
thestabilizing
solution
ofthe
Riccati equation is
X = T2T1-1
(10.42)
The stabilizing solution Xis real, symmetric, positive semi-definite, and unique.
Notice that the stabilizing solution exists only if the two conditions are satisfied: (1) The Hamil-tonian
matrix has no eigenvalues on the imaginary
axis, and (2) the matrix T1is invertible.
These
two conditions are satisfied under the two assumptions of the LQR problem formulation, which are: (1)
(A,B) is stabilizable, and (2) the weighting matrices Qand Rare positive semi-definite and positive
definite, respectively.
Example 10.22 (The Stabilizing
Solution of an Algebraic
Riccati Equation)
Considerthe system
x?(t)
which
was employed
in
= Ax(t)+Bu(t)
Examples
=
? ? ??
-2 -2 x(t)+
2
3
7.27 and 7.28 to demonstrate
-1
1
u(t)
the pole-placement
state-feedback
Here, we will usethe LQR Riccati equation approach to find a state feedback u(t)
that the closed-loop
system is stable and the performance
index
ap-proaches.
= Fx(t) so
10.5
State-Feedback
PI =
Control
via Linear
?8
Quadratic
Regulator
Design
383
? dt
?xT(t)Qx(t)+uT(t)Ru(t)
0
is minimized, wherethe weighting matricesare chosen as
Q =
??
10
,
01
R =1
Note
that
theopen-loop
system
has
poles
at-1and
2;hence,
itisunstable.
(A,B)
isstabilizable
since
it is controllable.
The Hamiltonian
matrix corresponding
Riccati
-2 -2 -1 1 ?
?
?-Q -AT ?=?-1 0 2 -2
0 -1 2 -3
10.39, is
A -BR-1BT 2 3 1 -1
H =
whose eigenvalues
to the algebraic
?
?
?
?
are
-2.4885,-0.89856,
2.4885,
0.89856
and the eigenvectors associated withthe stable eigenvalues are
?
??
=?
?
0.28724
T =
T1
T2
?
?
?
?
?
0.41001
-0.2049
-0.19405
-0.3267
0.59762
?
?
-0.87680.66112
hence, the stabilizing
Riccati solution is
X = T2T-1
1
=
6.5737
10.81
10.81
19.433
The state-feedback gain is thus
?
F =-R-1BTX
=? -4.2361
-8.6231
and the closed-loop system poles are now the eigenvalues of A+BF,
which are
-2.4885,-0.89856
exactly the same as the stable eigenvalues
These numerical
%
CSD
Hamiltonian
matrix.
results are obtained by running the following
Ex10.22
State
A=[-2
-2;2
H=[A
-B*inv(R)*B';
T2=V(3:4,1:2),
of the
3];
feedback
B=[-1;1];
Riccati
eig_A=eig(A),
-Q
-A'],
MATLAB code:
Hamiltonian
Q=eye(2);
[V,D]=eig(H),
R=1;
T1=V(1:2,1:2),
X=T2*inv(T1),
Riccati_Eq_Check=A'*X+X*A-X*B*inv(R)*B'*X+Q,
F=-inv(R)*B'*X,
eig_ABF=eig(A+B*F)
A MATLAB command:
lqr
can also be employed to find
an LQR solution:
equation,
Equation
384
>>
10
State Feedback
and Linear
Quadratic
Optimization
[K,S,E]=lqr(A,B,Q,R)
K
=
S
=
4.2361e+00
E
8.6231e+00
6.5737e+00
1.0810e+01
1.0810e+01
1.9433e+01
=
-2.4885e+00
-8.9856e-01
Note
thatK=-F,S=X,andEgives
theeigenvalues
ofA-BKorA+BF.
If only
thesolution
of the algebraic
follows:
>>
Riccati equation is of interest,
the
MATLAB command:
are can be applied as
X=are(A,B*inv(R)*B',Q)
X
=
6.5737e+00
1.0810e+01
1.0810e+01
1.9433e+01
10.5.3
Weighting
To employ
the
a nonlinear
Matrices
Q and
Rin the
Performance
control
design approach,
LQR state-feedback
dynamics
Index
model of the system to be controlled,
Integral
usually
the
we start
with the knowledge
desired operating
equilibrium,
and measurement noises, system model uncertainties, and actuator limitations.
the design
process,
demonstrate
we will use the simple
how to formulate
inverted
a state-feedback
pendulum
control
system
problem
discussed in
of
distur-bances
To demonstrate
Section
as an LQR optimization
4.4 to
problem.
The schematic of the simple inverted pendulum system is depicted in Figure 4.8, wherethe two state
variables x1 = ? and x2 =?? arethe angular displacement andthe angular velocity of the pendulum,
respectively, and the control input u = fa is the external control force perpendicular to the gravity. The
nonlinear
dynamics
model of the system is represented
????sinx1-b
x?1
x?2
=
by the following
nonlinear
state equation:
?
x2
(10.43)
m?2
x2 + m?
1 cosx1
u
g
Theequilibrium ofinterestis at x* =[0 0]T, whichrepresentsthe stick atthe upright position ? = 0
with zero angular velocity?? = 0. Assumeg = 9.8 m/s2,? = 1.089 m, m = 0.918 kg, b = 0.551 Nss
that
g
b
= 9,
?
1
m?2=0.6,
m?
= 1
hence,the state equation of the linearized modelatthe upright stick equilibrium x* =[0
found as
x?(t)
= Ax(t)+Bu(t)
Note that the uncompensated
=
?9 -0.6? ??
0
system is not stable
1
x(t)+
0
1
at this equilibrium
0]T can be
u(t)
(10.44)
since the eigenvalues
of the
Amatrix
are2.715
and-3.315.
Recall
thatin Section
4.4.3,
astate-feedback
controller
u =Fx=
[-34 -7.4]
x was
designed
based
onthepole
placement
approach
toplace
theclosed-loop
system
poles
at-4 j3sothatthedamping
ratio
and
thenatural
frequency
are? =0.8and?n=5rad/s,
respectively.
Asimulation
that
demonstrates the performance
was shown in Figure 4.11.
of this pole-placement
state-feedback
con-troller
10.5
In the following,
State-Feedback
we will employ the
Control
via Linear
Quadratic
Regulator
LQR approach to design a state-feedback
Design
controller
385
to stabilize
the system and minimize the performance index PI,
PI =
?8
? dt
?xT(t)Qx(t)+uT(t)Ru(t)
0
where Q and Rare positive semi-definite and positive definite, respectively. Aslong as (A,B) is sta-bilizable
and the Q and Rrequirements are satisfied, any state-feedback controller designed based on
the LQR approach theorem, Theorem 10.20, will stabilize the system. However, to achieve a desired
closed-loop system performance the weighting matrices Qand Rneed to be chosen carefully according
to the performance requirement and the control-input constraint.
In the following
examples, we will not only design an LQR state-feedback controller to stabilize
the system atthe originally unstableequilibrium, x* =[0 0]T. Wewouldlike the controller to be
able to bring the system from a perturbed state, say x(0) =x0 =[0.2618 0]T, back to the equilib-rium
x* =[0 0]T as quickly and smoothlyas possiblewithin control-input constraint. Notethat
x1 = ? = 0.2618radis 15?.
Since the controllability
matrix
[B
AB] =
?1 -0.6?
0
1
is of full rank, the system is controllable and thus stabilizable.
strategy, u(t)
complex
There exists a state-feedback control
= Fx(t), so that the closed-loop system poles orthe eigenvalues of A+BF,
plane as well as the control-input
we will consider
three
sets of Q and
constraints
R weighting
are satisfied.
In the following
matrices, respectively.
are in the left
three
examples,
For each set, there
a unique stabilizing optimal controller. By evaluating the performance of each controller,
see how the choice of Q and R would affect the performance of the system.
exists
we may
Fig. 10.19: Simulation diagram to evaluate LQR controller performance.
Although
the controller
is designed
based on the linearized
model, Equation
10.44, the
nonlinear
model Equation 10.43 is employed in the simulation, as shown in Figure 10.19. The function
Fcn is defined by
block
f(u)=9*sin(u(1))-0.6
*u(2)+cos(u(1))
*u(3)
where u(1), u(2), and u(3), respectively, represent the state variables x1, x2, and the control-input
u of Equation 10.43. Theinitial condition of the state variables are assigned as x20=0 an
386
10
State Feedback
x10=0.2618,
and Linear
respectively,
inside
Quadratic
Optimization
the integrator
stick has deviated from the equilibrium
blocks.
This initial
condition
meansthat the pendu-lum
by 15 degreesto the right with zero angular velocity.
Example 10.23 (Case 1 LQR State Feedback Controller for the Simple Inverted
Pendulum)
Considerthe state equation, Equation 10.44, whichis the linearized state-space model of the nonlin-ear
simpleinverted pendulum dynamics of Equation10.43atthe unstableequilibrium x* =[0
the weighting matrices Qand R be
Q =
??
q1 0
,
0 q2
0]T.Let
R = r1
then
xTQx
= q1x2
1 +q2x22,
and the performance index (also called cost function)
PI =
uTRu
= r1u2
will be
?8 2+r1u2?
?8 ?8 ?8
? q1x21 +q2x2
dt
x2
1 dt
= q1
0
+q2
0
x2
2 dt
0
+r1
u2 dt
0
The three terms on the right-hand side of the equation are the weighted total energies of the angular
displacement(x1 = ?),the angular velocity (x2 =??), andthe control-input force (u = fa), respectively.
For Qto be positive semi-definite and R positive definite, q1 and q2 haveto be greater or equal to zero
and r1 is required to be greater than zero. The control-input
weight r1 cannot be zero since r1 = 0
meansthat infinity feedback is allowed, which of course is not practically possible. Making r1 larger
will put
more constraint
to be made larger
on the control-input
if the reduction
energy consumption.
of the energy
Similarly,
of x1 would improve
the
weight
q1 may need
the performance.
We will begin with the selection of the weights q1 = q2 = 1 and r1 = 1, find the unique stabilizing
controller that minimizes the performance index associated with this particular weight selection, eval-uate
the performance of the closed-loop system, and then revise the weight selection according to the
performance evaluation.
Withthis Q, Rselection,
??
?
Q =
1 0
0 1
,
R =1
the unique stabilizing solution for the algebraic Riccati equation can be found as
X =
60.831 18.055
18.055 5.5213
The state-feedback gain is thus
?
F =-R-1BTX
=? -18.055
-5.5213
and
theclosed-loop
system
poles
areat-2.5018
and-3.6195.
Since
theinitial
state
isx(0)=x0=
[0.2618
0]T,the
minimal performance index is
0 Xx0 = 4.1693
PIopt =xT
Based on Equation
10.43 and the simulation
diagram in Figure 10.19, a Simulink
program is assem-bled
to conduct simulations to observe the time-domain responses dueto the initial conditions of x1 and
x2. Thetime-domain responsesof the two state variablesx1(t) = ?(t) and x2(t) =??(t) dueto the initial
10.5
State-Feedback
Control
via Linear
Quadratic
Regulator
Design
387
conditions x1(0) = 0.2618 rad and x2(0) = 0 rad/s, are shown on the left-hand side of Figure 10.20.
Meanwhile, the control-input action is recorded on the right graph of the figure. The pendulum initially
is tilted to the right by 15?(0.2618 rad). The deviation of the pendulum position from the equilibrium
x*=[0 0]Tprompted
thecontrol-input
u(t)
toreact
immediately,
changing
from0Nto-4.7Nand
then
gradually
reducing
to zero so that the angular
velocity
x2 and the angular
displacement
x1 can
change accordingly to bring the pendulum back to the equilibrium. It takes about 2.8 seconds to bring
the pendulum back to the equilibrium.
Comparing the simulation results of Case 1 LQR state-feedback controller with those of the pole
placement state-feedback controller shown in Figure 4.11, we observe that the latter only takes about
1.7seconds
toconverge
totheequilibrium
while
using
more
control-input,
-8.8N.
Inthenext
ex-ample,
we will reduce the control-input
weight r1 to allow
using
more control-input
energy and then
observe if this change will improve the performance.
Fig. 10.20: Simulation results of the Case1 LQR controller performance.
Before getting into the next example, it is interesting to see if multiplying Q and R by the same
constant would change the outcome of the LQR state-feedback controller design. Let the Q and Rin
Case 1 design be multiplied
by 10 to become
Q =
??
10 0
0 10
,
R = 10
and repeat the LQR controller design in Example 10.23. We will see the stabilizing solution Xto the
algebraic Riccati equation becomes 10 times of the solution obtained in Example 10.23, and thus, the
minimal performance index PIopt will increase 10 times accordingly to
0 Xx0 = 41.693
Jopt = xT
However,the stabilizing state-feedback gain matrix F remains the same as
?
F =-R-1BTX
=? -18.055
-5.5213
since the same variation in Rand X cancels each other. Therefore, only the relative
weights areimpor-tant,
and the value of the minimal performance index PIoptdoes not reflect the performance ofthe
closed-loop system.
Example 10.24 (Case 2 LQR State Feedback Controller for the Simple Inverted
Pendulum
388
10
State Feedback
In this example,
and Linear
we consider
Quadratic
Optimization
the same simple
inverted
pendulum
control
problem
with the same
LQR state-feedback control design as in Example 10.23. The only difference is in the selection of the
weighting
matrices
Q and R.
Let the weighting matrices Qand Rbe
Q =
in
which the
? ?? ?
q1 0
1 0
=
0 q2
Q matrix is the same as that in
usage of control-input
energy that
0 1
,
R =r1 = 0.001
Case 1, only the
may improve
weight r1 is reduced to 0.001 to allow
the closed-loop
system
performance
by shortening
more
the
convergence time to the equilibrium.
Withthe Q, Rthus selected, the unique stabilizing solution for the algebraic Riccati equation can be
found as
1.088 0.041879
X =
0.041879 0.032326
?
The state-feedback
?
gain is now
?
F =-R-1BTX
=? -41.879
-32.326
and
theclosed-loop
system
poles
areat-1.0308
and-31.895.
Since
theinitial
state
isx(0)=x0=
[0.2618
0]T,the minimal performance index is
0 Xx0 = 0.074568
PIopt =xT
Thetime-domainresponsesofthe two statevariablesx1(t) = ?(t), x2(t) =??(t), andthe control
input u(t) due to the initial
condition: x1(0)
= 0.2618 rad and x2(0)
= 0 rad/s, are shown in Figure
10.21.
It does
utilize
more
control-input
magnitude
with-11Nofforce
compared
with-4.7N
in
Case 1. However, the
converges
performance
to the equilibrium
within
Fig. 10.21: Simulation
is
not improved.
Actually
it
gets
worseneither
x1 nor x2
3 seconds.
results
of the Case 2 LQR controller
performance.
The choice of q1 (the weightfor x1 = ?) and q2(the weightfor x2 =??) to beidentical in Cases
1 and 2 may not be a good one. Although
the energy of x2 (angular
For this reason, q2 should
velocity)
be chosen
eventually
both x1 and x2 shall
converge
will slow down the convergence of x1 (angular
much smaller than q1. Withthe
to zero, limiting
displacement).
weights being chosen as q1 = 1000
10.5
q2 = 0, and r1
State-Feedback
= 1, we saw a great improvement
Control
via Linear
in performance.
Quadratic
After a few
Regulator
Design
minor revisions,
389
it seems
that the weight selection of q1 = 1,000, q2 = 20, and r1 = 1.2 will give an optimal performance. The
LQR controller
design associated
with this
Q =
and the performance
Example
evaluation
10.25 (Case
In this example,
q1 0
103
0
0
20
=
0 q2
of the closed-loop
3 LQR
matrices
? ?? ?
State Feedback
,
system
R = r1
= 1.2
will be given in the following
Controller
for the
Simple Inverted
example.
Pendulum)
we consider the same simple inverted pendulum control problem
LQR state-feedback
control
weighting
Q and R. Let the
matrices
Case 3 weighting
design as in
Q=
Example
weighting
10.23.
matrices
The only difference
103 0
,
0 20
=
of the
Q and R be
? ?? ?
q1 0
0 q2
with the same
is in the selection
R = r1 = 1.2,
in which the weight q1is muchlarger than the weight q2to emphasize the minimization of the x1 energy
while allowing
selected to
enough
energy for x2 to speed up the convergence
make sure the control-input
constraint
condition
is
to the equilibrium.
The weight r1 is
met.
Fig. 10.22: Simulation results of the Case3 LQR controller performance.
Withthe Q, Rthus selected, the unique stabilizing solution for the algebraic Riccati equation can be
found as
361.08 47.086
X =
The state-feedback
?
47.086
11.007
gain is now
?
F =-R-1BTX
=? -39.238
-9.1725
and
theclosed-loop
system
poles
areat -4.8863
j2.5224.
Since
theinitial
state
isx(0)=x0=
[0.2618
0]T,the minimal performance index is
PIopt =xT
0 Xx0 = 24.748
Thetime-domain responsesof the two state variables,x1(t) = ?(t), x2(t) =??(t), andthe control
input u(t) due to the initial
condition, x1(0)
= 0.2618 rad and x2(0)
= 0 rad/s, are shown in Figure
390
10
State Feedback
10.22.
Withthis
and Linear
Quadratic
LQR state-feedback
Optimization
controller,
it only takes
1.15 seconds
by using less than
control input to bring the perturbed
pendulum state at the 15-degree tilted
to the equilibrium
pendulum
statethe
upright
10.3N of
angular position back
position.
Withthis state-feedback control law u = Fx, the closed-loop system poles are the eigenvalues of
A+BF,
which
are-4.8863
j2.5224.
The
damping
ratio
and
thenatural
frequency
associated
with
the
closed-loopsystem polesare ? = 0.889and ?n = 5.499rad/s.
10.6 Exercise Problems
P10.1: In this exercise problem, we will review how to find equilibriums of a nonlinear system, and how
to obtain a linearized
state-space
study the system characteristics
lightly
model of the nonlinear
system at a chosen equilibrium.
at and around the equilibrium.
Assume the nonlinear
Then
we will
state equation
of a
damped pendulum system, whichis similar to the onein Equation 10.4, is given below,
x? =
???
x?1
x?2
=
?
x2
= f(x,u)
-a0sinx1-a1x2+b0u
where a1 = 0.1, a0 = 20, and b0 = 40. The state variables
=
? ?
f1(x1,x2,u)
f2(x1,x2,u)
x1 and x2 represent the angular
(10.45)
displacement
and the angular velocity of the pendulum system, respectively. The control input uis the applied torque
to drive the system.
Fig. 10.23: Identify equilibriums andfind the linearized
P10.1a: As a brief recap of the linearization
the equilibriums.
model at a chosen equilibrium.
process, Figure 10.23 shows that the first step is to find
Assumethe operating equilibrium is chosen to keep the angular displacement of the
pendulum at 18? = 0.1p rad, whichis x1 = x*1 = 0.1prad andx2 =x*
2 = 0 rad/s. Find the corresponding
control input u = u* Nm at the equilibrium.
P10.1b: Find the linearized state equation
x?(t) = x(t)+B
A
u(t)
(10.46
atthe equilibrium (x*1,x*2,u*) chosenin P10.1a.
P10.1c: Find the characteristic equation of the linearized model and the poles of the system, and com-ment
onthe system characteristics based on the damping ratio, natural frequency, andthe location of the
poles on the complex plane.
P10.1d: Build a simulation program according to the linearized state-space modelin Equation 10.46.
Conduct
the simulation withthefollowing conditions:
u(t)
= 0,x1(0) = 9? =0.05prad,andx2(0) =
10.6
Exercise
Problems
model in
Equation
391
0rad/s.
Plot
thesimulation
results
x1(t)and
x2(t)fort =0.
P10.1e:
Build a simulation
program
according
to the nonlinear
state-space
10.45.
Applythe equilibrium control input u(t) = u* Nmto the system, and observethe values of the state
variables x1(t) and x2(t) after the system reaches the steady state. Whatarethe values of x1(t) and x2(t)
at the equilibrium state?
P10.1f: Withthe nonlinearsimulationprogramatthe equilibrium(18?,0,u*), manuallymovex1from
18? to the 27? position, and release at t
= 0. Plot the state variables x1(t), x2(t), and the control input
u(t). Comparethe nonlinearsimulationresponsesx1(t), x2(t) withthe linear simulation responses
x1(t)
andx2(t) obtainedin P10.1d.
P10.2a: This exercise problem is a continuation of Problem P10.1. As revealed from the solutions of
P10.1, the time response of the uncompensated pendulum system is oscillatory with large overshoot. In
the following we would design a controller, also called a compensator, to improve the performance of
the system. First of all, let us check the controllability of the system. Show that the system represented
by Equation 10.46 is controllable.
P10.2b: Find a state feedback u(t)
= F
x(t)
so that
n = s2 +18s+100
|sI-(A+BF)|=s2+2??ns+?2
That
is,theclosed-loop
system
poles
willbeplaced
at-9 j4.359,
or,equivalently,
thedamping
ratio
andthe naturalfrequency are ? = 0.9 and ?n = 10rad/s, respectively.
Fig. 10.24:
Closed-loop
system
with state feedback.
P10.2c: Build a simulation program according to Figure 10.24(a), which includes the linearized state-space
model and the state-feedback
controller
obtained in P10.2b.
Conduct the simulation
with the initial
condition
x1(0)=9?=0.05prad,
and
x2(0)=0rad/s.
Plot
thestate
response
x1(t)and
x2(t)fort =0.
P10.2d:
Build a simulation
program
according
to
Figure 10.24(b),
which includes
the nonlinear
model and the state-feedback controller obtained in P10.2b. Conductthe simulation
state-space
withthe initial
condition:
x1(0)=27?
=0.15prad,
and
x2(0)
=0rad/s.
Plot
thestate
response
x1(t)
and
x2(t)
fort =0.
P10.2e: Comparethe nonlinearsimulation results x1(t) and x2(t) in P10.2d withthe linear simulation
resultsx1(t)
andx2(t)
obtained in P10.2c, and give your comments.
P10.3: Consider a set of linear algebraic equations, whichis represented in the following
matrixform
392
10
State Feedback
and Linear
Quadratic
Optimization
?2 1 -1?
Ax
=b,where
A=?34-10 -3-5?
1 2
A systematic
way of investigating
the null space of the
the solution
existence
1
?
?
?
?
and uniqueness is to find the range space and
matrix.
P10.3a: Findthe range spaceof the matrix A, Range(A).
P10.3b: Findthe null space ofthe matrix A, Null (A).
P10.3c: Find the condition on the vector b such that the equation has solutions.
P10.3d: Assume bT = ? 5 4 3 2 . Doesthe equation have a solution? Is the solution unique? If the equa-?
tion has morethan one solution, find the full set of solutions.
P10.4: Recallthat the controllable subspace is the range space of the controllability
matrix C, while the
uncontrollable subspace is the null space of CT. The null space of CTis the orthogonal complement of
the range space of C. Consider a system represented by the following state equation:
x?(t)
= Ax(t)+Bu(t)
=
?
?
? ??
?x(t)+
? ?u(t)
-14 -19-23
19
27
33
-1
0
1
-6 -9 -11
P10.4a:
Find the controllability
matrix
C, and determine if the system is controllable.
P10.4b: Find the controllable subspace of the system, which is the range space of the controllability
matrix, Range(C).
CT?
P10.4c: Find the uncontrollable subspace of the system, which is the null space of the transpose of the
controllability
matrix, Null ?
.
P10.4d: Usethe basis vectors of the range space of Candthe null space of CTto construct a similarity
transformation
matrix T = ? T1 T2 ? to transform the state equation into the controllability
decomposition
form, as shown in Theorem 10.9.
P10.4e: Is the system stabilizable?
Explain.
P10.4f: Design a state-feedback controller sothat the closed-loop system is internally stable.
P10.4g: Find the closed-loop system poles.
P10.4h: Assumethe initial state of the system is x(0)
closed-loop
=?
system.
?T
1-13 .
Plot the state response x(t) of th
P10.5: For a SISO system, the zeros can befound easily from its transfer function.
the case for
MIMO systems. In this exercise,
model to find the zeros of the system.
the system.
you
will apply
However,that is not
Theorem 10.17 to a simple
SISO state-space
Then verify the result using the zeros of the transfer
function
of
10.6
P10.5a:
Consider a SISO system described
by the following
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
state-space
Exercise
Problems
393
model
where
A =-1,B=1,C=-3,andD=1.Determine
thepole
and
zero
otfhesystem
directly
using
the
state-space model. That is, use Theorem 10.17to determine the zero, and compute the eigenvalue of the
A matrixto obtain the pole.
P10.5b: Find the transfer function of the system and use it to determine the zero and the pole of the
system.
P10.5c: Comment onthe relationship of the poles and zeros obtained in P10.5b and P10.5c.
P10.6: Consider a MIMO system described by the following state-space model
where
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
? ? ??
?100 -10
00 -248
? ? ??
04 0 00
1 0
A =
0
10
0 0
0 0
?
?
?
?
?
?
?
?
?
?
0 0
1
0 0
?
,
B =
?
00 0 10
?
0 1 ,
?
C =
0 1 0
1
3
?
0 0
?
00
P10.6a: Find the poles of the system by computing the eigenvalues of the A matrix, and determine the
zeros of the system by using Theorem 10.17.
P10.6b: Find the transfer function matrix of the system, and useit to determine the poles and zeros for
each entry of the transfer function matrix.
P10.6c: Comment onthe relationship of the poles and zeros obtained in P10.6b and P10.6c.
P10.7: In this exercise problem,
considered in Section 10.2.2.
wewould like to revisit the lightly
damped pendulum
positioning
system
Recall that at the equilibrium
(x*1,x*2,T*) = (15?,0?/s,0.19043Nm)
we havethe linearized state-space model
x?(t)
= x(t)+BT(t)
A
wherethe matrices A and B are given by Equation 10.6,
A =
?-21.32-0.135? ??
0
1
,
B =
0
30
A pole placementapproach wasemployedto improve the dampingratio from ? = 0.0146to ? = 1.0011.
In the following,
we will utilize the
index is optimized.
LQR approach to design a state-feedback
controller
so that a perfor-mance
394
10
State Feedback
and Linear
Quadratic
Optimization
P10.7a: For notational simplicity, the state vectorx(t) and the control input T (t) will bereplacedby
x(t) and u(t), respectively. Thatis, the linearized state-space model of the system to be controlled is
x?(t)
= Ax(t)+Bu(t)
=
?-21.32-0.135? ??
0
1
The objective is to design a state-feedback controller u(t)
stable and the following performance index
PI
=
0
x(t)+
30
(10.47)
u(t)
= Fx(t) so that the closed-loop system is
?8
?xT(t)Qx(t)+uT(t)Ru(t)
? dt
0
is
minimized.
The solution
of this optimization
problem is
u(t)=Fx(t) withF =-R-1BTX
where Xis the positive semi-definite stabilizing solution of the following
algebraic Riccati equation:
ATX+XA-XBR-1BTX+Q
=0
Let the weighting matricesin the performance index integral be
Q =
Find X, which is the positive semi-definite
??
1 0
0 1
,
stabilizing
R =1
solution
of the above algebraic
Riccati equation,
and
then
use
ittodetermine
theoptimal
state-feedback
gain
matrix
F =-R-1BTX.
P10.7b:
Find the closed-loop
damping ratio
system poles,
which are the eigenvalues
of A+BF,
and the corresponding
and natural frequency.
P10.7c: Letthe initial state be
x(0)T = ? 20? 0?/s ? = ? 0.349rad 0rad/s ?
Plot the state response
x(t)
and the control input
u(t).
P10.7d: Comment onthe pole location andthe simulation results of the closed-loop system.
P10.7e: Changethe weight matrices Q and R,repeat the above design and simulations, and comment on
how the weighting matrices affect the performance
11
ObserverTheory and Output Feedback Control
U
P to now, we have learned how to employ the state-space model and state-feedback control
with pole placement, linear quadratic optimization, and internal model principle to achieve sta-bilization,
tracking, regulation, and performance enhancement. The state-space approach can
also work together withthe classical root locus design approach to achieve precise pole placement per-formance.
However,the state-feedback approach requires direct accessto all the system state variables,
which maynot be possible in manyapplications. Consequently, we needto generate suitable estimates
of the states based on the available
observer structure to efficiently
the state-space model.
The main computation
gain
matrix,
information.
The observer
theory
provides
estimate the states using the inputs, the
a brilliant
involved in the design of the observer is the determination
which in fact is a dual
problem
of finding
full-state
measurable outputs, and
the state-feedback
of the ob-server
gain
matrix.
In
other words, the algorithm for computing the state-feedback gain matrix can be employed to find the
observer
gain
matrix. Furthermore,
the observer
can be integrated
seamlessly
together
with the state-feedback
gain matrix to construct the output feedback controller, and the overall closed-loop system
poles
will be the observer
we obtained
poles together
using state feedback.
with the regulator
poles,
which are the eigenvalues
Therefore, the observer and the state feedback
of
A+BF
can be designed sep-arately.
Similar to the state-feedback design, the observer can be designed using the pole placement
approach or the quadratic performance index minimization approach. The output feedback controller
consisting
of the observer
performance
index
and the state-feedback
minimization
gain is referred
approach is involved
as the
H2 controller
if the
quadratic
in the design process.
11.1 Observability
Duality is aninteresting property of nature. For example, humans left and right hands are dual to each
other. We will soon find out that observability and controllability
are a duality pair, and the observer
gain and the state-feedback gain are also a dual. Understanding either one of the dual will makethe
learning of its counterpart very easy. Withthe concept of controllability and the design of the state-feedback
gain we have covered the previous chapter, it will be a breezein learning their counterparts in
this chapter.
Although
our daily life.
observability,
controllability,
and stability
For example, assume a person has a tumor
are technical
terms, they actually
but it is not observable.
are related to
Without knowing
what
the problem is, there is no wayto fix it. If the tumor is growing, the system will become unstable and
eventually cease working. Onthe other hand,if it is observable, it is still required to be controllable fo
396
11
Observer
the problem
Theory and
to be fixed.
Output
The doctors
Feedback
Control
need to have
means, either
by surgical
procedure
or by injecting
medicine or through some other treatment to remove or contain the tumor.
For a control system to work,the system to be controlled is required to be stabilizable and detectable.
As defined in the previous chapter, a system is stabilizable if the uncontrollable
part of the system
is stable.
Similarly,
a system
is detectable
if the
unobservable
tumor
example, if the tumor is benign (stable) it is considered
11.1.1
Observability
Consider a system
part
of the system is stable.
In the
as detectable even it is unobservable.
Rank Test
with the linear
state-space
model
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
(11.1)
with
x?Rn,u
?Rm,y
?Rp.
The
basic
concept
ofobservability
istoidentify
thestate
xgiven
observation
of the output y over some time interval [0,t f] , t f > 0.
Definition 11.1 (Observability)
The linear
time-invariant
system described
by Equation
11.1 is said to be (completely)
observable
iftheinitialstate
x(0-)canbeuniquely
determined
using
themeasurements
oftheoutput
y(t)and
the
knowledge
ofthe
inputu(t),which
canbezero,
over
afinite
interval
oftime
0- =t =tf.
Note
thatonce
x(0-)isknown,
x(t)can
bedetermined.
Similar
tothecontrollability
rank
testof
Theorem 10.6, observability can be determined using the following theorem.
Theorem 11.2 (Observability
Rank Condition)
Consider
thefollowing
npnobservability
matrix:
? ?
? ?
C
CA
?
O=
?
?
(11.2)
?
...
?
?
CAn-1
The system described by Equation 11.1 or the matrix pair (A,C) is (completely) observable ifand
if
rank
only
O = n
Notice that the matrices B and D play no role in determining observability, thereby implying that the
control input u(t) is not relevant.
Example 11.3 (Observe the Initial
Considerthe following
State)
simple two-state one-output linear system:
x?(t) = Ax(t)+Bu(t)
=
?? ??
01
10
y(t)=Cx(t)
=? 1-1? x(t)
x(t)+
1
1
u(t
11.1
The observability
matrix of the system is
O =
Since rank
O =1
397
? ??1 -1?
C
=
CA
-1 1
< 2,the system is not observable based on Theorem 11.2.
The unobservability
response and the initial
initial stateis
Observability
of the system can also be verified
state according
to the definition
based on the relationship
of observability.
between the output
The output response
due to the
?
?x(0)
=e-t
-e-t
?
-e-t
e-t
?
? ??e-t -e-t?
?e-t -e-t?
et-e-t
? -1
? et+e-t
y(t)=CeAtx(0)
=0.51
et-e-tet+e-t
x(0
?
Taking a derivative of the equation, we obtain
y?(t)
=?
x(0)
Combine the two equations into a matrix form:
y(t)
=
y?(t)
Since the
matrix
from the information
-e-t e-t
-e-t
x(0)
e-t
is singular, the initial state x(0) cannot be uniquely determined
of y(t) and its derivatives. Therefore, according to the definition of observability
the system is not observable.
If a system is not observable
way to easily
it is necessary
make such an identification
to identify
is to transform
the
unobservable
physical
the system to diagonal form (or,
modes. One
more generally
into Jordan form). In the following, the system considered in the previous example will be employed to
illustrate
which part of the system is unobservable.
Example 11.4 (Observable
Modes)
Consider the system of Example 11.3. We will compute the eigenvalues ?1,?2 and corresponding
eigenvectors v1,v2. Fromthe latter define the similarity transformation matrix T =[v1,v2] and implement
thetransformation from the original statexto the newstatex, x
=T-1x to obtainthe transformed system
x? =? T-1AT ?x+ ? T-1B ? u
(11.3)
y =[CT] x
The eigenvalue-eigenvector pairs of A are:
1
?1=-1,v1= v2
? ?, ?2=1,v2= v ??
1
-1
1
2
Notethat there aretwo real eigenvalues and the first
the two eigenvectors
weset up the transformation
x? =
y =
Webegin with the knowledge (from
1
1
modeis stable. The second oneis unstable. Using
matrix T and obtain the transformed
? ?x+?v0?
?v
-10
0 1
2
u
2 0 ?x
Example 11.3) that there is one unobservable
states,x1,x2 denote the motion associated with modes1 and 2, respectively.
are decoupled.
They can be solved
state equations
one at atime,
each independent
mode. The new
Observethat the state equa-tions
of the other. Inspection
of the
transformed CT matrixshowsthat the output is not affected byx2, indicating thatx2 is the unobserv-able
mode.
398
11
Observer
Theory and
11.1.2
The Observability
Output
Feedback
Decomposition
Control
Form
The ability to identify the observable part of a system has control design implications
mode implications.
Here, we discuss the decomposition
that separates the observable
and the unobservable
Theorem 11.5 (Observability
introduced
by
beyond the inter-pretive
Kalman [Kalman,
1963]
parts of a linear system.
Decomposition Form)
Given a linear system in the form
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)
with
x ?Rn,u
?Rm,andy
?Rp.
Ifthesystem
isnotobservable
and
rank
then there exists a nonsingular
new statex, x
similarity
O =r <n
transform
matrix T that transforms
the original
state x to the
= T-1x such that the transformed system
x? = ? T-1AT ?x+ ? T-1B? u =Ax+ Bu
y =[CT] x
hasthe form
A =
=Cx
? ? ??
A11 0
A21A22
B =
,
B1
B2
(11.4)
C = ?C1 0 ?
?A11,
? is
where
A11
?Rrr,
B1?Rrm,
C1?Rpr,
and
thesubsystem
B1,
C1
Proof:
Hint: Choose T such that
T-1 =
observable.
??
U1
U2
(11.5)
wherethe rows ofU1 are any r linearly independent row vectors of the observability
Therows ofU2 together withthose ofU1 span the whole n-dimensional space.
Example 11.6 (Observability
Recall that the system
Decomposition
Form)
considered in Example
x?(t)
matrix O(C,A).
= Ax(t)+Bu(t)
11.3 is
=
?? ??
01
10
x(t)+
1
1
u(t
y(t) =Cx(t)=? 1-1? x(t)
The observability
matrix of the system is
O =
? ??1 -1?
C
CA
=
-1 1
Since rank O = 1 < 2,there is only onelinearly independent row vector in
O; U1is chosen as
11.1
Observability
399
U1=? 1-1?
and U2should
be chosen so that
the similarity
transformation
T-1 =
is nonsingular.
matrix
??
U1
U2
Hence, wechoose U2 = ? 0 1 ? so that we have
T-1 =
???1 -1?
U1
=
U2
0
1
??
? T=
1 1
0 1
Then the transformed state-space model will be
x? =
? ? ?? ? ? ??
A11 0
A21A22
B1
x+
-10 x+
u =
B2
1 1
0
1
y = ?C1 0?x =? 1 0?x
Notethatx2 has no effect onthe output y; hence,the dynamics associated withA22 = 1 are not
observable,
while
thedynamics
associated
with
A11=-1are
observable.
Remark 11.7 (Observable
and Unobservable Subspaces)
The observable subspace is the range space of the transpose
of the observability
matrix
OT, while the
unobservable subspace is the null space of O. The null space of Ois the orthogonal complement of the
range space of
OT.
Forthe system in Example 11.6, the observable subspace is a one-dimensional
the linearly
independent
column
vectors
of
space spanned by
OT, which is
??
OT?
Null(O)~ ??
1
Range?
The unobservable subspace is
~ -1
1
1
which is orthogonal to the range space of OT.
Note that the space spanned by the vector
subspace
although
its associated
If the similarity transformation
similarity
UT
2 =[0 1]T of Example 11.6 is not the unobservable
transform
achieves
observability
decomposition.
matrix
T-1 =
is chosen according
T still
to the basis vectors
??
U1
U2
of the range space of
T-1 =
OT and the null space of
???1 -1?
then the transformed state-space model will be
U1
U2
=
1 1
O as
400
11
Observer
Theory and
x? =
Output
Feedback
Control
? ??? ?-10? ??
B1
A11 0
A21A22 x+
B2
u =
0
1
x+
0
2
u
y = ?C1 0 ?x = ? 1 0?x
Although the numbers are slightly different, the structure of the observability decomposition form
remains
the sameandthe eigenvalues
ofA11andA22,respectively,
areidenticalto thoseobtained
in
Example 11.6.
Fig. 11.1: Observable and unobservable subspaces.
The observable
and unobservable
subspaces
of the system are illustrated
in Figure 11.1. Assume the
initial stateisxinitial is at [2 1]T position. Sincethe eigenvalue associated with the observable subspace
spanned byx1 is stable, the position of the state on the x1 axis will movefrom 2 to 0. Onthe other
hand,the eigenvalueassociated withthe unobservablesubspace spanned byx2 is unstable;hence,the
position of the state onthex2 axis will grow without bound. Furthermore, no control action can betaken
to convert the unstable statex2 to become stable becauseit is unobservable.
11.2 Dualityin State Space
For a feedback
control
system to
work, it
needs to
observe
and gather the relevant
information
for
decision making, and once a decision is made,it requires control actions to accomplish the objective.
The observation and control processes seem to be completely different; yet surprisingly they are
dual to each other, just like left and right hands! We will see that observability and controllability
are a duality pair, and the observer design and the state-feedback control design are also a dual.
Understanding either one will makeit easy to learn the counterpart.
11.2.1
Duality
of Controllability
and
Observability
Recallthat for the linear system represented bythe state-space modelin Equation 11.1
11.2
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)+Du(t)
Duality in
State Space
401
with
x?Rn,u
?Rm,y
?Rp.
The
controllability
and
observability
ofthesystem
are,
respectively,
deter-mined
by the rank of the controllability
matrix
C=[BABA2B An-1B]
and by the rank of the observability
(11.6
matrix
? ?
? ?
C
CA
O =
?
?
?
(11.7)
?
...
?
?
CAn-1
Since a matrix and its transpose
rank
which has the same form
OT
have exactly the same rank,
= rank [CT
we have
(AT)n-1CT]
ATCT (AT)2CT
as the controllability
matrix
C. Therefore,
observability
(11.8)
and controllability
is
a duality pair. The basic concept and theory developed based on controllability
can be easily ex-tended
for observability, and the computing resources available for analysis and design relevant to
controllability
can also be applied to those relevant to observability.
As we have learned from
matrix
Example
C and the uncontrollable
Controllable subspace
=
subspace is the range space of the control-lability
subspace is the null space of
Range (C) ,
Owing to duality, it is not surprising
Observable subspace
10.12, the controllable
Uncontrollable subspace
to have the following
= Range ?
CT,or
Null ?
CT?
(11.9)
deductions,
OT?
,
=
Unobservable subspace
=
Null (O)
(11.10)
as described in Remark 11.7.
11.2.2
State-Space
Models in
Controller
Form and
Observer
Form
Recallthat a given system can have manystate-space models depending on how the state variables are
defined. Usually physical variables like displacements, velocities, voltages, and currents are chosen as
state variables so that the state-space model can directly reflect the status of the physical variables of
interest. However, for computational reasons in the design process,the state-space model may need to
be transformed to a special form like a companion form or a diagonal form, discussed in Sections 6.6.2,
7.4.1, and 7.4.2.
The signal flow graph shown in Figure 11.2(a) is a state diagram of the following
???-a0-a1?????
??
x?1
x?2
=
0
x1
y = ? b0 b1?
x2
Note that
this state-space
state-space
1
x1
x2
0
1
u = Ax+Bu
(11.11)
=Cx
model is in companion
form,
modelin this form is always controllable.
model can be found
+
state-space model:
which is also called controller
form
since
Thetransfer function associated with the state-space
using either the state-space to transfer
function
formula,
402
11
Observer
Theory and
Output
Feedback
Control
? b0 b1 ?
G(s)=C(sI-A)-1B
=
? ??
?
s+a1
1
-a0s
0
1
s2 +a1s+a0
=
b1s+b0
s2 +a1s+a0
or Masons gain formula,
G(s) =
M1?1 +M2?2
b1s-1 +b0s-2
=
?
b1s+b0
=
1-(-a1s-1
-a0s-2)s2
+a1s+a0
Fig. 11.2: Duality between the controller form and observer form realizations.
Now let us take a look at the signal flow graph shown in Figure 11.2(b). Its shape is identical
to the one shown in Figure 11.2(a), but the signal flow directions are all reversed and the input u
and output y have exchanged positions. For ease of distinction, the state variables in Figure 11.2(b)
are marked with~x1 and~x2. These two signal flow graphs are dual to each other from a graphical
point of view. We would like to investigate the mathematical relationship between the two systems.
The state-space model associated with the state diagram in Figure 11.2(b) can befound in terms of
the state variables~x1 and~x2:
???01 -a0
?????
-a1
??
x?1
~
x~?2
x~1
x~2
=
x~1
y~ = ? 0 1 ?
x~2
Notice that the two
state-space
b0
+
b1
u =~A
~x+~B
~u
(11.12)
=~~
Cx
models have the following
relationship:
A~ = AT, B~ =CT, C~ = BT
Hence, the two state-space
models, Equations
11.11 and 11.12,
(11.13)
or the two
state diagrams in
Figure
11.2(a) and Figure 11.2(b), are dual to each other. Sincethe state-space modelin controller form is con-trollable,
its dualthe
referred
state-space
model in
Equation
11.12 or the state diagram in
Figure 11.2(b)is
asthe observer form.
The relationship
between the two state-space
models can be found in the following:
?T
CT=?
C(sI-A)-1B
?~B
-1
=BT?
sI-AT?-1
G~(s) =~C?sI-~A
= GT(s)
(11.14
11.3
For the SISO case, these two transfer
Minimal
functions
Realization
and Controllability
are identical,
but for
and
Observability
403
MIMO case, they in general are
different.
11.3
Minimal Realization and Controllability
A minimal state-space
model, also referred
or the dimension of the A matrix is
observable.
Theorem 11.8 (Minimal
and Observability
as minimal realization,
meansthat the number of state vari-ables
minimal. A minimal state-space
Realization and Controllability
modelis controllable
and
and Observability)
Giventhe state-space model (A,B,C),
x?(t)
= Ax(t)+Bu(t)
y(t)
=Cx(t)
the following three statements are equivalent:
(a) The state-space model (A,B,C) is controllable and observable.
(b) The state-space model (A,B,C) is a minimal realization.
(c) Thetransfer function N(s)D-1(s) or D-1(s)N(s) associated with(A,B,C) is irreducible.
Remark 11.9 (The Computation Involved in Obtaining an Irreducible
Transfer
and Numerically Unreliable for High-Order Large MIMO Systems)
Function Is Com-plicated
For SISOcases,the two fractional descriptions, N(s)D-1(s) and D-1(s)N(s), areidentical and can
be written as N(s)/D(s),
whichis irreducible if and only if N(s) and D(s) have no common polynomial
factors, or, equivalently, there is no possible pole-zero cancellation in N(s)/D(s).
However, for MIMO
cases, N(s)D-1(s) and D-1(s)N(s) are in general different, and they are called right MFD (matrix
fractional
description)
and left
MFD, respectively.
Anirreducible
left
or right
MFD can be obtained
by
removing the greatest common left or right divisor from N(s) and D(s), but the computation involved is
complicated and maybe numerically untrustworthy, especially for high-order, large MIMO systems. For
this reason, we will only employ the first two statements of Theorem 11.8 to determine if a given
MIMO state-space model is a minimal realizationin
other words, based on the controllability
and observability of the modelinstead of the irreducibility
of the MFD.
Example
11.10 (How
Controllability
and
Observability
Consider a system that is represented bythe following
Affect
Minimal
Realization)
state-space modelin diagonal form:
??? ?????
??
??=? ???+
??u
=Ax+Bu,
y= ??=Cx
x?1
x?2
x?3
?1 0 0
0 ?2 0
0 0 ?3
x1
b1
x2
b2
x3
b3
According to Theorem 11.8, the state-space
and observable. The controllability
x1
? c1 c2 c3 ?
x2
x3
model is a minimal realization
matrix of the system i
if and only if it is con-trollable
404
11
Observer
Theory and
Output
Feedback
Control
? ?
? 2b2
?
b1 ?1b1 ?2
1 b1
C
=
? B AB A2B? =
?
b2 ?2b2 ?2
?
b3 ?3b3 ?2
3 b3
The determinant of the controllability
matrix is
detC=b1b2b3
(?1-?2)
(?2-?3)
(?3-?1)
Hence,
thesystem
iscontrollable
if andonly
if detC
?=0,or,equivalently,
?1,?2,and?3are
distinct and b1, b2, and b3 are nonzero. Similarly,
we have
detO=c1c2c3
(?1-?2)
(?2-?3)
(?3-?1)
and so the system is observable if and only if ?1, ?2, and ?3 are distinct, and c1, c2, and c3 are nonzero.
Therefore, the state-space modelis a minimal realization if and only if ?1, ?2, and ?3 are distinct,
and b1, b2, b3, c1, c2, and c3 are nonzero.
These facts regarding the controllability, observability, and minimal realization
graphically in the following example using the state diagram.
Example
11.11 (Graphical
Interpretation
of Controllability,
Observability,
can also be inter-preted
Minimal
Realization)
The diagonal state-space model considered in the previous example can be graphically represented
by the state diagram shown in Figure 11.3.
Fig. 11.3: The state-space
model is a minimal realization
if and only if it is controllable
and observable.
It is clear that the system is composed of three subsystems in parallel connection, and the three
subsystems are decoupled. Thetransfer function of the system can be found as
Y(s)
U(s)
b1c1
= G(s) =
b2c2
+
b3c3
+
s-?1 s-?2 s-?3
(11.15
11.3
If
one of the input
branches,
Minimal
Realization
and Controllability
b1, b2, b3, is zero, say b1 = 0, then
and
Observability
the branch associated
405
with
b1 in the state diagram will disappear and the control input u will have no effect on the dynamics of
the subsystem with eigenvalue ?1; hence, the system becomes uncontrollable. Meanwhile, the transfer
function
of Equation
11.15
will become
G(s) =
b2c2
b3c3
+
s-?2 s-?3
which is a second-order system now, and therefore the third-order state-space modelis not a minimal
realization
anymore.
If one of the output
in the state diagram
branches, c1, c2, c3, is zero, say c2 = 0, then the branch associated with c2
will disappear and the state x2 will have no influence
on the output y, which
means
the state of the subsystem with eigenvalue ?2 becomes unobservable; hence,the system is unobservable.
Meanwhile, the transfer function of Equation 11.15 will become
G(s) =
b1c1
b3c3
+
s-?1 s-?3
which is a second-order system now, and sothe third-order state-space modelis not a minimal realiza-tion
any longer.
If the eigenvalues ?1, ?2, and ?3, are not distinct, say ?2 = ?1,then the top two subsystems in the
state diagram will collapse into one subsystem becausetheir eigenvalues are identical. Consequently, the
disappeared subsystem will become uncontrollable and unobservable. Meanwhile, the transfer function
will become
b1c1
b2c2
b3c3
b1c1 +b2c2
b3c3
G(s) =
+
+
=
+
s-?1 s-?1 s-?3
s-?1
s-?3
which is a second-order system now; hence, the third-order state-space modelis not a minimal realiza-tion.
Therefore,the state-space modelis a minimalrealization if and only if ?1,?2,and ?3 are distinct and
b1, b2, b3, c1, c2, and c3 are nonzero.
Remark 11.12 (Minimal
Realization and Pole-Zero
Cancellation)
Considerthe state-space model with the transfer function, Equation 11.15, shown in Example 11.11.
Let
?1=-1, ?2=-2, ?3=-3, b2=b3=c1=c2=c3=1
while keeping
b1 as afree parameter so that the transfer
Y(s)
U(s)
= G(s) =
b1
s+1
1
+
s+2
1
+
s+3
function
can be rewritten:
(b1 +2)s2 +(5b1 +7s)+(6b1
=
+5)
(s+1)(s+2)(s+3)
Then the system is controllable and observable and the associated state-space modelis a minimal
realization
if and
only
if b1
?=0orthe
transfer
function
G(s)
isirreducible,
which
means
that
there
isno
pole-zero cancellation in G(s). In casethat b1 = 0,the transfer function
2s2 +7s+5
G(s) =
Notice that the denominator
(s+1)(s+2)(s+3)
and numerator
function
will become
(s+1)(2s+5
(s+1)(s+2)(s+3)
have a common
function has a pole and a zero at the same location
the transfer
=
will become
factor
s + 1. In
other
words, the transfer
of complex plane. After the pole-zero cancellation,
406
11
Observer
Theory and
Output
Feedback
Control
G(s) =
which meansthat the transfer
function
2s+5
(s+2)(s+3)
before the pole-zero cancellation
was reducible.
Similarly, there will be a pole-zero cancellation at s = ?2 if c2 is zero. For another case that has
double eigenvalues at ?1 and the third eigenvalue at ?2,the pole-zero cancellation will occur at s = ?1.
ForSISO systems,theirreducibility ofthe transfer function N(s)D-1(s) can beeasily determinedby
inspecting if the denominator and numerator polynomials have common factors. However, for
systems, the denominator and numerator of the transfer function are matrices, and it
MIMO
usually
requires a moreinvolved procedure to determine if N(s)D-1(s) is irreducible.
11.4 State-Space Modelsand Minimal Realizations of MIMO Systems
One of the advantages
of the state-space
control
system
classical control is that they can be applied to
state-space
models are available.
knowledge
of assembling
regarding
Our emphasis
state-space
SISO systems
and
design approaches
up to now has been on SISO systems;
models, concept
certainly
analysis
over the
MIMO systems as easily as SISO systems if the
of controllability,
can be extended to
observability,
MIMO systems,
however, the
and
minimal re-alization
although there are some
differences.
In this section,
we will discuss how to assemble state-space
or interconnected
subsystems
with
multiple inputs
models based on transfer
and outputs.
The signal flow
function
ma-trices
graph shown in
Figure 11.4 represents a MIMO system with two inputs and two outputs consisting of four SISO sub-systems
G11(s),
The
G12(s), G21(s), and
G22(s).
MIMO system in Figure 11.4 can also be represented
? ??
y1(s)
y2(s)
11.4.1
Assume
Direct Realization
by the following
G11(s) G12(s)
=
G21(s) G22(s)
???
model for
each subsystem
function
matrix:
u1(s)
(11.16
u2(s)
Approach to Assemble a MIMO State-Space
we have a state-space
transfer
of the
Model
MIMO system in
Figure 11.4. Then
by stacking the state vectors of all the subsystems into one weintegrate the four subsystems into one
state-space model. The order of the overall model should be the sum of the orders of the four subsys-tems.
However, we will see that even though the subsystem models are all minimal realizations, the
overall modelis not necessarily a minimal realization.
Example 11.13 (Direct
Consider the
subsystems
Realization of a MIMO System)
MIMO system shown in Figure 11.4, where the transfer functions
of the four SISO
are given as
G11(s) =
-1 ,
s+3
G12(s) =
1
s+3
,
G21(s) =
1
s
,
G22(s) =
1
s(s+3)
Following the state-space model construction procedure in Section 6.6.2, each state-space modelof
the four
SISO subsystems
can be assembled
as follows:
11.4
State-Space
Models and
Minimal
Realizations
of
MIMO
Systems
407
-u1
s+3 ? (s+3)y11(s)
=-u1(s)? x?1=-3x1
y11 = x1
(i) G11(s) = -1
(ii)
G12(s) =
s+3
1
G21(s) = s
1
(iii)
? (s+3)y12(s)= u2(s) ?
? sy21(s) = u1(s)
y12 = x2
x?3 = u1
?
y21 = x3
???0 -3??
???
??
x?4
x?5
1
(iv)
x?2=-3x2
+u2
G22(s) = s(s+3)
? (s2 +3s)y22(s)= u2(s)
?
0
1
x4
=
0
+
x5
1
u2
x4
y22 = ? 1 0 ?
x5
Fig. 11.4: A MIMO system can be an interconnected system of several subsystems with multiple inputs
and outputs.
Withthese state-space
11.4,
models of the four
we have the following
state-space
SISO subsystems
model for the overall
and the interconnected
???0 -300 0 ???? ?
??
???0 0 00-3???? ?
??
??? ?
??
?
?
?
?
x?1
x?2
x?3
x?4
x?5
x1
-3 0 00 0
?
?
=
?
?
?
?
y1
y2
0
?
0
0
0 0 0
0
-10
x2
?
0 0 1
?
?
?
?
?
?
?
?
0 1
?
x3
x4
?
?
+
?
?
?
?
11000
00110
=
?
?
?
?
u1
?
1
?
0
?
u2
?
0
x5
x1
x2
x3
x4
x5
graph in Figure
MIMO system:
0
:=
Ax+Bu
?
0 1
(11.17
?
?
?
:=Cx
?
The poles of the MIMO model are simply the eigenvalues of the A matrix, while the zeros can be
computed based on the algorithm in Theorem 10.17. By the MATLAB codes:
% CSD
Ex11.13
A11=-3;
B=[-1
MIMO
A22=-3;
0;0
1;1
poles
A33=0;
0;0
0;0
zeros
A44=[0
1];
C=[1
1;0
1
-3];
0
0
A=blkdiag(A11,A22,A33,A44);
0;0
0
1
1
0];
D=[0
0;0
0];
408
11
Z=[A
Observer
-B;C
Theory and
-D];
Output
Feedback
E=blkdiag(eye(5),zeros(2,2));
poles_G=eig(A);
disp('Poles
poles_G',
Control
of
the
MIMO
state-space
model
are:'),
zeros_G=eig(Z,E);
disp('Zeros
of
the
MIMO
we will have the following
state-space
model
(ignore
Inf)
are:'),
zeros_G'
results,
>>
Poles
of
-3
the
MIMO
-3
Zeros
of
state-space
0
the
MIMO
-Inf
0
state-space
-4.0000
model
are:
model
(ignore
-3
-0.0000
Inf)
Inf
are:
Inf
Inf
-3.0000
Therefore,
theMIMO
state-space
model
has
fivepoles
ats =-3,-3,-3,0,0,
and
three
zeros
at
s =-4,-3,0.
Next, we will determine if the
state-space
MIMO state-space
model is a minimal realization
model is a minimal realization.
if and only if it is controllable
and observable
This MIMO
according
to
Theorem 11.8. In the casethat the modelis not minimal, the Kalman decomposition approach of Sec-tions
10.3.2 and 11.1.2
uncontrollable
will be employed
and unobservable
The controllability
to find
a minimal realization
of the system
by removing
the
modes.
matrix is
?0 10 -3 0 9 0 -270 81?
1 0 -30 9 0 -27
?00 00
10 -3 0 9 0 -270 81
-1 0 3 0 -9 0 27 0 -81 0
?
C = ? B AB A2B A3B A4B? =
?
1
?
0 0 0
0
0
0
0
0
0
?
?
?
?
?
and
rank
C=4 <5 ? system
isuncontrollable
The observability
matrix is
??
??
?
O =
?
C
CA
CA2
?
?
CA3
CA4
?
1
0 0
0
0
0
1 1
0
?
?
?
?
?
?
?
?
?
?
?
?
?
?
0
0
0 0
1
9
9
0 0
0
?
?
?
?
?
?
?
0 0 00 -3
-27 -2700 0
0
81
?0
?
?
?
-3 -3 00 0
?
?
=
1
0 00
81 0 0
9
0
?
?
?
?
?
?
?
?
0 00-27
and
rankO=3 <5 ? system
isunobservable
There are two unobservable modes and one uncontrollable
mode. We will first perform observ-ability
decomposition to remove the two unobservable modesso that the observable model will be
of third
order. The uncontrollable
mode may be overlapped
with the unobservable
ones. If that is the
case, the observable model will be also controllable, and, therefore the observable modelis a minimal
realization. Otherwise, controllability decomposition is neededto remove the uncontrollable mode.
11.4
Let U1 and U2 be
State-Space
Models and
Minimal
Realizations
of
MIMO
Systems
409
? ? ?10 -10
0 0?
0
1
-10
? ?,U=2
1 1 0 0 0
U1 =
0 0 1 1 0
00001
where the three row
vectors
of the
U1 matrix are any three linearly
independent
row
vectors of the
observability matrix O, or, in other words,the three column vectors of UT
1 span the three-dimensional
observable subspace or the range space of OT. Onthe other hand, the two row vectors ofU2 are chosen
so that the similarity
transformation
matrix
T-1 =
??
U1
U2
is nonsingular and U1UT
2 = 0.In other words,the column vectors of UT
2 form a basis of the null space
of O, which is the orthogonal complement ofthe observable subspace. This similarity transformation
x = T-1x
transforms
the state-space
x? =
modelinto the following
observability
decomposition form:
-30 0 0 0? ?
-1 1 ?
?
? ? ?? 00 00 -30 -30
00
?0 0 -1 0 0? ? ?
? ?
A11 0
0
B1
A21A22 x+
u =
B2
0 1
0
0
?
?
?
?
?
?
?
?
1
0
0
1
?
x+
?
?
?
?
-1 -1
1
y =?C1 0?x =
?
?
?
0
10000
x
01000
After deletingthe unobservablemodes,wehavethe third-order model(A11,B1,C1)for the
system,
which is observable.
minimal realization
Since this third-order
of the system.
observable
model is also controllable,
it is a
? ? ??
?0 0 -3?
xmin
+? ?u
??
x?min =A11xmin +B1u
=
-30 0
0
0
-11
1
1
0
0 1
y =C1xmin =
100
0 1 0
xmin
which
has
three
poles
ats =-3,0,-3
and
one
zero
ats =-4.These
three
poles
are
theeigenvalues
A11,
and
thezero
s =-4iscomputed
based
onthealgorithm
in Theorem
10.17.
of the matrix
In summary, the fifth-order
MIMO state-space model obtained by this direct realization ap-proach
is not a minimal realization even though the four individual subsystems state-space models
were all
minimal individually.
Apparently,
one of the two
Hence, after removing the unobservable
for the MIMO system.
11.4.2
MIMO State-Space
In addition
Modelsin Block Controller
to the direct realization
MIMO state-space
approach
unobservable
modes is also uncontrol-lable.
modes we obtained a third-order
and Block Observer Forms
discussed in Section
models. Recall that in Section 11.2.2,
11.4.1, there are
welearned
construction
procedures
can be extended to
many ways to con-struct
how to construct
modelin controller form or in observer form for a given SISOtransfer function.
and observer form
minimal realiza-tion
MIMO cases.
a state-space
These controller form
410
11
Block
Observer
Controller
Theory and
Output
Feedback
Control
Form
Consider
a MIMO
system
represented
bythep mtransfer
function
matrix
G(s)
shown
in Figure
11.5. Thetransfer function
description) as
matrix G(s) can be rewritten in the form of right
G(s) =
N(s)
d(s)
R (s)
=N(s)
D-1
where
MFD(matrix fractional
DR(s) = d(s)Im
(11.18)
Let ?(s) = DR(s)-1u(s). Then wehavethe following two equations:
DR(s)?(s)
= u(s)
(11.19a)
y(s)=N(s)
DR(s)-1u(s)
=N(s)?(s)
(11.19b)
Assume
d(s) = s2 +a1s+a0
and
N(s) = N1s+N0
Then Equations 11.19a and 11.19b will lead to the following:
DR(s)?
(s)=u(s) ? (s2+a1s+a0)?
(s)=u(s)?? =-a1
??-a0?
+u
y(s)=N(s)
?(s) ? y(s)=(N1s+N0)?
(s) ? y=N1
??+N0?
Fig. 11.5:
A MIMO transfer
function
Fig. 11.6:
A block controller
form realization
Based on Equations
function in right
11.20a and 11.20b,
in right
a block
MFD for block controller
in right
controller
of the
realization
(11.20b)
form realization.
of a MIMO transfer function
form
(11.20a)
MFD.
MIMO transfer
MFDis constructed and shown in Figure 11.6. Withthe state vectors x1 and x2 assigned
at the outputs of the integrators s-1Im, we havethe following
state-space model equations
11.4
State-Space
Models and
Minimal
Realizations
of
MIMO
Systems
??? ?????
??
x?1
x?2
0
=
x1
x2
-a0Im-a1Im
y = ? N0 N1 ?
Example 11.14 (Block
Im
Controller
x1
0
Im
+
411
u:= Ax+Bu
(11.21
:=Cx
x2
Form Realization of a MIMO System)
In this example, we will employ the same MIMO system considered in Example 11.13 to demon-strate
the block controller form realization approach. Thetransfer function matrix ofthe MIMO system
is rewritten in the following in the form of right MFD:
G(s) =
That is,
?-1/(s+3) 1/(s+3)??-s s ?? ?-1
:= N(s)
D-1
? ? ?-11?
1/s
a0 = 0,
Then according
to Equation
s+3
a1 = 3,
11.21,
0
s(s+3)
=
1/(s(s+3))
1
0
0 0
N0 =
31
we have a state-space
,
R (s)
s(s+3)
N1 =
1 0
model in block controller
form for the
MIMO
? ? ??
0
x?=?00-3
?u
:=Ax+Bu
00 0 -3?x+?
?00 -11?
system:
00 1
00 0
y =
0
1
?
?
?
?
?
?
31 1 0
x
0
0
1
0
0
0
0
1
?
?
=Cx
The
poles
oftheMIMO
model
are
theeigenvalues
oftheAmatrix,
which
are
s =-3,-3,0,0.
The
zeros
can
becomputed
based
onthealgorithm
in Theorem
10.17,
which
are-4,0.
Next, we will determine if the MIMO state-space modelis a minimal realization. This MIMO state-space
modelis a minimal realization if and only if it is controllable and observable according to Theorem
11.8. Since the state-space modelis in block controller form, whichis always controllable, only the ob-servability
of the model needsto be examined.
The observability
matrix is
?00 -1 1 ?
??
-9 9
O=? ?=00000027-27
0 -3
? ?
31 1
?
C
?
?
CA
?
?
CA2
?
0
00 3 -3
0 0 0
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
CA3
0 0 0
9
and
rankO=3 <4 ? system
isunobservable
There is one unobservable
mode. The Kalman decomposition approach of Section 11.1.2 will be em-ployed
to find a minimal realization
of the system
by removing
the unobservable
mode.
412
11
Observer
Theory and
Output
Feedback
Control
?00 -11? -1300
=
? ?,U2
Let U1 and U2 be
3 1
U1 =
1
0
?
?
00 0 1
where the three row
observability
observable
vectors
matrix
of the
U1 matrix are any three linearly
O, or, in other
words, the three column
subspace or the range space of
the similarity transformation
vectors
independent
of
row
vectors of the
UT
1 span the three-dimensional
OT. On the other hand, the row vector
ofU2 is chosen so that
??
matrix
U1
T-1 =
U2
2 = 0. In other words, the column vectors of UT
2 form a basis of the null space
is nonsingular and U1UT
of O, which is the orthogonal complement of the observable subspace. This similarity transform
x = T-1x
transforms the state-space modelinto the following
observability decomposition form:
-30 0 0? ?
-11?
?
? ? ?? 0 0-30?
u=? x+??
? ?
A11 0
x? =
x+
A21A22
B1
B2
0 0 1
0
?
?
?
?
?
?
1 0 2 0
1 0 0 0
y =?C1 0?x =
0 1 0 0
1
0
0
1
?
?
0 0
x
wherethethird-ordersubsystem(A11,B1,C1)is observable.Notethat thisthird-orderobservablemodel
is also controllable.
After removing the unobservable
system
of the
? ? ??
?0 0 -3?
xmin
+? ?u
??
x?min =A11xmin +B1u
y =C1xmin =
mode, we have a minimal realization
1 0 0
0 1 0
-30 0
=
-11
0 0 1
1 0
0 1
xmin
which
has
three
poles
ats =-3,0,-3
and
one
zero
ats =-4.These
three
poles
are
theeigenvalues
of the
matrix
A11,
and
the
zero
s =-4
iscomputed
based
on
the
algorithm
inTheorem
10.17.
In summary,
form
although the fourth-order
realization
approach.
approach
After removing
is still
the
not
MIMO state-space
model obtained
minimal, it has less order
unobservable
than that
mode, we have obtained
by the block con-troller
of the direct real-ization
a third-order
minimal
realization.
Block Observer Form
Consider
a MIMO
system
represented
bythepmtransfer
function
matrix
G(s)
shown
inFigure
11.7.
The transfer function
matrix G(s) can be rewritten
in the form
of left
MFD (matrix
fractional
description)
as
G(s) =
N(s)
d(s)
L (s)N(s) where
DL(s)
=d(s)Ip
= D-1
(11.22)
Recallthat the state-space modelin block controller form shown in Equation 11.21 was derived from
a MIMO transfer function
in right
MFD following
a procedure similar
to that of the SISO controller
form.
11.4
Now, to find the counterpart
left
State-Space
of the block
Models and
controller
form
Minimal
Realizations
associated
with a
of
MIMO
Systems
MIMO transfer
413
function
in
MFD, we wantto take advantage of the duality property instead of deriving it from scratch. In the
following,
we will use the short-hand
notation
?? ??? ?
0
G(s)
=N(s)
(d(s)Im)-1?
Im
,
-a0Im
-a1Im
0
, ? N0 N1?
Im
(11.23)
torepresent
therelationship
between
therightMFD,
N(s)(d(s)Im)-1,
and
thestate-space
model
in
block controller
form
shown in Equation
11.21.
Fig. 11.7: A MIMO transfer function in left
MFDfor block observer form realization.
Now,
thetranspose
ofG(s)
=(d(s)Ip)-1
N(s)is GT(s)
=NT(s)
(d(s)Ip)-1,
which
isintheform
of right
MFD. Hence, by using the relationship in Equation 11.23 we havethe following:
?? ??? ?
0
GT(s)
=NT(s)
(d(s)Ip)-1?
Then, taking
transpose
on both sides
Ip
0
,
-a0Ip-a1Ip
of the relationship
with
theleftMFD
G(s)
=(d(s)Ip)-1
N(s):
G(s)
=(d(s)Ip)-1
N(s) ?
Ip
, ? NT
0 NT
?
1
will give the block
observer form
0 -a0Ip
??
??? ?
Ip -a1Ip
,
N0
, ? 0 Ip ?
N1
(11.24)
associated
(11.25)
Therefore,
theMIMO
transfer
function
inleftMFD
(d(s)Ip)-1
N(s)has
astate-space
model
inblock
observer form
asthe following:
???Ip0 -a0Ip
?????
-a1Ip
??
x?1
x?2
x1
=
y = ? 0 Ip ?
x2
x1
x2
N0
+
N1
u = Ax+Bu
(11.26
=Cx
The state diagram associated with the state-space modelin block observer form is shown in Figure
11.8. Notice that the shape of the graphs in Figure 11.8 is the same as that of Figure 11.6, but the
directions
of the signal flow are reversed
and the positions
of the input
and the output have exchanged.
Thesetwo state diagrams are dual to each other.
Example 11.15 (Block
Observer Form Realization of a MIMO System)
414
11
Observer
Theory and
In this example,
Output
Feedback
we will employ the same
Control
MIMO system considered in Example
11.13 and Example
11.14 to demonstrate the block observer form realization approach. Thetransfer function
MIMO system is rewritten
G(s) =
in the following
in the form
?-1/(s+3) 1/(s+3)??
1/s
a0 = 0,
?-1
?-s s? L(s)N(s)
?? ? ?
0
That is,
a1 = 3,
N0 =
matrix of the
MFD:
0
s(s+3)
=
1/(s(s+3))
of left
s+3
s(s+3)
00
,
31
-11
N1 =
Then according to Equation 11.26, we have a state-space
MIMO system:
:= D-1
1
1 0
model in block observer form for the
? ?? ?
0 0
0
0
0
0
0 0
0
0
3
1
x?=? ?x+?
?u
:=Ax+Bu
?
?
y =
?
?
10 -3 0
01 0 -3
?
?
-11
?
?
1 0
? ?
0010
0001
x
=Cx
The
poles
oftheMIMO
model
are
theeigenvalues
oftheAmatrix,
which
are
s =-3,-3,0,0.
The
zeros
can
becomputed
based
onthealgorithm
in Theorem
10.17,
which
are-4,0.
Fig. 11.8: A block observer form realization of a MIMO transfer function in left
Next,
we will determine
if the
model is a minimal realization
MIMO state-space
model is a minimal realization.
if and only if it is controllable
11.8. Since the state-space modelis in block observer form,
of the model needsto be examined.
The controllability
matrixis
C
=?
?
0 00 0
3 10 0
MIMO state-space
which is always observable, only the con-trollability
0
0
0
0
0
0
0
0
?
?
This
and observable according to Theorem
BAB
A2B
A3B
=?1 00 1 0 -30 0
?
MFD.
?
?
-113 -3 -9 9 27-27
?
and
rank
C=3<4 ? system
isuncontrollable
There is one uncontrollable
mode. The Kalman decomposition approach of Section 10.3.2 will be em-ployed
to find a minimal realization
of the system
by removing
the uncontrollable
mode.
11.4
State-Space
Models and
Minimal
0 0
0
MIMO
Systems
415
1
10 0
?
?
?
?
?
?
0 0
where the three column
controllable
of
? ? ??
T1
=?13-3?,T2
=??
Let T1 and T2 be
controllability
Realizations
0
0
1
?
?
0
vectors of the T1 matrix are any three linearly
independent
column
vectors of the
matrix C, or, in other words,the three column vectors of T1 span the three-dimensional
subspace
or the range space of
C.
Onthe other hand, the column vector of T2is chosen so that the similarity transformation
matrix
T = ? T1 T2 ?
isnonsingular
andTTT2=0.Inother
words,
thecolumn
vectors
ofT2
form
abasis
ofthenull
space
1
of CT, whichis the orthogonal complement of the controllable subspace. This similarity transformation
x = T-1x
transforms
the state-space
model into the controllability
decomposition
form
?? ?
0 1/3
? ? ?? ?0-3
x+? ?u
u=?1 0 -3 0 ?
0 0
A11A12
x? =
x+
0 A22
0
0
3
?
?
?
?
?
?
0 0
y = ?C1 C2 ?x =
0
B1
0
?13 -30?
0 0
1
0
0
1
-1/30
1
0
0
0
?
?
x
where
thethird-ordersubsystem
(A11,B1,C1)is controllable.Notethatthis third-ordercontrollable
model is also observable.
After removing
the uncontrollable
mode, we have a minimal
realization
?0 -3 0 ? ? ?
?1 0 -3?
xmin
+? ?
?13 -3?
0
0
0
3
x?min =A11xmin +B1u =
0 0
1
1
-1/3 0
1
y =C1xmin =
of the system,
0
xmin
which
has
three
poles
ats =-3,0,-3
andone
zero
ats =-4.These
three
poles
are
theeigenvalues
A11,
and
thezero
s =-4iscomputed
based
on
thealgorithm
inTheorem
10.17.
of the matrix
In summary, although the fourth-order
MIMO state-space model obtained by this block ob-server
form realization approach is still not minimal, it has less order than that of the direct
realization approach. After removing the uncontrollable
mode, we have obtained a third-order
minimal realization.
11.4.3
MIMO State-Space
Gilbert diagonal realization
Modelsin Gilbert
is a natural
Diagonal Form
generalization
of the SISO diagonal realization.
apm MIMO
system
with
atransfer
function
matrix
Consider
416
11
Observer
Theory and
Output
Feedback
Control
G(s) =
N(s)
(11.27)
d(s)
where
d(s)=(s-?1)(s-?2)(s-??),??=?j,
i degN(s)
<?.LetG(s)
bewritten
as
R2
R1
G(s) =
+
s-?1
R?
s-?2+ +
where
s-??
Ri =lim
(s-?i)G(s)
(11.28)
s??i
and assume Rican be decomposed as
Ri=CiBiwhere
Ci?Rpri
, Bi?Rrimandri=rank
Ri
Then it is easy to verify that the state-space
model (A,B,C)
(11.29
with
?
? ??
, C= C1C2 C?
?00
0? ??
0 0 ??Ir?
?1Ir1 0 . . .
0 ?2Ir2
0
0
?
A =
...
?
?
B1
B2
?
?
,
?
B =
?
?
?
?
?
?
...
B?
(11.30)
?
?
?
is a state-space model of G(s), and the dimension of the state spaceis n = ? ri.
i=1
Example 11.16 (Gilbert
In this example,
we will employ the same
11.14, and Example
matrix of the
Diagonal Realization of a MIMO System)
11.15 to demonstrate
the
MIMO system considered
Gilbert diagonal
in
realization
Example
approach.
11.13,
Example
The transfer
func-tion
MIMO system is rewritten
G(s) =
?
? ? ?
? ???
?0 -1/3?
1/s
R1 = lim sG(s)
=
s?0
0
s?-3
model (A,B,C)
0
1 1/3
R2 = lim (s+3)G(s)
A =
=
1/(s(s+3))
where
Hence, the state-space
1
-1/(s+3) 1/(s+3)
=
-s s
s+3
s(s+3)
=
0
1
R1
=
1
s
R2
+
s+3
? 1 1/3 ?:=C1B1
-1 1
I2 :=C2B2
with
? ??? ?
?0 -3I2??
0 -3 0
?0 0 -3?,B= ? ?
0
0 0
0
0
B1
B2
=
C = ? C1 C2? =
=
1 ?
?01 -10 -1/3
1 1/3
1 0
0 1
isaGilbert
diagonal
realization,
which
has
three
poles
ats =-3,-3,
and
0a
, nd
one
zero
ats =-4.
The controllability
matrix is
?1 0 -3 0 90?
?0 1 0 -309?
1 1/3
C = ? B AB A2B? =
0
0 00
11.4
State-Space
Models and
Minimal
Realizations
of
MIMO
Systems
417
and
rank
C=3 ? system
iscontrollable
The observability
matrix is
O =
1 ?
?01 -10 -1/3
? ? 0 3 -3
? ?=0 -9 9
?0 0 -3
C
CA
CA2
?
?
?
?
?
?
0 0
?
1
?
?
?
?
?
and
rank
O=3 ? system
isobservable
Therefore, the
Gilbert diagonal realization is
Theorem 11.17 (Gilbert
Realization is
Gilbert realization is controllable
Proof:
Based on the
controllability
minimal.
Minimal)
and observable and therefore is a minimal realization.
Gilbert diagonal realization
defined by Equations 11.28, 11.29, and 11.30, the
matrix can be written as
?1B1?n-1?
?B1
B2
?2B2?n-1
C= BAB An-1B
=
?...
B???B??n-1?
Im?1Im?n-1?
0 0?
?B1
?
0 B2 0 Im?2Im ?n-1
:=BU
?00
...??Im?n-1?
0 0...0
B???
Im
B1
1
?
?
B2
2
?
?
?
?
...
?
...
?
B?
?
which can be decomposed
into a product
oftwo
matrices as
Im
1
C =
?
?
?
?
?
?
?
?
?
Im
2
?
?
...
...
?
Im
?
?
Note
thatthedimensions
ofBandUaren?mand
?m?m,
respectively,
where
n= ?ri.Itiseasy
i=1
to see that the rank ofU
is ?m; hence,
U is nonsingular.
By Sylvesters
inequality,
rank(B)+rank(U)-?m=
rank(C)
=min{rank(B),rank(U)}
we haverank(C)
= rank(B).
Bythe Gilbert construction,
?
rank(B)
1
Therefore, the realization is controllable.
?
?rank(Bi)
=
=?ri
=n
1
The observability can be proved similarly.
418
11
Remark
Observer
Theory and
11.18 (Comparison
So far in this
chapter,
Output
of
Feedback
MIMO
Control
State-Space
we have learned
a few
Model Construction
MIMO state-space
Approaches)
model construction
approaches to
obtain state-space modelsin direct realization, block controller form, block observer form, and Gilbert
diagonal form. The Gilbert diagonal realization is controllable and observable andtherefore is a minimal
realization. The restriction of this approach is that the least common denominator of all the entries
of the MIMO transfer function
matrix, like d(s) in Equation 11.27, is required to have no repeated
zeros
(i.e.,?i
?=?j).
The other approaches
do not have this restriction;
however, they
usually lead to nonminimal
realiza-tions.
The block controller form is always controllable, but usually unobservable with order equal to ?m,
where
mis the number
usually
uncontrollable
of inputs
and ? = degd(s).
The block
observer form is always
with order equal to ?p, where p is the number
observable
of outputs and ? = degd(s).
but
The
direct realization is usually uncontrollable and unobservable with order equal to the sum of the orders
of all SISO subsystems. For nonminimal realizations, the Kalman decomposition approach of Sec-tions
10.3.2 and 11.1.2 can be employed to find a minimal realization of the system by removing
the uncontrollable and unobservable modes.
There exist other approaches to construct
Kalman decomposition.
For example,
a
minimal realizations
MIMO transfer
function
directly
without the need of using the
can be represented
by a general right
MFD
asG(s)
=N(s)
D-1(s).Agreatest
common
right
divisor
R(s)
can
befound
and
extracted
from
1 (s),
D(s)=D1(s)R(s)
andN(s)=N1(s)R(s)
toreduce
G(s)
toan
irreducible
MFD,
G(s)=N1(s)
D-1
then a state-space modelconstructed based onthe irreducible MFD will be a minimal realization. Details
regarding the irreducible MFDand greatest common right divisors can be found in the book by [Kailath,
1980].
11.5 Full-Order
Observerand Output Feedback Control
Recall that we witnessedthe power of the state-space controller
state-feedback
employed
whenthe four
magic numbers in the
gain
matrix F = ? 139.31 24.452 30.138 30.67 ? of Equation 7.120 in Section 7.6.4 were
to successfully
stabilize a highly unstable cart-inverted
pendulum system. Later, in Sections
8.6.3, and 8.7, welearned more about the state-feedback control and utilized it to work together
with the pole placement approach, the root locus design, and the regulation/tracking
theory to
solve the lightly
damped system control
problem,
and achieve aircraft
altitude regulation
and flight
path
precision tracking. In Chapter 10, the linear quadratic performance optimization
with control-input
constraint wasincorporated in the state-feedback control system design process to achieve an optimal
control.
However,it is rare that all state variables are available and not contaminated in the measurementpro-cess.
The state feedback
concept
would be impractical
if not for the invention
of the ingenious
full-order
observer. The brilliant observer theory wasinvented to utilize not only the measured output but also
the actuation input, the noise model, and the system modelto determine an optimal estimate of all
state variables.
As mentioned earlier, there is a duality relationship
control
applied to the
they
theory.
The knowledge
design and implementation
work together
perfectly:
and tools
between the observer theory
for
of observers.
state-feedback
Furthermore,
the observer is responsible
control
as two
to provide
and the state-feedback
can immediately
subsystems
timely
be
of the con-troller,
accurate assess-ment
of the system states for the state-feedback controller to makethe right decision and achieve the
desired performance. In addition, we will see that the observer and the state-feedback controlle
11.5
can be designed
separately
without interfering
Full-Order
Observer and
Output
each other; this important
Feedback
Control
property is referred
419
as the
separation principle.
11.5.1
Brief Review of State-Feedback
Control
Fig. 11.9: A basic state-feedback control system.
In the following,
we will briefly review the structure of the state-feedback control system. Then we
will see the extension of it to the output feedback control system is fairly straightforward:
Simply
add a full-order
observer.
The block diagram shown in Figure 11.9 representing the basic state-feedback control system is
simple and can be easily implemented, yetit is effective if the state equation of the state-space model,
x? = Ax+Bu, is given and all the state variables are available for feedback. The control law is, u = Fx,
where Fis a constant gain matrix to be determined so that the closed-loop system is stable and has
a desired performance. By desired performance, it usually meansthat the specifications regarding the
following are satisfied: robust stability, steady-state error, transient response, control-input constraints,
and so on.
There are two main approaches available in designing the state-feedback gain matrix F. Oneis the
pole placement approach and the other is the linear quadratic optimization approach. The pole
placement approach can be effective if we know wherethe desired pole locations should be for the
closed-loop system. Although the system behavior and performance are mainly determined bythe pole
locations of the system, it is by no meansan easytask to find an optimal set of pole locations, especially
for high-order systems. Onthe other hand, for the linear quadratic optimization approach, the issue is in
the choice of the weighting matricesin the performance index integral. The guideline is to assign more
weight on the state variable
of interest,
but it
may not work if this state variable is implicitly
another state variable, which we wrongfully ignore.
matrix
need to
system
be properly
selected
according
affected
by
Therefore, the pole placement or the weighting
to the system
dynamics
to achieve
desired
closed-loop
performance.
Notethat the basic state-feedback control system structure shown in Figure 11.9 works for any fixed
operating condition control problem, like the cart-inverted pendulum control problem in Section 7.6 or
the simple inverted pendulum control problem considered in Section 10.5.3.
Onthe other hand, if the operating condition is required to follow a time-varying command or
a tracking signal, then the basic state-feedback control system structure needs to be modified to
include a tracking regulator like the one shown in Figure 10.8 of Section 10.2.4 for the pendulum
positioning
system
420
11
Observer
Theory and
11.5.2
Observer-Based
Fig. 11.10:
Output
Feedback
Control
Controller
A basic output feedback
control
system
with full-order
observer.
Forthe state-feedback control problem, all the state variables in x are assumed available for feedback.
However, in some practical applications, not all state variables are available for feedback. Instead, the
measuredoutput is given together withthe state equation asfollows:
x? = Ax+Bu
y
(11.31)
=Cx+Du
where
x?Rn,u
?Rm,y
?Rp.
Since only the measured output y(t), instead of the state vector, is available for feedback, a full-order
observer is constructed to generatex, which is an estimate of x, to serve as a substitute of x for
state-feedback
control.
That is,
(11.32)
u = F
x
Then, from Figure 11.10, it can be seen that the output feedback controller, also referred asthe observer-based
controller,
includes
the full-order
observer
and the state-feedback
gain F matrix inside the dash-line
enclosure.
Structure of the Full-Order
Observer
Fig. 11.11: Structure of the full-order
observer
11.5
The structure
full-order
observer
of the full-order
Full-Order
observer is shown
in
Observer and
Figure
Output
11.11.
Feedback
The dynamics
Control
421
equation
of the
observer, with u and y as the inputs andx asthe state vector of the observer, is given by the
equation:
x?=Ax+Bu+L(Cx+Du-y)
which can be rewritten as
x? =(A+LC)
x+ ? B+LD-L?
??
u
y
(11.33)
Notethat the state-space model,(A,B,C,D), the input u, andthe output y of the plant (i.e., the system to
be controlled), are employed to construct the observer. The only parameter matrix to be determined
in the design process is L, which is referred as the observer gain matrix. The observer itself is an
n-th order dynamic system represented by a state equation with state vectorx and input vector
stability
of the observer is determined
observer stable (i.e., the eigenvalues
of
by A+LC.
A+LC
??
u
y
The observer gain L can be chosen to
are on the left-hand
side of the complex
. The
makethe
plane) if and
only if (C,A) is detectable. The observer has to be stable since the steady-state reconstruction
error
e(t)=x(t)-x(t)can
only
bezero
iftheobserver
isstable,
asshown
inthefollowing
theorem.
Theorem 11.19 (Reconstruction
Error of the Full-Order
Observer)
Consider the plant of Equation 11.31 and its associated full-order
observer described by Equation
11.33.
Let
the
reconstruction
error
bee(t)=x(t)-x(t).Show
that
e(t)?0ast ?8forallinitial
error
e(0)
ifand
only ifthe
observer is stable, or the eigenvalues
ofA+LC
are on the left-hand
side ofthe complex
plane.
Proof:
Since
e?(t) =?x(t)?x(t)
=Ax+Bu-Ax-LCx-Bu-LDu+Ly
=Ax-Ax-LCx-LDu+L(Cx+Du)
we have the differential
equation,
e? (t)
=(A+LC)e(t)
whichyieldsthe solution e(t) = e(A+LC)te(0).
11.5.3
Now,
in
Design of
Observer-Based
we can combine the three
Equation
Output
Feedback
parts of the
11.31, the state-space
Controller
closed-loop
model of the full-order
system: the state-space
observer in
Equation
model of the
plant
11.33, and the state-feedback
controller in Equation 11.32 together and show them onthe same graph in Figure 11.12. The
plant
model is in red,
to generatex,
whose input
u and output y are fed into the full-order
observer,
which is in black,
which is an estimate of the state x. Then the state vector of the observer,x is fed back to
the state-feedback
gain
matrix block
F, which is in blue, to generate the control-input
vector
u to close
the loop. Notice that the observer-based output feedback controller is the combination of the full-order
observer and the state-feedback controller; hence,the state-space model of the observer-based
output feedback
controller
can be derived from the block diagram
Obviously, the output of the controller
observer
also serves as the state vector
of Figure 11.12.
is u and the input to the controller
of the output
feedback
controller.
is y. The state vector
of the
From the block diagram
of
the closed-loop system in Figure 11.12, we havethe state equation of the observer andthe state-feedback
equation in the following
422
11
Observer
Theory and
Output
Feedback
Control
x?=Ax+Bu+L(Cx+Du-y)
u = F
x
Replacing the control input uin the state equation by Fx so that the state equation only has y asits input,
then these two equations become
x?=(A+BF+LC+LDF)
x-Ly:= Ak
x+Bky
(11.34)
u = F
x :=Ck x
This is the state-space
model of the observer-based
output feedback controller,
(Ak,Bk,Ck),
where
Ak=A+BF+LC+LDF,
Bk=-L, Ck=F
Fig. 11.12: State-space
Theorem 11.20 (Regulator
model of the observer-based
output feedback
controller.
Poles and Observer Poles)
For the closed-loop system with the observer-based controller shown in Figure 11.12, the closed-loop
system poles are the regulator poles (the eigenvalues ofA+BF)
together withthe observer poles
(the eigenvalues ofA+LC).
Proof:
Fromthe state-space model ofthe plant in Equation 11.31 andthe state-space model ofthe observer-based
controller in Equation 11.34, we havethe state equation ofthe closed-loop system as
x? = Ax+Bu
= Ax+BF x
x? =(A+BF+LC+LDF)
x-Ly =(A+BF+LC+LDF)
x-L(Cx+DF
x)
whichis rewritten in matrix form as
???
x?
x?
A
=
BF
??
?
x
x
-LCA+BF+LC
11.5
Hence, the closed-loop
Observer and
Output
Feedback
Control
423
system poles are the roots of
det
Since
Full-Order
?
?sI-A
LCsI-(A+BF+LC)
-BF
=
sI-A
-BF ?
????sI-(A+BF)sI-(A+LC)
? ??
LCsI-(A+BF+LC)
I
-BF
0
-I I
we have
det
I 0
=
I I
0
? (sI-(A+BF))det(sI-(A+LC))
?sI-A
LCsI-(A+BF+LC)=det
-BF
Therefore, the closed-loop system poles are the eigenvalues of A+BF together
of A+LC.
with the eigenvalues
Remark 11.21 (Design of the Output Feedback Controller)
Owing to Theorem 11.20 and the separation principle, the regulator poles
can be independently chosen to meetthe closed-loop performance requirement.
in designing a state-space output feedback controller are to find the regulator
observer gain matrix L so that the regulator poles (the eigenvalues of A+BF)
(the eigenvalues
of
A+LC)
matrices can be obtained
are at the
using the
desired locations
pole placement
on the left
approach
and the observer poles
Therefore, the key steps
gain matrix F and the
and the observer poles
half complex
given in
plane.
Sections
These two gain
7.5 and 7.6.4 or the
performance index optimization approach discussed in Sections 10.5.1 and 10.5.2.
Example 11.22 (An
Observer-Based
Controller
Design Using Pole Placement)
Considerthe system
x?(t)
y(t)
To find the regulator
weshould have
= Ax(t)+Bu(t)
=Cx(t)
that gives
?
|?I-(A+BF)| =
?
?
?
?? ??
00
10
x(t)+
1
0
u(t)
(11.35)
= ? 1 1 ? x(t)
gain matrix F = ? f1 f2 ?
sothattheeigenvalues
ofA+BF
areat-1 j,
? ??? ? ?
??-f1 -f2??
?=?2
-f1?
-f2=?2
+2?
+2
=(?+1)2
+1
A+BF
Hence,
=
=
00
1 0
+
1
0
? f1 f2 ? =
f1 f2
1 0
?
-1
?
?
f1 =-2 andf2 =-2
vNext,
wewould
liketofindtheobserver
gainmatrix
Lsothattheeigenvalues
ofA+LC
areat
- 3 j. LetL=[?1?2]T,
then
424
11
Observer
Theory and
Output
Control
? ??? ? ?
?-?1 -?1??
|?I-(A+CL)|= ?
?=?2
-(?1
+?2)?
+?1(2?2
-1)
=?2
+2
A+LC
Thus,
Feedback
00
10
=
+
?1
?1
?1 1 ? =
?2
?1
1+?2 ?2
v3?
?
?
?
-1-?2? -?2
?
+4
?
?
which gives
?1=-4 and?2=0.5359
Now, according to Equation 11.34 the observer-based output feedback controller is
x? = Ak
x+Bky
(11.36
u =Ck
x
where
Ak = A+BF+LC
=
?
? ?, Ck=F= -2 -2
?
-6
4
-6
1.5359 0.5359 , Bk=-L=
?
-0.5359
Now the plant model, Equation 11.35, andthe observer-based controller,
construct a closed-loop system with the following state equation:
?
Equation 11.36, together
?
??? ?? ?? ??
??? ??
?
-6-6?
=?
x?
Ax+Bu
x?
=
Ax+BCk x
Ak
x+BkCx
=
Ak
x+Bky
0
0
4
0
1
4
?
?
A
=
-2
-2
0
BCk
x
BkC Ak
0
?
?
x
x
:=
x
ACL
x
x
-0.5359
-0.5359
1.5359
0.5359
v
The
eigenvalues
ofACL,
the
closed-loop
system
poles,
can
be
f
ound
as-1
j
and
3 j,
which are the eigenvalues of A+BF together with the eigenvalues of A+LC, asshown in Theorem
11.20.
The numerical
% CSD
%
Ex11.22
Find
A=[0
results
F
0;1
can be verified
Observer-based
to
0];
place
the
B=[1;0];
F=place(A,B,P);
by running
control
regulator
C=[1
the following
Pole
poles,
1];
MATLAB
placement
eigenvalues
of
A+BF
P=[-1+i,-1-i];
F=-F,
eig_ABF=eig(A+B*F),
%
Find
At=A';
L
to
Bt=C';
place
Ct=B';
Ft=place(At,Bt,Pt);
the
observer
poles,
eigenvalues
of
Pt=[-sqrt(3)+i,-sqrt(3)-i];
Ft=-Ft;
L=Ft',
eig_ALC=eig(A+L*C),
%
Find
the
Ak=A+B*F+L*C,
eig_cl=eig(A_cl)
closed-loop
Bk=-L,
system
Ck=F,
poles
A_cl=[A
B*Ck;Bk*C
Ak],
A+LC
code:
11.6
LQG
Control
Problem and the
H2 Control
Theory
425
11.6 LQG Control Problem and the H2 Control Theory
The H2 control
problem formulation
considers
a more general set of the
H2 norm optimization
problems
than the LQG (linear quadratic Gaussian) control problem. We will start from a brief description of the
LQG control problem, of whichthe solution requires an observer to provide an optimal estimate of all
state variables
and an optimal
state-feedback
control
to
minimize
a performance
index.
Then we will
describe the standard H2control problem formulation and solutions according to the celebrated Doyle,
Glover, Khargonekar, and Francis paper [Doyle et al., 1989].
11.6.1
The LQG
Control
Problem
Considerthe state-space model of the system P,
x?(t) = Ax(t)+Bu(t)+Wdd(t)
(11.37)
P:
y(t)
=Cx(t)+Wnn(t)
where
A,Wd
? Rnn,B?Rnm,
C?Rpn,Wn
?Rpp,and
thevariables
x(t),y(t),u(t),d(t),and
n(t)
are the state vector,
respectively.
measured output,
The pairs (A,B)
input
d(t)
and the
and (C,A)
control
input,
disturbance
input,
and
measurement
are stabilizable and the detectable, respectively.
measurement
noise n(t)
are assumed
white noises
noise,
The dis-turbance
with the covariances
E[d(t)dT(t +t)] =Ind(t) and E[n(t)nT(t +t)] =Ipd(t), respectively.
The objective
of the LQG (Linear
Quadratic
Gaussian) control
problem is to design an output feed-back
controller (Ak,Bk,Ck),
x?K(t) = AKxK(t)+BKy(t)
K:
(11.38)
u(t)
=CKxK(t)
so that the closed-loop system is stable and the performance index
PI = E
??8
?
u Wuu(t) ? dt
? xT(t)WT x Wxx(t)+uT(t)WT
0
?
?
(11.39)
is minimized,
where
E[] stands
fortheexpected
value.
Next,
we will describe the standard
H2 control
problem
formulation
and solutions.
Then
we will
rephrase the LQG control problem in the standard H2 control problem format so that the standard H2
control solution formula can be applied to the LQG control problem.
11.6.2
The Standard
H2 Control Problem
The standard H2control problem formulation
starts from a generalized plant G with state-space model
given by the equation,
x?(t) = Ax(t)+B1w(t)+B2u(t)
G:
z(t)
=C1x(t)+
y(t)
=C2x(t)+D21w(t)
which can be rewritten in matrix form as follows:
? ??
? ?=?
x?(t)
G:
?
z(t)
y(t)
A B1 B2
?
?
C1 0
(11.40)
+D12u(t)
D12
C2 D21 0
?? ?
???
x(t)
?
?
w(t)
u(t)
?
(11.41
426
11
Observer
Theory and
Output
Feedback
Control
This state-space model hastwo output vectors: the measuredoutput vector y(t), which consists of
all the measurable signals available for feedback, and the controlled output vector z(t), which is com-posed
of tracking
error, disturbance response
and control-input
costs. The generalized
input vectors: the control-input vector u(t) as usual andthe exogenous input
disturbance input, measurement noise, and reference input/command.
plant also has two
w(t) that mayinclude the
Thematrices
A?Rnn,
B1?Rnm1
, B2?Rnm2
, C1?Rp1n,
C2?Rp2n,
D12
?Rp1m2
, D21
?
Rp2m1are
assumed to satisfy the following
(i)
(ii)
Fig. 11.13:
conditions:
(A, B2) is stabilizable
and (C2, A) is detectable.
DT
C1 D12? = ? 0 I ?
12 ?
Closed-loop
system
and
? ? ??
B1
DT
21 =
D21
with generalized
plant
(11.42)
0
I
G and controller
K.
The objective of the H2control problem is to find a controller
x?K(t)
= AKxK(t)+BKy(t)
K:
(11.43)
u(t)
=CKxK(t)
so that the closed-loop system shown in Figure 11.13is internally stable and the H2 norm of the closed-loop
system
transfer
function
fromwtoz,?Tzw?2,
is minimized.
Now, before weget into the solution of the problem we need to digress a little bit to review some
background materials,like how to find a state-space modelfor the closed-loop system transfer function
Tzw,the physical meaning of the H2 norm, and the relevant computation algorithms. We will organize
these review materialsin the form of exercises, definitions, theorems, and remarks.
Exercise 11.23 (The
Closed-Loop
Since the generalized
plant
Transfer
G has two input
be broken down into four subsystems,
Function
vectors,
Tzwin the
w and u, and two
y
= G
w
u
Shown in
=
G11 G12
w
G21 G22
u
Figure
11.13)
output vectors, z and y, it can
G11, G12, G21, and G22. Thus,
?? ??? ??
?
z
System
z = G11w+G12u
?
y = G21w+G22
11.6
With the feedback
z(s)
Proof:
=
control,
u
LQG
=Ky, the closed-loop
Control
transfer
Problem and the
function
from
H2 Control
Theory
427
wto z can be found as follows:
?G11(s)+G12(s)K(s)(I-G22(s)K(s))-1G21(s)
?w(s):=Tzw(s)w(s)
(11.44
Left as an exercise.
Exercise 11.24 (The State-Space
Model of Tzwin the System Shown in Figure 11.13)
For ease of computation, we will usea state-space model ofTzwinstead of the s-domain transfer func-tion
obtained in Exercise 11.23. The state-space model ofTzw can be found by combining the generalized
plant modelin Equation 11.40 and the controller modelin Equation 11.43 asfollows:
???
x?
x?K
=
B2CK
x
AK
xK
z = ? C1 D12CK?
Proof:
???? ?
??
A
BKC2
B1
+
BKD21
w
(11.45)
x
xK
Left as an exercise.
Remark 11.25 (H2 Norm of a SISO System P(s))
Assume Z(s)
= P(s)V(s),
where the output z(t), input
v(t),
and the impulse
response
p(t)
are the
inverse Laplace transforms of Z(s), V(s), and P(s), respectively. The definition of the square ofthe H2
norm of P(s) is defined as
1
?8
2:=
?P?2
|P(j?)|2d?
2p
(11.46)
-8
which by Parsevals theorem,
wehave
?8
2:= |p(t)|2dt
?P?2
2 =?p?2
(11.47)
0
2 =
Assume
theinput
isaunit
impulse
function
v(t)=d(t);then,
Z(s)=P(s),
which
implies
?P?2
2.
Therefore,
the
H2
norm
of
the
system
is
the
square
root
of
the
total
energy
of
the
?p?2
2 =?z?2
impulse response of the system, or is the square root of the output energy of the system driven by
a unit impulse
total
input.
For
MIMO systems, the square of the
output energies at the outputs
Definition 11.26 (Controllability
of the system
and Observability
For a stable system G(s) =(A,B,C),
Lc =
?8
respectively.
function
at all inputs.
Grammians)
the controllability
eAtBBTeATtdt and
0
H2 norm of the system is the sum of the
driven by unit impulse
and observability grammians are defined as
Lo =
?8
eATtCTCeAtdt
0
428
11
Theorem
Observer
Theory and
11.27 (Computation
The controllability
Output
Feedback
Control
of Controllability
and
Observability
Grammians)
?8
grammian Lc = eAtBBTeATtdt satisfies the Lyapunov equation
0
ALc +LcAT +BBT = 0
?8
The observability grammian Lo = eATtCTCeAtdt satisfies the Lyapunov equation
0
ATLo +LoA+CTC
Theorem
11.28 (Computation
For a stable system
=0
of H2 Norm)
G(s) = (A,B,C),
the
H2 norm ofthe system is
?
?
?G?2
= trace(CLcCT)
= trace(BTLoB).
11.6.3 Solutions to the Standard
H2 Control Problem
The key steps to obtain the solution
gain
for the standard
matrix F and the optimal
The optimal state-feedback
H2 control
observer gain
problem
are to find the optimal
gain matrix Fis
2X
F =-BT
where Xis the stabilizing
solution
of the algebraic
(11.48)
Riccati equation [Zhou
et al., 1995,
2 X+CT1 C1 = 0
ATX+XA-XB2BT
with the corresponding
observer
gain
matrix
-B2BT
?-CTA1C1-AT
?
2
solution
(11.50)
(11.51)
2
of the algebraic
Riccati equation
2 C2Y+B1BT1
AY+YAT
-YCT
with the corresponding
(11.49)
L is
L =-YCT
where Y is the stabilizing
Kailath, 1980],
Hamiltonian matrix [Zhou et al., 1995, Kailath, 1980],
H =
The optimal
state-feedback
matrix L.
=0
(11.52)
Hamiltonian matrix
J =
?-B1BT-CT-A ?
AT
2 C2
1
(11.53
11.6
Then the optimal
H2 controller
LQG
Control
Problem and the
H2 Control
Theory
429
is
x?K(t)=(A+B2F+LC2)xK(t)-Ly(t)
Kopt(s) :
u(t)
(11.54)
= FxK(t)
and the minimum H2 norm is
?FGf
?2 ?C1Gf
?2
2 +
2 +
min?Tzw?2
2 =?GcB1?2
2 =?GcL?2
?
where
?
?
Gc = (A+B2F,
I, C1 +D12F)
Gf = (A+LC2,
B1 +L2D21,I)
(11.55)
?
2
(11.56)
11.6.4
Application
In the following,
the
of the Standard
H2 Control Formula
H2 control solution
formulas
shown in the previous subsection,
Section 11.6.3,
will
be employed to solve the LQG control problem described in Section 11.6.1.
Recallthat the LQG control problem is to minimize the performance index of Equation 11.39 and one
ofthemain
objectives
oftheH2
control
problem
isto minimize
?Tzw?2,
theH2
norm
otfheclosed-loop
system transfer
function
Tzw. To equalize these two
output of the
? ?? ?
plant has to be defined as
z(t)
measures, the controlled
Wxx(t)
=
:=
Wuu(t)
e(t)
generalized
(11.57)
v(t)
where e(t) =Wxx(t) and v(t)
=Wuu(t) represent the weighted error and the weighted control input
constraint. The weighting matrices here play the same roles as Q =WT
x Wxand R =WT
u Wuin the LQR
control problem westudied in Chapter 10. Meanwhile, the disturbance input d(t) and the measurement
noise n(t) are grouped together asthe exogenous input vector:
w(t) :=
??
d(t)
(11.58)
n(t)
Now, with the definition of z and w, and together with the LQG plant model P given in Equation
11.37, we havethe following generalized plant modelfor the LQG problem:
?? ????? ????? ??
Wu
?? ?
? ??
x?(t)
GLQG
:
?
?
?
e(t)
v(t)
?
?
?
=
?
?
?
y(t)
A
? Wd 0 ?
B
Wx
0
00
00
0
C
? 0 Wn?
0
x(t)
?
?
?
?
?
?
d(t)
n(t)
?
?
(11.59)
?
u(t)
To satisfy the generalized plant assumptions in Equation 11.42, werescale the control input u(t) and
the noiseinput n(t),
u(t)?u(t) =Wuu(t),
n(t)?n(t) =Wnn(t)
(11.60)
so that the generalized plant becomes
?? ????? ????? ??
? ??
??
n(t)
?
x?(t)
GLQG
:
?
?
?
e(t)
v(t)
y(t)
A
?
?
?
?
=
?
?
? Wd0 ? BW-1
u
Wx
0
00
00
0
I
C
?0 I ?
0
x(t)
d(t)
?
?
?
?
?
?
?
?
u(t)
?
(11.61
430
11
Observer
Theory and
Output
Feedback
Control
According to the standard H2control notations, we havethe following
matricesfor the generalized
plant:
A = A, B1 =? Wd0?,
C1 =
?? ??
Wx
,
0
C2 =C,
0
I
D12 =
(11.62)
D21 = ? 0 I ?
Now, weare ready to work on the following
Example 11.29(LQG/H2
B2 = BW-1
u
demonstration example.
Output Feedback Controller for the Simple Inverted Pendulum)
InSection
4.4.3,
astate-feedback
controller
u =Fx=[-34 -7.4]
x was
designed
based
onthe
pole
placement
approach
toplace
theclosed-loop
system
poles
at-4 j3sothat
thedamping
ratio
and
the natural frequency are ? = 0.8 and ?n = 5rad/s, respectively. The same simple inverted pendulum
system wasemployed again in Section 10.5.3 to demonstrate how to formulate a state-feedback control
problem as an LQR optimization problem, especially on the selection of the Qand R weighting matrices
in the performance index.
In this LQG/H2 control problemformulation, the selection ofWx(part ofC1in the generalizedplant)
andWu (part of D12before changing u(t) to u(t)) is equivalent to that of Q and Rsince Q =WT
x Wxand
R = WT
u Wu. For ease of comparison, we will employ the same simple inverted pendulum dynamics
modellinearized
model for design and nonlinear modelfor simulationto
validate the output feed-back
LQG/H2 control system design.
The linearized
state-space
model of the nonlinear
10.43 atthe unstableequilibrium x* =[0
x?(t) = Ax(t)+Bu(t)+Wdd(t)
y(t)
=Cx(t)+Wnn(t)
simple inverted
=
dynamics
?9 -0.6? ??
0
1
x(t)+
of Equation
0
1
u(t)+Wdd(t)
(11.63
= ? 1 0 ? x(t)+Wnn(t)
Noticethat the output y(t) consists of only one state variable, x1(t)
the pendulum stick, contaminated
pendulum
0]Tis
= ?(t), the angular displacement of
with the measurement noise n(t).
Owing to the separation principle, the designs of the state feedback andthe observer do not interfere
with each other. Sincein Example 10.25 wehad already designedthe state-feedback part using the LQR
approach, whichis actually half ofthe LQG/H2 control design, we willselectthe equivalent weighting
matrices in the performance
index
as
Wx =
Notice that the
For the
WxandWu
weighting
1000
0
are equivalent
matrices Wd and
very similar to that betweenWx
the trade-off
?v ?
andWn.
to the
0
v20
and
Wu =
Q and R selected in
Wnin the observer
v1.2
Example
10.25.
design, the trade-off
Recall that only the relative
between
weights are important
Wd and
Wnis
in determining
of the weighting matrices,so we will fixWn atWn = 1 and havethree cases of design based
on the selection
of the
Case A:
weighting
Wd =
matrixWd:
??
100
,
0
Case B:
Wd =
??
10
,
0
Case C:
Wd =
??
1
0
(11.64)
11.6
Now, with these
A =
matrices in the generalized
?9 -0.6?
v
?
???
C1 =
0
1
Wx
0
,
=
0
0
v20
0
0
we can solve the two algebraic
Control
Problem and the
H2 Control
Theory
431
plant,
??v?
????
? ??
=C
=
?,D12
=??,C2
B1 = ? Wd 0 ? =
1000
LQG
100
0
0
0
,
B2 = BW-1
u
0
0
1
=
? 1 0 ?,
0
1
1.2
D21 = ? 0 1 ?
Riccati equations, compute the state-feedback
the observer gain matrix L, and complete the LQG/H2 controller
(11.65)
gain matrix F and
design.
The stabilizing solution of the state-feedback Riccati equation
2 X+CT1 C1 =
ATX+XA-XB2BT
is
X =
?
361.08
47.086
47.086
11.007
?
and the state-feedback gain is
?
F =-BT
2X=? -42.983
-10.048
Notethat this F matrix hereis a factorWu different from the F obtained in Example 10.25. Actually the
design is exactly identical;
regulator
the difference
poles (i.e., the eigenvalues
is the
of A+B2F),
mathematical
scaling
defined
by Equation
11.60.
The
are
-4.8863j2.5224
which are exactly the same asthose obtained in Example 10.25.
For the observer design in
Case A,the stabilizing solution of the observer Riccati equation
2 C2Y+B1BT1
AY+YAT
-YCT
=0
?
L =-YCT? ?
is
Y =
and the observer gain is
?
100.10 9.6136
9.6136 67.186
2 =
-100.10
-9.6136
The observer poles (i.e., the eigenvalues of A+LC2) are
-100.09and -0.60617
The Simulink
simulation
diagram is shown in Figure 11.14,
in Figure 10.19 except the controller
now is an output feedback
which is almost the same as that shown
controller
instead
of the state feedback
one.
Three LQG/H2 controllers are designed according to different selections of the weighting matrices
Wdand Wn. The simulation results of Case A with Wd = [100 0]T, Wn = 1 are shown in blue on
Figure 11.15. The blue initial state responses, x1(t), x2(t), and u(t), are almost as good as their
432
11
counterparts
Observer
shown
Case A LQG/H2
Theory and
in
Figure
controller
Output
Feedback
10.22,
Control
which used the same state-feedback
places the two regulator
gain
matrix
poles, eigenvalues of A+B2F,
F. Thus, the
at exactly the
same
locations:
-4.8863
j2.5224.
The
onlydifference
isthattheLQG/H2
controller
adds
two
observer
poles,
-100.09
and-0.60617,
tothesystem
astheresult
ofselecting
weighting
matrix
Wd =[100 0]T, Wn = 1in Case A observer design. For Case A design, the addition of the observer
poles has only slightly reduced the performance compared with the perfect state feedback case.
Fig. 11.14: The diagram of the Simulink program, SIP SSNLmodel.mdl,
usedin the simulation.
Fig. 11.15: Simulationresults ofthe three casesLQG/H2 controller performance.
For
Case B design, the
design
of the state-feedback
part remains
the same, still
placing
the
A+B2F
eigenvalues
atthesame
-4.8863
j2.5224.
Buttheobserver
weighting
matrices
are
changed to Wd = [1 0]T, Wn = 1, which leads the observer poles (i.e., the eigenvalues of A+LC2),
to -10.833
and0.99845.
The
initialstate
response
associated
withCase
Bsdesign
isalso
shown
bit worse than that of Case A
in Figure 11.15, but in red. The response of Case B design is alittle
design
11.6
In Case C design, the weighting
LQG
Control
matrixWd is further
Problem and the
reduced toWd
H2 Control
=[1
Theory
433
0]T, Wn = 1. The reg-ulator
parts
are
stillunchanged,
but
thetwoobserver
poles
have
moved
tonew
locations:
-3.6322
and-2.4833.
The
initial
state
response
associated
with
Case
Cdesign
isalso
shown
in Figure
11.15,
but in brown.
It is obvious that
From the simulation
larger than
results,
Wn,the control
effect reduction.
the brown
response is not as good as the other two.
we can see that
design
when the disturbance
will emphasize
disturbance
weight
Wdis chosen to be much
response recovery
Onthe other hand,if the noise reduction is
more than the noise
more urgent, the designer would increase
Wnor reduceWd.
The MATLAB codes usedto conduct the computation
% CSD_ex11p29.m
% This
May
program,
% automatically
% and
10,
2020
to
to
plot
1;
9
0;
0
-0.6],
SIP_SSNLmodel.mdl
SIP_plot3A.m,
x10=x10_deg*d2r,
B=[0;
1],
x20=0,
eig(A),
C=[1
Wn=1,
D12=[zeros(2,1);1],
F=-B2'*X,
SIP_plot3B.m,
results.
0],
B2=B*inv(Wu),
D21=[0
1],
eig_ABF=eig(A+B2*F),
A
Wd=[100;
0],
B1=[Wd
Ak=A+B2*F+L*C2,
zeros(2,1)],
Bk=-L,
Bcl=[B1;Bk*D21],
Ck=F,
Ccl=[C1
Y=are(A',C2'*C2,B1*B1'),
L=-Y*C2',
eig_ALC=eig(A+L*C2),
D12*Ck],
sys_cl=ss(Acl,Bcl,Ccl,Dcl),
%
file:
call
Wu=sqrt(1.2),
C2=C,
X=are(A,B2*B2',C1'*C1),
Design
simulation
sqrt(20)],
C1=[Wx;zeros(1,2)],
Simulink
and
x10_deg=15,
A=[0
Wx=[sqrt(1000)
the
are given in the following:
Cntrl
Call
simulations
r2d=1/d2r;
x0=[x10;x20],
H2
will
conduct
SIP_plot3C.m,
d2r=pi/180;
%%
SIP
CSD_ex11p29.m,
and simulation
Acl=[A
B2*Ck;Bk*C2
Ak],
Dcl=zeros(3,2),
damp(sys_cl),
eig_Acl=eig(Acl),
Simulation
sim_time=2,
sim_options=simset('SrcWorkspace',
'current',
open('SIP_SSNLmodel'),
'DstWorkspace',
sim('SIP_SSNLmodel',
[0,
'current');
sim_time],
sim_options);
run('SIP_plot3A')
%%
Design
B
Wd=[10;
0],
B1=[Wd
Ak=A+B2*F+L*C2,
zeros(2,1)],
Bk=-L,
Bcl=[B1;Bk*D21],
Y=are(A',C2'*C2,B1*B1'),
Ck=F,
Ccl=[C1
D12*Ck],
sys_cl=ss(Acl,Bcl,Ccl,Dcl),
L=-Y*C2',
eig_ALC=eig(A+L*C2),
Acl=[A
B2*Ck;Bk*C2
Ak],
Dcl=zeros(3,2),
damp(sys_cl),
eig_Acl=eig(Acl),
% Simulation
sim_time=2,
sim_options=simset('SrcWorkspace',
'current',
open('SIP_SSNLmodel'),
'DstWorkspace',
sim('SIP_SSNLmodel',
[0,
'current');
sim_time],
sim_options);
run('SIP_plot3B')
%%
Design
Wd=[1;
C
0],
B1=[Wd
Ak=A+B2*F+L*C2,
zeros(2,1)],
Bk=-L,
Bcl=[B1;Bk*D21],
Ccl=[C1
sys_cl=ss(Acl,Bcl,Ccl,Dcl),
Y=are(A',C2'*C2,B1*B1'),
Ck=F,
L=-Y*C2',
eig_ALC=eig(A+L*C2),
D12*Ck],
Acl=[A
B2*Ck;Bk*C2
Ak],
Dcl=zeros(3,2),
damp(sys_cl),
eig_Acl=eig(Acl),
% Simulation
sim_time=2,
sim_options=simset('SrcWorkspace',
open('SIP_SSNLmodel'),
'current',
sim('SIP_SSNLmodel',
'DstWorkspace',
[0,
'current');
sim_time],
sim_options);
run('SIP_plot3C')
where SIP plot3A.m
different colors.
%Filename:
SIP_plot3A
is given below, andits copies SIP plot3B.m,
SIP
plot3C.m
will print in
434
11
Observer
figure(10),
grid
Theory and
Output
Feedback
subplot(1,2,1),
minor,
plot(t,x1,'b-',t,x2,'b--'),
title('x1
and
plot(t,cntrl,'b-'),
Control
grid
x2'),
on,
hold
grid
grid
on,
on,
subplot(1,2,2),
minor,
title('cntrl'),
hold
on
11.7 Exercise Problems
P11.1: Recallthat the observable subspace is the range space of OT,the transpose of the observability
matrix, while the unobservable
Ois the orthogonal
complement
subspace is the null space ofO, the observability
of the range space ofOT. In this exercise,
matrix. The null space of
we will identify
the observable
and the unobservable subspaces of a state-space model, and usethe basis vectors of these subspaces to
construct a similarity transformation to transform the modelinto an observability decomposition form.
The state-space model of the system to be considered is given as,
?-8 -9 -12? ??
?-1 -1 -3 ?x(t)+
??u(t
1
x?(t)
= Ax(t)+Bu(t)
=
6
7
12
0
0
y(t)
=Cx(t)+Du(t)
P11.1a: Find the observability
= ? 4 6 8 ? x(t)+u(t)
matrix O,and determine if the system is observable.
P11.1b: Find the observable subspace of the system, which is the range space of OT,the transpose of
the observability
matrix.
P11.1c: Find the unobservable subspace of the system, which is the null space of O,the observability
matrix.
P11.1d: Usethe procedure shown in Theorem 11.5 to transform the state-space modelinto the observ-ability
decomposition form.
P11.1e: Is the system detectable? Explain.
P11.1f: Deletethe unobservable modeof the state-space modelin observability decomposition form to
obtainanobservable
second-order
subsystem
(A11,B1,C1,D).Isthesubsystem
controllable?
P11.1g:Showthatthe observable
second-order
subsystem(A11,B1,C1,D)is controllable.
P11.1h:
Isthesystem(A11,B1,C1,D)
in P11.1g
a minimal
realization?
Explain.
P11.1i:Findthe polesofthe system(A11,B1,C1,D).
P11.1j:Compute
thezerosofthesystem(A11,B1,C1,D)usingTheorem
10.17.
P11.1k:
Find
the
transfer
function
Gmin(s)
=C1(sIA11)
B1+D,
and
verify
the
zeros
ofthe
system
obtained in P11.1j.
P11.2: Consider a MIMO system described by the following transfer function
matrix:
11.7
G(s) =
?
?
Problems
435
?
?
s
-s
Exercise
(s-1)(s+3)2
(s-1)(s+3)2
s+2
1
(s-1)(s+3)2
(s-1)(s+3)2
P11.2a: Find a state-space modelin block observer form for the
MIMO system.
P11.2b: Find the poles and zeros of the system using the state-space modelobtained in P11.2a.
P11.2c: Is the state-space model controllable? If it is not controllable, find a similarity transformation to
transform it into the controllability decomposition form, andthen check if the system is stabilizable.
P11.2d: If the transformed state-space modelis stabilizable, delete the uncontrollable
the controllable subsystem.
modesto obtain
P11.2e: Show that the controllable subsystem is observable, and, therefore, the controllable subsystem
obtained in P11.2dis a minimal realization.
P11.2f:
Find the poles and zeros of the
minimal realization.
P11.2g: Comment on the difference between the results obtained in P11.2b and P11.2f.
P11.3: Consider a third-order
SISO system described by the following state-space model,
0 -1 0 ? ? ?
??? ????? ?
?
?
?101 -2? ? ?u
?? ??
x?1
x?2
=
A11 A12
x1
A21 A22
x2
x1
y = ? C1 C2?
B1
+
= ? 1 0 0?
x2
u =
B2
-3 -2 0
x1
x2
+
1
1
-2
(11.66
x1
x2
where
x1isa21 vector.
Note
that
thestate-space
model
isinobservability
decomposition
form.
P11.3a:
Show that the system is not observable,
P11.3b:
Show that the subsystem
but it is detectable.
x?1 = A11x1 +B1u =
y
=C1x1
?0 -1? ??
-3 -2
x1 +
1
u
1
(11.67)
= ? 1 0 ? x1
is a minimal realization of the system.
P11.3c: Find a state-feedback gain matrix F1 so that the eigenvalues of A11 +B1F1(i.e., the regulator
poles)
are-1 j.
P11.3d:
Find an observer
gain
matrix L1 so that the eigenvalues
of A11 +L1C1 (i.e., the observer poles)
are-1.414
j.
P11.3e:
Construct the observer-based
output feedback
controller,
as shown in Figure 11.12.
P11.3f: Show that the observer-based output feedback controller can be represented by the following
state-space
model:
436
11
Observer
Theory and
Output
Feedback
Control
x?K(t)=AKxK(t)+BKy(t)
=(A11
+B1F1
+L1C1)xK(t)-L1y(y)(11.68)
u(t)
=CKxK(t)
= F1xK(t)
P11.3g: Combine Equation 11.66, the state-space model of the plant, together with Equation 11.68,
the state-space model of the observer-based output feedback controller, to form the state-space model
of the closed-loop system. Then compute the eigenvalues of the closed-loop system, which should in-clude
the regulator poles(i.e., the eigenvalues of A11 +B1F1),the observer poles (i.e., the eigenvalues of
A11 +L1C1), and the unobservable pole of the plant (i.e., the eigenvalues of the matrix A22).
P11.4: In this exercise problem, we will revisit the lightly damped pendulum positioning system consid-ered
in Section 10.2.2. The LQR approach wasemployed to design a state-feedback controller in P10.7,
an exercise
problem in
Chapter 10, to
minimize
a performance
index.
Since not all the state variables
can be measured without contamination in practice, the more versatile LQG/H2
approach is employed
here to address the issue.
The linearized
state-space
output feedback design
model of the lightly
damped
pendulum positioning system is described by
x?(t) = Ax(t)+Bu(t)+Wdd(t)
y(t)
=Cx(t)+Wnn(t)
?
0
=
1
? ??
x(t)+
-21.32-0.135
0
u(t)+Wdd(t)
30
(11.69
= ? 1 0 ? x(t)+Wnn(t)
Notice that the output y(t) consists of only one state variable x1(t) = ?(t), the angular displacement
of the pendulum, contaminated with the measurement noise n(t). As discussed in Section 11.6.4, the
generalized plant can be set up as shown in Equations 11