Basisc of Automation and Control
Lecture 1 - Basic concepts / Control system design
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basisc of Automation and Control
Programme
Information about lectures and laboratories
Automation Systems, lecture - Jakub Możaryn, PhD, Eng.,
Faculty of Mechatronics, room. 341, e-mail:
jakub.mozaryn@pw.edu.pl, webpage: http://jakubmozaryn.esy.es
Automation Systems, laboratories - Jakub Możaryn, PhD, Eng.,
Faculty of Mechatronics, room. 341, e-mail:
jakub.mozaryn@pw.edu.pl, webpage: http://jakubmozaryn.esy.es
Jakub Możaryn
Basisc of Automation and Control
Programe
Lecture information
Lecture: 20 hours
Laboratories: 10 hours
Work at home: 20 hours
Exam preparation: 10 hours
Conditions to pass the lecture: pass the writing exam and
attend all laboratories
Jakub Możaryn
Basisc of Automation and Control
Literature
R.C. Dorf, R.H.Bishop, Modern Control Systems, Prentice Hall,
2008
G.F. Franklin, J.D. Powell, A. Emami-Naeini, Feedback Control of
Dynamic Systems, Addison-Wesley, 1994
N.S. Nise, Control Systems Engineering, Wiley, 2015
Massive open on-line courses (MOOC): Coursera, EdX, Iversity
Jakub Możaryn
Basisc of Automation and Control
Introduction
Today, many devices are equipped with what is generally called automation. Ranging from household appliances like iron (temperature controller), washing-machine (programmer) to devices with the most advanced
technology as the aircraft (autopilot).
One of the first controllers, which has been applied in practice was Watt’s
centrifugal governor to stabilize the rotation of the steam engine (1784).
Since then, automation has become a type of science, and the number
of its practical application is constantly growing. The theory of automatic
control now includes:
theory of linear systems,
feedback control,
theory of nonlinear systems,
optimal control,
theory of discrete systems (logical automation systems),
robotics.
Jakub Możaryn
Basisc of Automation and Control
Steam engine
Figure: Steam engine with the centrifugal speed governor
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Basisc of Automation and Control
Centrifugal speed governor
The valve starts fully open at zero speed, but as the balls rotate and rise
on the rods. The central valve stem is forced downward and closes the
valve.
Figure: The detailed view of the centrifugal speed governor
Jakub Możaryn
Basisc of Automation and Control
Aims of the lecture
Aims of the lecture
Acquiring the ability to recognize and assess the problems of
automation and control.
Understanding the basic concepts of automation of different
processes, methods to determine the nature and elements of
automation with continuous and discrete action.
Understanding the basic principles of operation of control systems
and functions of the elements of these systems.
Understanding the requirements for control systems and methods of
ensuring the fulfillment of these requirements.
Jakub Możaryn
Basisc of Automation and Control
List of lectures
Introduction: basic concepts, classification of control systems of
continuous processes, examples.
Description methods of dynamical systems: differential equations,
transfer functions, time and frequency domains.
Static and dynamic features of the basic elements of the automation
system.
Block diagrams.
Process / plant - model identification.
PID controllers.
Stability of control systems, stability criteria.
Static and dynamic indexes of control quality, selection of
parameters of controllers.
Design of the control system.
Complex control systems.
Jakub Możaryn
Basisc of Automation and Control
Natural and technological processes
Natural processes
Physical and chemical transformations of the state of matter that takes
place without human intervention. Examples: weather changes, water
movement in rivers, tectonic movements, chemical processes in the
human body (eg changes in the level of insulin and glucose).
Technological processes
Processes carried out by a man with the use of appropriate devices
constructed by him in order to obtain the intended changes in the
state of matter. Example: changing the temperature in the furnace,
changing the water level in tanks in petrochemical installations.
During the lecture, issues related to technological processes and
their control will be discussed.
Jakub Możaryn
Basisc of Automation and Control
Basic concepts
Figure: Block diagram of the process
Control System
Control System is an interconnection of components constituting a
system configuration that shall provide the desired system response
(behavior).
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Basisc of Automation and Control
Basic concepts
Open-loop control system
Jakub Możaryn
Basisc of Automation and Control
Basic concepts
Open-loop control system
Examples:
toaster,
cofee vending machine.
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Signals
yr (t) - desired output
response,
u(t) - control signal,
x(t) - input signal,
y (t) - output signal.
Basisc of Automation and Control
Example - filling the glass with water
Figure: Example of the process control - the control of the level in the tank
(glass).
Jakub Możaryn
Basisc of Automation and Control
Example - filling the glass with water
Aim: Fill half of the glass with
water.
Figure: Example of the process
control - the control of the
level in the tank (glass).
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Function:
sensors - eyes, force estimation
(weight of glass),
actuators - hand, valve,
controller - brain.
Basisc of Automation and Control
Basic concepts
Closed-loop control system
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Basisc of Automation and Control
Basic concepts
Closed-loop control system
Examples:
The temperature control in
the greenhouse.
The water level control in
the tank.
The autopilot.
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Signals
u(t) - control signal,
x(t) - input signal,
y (t) - output signal,
yr (t) - desired output
response,
ym (t) - measured output,
e(t) = yr (t) − ym (t) - error
signal.
Basisc of Automation and Control
Basic concepts
Closed-loop feedback control system
The closed-loop feedback control system tends to maintain a
prescribed relationship of one system variable (measured process output
value) to another variable (desired process output response) by
comparing functions of these variables and using their difference as a
means of control.
Error signal e(t) is amplified.
The controller causes the actuator to modulate the process in order
to reduce the error e(t).
Feedback concept
The closed-loop control system uses a measurement of the output and
feedback of this signal to compare it with the desired output value
(reference value or command).
Jakub Możaryn
Basisc of Automation and Control
Basic concepts
Closed-loop control system
Elements
C - controller, A - actuator, O - process/plant, S - sensor
Main Path indicates always an essential input value of the system and
output value. This path typically describes the main flow of material or
energy in the system.
Feedback Path is used to transmit information. Energy requirements of
this path are usually small in comparison with the main path, and can be
omitted.
Jakub Możaryn
Basisc of Automation and Control
Basic concepts
Closed-loop control system - disturbances and noise
Advantages of the closed-loop control system over the open-loop
control system
Rejection of external disturbances d(t),
Improvement of the measurement noise n(t) attenuation.
Jakub Możaryn
Basisc of Automation and Control
Basic concepts
Closed-loop control system - simplified scheme
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Basisc of Automation and Control
Basic concepts
Multi-loop Feedback Control System
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Basisc of Automation and Control
Basic concepts
Single Input Single Output (SISO) system
Multiple Input Multiple Output (MIMO) system
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Basisc of Automation and Control
Control System Design
R.C. Dorf, R.H. Bishop, Modern Control Systems, Prentice Hall, 2009
Jakub Możaryn
Basisc of Automation and Control
Basics of Automation and Control
Lecture 2 - Mathematical Models of Dynamical Systems
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Mathematical modelling for control systems
Real processes, and thus control systems, have nonlinear properties:
turbulences,
multiple stable states,
hysteresis,
energy losses due to friction.
In practice, to simplify the mathematical description, there is carried linearization, enabling the formulation of the approximate description of a
linear phenomenon, in vicinity of the operating point (this point corresponds to the nominal or average operating conditions of the system).
Linearization steps
1
2
3
description of the phenomenon in the form of differential equations,
linearization,
operational calculus: differential equations → algebraic equations.
Jakub Możaryn
Basics of Automation and Control
6 step approach to modelling
STEP 1: Define the system and its components.
STEP 2: Formulate the mathematical model and fundamental
necessary assumptions based on basic physical principles.
STEP 3: Obtain differential equations representing the
mathematical model.
STEP 4: Solve equations for the desired output variables.
STEP 5: Examine the assumptions and solutions.
STEP 6: If necessary, reconsider and redesign the system.
Jakub Możaryn
Basics of Automation and Control
Description of linear models
The basic forms of mathematical description of the system dynamical properties are:
Equations of Motion: equations of system dynamics in form of
differential equations.
Transfer function.
State Space Equations (not covered in the course).
In the case of dynamical system (process) with one input signal x(t) and
one output signal y (t) equation of motion describes the relationship
between the output signal y (t) and the input signal x(t) in a following
form:
y (t) = f (x(t)) = f (x, t) = f (x)
(1)
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Basics of Automation and Control
Description of linear models / systems
Principle of superposition:
f (x1 + x2 ) = f (x1 ) + f (x2 ), and f (0) = 0.
(2)
Space of solutions of the equation that satisfies (2) is a linear space.
Homogeneity (implies scale invariance):
Function f (x, y ) is said to be homogeneous of degree k if
f (βx, βy ) = β k f (x, y ), and f (0) = 0,
where: β - constatnt coefficient.
Linear system
Homogenous system, which preserve the principle of superposition.
Nonlinear system
The system, which does not preserve the principle of superposition
and/or is not homogenous.
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Basics of Automation and Control
(3)
Description of linear models / systems
General form of the differential equation describing linear system:
an
d ny
d n−1 y
d mx
d m−1 x
+an−1 n−1 +· · ·+a0 y = bm m +bm−1 m−1 +· · ·+b0 x (4)
n
dt
dt
dt
dt
where: y - output signal, x - input signal, ai , bi - constant coefficients.
Jakub Możaryn
Basics of Automation and Control
Proportional elements
Dynamics equation relationship between input and
output signal:
Figure 1: Proportional element voltage divider
Input signal x(t) - voltage
U1 (t).
Output signal y (t) - voltage
U2 (t).
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U2 (t) =
R2
U1 (t)
R1 + R2
(5)
General equation of proportional
element
y (t) = kx(t)
Basics of Automation and Control
(6)
First order lag elements
Dynamics equation relationship between input and
output signal:
L dU2 (t)
+ U2 (t) = U1 (t) (7)
R dt
Figure 2: First order lag element RL filter
Input signal x(t) - voltage
U1 (t).
Output signal y (t) - voltage
U2 (t).
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General equation of first order
lag element
T
dy (t)
+ y (t) = kx(t)
dt
Basics of Automation and Control
(8)
Static characteristic
Static characteristics
Static characteristic fs describes the
dependence of the output signal y of
the system from the input signal x
in steady state.
Steady state
Steady state is a state in which all
derivatives of the input signal
and output signal are equal to
zero. In such a situation the output
signal has a steady value.
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Figure 3: Static characteristics
of linear system.
Basics of Automation and Control
Linearization
Creation of linear description of the system, based on nonlinear description
of this system is called linearization.
Linearization of nonlinear description in the form of nonlinear algebraic
equations is called static linearization. (There are no derivatives)
Linearization of nonlinear description in the form of nonlinear differential
equations is called the dynamic linearization.
Methods of static linearization
tangent method: obtain the best relation between the linear and
nonlinear description of a system for a given value of the
independent variable (input), and hence a particular value of the
dependent variable (output).
secant method: obtain the best relation between the linear and
nonlinear description of a system in the specified range of
changes of the independent variable (input).
Jakub Możaryn
Basics of Automation and Control
Static linearization
Figure 4: Static linearization; a) secant method, b) tabgent method.
In control system design, there is considered the behavior of plant/system
in a vicinity of a specified operating point. Therefore in practical applications tangent linearization method is much useful.
Jakub Możaryn
Basics of Automation and Control
Tangent method
Process of linearization using tangent method involves:
replacement of the curve representing nonlinear relationship
y = f (x) with its tangent at operating point,
transfer the origin to operating point,
replacement in mathematical model absolute variables x and y
with deviations of these variables from operating point incremental variables ∆x and ∆y .
Static characteristics obtained using linearized equation, in terms of
the specified operating point, is a linear function. It can be also
obtained by linearization of real characteristics in terms of the same
operating point.
Jakub Możaryn
Basics of Automation and Control
Dynamic linearization
An example of differential equation, which describes linear relationship
between functions x(t), y (t) and their derivatives.
F [y (t), ẏ (t), ÿ (t), . . . , y (n) (t), x, ẋ(t), ẍ(t), . . . , x (m) (t)] = 0
(9)
During dynamic linearization, functions x(t), y (t) and their derivatives
are treated analogously to variables of implicit function.
)
(
)
(
m
n
X
X
∂F
∂F
(i)
(j)
∆y
+
∆x
=0
(10)
∂y (i) y (i)
∂x (j) x (j)
j=0
i=0
0
0
where:
∆y = y (t) − y0 , ∆y (1) =
d∆y
d n ∆y
, . . . , ∆y (n) =
dt
dt n
∆x = x(t) − x0 , ∆x (1) =
d∆x
d m ∆x
, . . . , ∆x (m) =
dt
dt m
Jakub Możaryn
Basics of Automation and Control
Laplace transform
Replacing differential equation with transfer function (algebraic equation)
needs transition from the time domain (t) to the complex plane (s).
f (t) ⇔ f (s), where s = c + jω
(11)
where: c - real part coefficient, ω - conjugate part coefficient.
Laplace transform
Z∞
f (s) = L[f (t)] =
f (t)e −st dt
(12)
0
Inverse Laplace transform - Riemann-Mellin integral
−1
f (t) = L
1
[f (s)] =
2πj
c+jω
Z
F (s)e st ds
c−jω
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Basics of Automation and Control
(13)
Properties of Laplace transform
Linearity property
L [f1 (t) + f2 (t)] = L [f1 (t)] + L [f2 (t)]
(14)
L[k · f (t)] = k · L[f (t)]
(15)
L−1 [F1 (s) + F2 (s)] = L−1 [F1 (s)] + L−1 [F2 (s)]
(16)
L
−1
[k · F1 (s)] = k · L
−1
[F (s)]
Transform of derivative
n
0
X
d f (t)
n
=
s
·
F
(s)
−
s i · f n−1−i (0+)
L
dt n
i=n−1
where: f (n−1) =
d n−1 f (t)
.
dt n−1
In special case of zero initial conditions
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P0
i=n−1
s i · f n−1−i (0+) = 0.
Basics of Automation and Control
(17)
(18)
Properties of Lapace transform
Transform of integral
Z t
F (s)
1
L
f (τ )dτ = · L[f (t)] =
s
s
0
(19)
Shift in the complex plane (frequency shift)
L [e −α·t · f (t)] = F (s + α)
L [e +α·t · f (t)] = F (s − α)
(20)
Shift in the real plane (time shift)
L [f (t − τ )] = e −τ s · L[f (t)] = e −τ s · F (s)
L [f (t + τ )] = e +τ s · L[f (t)] = e +τ s · F (s)
(21)
Initial and final value theorems
if there exists limt→0+ f (t) = f (0+ )
lim f (t) = f (0+ ) = lim s · F (s)
t→0+
s→∞
(22)
if there exists limt→∞ f (t) = f (∞), to
lim f (t) = f (∞) = lim s · F (s)
t→∞
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s→0
Basics of Automation and Control
(23)
Laplace transform of the linear systems
Laplace transform is used for an analysis of control systems. As a tool
for graphical analysis, complex plane S is used, where multiplication by
s has the effect of differentiation and division by s has the effect of
integration.
Analysis of complex roots of a linear equation, may disclose information
about the frequency characteristics and the stability of the system.
To determine the function’s Laplace transform the following conditions
must be met:
f (t) has a finite value in any finite interval,
f (t) has a derivative
df (t)
dt
in any finite interval,
there exists a set of real numbers X for which the integral
R∞
0
absolutely convergent.
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Basics of Automation and Control
e −ct is
Laplace transform of the linear systems
Linear system is described by following differential equation
an
d n−1 y
d mx
d m−1 x
d ny
+a
+·
·
·+a
y
=
b
+b
+· · ·+b0 x (24)
n−1
0
m
m−1
dt n
dt n−1
dt m
dt m−1
Using the n-th derivative property of the Laplace transform
n d y
= s n y (s) − s n−1 y (0+ ) − · · · − y n−1 (0+ )
L
dt n
and assuming that initial conditions are zero, one obtains
n d y
= s n y (s)
L
dt n
(25)
(26)
Laplace transform of the linear dynamic system (22) with zero initial conditions take the following form
y (s)(an s n +an−1 s n−1 +· · ·+a0 ) = x(s)(bm s m +bm−1 s m−1 +· · ·+b0 ) (27)
Jakub Możaryn
Basics of Automation and Control
Transfer function
Transfer function
For continuous-time input signal x(t) and output y (t), the transfer function G (s) is the linear mapping of the Laplace transform of the input,
X (s) = L[x(t)], to the Laplace transform of the output Y (s) = L[y (t)] at
zero initial conditions:
y (s)(an s n +an−1 s n−1 +· · ·+a0 ) = x(s)(bm s m +bm−1 s m−1 +· · ·+b0 ) (28)
G (s) =
bm s m + bm−1 s m−1 + · · · + b0
y (s)
=
x(s)
an s n + an−1 s n−1 + · · · + a0
(29)
Numerator
M(s) = bm s m + bm−1 s m−1 + · · · + b0
(30)
Denominator - characteristic equation
N(s) = an s n + an−1 s n−1 + · · · + a0
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Basics of Automation and Control
(31)
Determination of static characteristics from transfer
function
x0 = lim x(t),
t→∞
y0 = lim y (t),
t→∞
(32)
using the final value theorem
y0 = lim y (t) = lim sy (s) = lim sG (s)x(s)
t→∞
s→0
s→0
(33)
For the input signal in the form of the unit step
x0 = const ⇒ x(s) =
1
x0
s
y0
= lim G (s).
s→0
x0
(34)
(35)
Finally, the static characteristcs has a form
y0 =
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b0
x0
a0
Basics of Automation and Control
(36)
Methods for determining the transient response of the
system
an
d ny
d n−1 y
d mx
d m−1 x
+an−1 n−1 +· · ·+a0 y = bm m +bm−1 m−1 +· · ·+b0 x (37)
n
dt
dt
dt
dt
Classic:
Assumption of the initial conditions x(0), y (0).
Solution of differential equations.
Using transfer function:
f (t) = L−1 [y (s)] = L−1 [G (s)x(s)]
(38)
To perform Laplace transform and its reverse, which are the basic operations of a transfer function calculus, it is often sufficient to know basic
properties of transfer fuctions and tables of transfer fuctions.
Jakub Możaryn
Basics of Automation and Control
Typical input signals
Step with constant value
x(t) =
for t ≥ 0
for t < 0
xst 1(t)
0
x(s) = xst
1
s
Impulse - Dirac delta function
x(t) = δ(t) =
0 for t ̸= 0
∞ for t = 0
x(s) = 1
Ramp
x(t) = at
x(s) =
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Basics of Automation and Control
a
s2
System properties
Changes of output signal y (t) as a response to a specific change of an
input signal x(t)
Figure 5: Example of a transient response of the dynamical system
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Basics of Automation and Control
Table of transfer functions
...
Figure 6: Table of transfer functions
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Basics of Automation and Control
Basics of Automation and Control
Lecture 3 - Frequency response methods / Basic linear dynamical
elements
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Frequency response
Frequency response methods
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Basics of Automation and Control
Frequency response
Frequency response
The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal.
The sinusoidal signal is a unique input signal, and a resulting output signal
for linear time-invariant (LTI) system, as well as signals throughout the
system, is sinusoidal in the steady-state; it differs from the input waveform
only in amplitude and phase angle.
In the analysis of linear systems, frequency response methods are used
to examine the stability, bandwidth, and robustness of the LTI system.
Jakub Możaryn
Basics of Automation and Control
Frequency response
Figure 1: Creation of the frequency response
u(t) = A1 sin[ωt]
(1)
y (t) = A2 sin[ω(t − tφ )]
(2)
ω=
2π
Tosc
(3)
where: Ai , i = 1, 2 - the amplitude, ω - the angular frequency (constant
for input and output signals in case of LTI systems), tφ - the delay between
the output signal phase and the input signal phase, Tosc - the period of
oscillations.
tφ < 0 - the negative phase shift,
tφ > 0 - the positive phase shift.
Jakub Możaryn
Basics of Automation and Control
Frequency response
There are defined following quantities (as a function of a frequency):
Magnitude: an amplitude ratio of the output signal (response) to
the input signal (cause),
Phase angle shift: between the output signal and input signal
There are widely used following plots of the system in frequency domain:
an amplitude-phase polar plot - Nyquist plot,
amplitude and phase logarithmic plots - Bode plots (diagrams).
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Basics of Automation and Control
Frequency response
Figure 2: Input signal
Figure 3: Output signal (negative phase shift)
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Basics of Automation and Control
Frequency response
A phase shift of an output signal relative to an input signal can be expressed
as a shift in time by tφ - output signal is described by the following
function
y (t) = A2 sin[ω(t − tφ )]
The phase shift of the output signal relative to the input signal can be
expressed as a shift angle
φ(ω) = ωtφ
Therefore
y (t) = A2 sin[ωt − φ]
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Basics of Automation and Control
Frequency response
To describe systems, where the signals are in the form of sinusoidal functions, there is used transform in frequency domain G (jω).
The transform in frequency domain is connected with Fourier transform
which assigns the transform F (jω) in the frequency domain to the function in time domain f (t) according to the following equation (also called
Fourier Integral)
Z∞
F (jω) =
f (t)e −jωt dt
(4)
−∞
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Basics of Automation and Control
Spectral transfer function
Spectral transfer function
Spectral transfer function is the ratio of the Fourier transform of the
output signal to the Fourier transform of the input signal.
Gjω =
y (jω)
x(jω)
Between spectral transfer function, and transfer function there is formal relation
G (jω) = G (s)|s=jω
resulting from the relation between transforms of Laplace and Fourier
(Fourier transform is special case of Laplace transform)
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Basics of Automation and Control
Spectral transfer function
Using property of Laplace transform - theorem of the shift in the real
variable domain
L{f (t + τ )} = L{f (t)}e τ s
The spectral transfer function of the object in the case of a sinusoidal
signal at an input has the following form
A2 (ω) L {sin[ω(t)]} e tφ s
A2 (ω) tφ s
L {A2 (ω)sin[ω(t + tφ )]}
=
=
e
L {A1 sin[ω(t)]}
A1
L {sin[ω(t)]}
A1
G (s) =
Because
G (jω) =
Y (jω)
,
U(jω)
G (jω) = G (s)|s=jω ,
tφ =
φ(ω)
ω
therefore
G (jω) =
A2 (ω) tφ s
A2 (ω) tφ jω
A2 (ω) jφ(ω)
e |s=jω =
e
=
e
A1
A1
A1
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Basics of Automation and Control
Spectral transfer function
The spectral transfer function can be presented as follows
G (jω) =
A2 (ω) jφ(ω)
e
= M(ω)e jφ(ω)
A1
where:
M(ω) =
A2 (ω)
A1
- the magnitude of the spectral transfer function
φ(ω) - the argument of the spectral transfer function
The spectral transfer function can be divided into two components
G (jω) = M(ω)e jφ(ω) = P(ω) + jQ(ω)
where:
P(ω) - the real part of the spectral transfer function,
Q(ω) - the imaginary part of the spectral transfer function.
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Basics of Automation and Control
Polar plot - Nyquist plot
Polar plot - Nyquist plot
A plot of a function expressed in polar coordinates, which is a locus of
the terminal point of the vector of spectral transfer function G (jω)
with changes
ω=0→∞
p
[P(ω)]2 + [Q(ω)]2
Q(ω)
φ(ω) = arctg
P(ω)
M(ω) =
P(ω) = M(ω) cos[φ(ω)]
Figure 4: Example of polar plot
Q(ω) = M(ω) sin[φ(ω)]
M(ω) = P(ω) cos[φ(ω)] + Q(ω) sin[φ(ω)]
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Amplitude and phase logarythmic plots - Bode diagrams
Frequency characteristics
The phase and magnitude in terms
of the frequency can be presented on
two separate plots:
the amplitude plot
L(ω) = |G (jω)| as the function
of the frequency ω ,
phase plot φ = arg G (ω) as the
function of the frequency ω.
Logarithmic gain (unit - decibel)
L(ω) = 10log10 M 2 (ω)
Figure 5: Logarythmic plots
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= 20 log M(ω)[dB]
Basics of Automation and Control
Basic linear dynamical elements
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Basics of Automation and Control
Basic linear dynamical elements
In complex automation systems, it’s often possible to extract a number
of the simple, indivisible, functional elements, connected together. Their
properties can be assigned with certain accuracy to few basic linear mathematical models. Abstract elements with properties corresponding to those
models are basic (or elementary) linear dynamical elements.
Methods of the description of basic elements:
equations of motion,
transfer functions,
static characteristics,
transient responses,
transfer functions in the frequency domain i.e spectral transfer
function,
polar plot in the frequency domain (Nyquist plot),
logarithmic plots in the frequency domain (Bode plots).
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Basic dynamical elements
Proportional
(non-inertial)
y (t) = ku(t)
T
T
dy (t)
+ y (t) = ku(t)
dt
dy (t)
dy (t)
= u(t), lub
= ku(t)
dt
dt
y (t) = T
T
T2
First-Order Lag
du(t)
dt
Differentiator (ideal)
dy (t)
du(t)
+ y (t) = Td
dt
dt
Differentiator (real)
d 2 y (t)
dy (t)
+ 2ξT
+ y (t) = ku(t)
dt
dt
y (t) = u(t − T0 )
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Integrator
Second Order Lag, if
0<ξ<1
Delay
Basics of Automation and Control
Proportional element
Figure 6: Examples of proportional elements: a) the two-port network, b) the
lever, c) the hydraulic lever
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Proportional element
Figure 7: Examples of proportional elements: a) the two-port network, b) the
lever, c) the hydraulic lever
a)
U2 (t) =
R2
U1 (t)
R1 + R2
Equation of motion
b)
b
y (t) = x(t)
a
y (t) = ku(t)
c)
where: k - gain.
d2
F2 (t) = 22 F1 (t)
d1
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Proportional element
Dynamic equation
y (t) = ku(t)
Static characteristic
yst = kust
Transfer fuction
G (s) =
Figure 8: Static characteristic of the
proportional element
Y (s)
=k
U(s)
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Proportional element
Transient response
1
y (t) = L−1 [ust k] = kust
s
Figure 9: Transient response of the proportional element
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Proportional element
Spectral transfer function
Gjω = G (s)|s=jω = k
P(ω) = k,
Q(ω) = 0
M(ω) = k
L(ω) = 20 log k[dB]
φ(ω) = 0
Figure 10: Nyquist plot
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Basics of Automation and Control
Proportional element
Amplitude diagram
L(ω) = 20 log k[dB]
Phase diagram
φ(ω) = 0
Figure 11: Bode plots
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First Order Lag Element - Example 1
Figure 12: First Order Lag Element - spinning disc on a shaft
dω(t)
+ Rω(t) = M(t)
dt
J dω(t)
1
+ ω(t) = M(t)
R dt
R
where: R - the viscosity coefficient in bearings, J - the moment of inertia,
M - the torque, ω - the angular velocity.
J
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First Order Lag Element - Example 2
Figure 13: First Order Lag Element - RL network
dI (t)
+ U2 (t)
dt
U2 (t)
I (t) =
R
L dU2 (t)
+ U2 (t) = U1 (t)
R dt
where: L - the inductance, R - the resistance, U1 (t) - the input voltage,
U2 (t) - the output voltage.
U1 (t) = L
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First Order Lag Element
Equations of motion
a) disc rotating on the shaft
1
J dω(t)
+ ω(t) = M(t)
R dt
R
Equation of a motion
T
dy (t)
+ y (t) = ku(t)
dt
b) RL network
where: T - time constant.
L dU2 (t)
+ U2 (t) = U1 (t)
R dt
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Basics of Automation and Control
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Basics of Automation and Control
First Order Lag Element
Dynamic equation
T
dy (t)
+ y (t) = ku(t)
dt
Static characteristic
yst = kust
Figure 14: Static characteristic of the
First Order Lag Element
Transfer function
G (s) =
k
Y (s)
=
U(s)
Ts + 1
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First Order Lag Element
Transient response
1 k
]
y (t) = L−1 [ust
s Ts + 1
−t
= ust k 1 − e T
Figure 15: Transient response of the First Order Lag Element
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First Order Lag Element
Spectral transfer function
Gjω = G (s)|s=jω =
P(ω) =
k
k
|s=jω =
= P(ω) + jQ(ω)
Ts + 1
Tjω + 1
k
T 2 ω2
+1
,
Q(ω) =
−kT ω
T 2 ω2 + 1
Figure 16: Nyquist plot, ωs - the corner frequency
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First Order Lag Element
Amplitude diagram
M(ω) = √
k
T 2 ω2 + 1
p
L(ω) = 20 log k − 20 log T 2 ω 2 + 1[dB]
for
1
= ωs
T
L(ω) = 20 log k[dB]
ω≪
for
ω≫
1
= ωs
T
L(ω) = (20 log k−20 log
p
T 2 ω 2 + 1)[dB]
Phase diagram
φ = −arctg(T ω)
Figure 17: Bode plots
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Integrator
Figure 18: Integrators: a) the hydraulic integrator, b) the gearbox
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Integrator
a) Hydraulic integrator
( s
)
2
Q = αb
(pz − ps ) x(t) = Bx(t)
ρ
Equation of motion
Q1 = Q2 = Bx(t) = A
dy (t)
dt
T
or
A dy (t)
= x(t)
B dt
b) Gearbox
dy (t)
= u(t)
dt
dy (t)
= ku(t)
dt
dφ(t)
ω
= x(t)
dt
r
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Integrator
Dynamic equation
T
dy (t)
= u(t)
dt
Static characteristic
ust = 0
Trasfer function
Figure 19: Static characteristic of the
integrator
1
Y (s)
G (s) =
=
U(s)
Ts
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Integrator
Transient response
y (t) = L−1 [ust
t
1 1
] = ust
s Ts
T
Figure 20: Transient response of the integrator
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Integrator
Spectral transfer function
1
1
1
|s=jω =
= −j
Ts
Tjω
Tω
1
P(ω) = 0,
Q(ω) = −
Tω
Gjω = G (s)|s=jω =
Figure 21: Nyquist plot
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Integrator
Amplitude diagram
M(ω) =
1
Tω
1
Tω
= −20 log T ω[dB]
L(ω) = 20 log
Phase diagram
− T1ω
0
π
= arctg(−∞) = −
2
φ(ω) = arctg
Figure 22: Bode plots
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Differentiator (ideal)
a) Tachometric generator
Figure 23: Differentiator - the tachometric generator
Uy (t) =
dθ(t)
dt
b) Liquid feeder (syringe)
Figure 24: Differentiator - the liquid feeder
Q(t) = A
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dx(t)
dt
Basics of Automation and Control
Differentiator (ideal)
Dynamic equation
y (t) = Td
du(t)
dt
Static characteristic
yst = 0
Figure 25: Static characteristic of the
differentiator (ideal)
Transfer function
G (s) =
Y (s)
= Td s
U(s)
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Differentiator (ideal)
Transient response
1
y (t) = L−1 [ust Td s] = ust Td δ(t)
s
Figure 26: Transient response of the differentiator (ideal)
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Differentiator (ideal)
Spectral transfer function
Gjω = Td s|s=jω = jTd ω
P(ω) = 0,
Q(ω) = Td ω
Figure 27: Nyquist plot
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Differentiator (ideal)
Amplitude diagram
M(ω) = Td ω
L(ω) = 20 log Td ω[dB]
Phase diagram
Td ω
0
π
= arctg(∞) =
2
φ(ω) = arctg
Figure 28: Bode plots
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Differentiator (real)
a) Absorber
A
du(t) dy (t)
−
= Q = k∆p
dt
dt
∆pA = Cy (t),
∆p =
C
y
A
A2 dy (t)
A2 du(t)
+ y (t) =
kC dt
kC dt
dy (t)
du(t)
T
+ y (t) = Td
dt
dt
Figure 29: Differentiator - the absorber
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Differentiator (real)
b) RC network
Figure 30: Differentiator - the RC network
dU2 (t)
dU1 (t)
+ U2 (t) =
dt
dt
dy (t)
du(t)
T
+ y (t) = Td
dt
dt
RC
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Basics of Automation and Control
Differentiator (real)
Dynamic equation
T
dy (t)
du(t)
+ y (t) = Td
,
dt
dt
Td
T
Static characteristic
kd =
yst = 0
Figure 31: Static characteristic of the
differentiator (real)
Transfer function
G (s) =
Y (s)
Td s
=
U(s)
Ts + 1
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Basics of Automation and Control
Differentiator (real)
Transient response
y (t) = L−1 [ust
t
Td − t
1 Td s
] = ust
e T = ust kd e − T
s Ts + 1
T
Figure 32: Transient response of the differentiator
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Basics of Automation and Control
Differentiator (real)
Spectral transfer function
Gjω =
P(ω) =
Td jω
Td s
|s=jω =
Ts + 1
Tjω + 1
Td T ω 2
,
T 2 ω2 + 1
Q(ω) =
Td ω
T 2 ω2 + 1
Figure 33: Nyquist plot
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Basics of Automation and Control
Differentiator (real)
Amplitude diagram
Td ω
T 2 ω2 + 1
p
L(ω) = [20 log Td ω−20 log T 2 ω 2 + 1]
M(ω) = √
Phase diagram
φ(ω) = arctg
1
π
= − arctg(T ω)
Tω
2
Figure 34: Bode plots
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Basics of Automation and Control
Second Order Lag Element
Figure 35: Second Order Lag: a) the pneumatic positioner, b) the RLC network
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Basics of Automation and Control
Second Order Lag Element
a) Pneumatic positioner
m
dy (t)
d 2 y (t)
+ Cy (t) = Ap(t)
+B
2
dt
dt
m d 2 y (t) B dy (t)
A
+ y (t) = p(t)
+
2
C dt
C dt
C
Equation of motion
T2
d 2 y (t)
dy (t)
+ 2ξ
+ y (t) = ku(t)
dt 2
dt
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Basics of Automation and Control
Second Order Lag Element
b) RLC network
U3 (t) = I (t)R
dI (t)
dt
dU2 (t)
I (t) = C
dt
U1 (t) = U2 (t) + U3 (t) + U4 (t)
U4 (t) = L
LC
d 2 U2 (t)
dU2 (t)
+ RC
+ U2 (t) = U1 (t)
dt 2
dt
Equation of motion
T2
d 2 y (t)
dy (t)
+ 2ξ
+ y (t) = ku(t)
dt 2
dt
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Basics of Automation and Control
Second Order Lag Element
Equation of motion
T2
dy (t)
d 2 y (t)
+ 2ξ
+ y (t) = ku(t)
2
dt
dt
1 d 2 y (t) 2ξ dy (t)
+ y (t) = ku(t)
+
ω0 dt
ω02 dt 2
d 2 y (t)
dy (t)
+ 2ξω0
+ ω02 y (t) = kω02 u(t)
2
dt
dt
where: 0 < ξ < 1 - the damping ratio, ω0 - the natural frequency.
Static characteristic
y = ku
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Basics of Automation and Control
Second Order Lag Element
Transfer function
G (s) =
k
Y (s)
= 2 2
U(s)
T s + 2ξTs + 1
G (s) =
kω02
Y (s)
= 2
U(s)
s + 2ξω0 s + ω02
Transient response
1
kω02
y (t) = L
ust 2
s s + 2ξω0 s + ω0
"
#
p
1
−ξω0 t
= kust 1 − p
e
sin ω0 1 − ξ 2 t + ϕ
1 − ξ2
−1
p
1 − ξ2
ϕ = arctg
ξ
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Basics of Automation and Control
Second Order Lag Element
Figure 36: Transient response of Second Order Lag element
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Basics of Automation and Control
Second Order Lag Element
Figure 37: Influence of values of the damping ratio ξ on the transient response
of the Second Order Lag element
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Basics of Automation and Control
Second Order Lag
Spectral transfer function
G (jω) =
kω02 [(ω02 − ω 2 ) − j2ξω0 ω]
(ω02 − ω 2 )2 + (2ξω0 ω)2
P(jω) =
kω02 [(ω02 − ω 2 )]
(ω02 − ω 2 )2 + (2ξω0 ω)2
Q(jω) = −
(ω02
k[2ξω03 ω]
− ω 2 )2 + (2ξω0 ω)2
Amplitude diagram
kω02
(ω02 − ω 2 )2 + (2ξω0 ω)2
q
L(ω) = 20 log kω02 − 20 log (ω02 − ω 2 )2 + (2ξω0 ω)2
M(ω) =
Phase diagram
φ = −arctg
2ξω0 ω
ω02 − ω 2
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Second Order Lag
Figure 38: Nyquist plot
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Figure 39: Bode plots
Basics of Automation and Control
Delay
Figure 40: Delay - the belt conveyor system
where: Q1 , Q2 - streams of mass, at the beginning and at the end of the
belt conveyor system.
Q2 (t) = Q1 (t − T0 ),
T0 =
L
v
Equation of motion
y (t) = u(t − T0 )
where: T0 - the transport delay.
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Basics of Automation and Control
Delay
Dynamics equation
y (t) = u(t − T0 )
Static characteristic
yst = ust
Transfer function
G (s) =
Y (s)
= e −T0 s
U(s)
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Figure 41: Static characteristic of the
delay
Basics of Automation and Control
Delay
Transient response
1
y (t) = L−1 [ust e −T0 s ] = ust 1(t − T0 )
s
Figure 42: Transient response of the delay
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Basics of Automation and Control
Delay
Spectral transfer function
G (jω) = e −jT0 ω
P(ω) = cos (−T0 ω)
Q(ω) = sin (−T0 ω)
Figure 43: Nyquist plot
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Delay
Amplitude diagram
M(ω) = 1,
L(ω) = 0
Phase diagram
φ(ω) = −T0 ω
Figure 44: Bode plots
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Basics of Automation and Control
Basics of Automation and Control
Lecture 4 - Block Diagram Models
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
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Basics of Automation and Control
Introduction
Block diagram model
Block diagram model (structural): Graphical representation of interrelationships between the parts of analyzed system, i.e. there are given
directions of signal flow and the relationships between input and output
signals of all components of the analyzed system.
A block diagram, of either a single element or a complex system, is a
form of a mathematical description of the systems function. It clearly
expresses the dependence of the output signals from the input signal, if
there is known information about properties (the transfer functions) of its
components.
Block diagrams consists of unidirectional, operational blocks that represent the transfer function.
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Introduction
Figure 1: Example of block diagram model
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Elements of block diagrams
Block: A rectangle with arrows representing
input and output signals. Inside rectangle
the transfer function is written.
y (s) = G (s)u(s)
(1)
Pickoff point (information point): Represents device that allow to retrieve the information and send it to several branches of the
system.
Summary junction: represents the device
that allow an algebraic summation of signals
and the signs of signals are distinguished.
z =u−y
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Basics of Automation and Control
Types of connections in the block diagram models
Using appropriate transformations, the block diagram representation
can be often reduced to a simplified block diagram with fewer blocks
than a original one, in which there are only 4 types of connections,
called elementary connections.
Elemetary connections are:
1
serial connection (chain, cascade),
2
parallel connection,
3
negative feedback loop,
4
positive feedback loop.
There are also several rules that allow to trasform a complex block diagram
to a simpler one.
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Types of connections in the block diagram models
Connection type
Serial connection
(chain)
Transfer function
Block diagram
G (s) = G1 (s)G2 (s)
Parallel connection
G (s) = ±G1 (s)±G2 (s)
Negative feedback
loop
G (s) =
±G1 (s)
1 + G1 (s)G2 (s)
Positive feedback
loop
G (s) =
±G1 (s)
1 − G1 (s)G2 (s)
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Block diagram transformations - pickoff points
Moving pickoff point
ahead of the block
Changing the order of
pickoff points
Moving pickoff point
behind the block
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Block diagram transformations - summary junctions
Moving a summary
junction behind a
block
Moving a summary
junction ahead of a
block
Separation of a multiinput summary junction
Changing the order of
summary junctions
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Block diagram transformations - pickoff point and
summary junctions
y (s) = u1 (s) − u2 (s)
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Basics of Automation and Control
(3)
Block diagram transformations - example 1, solution 1
Simplify the following block diagram
where: 1 and 2 - summary junctions.
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Basics of Automation and Control
Block diagram transformations - example 1, solution 1
The block diagram can be simplified using the following rules: a) moving
summary junction (2) behind the block, b) changing the order of summary
junctions (1) and (2).
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Block diagram transformations - example 1, solution 1
where
G′ (s) = 1 +
G′′ (s) =
1
G1 (s)
G1 (s)
1 − G1 (s)G2 (s)
(4)
(5)
finally
G( s) = 1 +
1
G1 (s)
1 + G1 (s)
=
G1 (s) 1 − G1 (s)G2 (s)
1 − G1 (s)G2 (s)
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(6)
Block diagram transformations - example 1, solution 2
The block diagram can be simplified using the following rules: a) moving
summary junction (1) ahead of the block, b) changing the order of summary junctions (1) and (2).
G( s) = [1 + G1 (s)]
1
1 + G1 (s)
=
1 − G1 (s)G2 (s)
1 − G1 (s)G2 (s)
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(7)
Multi-input components - Lever
Where: x1 , x2 , y - displacements.
Equation of motion:
y (s) =
b
a
x1 (s) +
x1 (s)
a+b
a+b
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(8)
Multi-input components - Hydraulic servodrive
Figure 2: Hydraulic servodrive - with spool valve
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Multi-input components - Hydraulic servodrive
Figure 3: Hydraulic servodrive - with spool valve
Where: x1 , x2 , y - displacements.
Equation of motion:
y (s) =
1
(x1 (s) + x2 (s))
Ts
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(9)
Multi-input components - Absorber
Figure 4: Absorber/Damper
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Multi-input components - Absorber
Equation of motion:
y (s) =
Ts
1
x1 (s) +
x2 (s)
Ts + 1
Ts + 1
Figure 5: Absorber: A - surface, Q flow, x1 , x2 , y , - displacements, C spring constant, α - valve constant
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Construction of block diagram models
The block diagram enables to determine the role and place of each element
present in the system and how this element influences the processing of
information.
In order to construct the block diagram model, the following steps should
be taken:
1
Identify interactions, caused by changes in the value of the input
signal.
2
Distinguish the elemets that process these interactions (blocks
in the block diagram).
3
Determine the transfer fuctions of distinguished elements.
REMARK: The number of elements present in the block diagram may be
larger than the number of structural elements in the block diagram - since
some components may be influenced by more than one input.
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Basics of Automation and Control
Construction of block diagram model - Mechanical
feedback 1
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Basics of Automation and Control
Construction of block diagram models - Mechanical
feedback 1
Transfer function
G (s) =
=
b
a
1 b
Ts a + b
1
=
a 1
1+
a + b Ts
1
a+b
T
s +1
a
Static characteristic
y=
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a
x
b
Basics of Automation and Control
Construction of block diagram models - Mechanical
feedback 2
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Basics of Automation and Control
Construction of block diagram models - Mechanical
feedback 2
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Basics of Automation and Control
Construction of block diagram models - Mechanical
feedback 2
Substitution
A=
a
e
−
a+b e +b
(10)
Transfer function
G (s) =
1
b
b
1
Ts
1 = a + b Ts + A
a + b 1 + A Ts
Static characteristic
y=
b
x
A(a + b)
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(11)
(12)
Automation Systems
Lecture 5 - Stability of linear dynamic systems
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Mozaryn
Automation Systems
Introduction
Stability
A stable system is a dynamic system with a bounded response to a bounded
input
Stable system - if pushed out a state of equilibrium (considered operating
point P) returns to the state of equilibrum (to some state K ) after the
termination of factors (disturbances d) that pushed the system from a
state of equilibrum.
In the case of linear time-invariant systems - the behavior of the system
after the termination of an action, that pushed the system out of the
equilibrum state, is a characteristic feature of the system and does
not depend on the type of the action before it’s termination. (simple
analysis)
In the case of nonlinear systems - their behavior, as an effect of an
action (signal), that pushed the system out of the equilibrum point, can
depend on the type and the magnitude of the action before it’s termination.
(complicated analysis)
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Stability
There are three types of behavior after pushing a system out of the equilibrum point:
1
The system returns to the equilibrium state in the operating point
prior to action that pushed system out of balance - asymptotical
stability,
2
The system returns to equilibrium state in the operationg point
other than one present, when action pushed system out of balance non-asymptotical stability, neutral, marginal
3
A system doesn’t reach a state of equilibrum - unstability,
instability; (IMPORTANT: a special case of such behavior are
sustained oscillations with constant amplitude - system on the
border of stability).
Figure 1: a) unstability, b) stability, c) neutral
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Automation Systems
Stability
Figure 2: a) unstability, b) stability, c) neutral
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Impulse response
Impulse response is the simplest case, that allows to determine the stability of the linear time invariant dynamical system (LTI).
Figure 3: Impulse response examples for: 1, 2 - non-asymptoticaly stable
systems, 3, 4 - asymptoticaly stable systems, 5, 6 - unstable systems, 7 system on the border of stability (sustained oscillations)
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Automation Systems
Impulse response
Impulse response is the simplest case, that allows to determine the stability of the linear time invariant dynamical system (LTI).
Using the following formula
y (t)|u(t)=δ = g (t) = L−1 {G (s)}
(1)
Some inverse Laplace transforms for impulse responses, useful for further
analysis are: 1
1
−1
−1
L
= 1(t)
(2)
L
= e ∓αt
(4)
s
s ±α
1
1
−1
−1
L
=t
(3)
L
= te ∓αt
(5)
s2
(s ± α)2
L
−1
As + B
s 2 + Cs + D
r
= Ae
C
2
t
cos(t
C2
2B − AC C t
D−
)+ √
e 2 sin(t
4
4D + C 2
r
D−
(6)
if: C 2 − 4D < 0.
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Automation Systems
C2
)
4
Impulse response - expotential functions
Figure 4: a) y = exp(t), b) y = exp(−t)
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Stability
Equation of motion of a LTI system:
an
dy (t)
d (m) tu(t)
du(t)
d (n) y (t)
+ · · · + a1
+ a0 = bm
+ · · · + b1
+ b0 (7)
(n)
dt
dt
dt
dt (m)
Transfer function:
G (s) =
bm s m + · · · + b2 s 2 + b1 s + b0
Y (s)
=
U(s)
an s n + · · · + a2 s 2 + a1 s + a0
(8)
y (t)|u(t)=δ = g (t) = L−1 {G (s)}
(9)
Impulse response:
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Automation Systems
Stability - asymptotical stability
Case 1: Characteristic equation has only single roots with nonzero negative real parts.
G (s) =
L(s)
an (s − s1 )(s − s2 ) . . . (s − sn )
(10)
Using the partial fraction decomposition
G (s) =
C2
Cn
C1
+
+ ··· +
s − s1
s − s2
s − sn
(11)
Impulse response
g (t) = L−1 {G (s)} = C1 e s1 t + C2 e s2 t + ... + Cn e sn t
(12)
Convergence check
lim g (t) = 0 if s1 , . . . , sn < 0
t→∞
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Automation Systems
(13)
Stability - asymptotical stability
Figure 5: Placement of roots of the system on the s-plane
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Stability - asymptotical stability
Case 2: Characteristic equation has one double root with nonzero real
part, and the rest of roots are unique with nonzero negative real parts.
G (s) =
an (s − s1
)2 (s
L(s)
− s3 ) . . . (s − sn )
(14)
Calculation of the step response and the convergence check
G (s) =
C1
C2
C3
Cn
+
+
+ ··· +
s − s1
(s − s1 )2
s − s3
s − sn
g (t) = L−1 {G (s)} = C1 e s1 t + C2 te s1 t + C3 e s3 t + ... + Cn e sn t
lim g (t) = 0 if s1 , . . . , sn < 0
t→∞
(15)
(16)
(17)
because, in case of s1 < 0, expotential function e s1 t decreases faster than
coefficient t increases.
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Stability - asymptotical stability
Figure 6: Placement of roots of the system on the s-plane
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Stability - neutral
Case 3: Characteristic equation has one zero root, and the rest of roots
are unique with nonzero real parts.
G (s) =
L(s)
an (s)(s − s2 ) . . . (s − sn )
(18)
Calculation of the step response and the convergence check
G (s) =
C1
C2
Cn
+
+ ··· +
s
s − s2
s − sn
(19)
g (t) = L−1 {G (s)} = C1 + C2 e s2 t + ... + Cn e sn t
(20)
lim g (t) = C1 if s2 , . . . , sn < 0
(21)
t→∞
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Stability - asymptotical stability
Figure 7: Placement of roots of the system on the s-plane
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Stability - instability
Case 4: Characteristic equation has two (or more) zero roots, and the rest
of roots are unique with nonzero real parts.
C1
C2
Cn
C3
L(s)
=
+
+ ··· +
+
an (s)2 (s − s3 ) . . . (s − sn )
s
(s)2
s − s3
s − sn
(22)
Calculation of the step response and the convergence check
G (s) =
G (s) =
L(s)
C1
C2
C3
Cn
=
+
+
+ ··· +
2
an
− s3 ) . . . (s − sn )
s
(s)
s − s3
s − sn
(23)
g (t) = L−1 {G (s)} = C1 + C2 t + C3 e s3 t + ... + Cn e sn t
(24)
(s)2 (s
lim g (t) = ∞
t→∞
system is instable
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(25)
Stability - asymptotical stability
Figure 8: Placement of roots of the system on the s-plane
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Stability - asymptotical stability
Case 5: Characteristic equation has unique real roots with nonzero parts,
and complex roots with negative nonzero imaginary parts.
s1 = a + jb,
G (s) =
s2 = a − jb
L(s)
an (s 2 − 2as + a2 + b 2 )(s − s3 ) . . . (s − sn )
(26)
(27)
Calculation of the step response and the convergence check
G (s) =
C1 s + C2
C3
Cn
+
+ ··· +
(s 2 − 2as + a2 + b 2 ) s − s3
s − sn
(28)
C2 + aC1 at
e sin(bt)+C3 e s3 t +...+Cn e sn t
b
(29)
lim g (t) = 0 if a < 0, and Re(s3 ), . . . , Re(sn ) < 0
(30)
g (t) = L−1 {G (s)} = C1 e at cos(bt)+
t→∞
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Stability - asymptotical stability
Figure 9: Placement of roots of the system on the s-plane
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Stability - sustaining oscillations
Case 6: Characteristic equation has unique real roots with nonzero parts,
and complex roots with zero real parts.
s1 = jb,
G (s) =
(s 2
+
s2 = −jb
b 2 )(s
L(s)
− s3 ) . . . (s − sn )
(31)
(32)
Calculation of the step response and the convergence check
G (s) =
C1 s + C2
C3
Cn
+
+ ··· +
(s 2 + b 2 ) s − s3
s − sn
(33)
C2
sin(bt) + C3 e s3 t + ... + Cn e sn t (34)
b
If s3, . . . , sn < 0, then system is on the boundary of stability and has
sustainig oscillations.
g (t) = L−1 {G (s)} = C1 cos(bt) +
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Stability - asymptotical stability
Figure 10: Placement of roots of the system on the s-plane
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Stability
Summary:
System is asymptoticaly stable, if it’s chatacteristic equation has
roots with nonzero negative real parts.
System is neutral, if it’s chatacteristic equation has one zero root
and the rest of roots with nonzero negative real parts.
System is unstable if it’s chatacteristic equation has more than one
zero root or unique roots with positive real parts.
System is on the boundary of stability (generates sustaining
oscillations) if it’s chatacteristic equation has one zero root and
doesn’t have unique roots with positive real parts, but has complex
roots with zero real parts.
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Stability - stability criteria
To determine the stability of linear (LTI) systems there is sufficient knowledge of the distribution of roots of characteristic equation on the
complex plane (s-plane).
The problems that arise within this method:
Complicated calculations of roots of higher order algebraic
equations.
Characteristic equation of the system is sometimes unknown.
Other methods of determining stability - the so-called: stability criteria
do not require the determination of the roots of the characteristic equation:
analytic criteria (Routh-Hurwitz, Routh),
graphical-analytical criterion: (Nyquist).
graphic criteria ( Michajlov,Evans-root locus),
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Routh-Hurwitz criterion
Routh-Hurwitz criterion allows to verify that algebraic equation of any
degree has only the roots with negative real parts. The use of RouthHurwitz criterion is limited to LTI systems with the transfer function in
the analytical form.
Routh-Hurwitz criterion
Algebraic equation of the degree n with constant, real coefficients
an
d (n) y (t)
d (n−1) y (t)
dy (t)
+
a
+ · · · + a1
+ a0
n−1
(n)
(n−1)
dt
dt
dt
(35)
has only the roots with negative real parts, when there are fulfilled folowing
conditions (Hurwitz conditions).
1
CONDITION I: All coeficients a0 , a1 , ..., an , of this equation are
non-zero, and have the same signs,
2
CONDITION II: All determinants of principal minors of so-called
Hurwitz matrix ∆n are positive.
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Routh-Hurwitz criterion
Hurwitz amtrix
Hurwitz matrix has following form
∆n =
an−1
an−3
an−5
−
0
0
0
an
an−2
an−4
−
0
0
0
0
an−1
an−3
−
0
0
0
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− 0
− 0
− 0
− −
− a2
− a0
− 0
0
0
0
−
a3
a1
0
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0
0
0
−
a4
a2
a0
(36)
n×n
Routh-Hurwitz criterion
Example: determine Hurwitz matrix for the 4th degree equation
a4 s 4 + a3 s 3 + a2 s 2 + a1 s + a0 = 0
a3
a1
0
0
a4
a2
a0
0
0
a3
a1
0
∆2 =
a3
a4
a1
a2
∆4 =
0
a4
a2
a0
(37)
(38)
It’s primary determinants are:
∆3 =
a3
a4
0
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a1
a2
a3
0
a0
a1
Automation Systems
(39)
(40)
Routh-Hurwitz criterion - example H1
Determine the value of gain kp that will ensure the stability of the
control system.
Figure 11: Block diagram - example H1
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Automation Systems
Routh-Hurwitz criterion - example H1
Gs =
1
1
(Ts+1)4
1
+ (TS+1)
4
=
1
(Ts + 1)4 + kp
(41)
Characteristic equation:
(Ts + 1)4 + kp = 0
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(42)
Routh-Hurwitz criterion - example H1
Characteristic equation:
(Ts + 1)4 + kp = 0
(43)
T 4 s 4 + 4T 3 s 3 + 6T 2 s 2 + 4Ts + 1 + kp = 0
(44)
therefore
4
a4 = T ,
3
a3 = 4T ,
2
a2 = 6T ,
a1 = 4T ,
a0 = 1 + kp
(45)
I. Hurwitz condition will be fulfilled if
a0 = 1 + kp > 0, which gives kp > −1
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(46)
Routh-Hurwitz criterion - example H1
Hurwitz matrix
∆4 =
a3
a1
0
0
a4
a2
a0
0
0
a3
a1
0
0
a4
a2
a0
=
4T 3
4T
0
0
II. Hurwitz condition will be fulfilled if
4T 3
det(∆2 ) = det
4T
4T 3

4T
det(∆3 ) = det
0

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4T
6T 2
1 + kp
0
4T
6T 2
4T
6T 2
1 + kp
0
4T 3
4T
0
0
T4
6T 2
1 + kp
(47)
>0
(48)

0
4T 3  > 0
4T
(49)
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Routh-Hurwitz criterion - example H1
det(∆2 ) = 24T 5 − 4T 5 = 20T 5 > 0
6
6
6
6
6
(50)
6
det(∆3 ) = 96T − 16T − 16T kp − 16T = 64T − 16T kp > 0 (51)
therefore, from II Hurwitz condition
kp < 4
(52)
I. and II. Hurwitz conditions will be fulfilled if
−1 < kp < 4
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(53)
Routh-Hurwitz criterion
Remaks to Routh-Hurwitz criterion
REMARK 1: Non-asymptotic stability occurs when in the characteristic
equation of degree n coefficient a0 = 0 (equation has one zero root), while
the remaining coefficients are positive.
After dividing both sides of the equation by s yields the equation of order
n − 1, for which the Routh-Hurwitz criterion should be applied, in order
to check the sign of the other elements.
If this equation satisfies the Hurwitz conditions it will mean that the system
has one zero root and the remaining roots have negative real parts and the
system is non-asymptotically stable - neutral.
REMARK 2: Routh-Hurwitz criterion can’t be used to determine the
stability of system with delays.
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Nyquist criterion
Nyquist criterion enables to determine the stability of a closed system
(with feedback loop) based on the frequency characteristics of the open
system.
Closed system transfer function
GZ (s) =
GR (s)G (s)
1 + G1 (s)G2 (s)
(54)
Open system transfer function
G0 (s) = GR (s)G (s)
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(55)
Simplified Nyquist criterion
Simplified Nyquist criterion
In the case when the characteristic equation of the open system does
not have roots with positive real parts (may have any number of zero value
roots), a closed system is stable where amplitude-phase characteristic of
open system does not include point {−1, j0}.
’does not include’ means, that moving along the characteristics toward
higher frequencies, point {−1, j0} stays to the left side of the characteristics.
REMARK: Simplified Nyquist criterion doesn’t include cases when characteristic equation of open system, besides negative or zero value roots,
has roots with positive real parts.
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Nyquist criterion
Nyquist criterion properties
frequency response of the open system, on the basis of which
stability of the closed system is determined, can be easily determined
analytically or experimentally.
criterion allows not only a stability check, but also allows designing
the system with specified dynamic properties,
criterion allows stability check of systems with delays.
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Nyquist criterion
Figure 12: The amplitude-phase characteristics of open system for 1) stable
closed-loop system, 2) unstable closed-loop system
Nyquist conditions (frequencies ω−π and ωp )
M(ω−π ) < 1; where ω−π : φ(ω−π ) = −π
(56)
φ(ωp ) > −π; where ωp : M(ωp ) = 1
(57)
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Nyquist criterion - characteristics examples
Figure 13: The amplitude-phase characteristics of open-loop systems,
corresponding to: stable closed-loop systems - characteristics doesn’t include
point {−1, j0}
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Automation Systems
Nyquist criterion - characteristics examples
Figure 14: The amplitude-phase characteristics of open-loop systems,
corresponding to: unstable closed-loop systems - characteristics include point
{−1, j0}
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Nyquist criterion - Bode charakteristics
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Nyquist criterion - Bode charakteristics
Nyquist conditions for amplitude and
phase characteristics
L(ω−π ) = 20 log M(ω−π ) < 0;
(58)
φ(ωp ) > −π; where L(ωp ) = 0
(59)
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Nyquist criterion - gain margin and phase margin
Gain margin
∆M =
1
M(ω−π )
∆L = −20 log M(ω−π )
(60)
(61)
Phase margin
∆φ = π + φ(ωp )
(62)
Gain margin and phase margin of
the stable system have positive
values.
INDUSTRIAL PRACTICE
30 deg < ∆φ < 60 deg
(63)
2 ≤ ∆M ≤ 4 → 6dB ≤ ∆L ≤ 12dB
(64)
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Nyquist criterion - example N1
Using Nyquist criterion determine the stability of the following system:
G0 (s) =
1
s 3 + 3s 2 + s + 1
1
=
− 3ω 2 + jω + 1
2
1
1 − 3ω − j(ω − ω 3
=
1 − 3ω 2 + j(ω − ω 3 ) 1 − 3ω 2 − j(ω − ω 3 )
2
1 − 3ω
−(ω − ω 3 )
+
j
(1 − 3ω 2 )2 + (ω − ω 3 )2
(1 − 3ω 2 )2 + (ω − ω 3 )2
G0 (jω) =
(65)
−iω 3
(66)
Real part and imaginary part
P(ω) =
1 − 3ω 2
;
(1 − 3ω 2 )2 + (ω − ω 3 )2
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Q(ω) =
−(ω − ω 3 )
(1 − 3ω 2 )2 + (ω − ω 3 )2
(67)
Automation Systems
Nyquist criterion - example N1
Real part
P(ω) =
1 − 3ω 2
+ (ω − ω 3 )2
(68)
−(ω − ω 3 )
(1 − 3ω 2 )2 + (ω − ω 3 )2
(69)
(1 −
3ω 2 )2
Imaginary part
Q(ω) =
ω
P(ω)
Q(ω)
0
1
0
p
1/3
0
-2.6
1
-0.5
0
∞
0
0
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Real shape of Nyquist diagram - MATLAB
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Nyquist criterion - example N2
Using Nyquist criterion determine the stability and gain/phase margins of
the following system:
G0 (s) =
10
1
(0.1s + 1)(0.001s + 1) 0.3s
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(70)
Nyquist criterion - example N2
G0 (s) = G1 (s)G2 (s)G3 (s)G4 (s)
(71)
therefore
1
1
1
0.1s + 1 0.01s + 1 0.3s
(72)
In the case of serial connection the
Bode characteristics of each
elements are summed.
G0 (s) = 10
L0 (ω) = L1 (ω) + Lω + L3 (ω) + L4 (ω)
(73)
φ0 (ω) = φ1 (ω)+φω +φ3 (ω)+φ4 (ω)
(74)
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Automation Systems
Lecture 6 - Controller and single-loop control system
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
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Controller
Figure 1: Place of the controller in the control system (the flow process with
the valve eg. gas pipelines, wastewater plants)
controlled variable: y (t),
process variable (PV ) : ym (t),
set point (SP): w (t),
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control variable (sym. CV ):
u(t) = f (e(t)),
control error: e(t) = SP − PV ,
Automation Systems
Controller
Role of the controller
Controller generates the control signal u(t) (CV, Control Variable) based
on the comparison of output signal ym (t)(PV, Process Variable) generated by the sensor that represents the output signal y (t) (controlled
variable) with the reference signal yr (t) (SP, Set Point).
The result of this comparison is error signal, called Deviation of the
Process Variable, defined as:
e(t) = ym (t) − yr (t); ⇒ e = PV − SP
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(1)
Structures of the control systems single loop
Figure 2: Structure of the control system - version 1 (w (t) = yr (t))
Figure 3: Structure of the control system - version 2 (w (t) = yr (t))
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Role of the controller - Steady State
Figure 4: Structure of the control system - version 2 (w (t) = yr (t))
In the steady state, when e(t) = 0, the controller should generate a control signal which causes activation of the actuator ensuring achievement of the predetermined value of the Controlled Variable - the operation
point.
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Role of the controller - Set Point
Figure 5: Structure of the control system - version 2
Set Point following: The occurence of the positive deviation e(t) > 0
(by increasing the setpoint yr (t)) causes an increase of the control value
u and, consequently, the expected increase in the value of the controlled
variable (y (t)).
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Role of the controller - Disturbances
Figure 6: Structure of the control system - version 2
Disturbance minimization: Occurence of the positive deviation e(t) > 0
(by decreasing the controlled variable y (t) due to disturbances) increase
the value of the controlled variable u(t) that compensates the impact of
dusturbances d(t) on the process.
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Structures of the control systems - normal and reverse
modes
Normal mode
In the case of the process, where an increase of the control signal u(t) is
connected with a decrease of the output signal (the transfer function Gr (s)
is negative), the negative feedback can be obtained as follows:
enormal (t) = ym (t) − yr (t).
(2)
Reverse mode
In the case of process, in which an increase of the control signal u(t)
causes an increase in the output (the transfer function Gob (s) is positive),
the following action of the controller shall be used to obtain the negative
feedback:
ereverse (t) = yr (t) − ym (t).
(3)
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Structures of the control systems - normal and reverse
modes
The increase in the signal from the The increase in the signal from the
controller (u(t)) closes the valve - controller (u(t)) opens the valve normal mode.
reverse mode.
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Technical realization of controllers
Figure 7: Technical realization of controllers
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Classification of the controllers - pt. 1
Controller type
Ctiteria
Type of the processed signals:
analogous
digital
The way of influence on the object:
continous
non-continous
Compliance with the law of
superposition:
linear
nonlinear
Destination:
specialized
universal
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Automation Systems
Classification of the controllers - pt. 2
Controller type
Ctiteria
Type of implementation:
mechanical
pneumatic
hydraulic
electrical
Algorithm of control action:
PID controllers
other (LQR, state-space,
predictive)
The energy required for operation:
direct action
indirect action
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PID controllers
Control Algorithm
Dynamic properties of controllers are realized by the control algorithm.
The most commonly used control algorithm (95 %) is called PID algorithm
(Proportional - Integral - Derivative). It is possible to realize simpler
algorithms: P, PI, PD, by setting gains of PID controller (kP , Ti , Td ).
P controller
∆u(s)
= kp
e(s)
(4)
∆u(s)
1
=
e(s)
Ti s
(5)
∆u(s)
1
= kp 1 +
e(s)
Ti s
(6)
Gr (s) =
I controller
Gr (s) =
PI controller
Gr (s) =
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Transfer functions of PID controllers
PD controller - ideal
Gr (s) =
∆u(s)
= kp (1 + Td s)
e(s)
(7)
PD controller - real (the ideal derivative part is substituted with the real
derivative part)


Gr (s) =

∆u(s)
Td s 

= kp 
1+


T
e(s)
d
s +1
kd
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(8)
Transfer functions of PID controllers
PID controller - ideal
Gr (s) =
1
∆u(s)
= kp (1 +
+ Td s)
e(s)
Ti s
(9)
PID controller - real (the ideal derivative part is substituted with the real
derivative part)


Gr (s) =

∆u(s)
1
Td s 

= kp 
1+
+


T
e(s)
Ti s
d
s +1
kd
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Automation Systems
(10)
Block diagram of PID controller
PID controller - real

Gr (s) =


Td s 
1
∆u(s)

= kp 
+
1
+


Td
e(s)
Ti s
s +1
kd
Figure 8: Block diagram of PID controller - paralell realization
Jakub Mozaryn
Automation Systems
(11)
P controller
Dynamics equation of P controller
∆u(t) = kp e(t)
(12)
∆u(t) = u(t) − up
(13)
u(t) = kp e(t) + up
(14)
where: kp - proportionl gain, up - operating point.
Proportional range
xp =
1
100%
kp
(15)
The proportional range describes the percentage (in relation to the full
range of the signal) of the change in the deviation e(t) that is required to
induce changes of the control signal u(t) of the full range.
Jakub Mozaryn
Automation Systems
I controller
Transfer function
Gr (s) =
Ti
1
∆u(s)
=
e(s)
Ti s
(16)
d∆u(t)
= e(t)
dt
(17)
where
∆u(t) = u(t) − u(0)
(18)
Step response (the response to the step function, 2 components)
u(t)|e(t)=e0 1(t)
1
= ∆u(t)+u(0) = u(0)+
Ti
Zt
e(τ )dτ = u(0)+e0
t
(19)
Ti
0
Static characteristic
e=0
Jakub Mozaryn
Automation Systems
(20)
I controller
Figure 9: Step response of I controller
Jakub Mozaryn
Figure 10: Static characteristic of I
controller - astatic algorithm
Automation Systems
PI controller
Transfer function
Gr (s) =
1
∆u(s)
= kp (1 +
)
e(s)
Ti s
1
∆u(t) = u(0) + kp e(t) +
Ti
(21)
Zt
e(τ )dτ
(22)
0
Step response (the response to the step function, 2 components)
∆u(t)|e(t)=e0 1(t) = e0 kp 1(t) + e0 kp
t
Ti
u(t)|e(t)=e0 1(t) = ∆u(t) + u(0) = e0 kp 1(t) + e0 kp
(23)
t
+ u0
Ti
(24)
Static characteristics
e=0
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Automation Systems
(25)
PI controller
Figure 11: Step response of PI controller
Integral time constant Ti (double time)
The component that describes an integral action increases with time from
an initial value, reaching in time t = Ti a value of the proportional component, which means double the gain in the output signal relative to
the proportional component.
Jakub Mozaryn
Automation Systems
PD controller - ideal
Transfer function
∆u(s)
= kp (1 + Td s)
e(s)
(26)
Step response (the response to
the step function, 2 components)
Gr (s) =
∆u(t)|e(t)=e0 1(t) = kp e0 [1 + δ(t)]
(27)
REMARKS:
PD algorithm doesn’t have
technical realisation because
1
kd =
→ ∞.
Td
The ideal PD controller is not
used, because the dynamics of
the actual devices requires a
specific signal duration to be
able to react to change (delay).
Jakub Mozaryn
Figure 12: Step response of PD
controller (ideal)
Automation Systems
PD controller - real
Transfer function


Td s 


Td
s +1
kd
(28)
Step response (respone to the
step function, 2 components)

Gr (s) = kp 
1 +
−kd
t
∆u(t)|e(t)=e0 1(t) = kp e0 [1+kd e Td ]
(29)
Jakub Mozaryn
Figure 13: Step response of PD
controller (real)
Automation Systems
PD controller
Figure 14: Ramp response of PD controller - (a) ideal and (b) real
Derivative time constant - Td (lead time)
The ramp response of the PD controller (ideal / real) explains the name
of the lead time of Td - in the case of ramp input, value of the control
variable as the sum of the components P and D is achieved earlier by
the time Td than value of the component P.
Jakub Mozaryn
Automation Systems
PID controller - ideal
Transfer function
1
∆u(s)
= kp 1 +
+ Td s
Gr (s) =
e(s)
Ti s
(30)
Step response (the response to
the step function, 3 components)
∆u(t)|e(t)=e0 1(t) = kp e0 [1+
t
+δ(t)]
Ti
(31)
Jakub Mozaryn
Figure 15: Step response of PID
controller (ideal)
Automation Systems
PID controller - real
Transfer function



Td s 
t

Gr (s) = kp 
1 + Ti + Td

s +1
kd
(32)
Step response (the response to
the step function, 3 components)
∆u(t)|e(t)=e0 1(t) = kp e0 [1+
−kd
t
+kd e Td ] Figure 16: Step response of PID
Ti
controller (real)
(33)
Jakub Mozaryn
Automation Systems
PID controller - real
Jakub Mozaryn
Automation Systems
Automation Systems
Lecture 7 - Quality of Control System and PID Controller Tuning
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Mozaryn
Automation Systems
Quality of the control system
Apart from the most important requirement of asymptotic stability,
there are imposed additional requirements on control systems, concerning the transient (dynamic) response and steady states. They are generally
referred to as quality requirements of the control system.
The requirements related to the steady state are formulated by determining
the so-called static accuracy of the control system - permissible values of
deviations of the system output from the set point in steady states (steady
state errors) in the case of disturbances or setpoint changes.
Requirements related to the transient response in the control systems are
determined by a number of indices, generally called dynamic quality indices of the control system.
Jakub Mozaryn
Automation Systems
Quality of the control system
The task of the control system is to minimize the deviation from the
setpoint (when the time approaches infinity) described as an error in
steady state:
e(t) = ez (t) + ew (t),
(1)
where
ez (t) - error caused by disturbance,
ew (t) - error caused by change of the set point.
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Automation Systems
Quality of the control system
When rating the quality of the control LTI system, because of the
superposition property, both components of the steady state error
e(t) = ez (t) + ew (t), can be analyzed separately.
Jakub Mozaryn
Automation Systems
Steady state error caused by disturbance
Transfer function
Gz (s) =
∆ym (s)
ez (s)
±Gz (s)Gob (s)
=
=
z(s)
z(s)
1 + Gob (s)Gr (s)
ez (s) = ∆ym (s) =
±Gz (s)Gob (s)
· z(s)
1 + Gob (s)Gr (s)
(2)
(3)
Steady state error caused by disturbance
ezst. = lim ez (t) = lim s · ez (s)
t→∞
ezst. = lim s ·
s→0
s→0
±Gz (s)Gob (s)
· z(s)
1 + Gob (s)Gr (s)
Jakub Mozaryn
Automation Systems
(4)
(5)
Steady state error caused by change of the set point
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Automation Systems
Steady state error caused by change of the set point
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Automation Systems
Steady state error caused by change of the set point
Transfer function
Gew (s) =
ew (s)
−1
=
∆w (s)
1 + Gob (s)Gr (s)
(6)
−1
∆w (s)
1 + Gob (s)Gr (s)
(7)
ew (s) =
Steady state error caused by change of the set point
ewst. = lim ew (t) = lim s · ew (s)
t→∞
ewst. = lim s ·
s→0
s→0
−1
∆w (s)
1 + Gob (s)Gr (s)
Jakub Mozaryn
Automation Systems
(8)
(9)
Steady state error - example
Determine the steady state error of the control system shown in the figure,
caused by a step change of disturbances z(t) = 2 and a step change of the
setpoint ∆w (t) = 5. Assume, that in the control system there is used:
P controller,
PD controller,
PI controller.
Jakub Mozaryn
Automation Systems
Steady state error - example
Transfer function
Gob (s) =
kob
(Ts + 1)4
(10)
P controller
Gr (s) = kp
(11)
Gr (s) = kp (1 + Td s)
(12)
1
Gr (s) = kp 1 +
Ti s
(13)
PD controller
PI controller
Disturbance
z(t) = 2 → z(s) =
2
s
(14)
Change of the set point
∆w (t) = 5 → ∆w (s) =
Jakub Mozaryn
5
s
Automation Systems
(15)
Steady state error caused by the disturbance - example, P
controller
ezst. = lim ez (t) = lim s
t→∞
s→0
Gob (s)
z(s)
1 + Gob (s)Gr (s)
(16)
P controller
ezst.P = lims→0 s
lims→0
2
Gob (s)
=
1 + Gob (s)Gr (s) s
kob
kob · 2
(Ts + 1)4 · 2
= lims→0
kob
(Ts + 1)4 + kob · kp
1+
kp
4
(Ts + 1)
(17)
Steady state error caused by the disturbance - P controller
ezst.P =
Jakub Mozaryn
kob · 2
1 + kob kp
Automation Systems
(18)
Steady state error caused by the disturbance - example,
PD controller
Gob (s)
z(s)
1 + Gob (s)Gr (s)
(19)
kob
(Ts + 1)4 · 2
ezst.PD = lims→0
=
kob
1+
k
(1
+
T
s)
p
d
(Ts + 1)4
kob · 2
= lims→0
(Ts + 1)4 + kob · kp (1 + Td s)
(20)
ezst. = lim ez (t) = lim s
t→∞
s→0
PD controller
Steady state error caused by the disturbance - PD controller
ezst.PD =
Jakub Mozaryn
kob · 2
1 + kob kp
Automation Systems
(21)
Steady state error caused by the disturbance - example, PI
controller
Gob (s)
z(s)
1 + Gob (s)Gr (s)
(22)
kob
(Ts + 1)4 · 2
ezst.PI = lims→0
=
kob
1
1+
k
(1
+
)
p
(Ts + 1)4
Ti s
kob · 2
= lims→0
=0
1
(Ts + 1)4 + kob · kp (1 +
)
Ti s
(23)
ezst. = lim ez (t) = lim s
t→∞
s→0
PI controller
Steady state error caused by the disturbance - PI controller
ezst.PI = 0
Jakub Mozaryn
Automation Systems
(24)
Steady state error caused by the disturbance - summary
P controller
ezst.P =
kob · 2
1 + kob kp
(25)
ezst.PD =
kob · 2
1 + kob kp
(26)
PD controller
PI controller
ezst.PI = 0
Jakub Mozaryn
Automation Systems
(27)
Steady state error caused by change of the set point example, P controller
ewst. = lim ez (t) = lim s
t→∞
s→0
−1
∆w (s)
1 + Gob (s)Gr (s)
(28)
P controller
ewst.P = lim s
s→0
5
−1
= lim
s→0
1 + Gob (s)kp s
−5
−5
=
kob
1 + kob kp
1+
kp
(Ts + 1)4
(29)
Steady state error caused by change of the set point - P controller
ewst.P =
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−5
1 + kob kp
Automation Systems
(30)
Steady state error caused by change of the set point example, PD controller
−1
∆w (s)
1 + Gob (s)Gr (s)
(31)
−1
5
1 + Gob (s)kp (1 + Td s) s
−5
−5
=
kob
1 + kob kp
1+
kp (1 + Td s)
(Ts + 1)4
(32)
ewst. = lim ez (t) = lim s
t→∞
s→0
PD controller
ewst.PD = lims→0 s
= lims→0
Steady state error caused by change of the set point - PD controller
ewst.PD =
Jakub Mozaryn
−5
1 + kob kp
Automation Systems
(33)
Steady state error caused by change of the set point example, PI controller
ewst. = lim ez (t) = lim s
t→∞
s→0
−1
∆w (s)
1 + Gob (s)Gr (s)
(34)
PI controller
−1
5
1
s
1 + Gob (s)kp 1 +
Ti s
−5
=0
kob
1
1+
kp 1 +
4
(Ts + 1)
Ti s
ewst.PI = lims→0 s
= lims→0
(35)
Steady state error caused by change of the set point - PI controller
ewst.PI = 0
Jakub Mozaryn
Automation Systems
(36)
Steady state error caused by change of the set point summary
P controller
ewst.P =
−5
1 + kob kp
(37)
ewst.PD =
−5
1 + kob kp
(38)
PD controller
PI controller
ewst.PI = 0
Jakub Mozaryn
Automation Systems
(39)
Conclusions about steady state errors
In a control system with a static object and P or PD control
algorithm there are non-zero steady state errors in relation to the
disturbances or setpoint changes respectively.
Increasing the proportional gain of the P or PD controller reduces
the value of static error. Reducing the static deviation by increasing
the gain kp is usually limited due to the stability of the system. (The
system with PD controller reaches the border of stability at a higher
gain than in the case of the regulator P).
Integral action in the controller (PI, PID) provides zero steady
state errors in relation to the disturbances or setpoint changes
respectively.
Jakub Mozaryn
Automation Systems
Dynamical quality of control system
Requirements related to the transient response in the control systems are
determined by a number of indices, generally called dynamic performance
quality indicies of the control system.
Groups of such indices are:
transient response indices,
indices descibing the frequency plots of the control system magnitude and phase margins,
integral indices.
Jakub Mozaryn
Automation Systems
Transient response indices
To evaluate the transient response following indices are used:
Maximum error (dynamical): em - the maximum value of error
after the step change of disturbance or setpoint.
Settling time: tr - it is the time between the moment of change of
the set point w (t), or introduction of disturbances z(t) , and the
moment when the error e(t) reaches a fixed value inside a boundary
∆e(t) (eg.∆e(t) = |0.05emax |).
Overshoot:
κ=
e2
· 100%
e1
where e1 and e2 are the first 2 consecutive biggest errors with
opposite signs, assuming steady state value of output y (t) after
transient response as the zero level (baseline).
Jakub Mozaryn
Automation Systems
(40)
Oscillatory transient response - disturbances
Rysunek: Oscillatory transient response of the control system to disturbances:
a) with non-zero steady state error, b) with zero steady state error
Jakub Mozaryn
Automation Systems
Aperiodic transient response - disturbances
Rysunek: Aperiodic transient response of the control system to disturbances: a)
with non-zero steady state error, b) with zero steady state error
Jakub Mozaryn
Automation Systems
Oscillatory transient response - setpoint
Rysunek: Oscillatory transient response of the control system to setpoint
change: a) with non-zero steady state error, b) with zero steady state error
Jakub Mozaryn
Automation Systems
Aperiodic transient response - setpoint
Rysunek: Aperiodic transient response of the control system to setpoint
change: a) with non-zero steady state error, b) with zero steady state error
Jakub Mozaryn
Automation Systems
Selection of controllers
The basic premise when choosing the type of controller is dynamic characteristics of the controlled process.
Rysunek: Control system
Basic equations, describing the properties of the controlled processes
Gob (s) =
kob
∆ym (s)
=
e −T0 s ,
∆u(s)
Tz s + 1
Jakub Mozaryn
Gob (s) =
Automation Systems
∆ym (s)
1 −T0 s
=
e
∆u(s)
Tz s
Selection of controllers
T0
< 0, 1 ÷ 0, 2 → - switch controllers (two- three- gain
Tz
controllers),
for
for 0, 1 ¬
dla
T0
< 0, 7 ÷ 1 ÷ 0, 2 → continuous controllers,
Tz
T0
> 1 → impulse controllers (impulse output signals).
Tz
T0
is in the range of
Tz
0, 2 ÷ 0, 7. Therefore, in industrial control systems the most common
controllers are continuous, with typical control algorithms P, PI, PD and
PID.
In the case of industrial processes common ratio of
Jakub Mozaryn
Automation Systems
Selection of controllers
An analysis of the controller algorithm with the process model leads to the
following conclusions concerning the selection of the control algorithm:
PI algorithm provides good control only for the low frequencies
of setpoint changes or disturbancs. Integral action is necessary to
obtain zero error in steady state.
PI algorithm provides wider bandwidth than PID algorithm, but
poorer performance for the low frequencies of setpoint changes or
disturbances.
Derivative action is recommended for objects with higher order
lag (such as thermal processes), because it allows the strong
interaction of control even at small deviations. PD controller does
not ensure the achievement of zero deviation in steady state .
PID algorithm merges to a certain extent the advantages of
PI and PD algorithms.
Jakub Mozaryn
Automation Systems
Selection of controllers
In practice, industrial controllers with the continuous algorithm are commonly used. Their parameters (settings) can be changed (adjusted) within
a wide range, so they can control properly processes with different dynamical properties.
Depending on the requirements of the stability and quality, the controller settings are selected using different selection procedures.
There are following settings of PID controller:
proportional gain kp = 0, 1 ÷ 100
integral gain Ti = 0, 1 ÷ 3600s
derrivative gain Td = 0 ÷ 3600s
Jakub Mozaryn
Automation Systems
Selection of controllers
Methods of PID controllers tuning:
Experimental methods - usually do not allow to achieve certain
quality of the control system, eg. Ziegler – Nichols, Pessen, Hassen
and Offereissen, Cohen-Coon, Äström – Hagglund .
Tabular methods - determining the set of controller parameters
based on the parameters of a mathematical model of the controlled
process and the required quality criterion of the control system (like
the smallest overshoot, shortest settling time. Problem: often the
minimization of different quality indices base on the contrary
requirements).
Autotuning, eg. relay method.
Jakub Mozaryn
Automation Systems
Tuning of the controllers
Ziegler-Nichols method
Type 1:
controller settings are selected on the basis of the parameters of the
closed-loop control system, brought to the border of stability (by
experimental excitation of the system).
It can be used to controller tuning in the control systems where
processes are described by static and astatic, higher order lag
elements.
Type 2:
It can be used to controller tuning in the control systems where
processes are described by static higher order lag elements,
controller settings are selected based on the transient response of
the controlled process.
Jakub Mozaryn
Automation Systems
Ziegler-Nichols method
Rysunek: Functional scheme of real control system
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Automation Systems
Ziegler-Nichols method, steps 1-3 / 6
Step 1: In the manual mode (M) by changing control variable (CV),
adjust the process variable ym (PV) to a state in which it is equal
with the required setpoint
Step 2: Set the controller to the proportional action (switch off
integral and derivative actions), set the operation point control value
of the controller equal to the setting obtained in Step 1 and set the
initial value of the controller gain kp > 0.
Step 3: Switch the system to automatic control (A) and if the
system maintains equilibrium, by changing SP produce an impulse
with some amplitude and pulse duration depending on the expected
dynamics of the process; observe or record the change in the
controlled variable. It is recommended to use a pulse with the
amplitude of 10 % of the process value changes ym (PV) and the
pulse duration of about 10 % of the estimated value of the time
constant of the controlled process.
Jakub Mozaryn
Automation Systems
Ziegler-Nichols method, steps 4-5 / 6
Step 4: If the transient response is underdamped, set higher values
of the proportional gain (Steps 1-3) until the system be on the
border of stability (constant oscillations).
Step 5: From the steady oscillations read ’critical’ proportional gain
kpkryt. and oscillation period Tosc .
Step 6: Set the patameters according to the table of setings
developed by Ziegler-Nichols.
Rysunek: Changes of the process variable (PV) obtained during Ziegler –
Nichols experiment
Jakub Mozaryn
Automation Systems
Ziegler-Nichols method
PID controller setting according to Ziegler-Nichols
Controller type
kp
Ti
Td
P
0, 50kpkryt.
PI
0, 45kpkryt. 0, 8Tosc
•
PID
0, 60kpkryt. 0, 5Tosc 0, 12Tosc
Jakub Mozaryn
Automation Systems
Basics of Automation and Control
Lecture 8 - Discrete process automation
dr inż. Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
dr inż. Jakub Możaryn
Basics of Automation and Control
Bibliography
Holdsworth, B., Woods V.: Digital logic design. Newnes, 2002
Potton, A.: An introduction to digital logic. Macmillan Education, 1973
Donzelli, G., Oneto, L., Ponta, D., Aguita, D.: Introduction to digital system
design. Springer, 2019
Hruz, B., Zhou, M.C.: Modelling and control of discrete event dynamics systems
with Petri nets and other tools, Springer, 2007
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Discrete processes
Discrete processes are processes that are described with variables with a
finite number of values; usually they are binary variables.
Binary processes
The processes that are described wit binary variables (two value;
numerals 0 and 1) are called binary processes. Information about the
state of such processes is conveyed by means of binary signals.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Figure: An example of a device for the implementation of a discrete process steel sheet bending. Marking: A - work-piece clamping, B - initial bend, C final bend.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Discrete process areas
technological processes related to the production of elements/pieces,
machine assembly,
assembly of electronic components,
packaging, dosing,
orientation and feeding systems,
handling systems, robotics,
of the inter-operational transport devices,
signaling, security, locks,
flexible production systems,
building automation,
service.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Automation of discrete processes as a field of technology deals with the
following issues:
technical implementation of discrete technological processes and
construction of technological equipment for individual processes,
selection of drives, actuators and sensory elements,
designing elementary process control systems (logic circuits, systems
of medium scale of integration - functional blocks, computer control
- programmable controllers),
for controlling complex production systems (concurrent control,
communication networks),
planning and management (e.g. production).
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 1
Example 1 - ventilation control
Binary output signal y of the room ventilation control system
y = 0, the fan motor is not running,
y = 1, the fan motor is running.
(1)
Signal y is generated from binary input signals
x1 , x2 , and x3 T temperature relays located in this room with the same
switching threshold:
xi = 0 if T < Ti ,
(2)
xi = 1 if T ≥ Ti
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 1
There are various variants of the dependence of the system output signal
on the input signals - the truth table.
State No.
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y1
0
0
0
0
0
0
0
1
y2
0
0
0
1
0
1
1
1
y3
0
0 or
0 or
0 or
0 or
0 or
0 or
1
1
1
1
1
1
1
y4
0
0 or 1
0 or 1
1
0 or 1
1
1
1
Table of logic values (Truth table)
Table of logic values (Truth table) specify the output values of the system
for all combinations of input signal values.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 1
The description of variants marked as y3 and y4 is ambiguous and
requires additional explanation.
State No.
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y3
0
0 or
0 or
0 or
0 or
0 or
0 or
1
1
1
1
1
1
1
y4
0
0 or 1
0 or 1
1
0 or 1
1
1
1
The operation of the system with the output signal y3 :
if the following
the next states
system y3 = 0;
if the following
the next states
system y3 = 1;
state of inputs appeared: x1 = 0, x2 = 0, x3 = 0, then in
the ventilation is turned off - the output signal of the
state of inputs appeared: x1 = 1, x2 = 1, x3 = 1, then in
the ventilation is turned on - the output signal of the
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 1
The description of variants marked as y3 and y4 is ambiguous and
requires additional explanation.
State No.
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y3
0
0 or
0 or
0 or
0 or
0 or
0 or
1
1
1
1
1
1
1
y4
0
0 or 1
0 or 1
1
0 or 1
1
1
1
In the case of a system with an output signal y4 , it turns on the ventilation
when any two relays indicate that the set-point temperature is exceeded,
and turns ventilation off when all relays have a zero signal.
dr inż. Jakub Możaryn
Basics of Automation and Control
Combinatorial systems
State number
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y1
0
0
0
0
0
0
0
1
y2
0
0
0
1
0
1
1
1
Combinatorial systems
In systems that implement dependencies y1 = f1 (x1 , x2 , x3 ) i y2 =
f2 (x1 , x2 , x3 ) the current state of the output signal depends only on current state of the input signals
dr inż. Jakub Możaryn
Basics of Automation and Control
sequential systemse
State number
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y3
0
0 or
0 or
0 or
0 or
0 or
0 or
1
1
1
1
1
1
1
y4
0
0 or 1
0 or 1
1
0 or 1
1
1
1
Sequential systems (Sytems with memory)
In the case of systems with the output signals y3 and y4 , certain states of
the input signals cause a change in the state of the output signal, after
which this new state of the output signal lasts (it is ’stored’) until the state
of the inputs appears which should result in another change of the output
signal.
Such systems are called systems with memory or sequential systems
(Latin sequentia - sequence).
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Table of values (Truth table), is used to define the operation of combinational systems, but is not suitable for describing the operation of
sequential systems; other methods are needed to determine how sequential systems operate.
Process dependent sequential systems
In the case of the discussed systems with the output signals y3 and y4 , the
desired changes in the output signals are made on the basis of information
about the state of the realized process (signals x1 , x2 and x3 ).
Time dependent sequential systems
These are systems without input signals - the desired changes in output
signals are caused by an appropriately programmed timer.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation
Figure: System a) combinatorial or sequential (process dependent), b)
sequential (time dependent)
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Basics of Automation and Control
Discrete process automation - Example 2
Przyklad 2
The sheet is folded in the pneumatic device. The A actuator clamps the
sheet which is initially bent by the B actuator and finally bent by the C
actuator.
After placing the sheet, the operator calls series of actuators movements
by pressing the appropriate ’START’ button. The course of these movements is presented in the so-called step diagram.
The actuator control system can be implemented as a process-dependent
or time-dependent system.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 2
Figure: Diagram of the actuator system for example 2 - sequential system
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - Example 2
In the case of process-dependent system, it is necessary to equip the
actuators with sensors that detect the extreme positions of the actuator
pistons. The signals of these sensors inform about the end of the appropriate movement of a given actuator and initiate the start of the next activity.
Figure: Reed sensor detecting the position of the piston
Figure: The location of the sensors detecting the position of the piston
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - process-dependent
sequential systems
In the case of implementing the control system as a process-dependent
system, its input signals are the signal from the ’START’ button and the
signals of sensors detecting positions of piston rods; output signals - signals
causing actuator movements.
Figure: Signals in process-dependent sequential systems - Example 2
A characteristic feature of the process is that the sequence of input signal
changes is defined - it results from assumptions about the process flow. The
sequential systems controlling such processes are linear program systems.
dr inż. Jakub Możaryn
Basics of Automation and Control
Discrete process automation - time-dependent sequential
systems
Mechanical or electronic programmers with internal time measurement are
used as time-dependent controlers.
Sequential time-dependent systems are systems with no input signals;
work without process supervison.
Time-dependent systems implement only linear programs.
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
The numeric codes are divided into
analytical
non-analytical (symbolic)
Analytical codes are the notation of an algebraic expression representing
the given number.
Any integer number can be expressed as
0
X
ai 10i = an 10n + an−1 10n−1 + ... + a0 100
(3)
i=n
The meaning of individual variables depends on the position taken in the
code notation; the decimal code is said to be positional code. The following
notation is used:
i – position number,
10i – i-th position weight,
ai – i-th position code variable,
10 – base of the positional code (base of expansion).
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
For example
1989 = 1 · 103 + 9 · 102 + 8 · 101 + 9 · 100
(4)
’1989’ is the decimal notation of a number
1 · 103 + 9 · 102 + 8 · 101 + 9 · 100
(5)
Therefore, notation of the number
L=
0
X
ai 10i = an 10n + an−1 10n−1 + ... + a0 100
(6)
i=n
In the decimal code: L10 = an an−1 ...a1 a0
coefficients an , ..., a0 (so-called code variables) can take values from 0 to
9.
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
Similarly, one can create analytical codes with different bases
L=
0
X
ai P i = an P n + an−1 P n−1 + ... + a0 P 0
(7)
i=n
If the analytical code is based on the number P (it can be an integer ≥ 2),
then the code variables ai can take the values from 0 to P − 1 .
A particularly important code is the so-called natural binary code with the
base P = 2, in which code variables can only take two values: 0 and 1.
A number expressed in natural binary code L2 = an an−1 ...a1 a0 is a notation
of a number
L2 =
0
X
ai 2i = an 2n + an−1 2n−1 + ... + a0 20
i=n
dr inż. Jakub Możaryn
Basics of Automation and Control
(8)
Non-negative integer codes
L10
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
a3
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
a2
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
a1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
a0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
The disadvantage of the natural binary code is occurrence of changes
of several code variables (say: a
few bits) when going to an adjacent numerical value, e.g. when going from 7 to 8 all bits are changing.
Table: Numbers 0 to 15 in natural
binary code
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Basics of Automation and Control
Non-negative integer codes
In technical devices, information about the values of individual code variables is transferred by means of binary signals. Since it is not possible to
force several signals to change exactly simultaneously, erroneous information appears when changing the transmitted numerical values.
Figure: Illustration of the so-called reading ambiguity
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
Reading ambiguity does not occur in the case of the so-called unit-distance
codes where a change in numerical value by 1 is always associated with a
change in value of only one bit.
Elementary unit distance codes are:
Gray code,
Gray +3 code,
ring codes (eg. Johnson counter).
Unit-distance codes are non-analytical (symbolic) codes - writing a number in such a code is not conventional notation of one mathematical
formula for the encoded number.
To read a number encoded in non-analytical code, one can Use code table
or specific rule.
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
Numbers 0 to 15 in the Gray code
The Gray code table has the characteristic axes of symmetry (blue
lines); hence the names of the codes
with this property - reflected codes
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
Numbers 0 to 15 in the Gray code
and the Gray +3 code
A) Gray code table,
B) Gray +3 code table.
dr inż. Jakub Możaryn
Basics of Automation and Control
Non-negative integer codes
Gray code
a3 , a2
00
Alternate form of Gray code table
01
11
10
00
0
7
8
15
a1 , a0
01 11
1
2
6
5
9 10
14 13
Figure: Graphical representation of the Gray code
dr inż. Jakub Możaryn
Basics of Automation and Control
10
3
4
11
12
Non-negative integer codes
Gray code decoding
One can use the following formula determining the absolute value of the
weight Wk of the k -th position:
|Wk | =
k
X
2i = 2k+1 − 1
(9)
i=0
In the Gray code, the weights of odd ones from the left are positive, and
the weights of even ones are negative.
For example:
(1101)g = (24 –1)–(23 –1) − (21 –1) = 15–7 + 1 = 9
(10)
Converting the natural binary code to Gray’s code: values should be
changed to opposite values for those items for which the higher position
(in binary code) is 1.
For example:
(1000110)2 = (1100101)g
dr inż. Jakub Możaryn
Basics of Automation and Control
(11)
Ring codes (Johnson counter)
Ring (and pseudo-ring) codes enable the encoding of even number sets.
To encode a set containing n numbers, n2 bits are needed.
For example
Decimal number
0
1
2
3
4
5
6
7
dr inż. Jakub Możaryn
a3
0
0
0
0
1
1
1
1
Code
a2 a1
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
a0
0
1
1
1
1
0
0
0
Basics of Automation and Control
’1-of-n’ code
Codes of the type ’1-z-n’ enable encoding of any set of numbers (n - means
the number of encoded elements). Therefore n binary variables are used
to encode n elements. In each notation of a number, one variable has the
value 1.
Decimal number
0
1
2
3
4
5
a5
0
0
0
0
0
1
a4
0
0
0
0
1
0
Code
a3 a2
0
0
0
0
0
1
1
0
0
0
0
0
a1
0
1
0
0
0
0
a0
1
0
0
0
0
0
Ring codes and ’1-of-n’ codes are non-minimal codes - they require more
variables (bits) than the natural binary code or Gray code.
dr inż. Jakub Możaryn
Basics of Automation and Control
Two-valued logical functions
Logical functions
Logical functions are functions whose independent variables and the dependent variable can have only a finite number of values.
Two-valued logical functions
Boolean functions whose independent and dependent variables can only
take two values are called two-valued or bivalent logical functions.
Two-value logical functions are used to describe the operation of discrete
control systems.
n
Various binary logical functions with the number of arguments n are 22 .
So there are only 4 binary unary functions, 16 binary functions, 256 ternary
functions, etc.
dr inż. Jakub Możaryn
Basics of Automation and Control
Unary logical functions
The basic forms of describing logical functions are:
tabular forms,
algebraic description.
Unary logical functions y = f (x)
x
0
1
y
0
0
Constant zero
y =0
x
0
1
y
0
1
x
0
1
Repetition
(12)
y =x
y
1
0
x
0
1
Negation
(13)
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y =x
y
1
1
Constant one
(14)
Basics of Automation and Control
y =1
(15)
Binary logical functions
Binary logical
x1 x2 y0
0
0
0
0
1
0
1
0
0
1
1
0
functions
y1 y2
0
0
0
0
0
1
1
0
y = f (x1 , x2 )
y3 y4 y5
0
0
0
0
1
1
1
0
0
1
0
1
Constant zero function, contradiction : y0 = 0
Conjunction, AND : y1 = x1 · x2
Prohibition by x2 , converse nonimplication :
y2 = x1 ∆x2 = x1 → x2 = x1 · x2
Repetition x1 : y3 = x1
Prohibition by x1 , material nonimplication, abjuction :
y4 = x2 ∆x1 = x2 → x1 = x1 · x2
Repetition x2 : y5 = x2
dr inż. Jakub Możaryn
Basics of Automation and Control
Binary logical functions
Binary logical functions y = f (x1 , x2 )
x1 x2 y6 y7 y8 y9 y10
0
0
0
0
1
1
1
0
1
1
1
0
0
0
1
0
1
1
0
0
1
1
1
0
1
0
1
0
Exclusive OR, exclusive disjunction, XOR :
y6 = x1 ⊕ x2 = x1 x2 + x1 x2
Alternation, disjunction : y7 = x1 + x2
Peirce function, alterative denial,NOR function:
y8 = x1 ↓ x2 = x1 + x2 = x1 · x2
Logical biconditional, NXOR : y9 = x1 ≡ x2 = x1 · x2 + x1 · x2
Negation, logical complement, NOT x2 : y10 = x2
dr inż. Jakub Możaryn
Basics of Automation and Control
Binary logical functions
Binary logical functions y = f (x1 , x2 )
x1 x2 y11 y12 y13 y14 y15
0
0
1
1
1
1
1
0
1
0
1
1
1
1
1
0
1
0
0
1
1
1
1
1
0
1
0
1
Material conditional : y11 = x2 → x2 = x1 + x2
Negation, logical complement, NOT x1 : y12 = x1
Converse : y13 = x1 → x2 = x1 + x2
Sheffer function, alternative denial, NAND :
y14 = x1 /x2 = x1 · x2 = x1 + x2
Constant one, tautology: y15 = 1
dr inż. Jakub Możaryn
Basics of Automation and Control
textitBoole algebra
Boole algebra deals with the dependencies between functions: alternative,
conjunction, and negation.
The functions alternative, conjunction and negation create the so-called
functionally complete system.
functionally complete system
A functionally complete system is a set (set) of logical functions that
enables the creation of algebraic notations of arbitrarily complex logical
functions.
Creating an algebraic notation of a logical function defined, for example,
in the form of a verbal description, in a tabular form or otherwise, is
called synthesis of this function, for which the knowledge of the Boolean
algebra is necessary.
The dependencies between the functions: alternative, conjunction and
negation are expressed by a set of theorems (laws) called Boole’s algebra axioms.
dr inż. Jakub Możaryn
Basics of Automation and Control
textitBoole algebra
Boole’s algebra axioms
Conjunction
Alternation
0=1
(16)
1=0
(21)
x ·0=0
(17)
x +0=x
(22)
x ·1=x
(18)
x +1=1
(23)
x ·x =x
(19)
x +x =x
(24)
x ·x =0
(20)
x +x =1
(25)
x1 · x2 = x2 · x1
Commutativity
(26)
x1 + x2 = x2 + x1
(27)
Associativity
x1 · (x2 · x3 ) = (x2 · x1 ) · x3
(28) x1 + (x2 + x3 ) = (x2 + x1 ) + x3 (29)
dr inż. Jakub Możaryn
Basics of Automation and Control
Algebra Boole’a
Distributivity of conjunction over disjunction
(x1 + x2 ) · x3 = x1 · x3 + x2 · x3
(30)
Distributivity of disjunction over conjunction
(x1 · x2 ) + x3 = (x1 + x3 ) · (x2 + x3 )
(31)
De Morgan laws
x1 · x2 = x1 + x2
(32)
x1 + x2 = x1 · x2
(33)
Double negation law (involution law)
x =x
(34)
On the basis of the above theorems, a number of other dependencies can
be created that are useful in transforming logical functions. The symbols
x0 , x1 , x2 , x3 in these statements can represent either a single argument
or any complex logical function.
dr inż. Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 8 - Discrete process automation
dr inż. Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
dr inż. Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 9 - Synthesis and minimization of the logical functions
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Synthesis of the logical functions
Definitions - based on the function with three arguments - y = f (x1 , x2 , x3 )
Elementary logical product
Any product of simple or negated arguments,eg. x1 · x3 , x1 · x2 · x3
Minterm
Elementary logical multiplication, with all arguments of the given function, eg.
K7 = x1 x2 x3
Elementary logical sum
Any sum of simple or negated arguments, eg. x1 + x3 , x1 + x2 + x3
Maxterm
Elementary logical sum, with all arguments of the given function, eg.
D0 = x1 + x2 + x3
Successive states of arguments of a given function, e.g. state 011 (x1 = 0,
x2 = 1, x3 = 1) form binary notations of decimal numbers, which we call state
numbers (eg. the state number of binary number 011 is 3).
Jakub Możaryn
Basics of Automation and Control
Synthesis of the logical functions
State No
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
Minterms
K0 = x1 x2 x3
K1 = x1 x2 x3
K2 = x1 x2 x3
K3 = x1 x2 x3
K4 = x1 x2 x3
K5 = x1 x2 x3
K6 = x1 x2 x3
K7 = x1 x2 x3
D0
D1
D2
D3
D4
D5
D6
D7
Maxterms
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
= x1 + x2 + x3
In table:
Minterm K is denoted by the index i, if for i-th state of arguments
it takes the value of 1.
Maxterm D is denoted by the index i, if for i-th state of arguments
it takes the value of 1.
Jakub Możaryn
Basics of Automation and Control
Synthesis of the logical functions
In table:
Minterm K is denoted by the index i, if for i-th state of arguments
it takes the value of 1.
Maxterm D is denoted by the index i, if for i-th state of arguments
it takes the value of 1.
It should be noted that for the adopted method of numbering the minterms
and the maxterms:
minterm Ki takes the value of 1 only for the i-th state of arguments;
is 0 for the remaining argument states,
maxterm Di takes the value of 0 only for the i-th state of
arguments; is 1 for the remaining argument states,
The number of minterms and the number of maxterms are equal to the
number of argument states.
Jakub Możaryn
Basics of Automation and Control
Synthesis of the logical functions
Any three-argument function (and similar functions with a different number
of arguments) can be written in the so-called minterm canonical form :
y = (x1 , x2 , x3 ) = y0 ·K0 +y1 ·K1 +y2 ·K2 +y3 ·K3 +y4 ·K4 +y5 ·K5 +y6 ·K6 +y7 ·K7
(1)
where: yn is the value of the function’s dependent variable at the n-th
argument state.
Any three-argument function (and similar functions with a different number
of arguments) can be written in the so-called maxterm canonical form :
y = (x1 , x2 , x3 ) = (y0 + D0 )(y1 + D1 )(y2 + D2 )(y3 + D3 )
(y4 + D4 )(y5 + D5 )(y6 + D6 )(y7 + D7 )
Jakub Możaryn
Basics of Automation and Control
(2)
Example: Synthesis of the logical functions
Minterm canonical form:
y = f (x1 , x2 , x3 ) = 1 · K0 + 1 · K1 + 0 · K2 + 0 · K3 +
+1 · K4 + 1 · K5 + 1 · K6 + 1 · K7
(3)
After the removal of the components with a
value of 0
Figure: Truth table of
the three-argument
function
y = f (x1 , x2 , x3 ) = K0 +K1 +K4 +K5 +K6 +K7
(4)
This function can be represented in the
symbolic (numerical) form:
X
y=
0, 1, 4, 5, 6, 7
(5)
The proper notation of the minterm canonical form of a given function is:
y = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
Jakub Możaryn
Basics of Automation and Control
(6)
Example: Synthesis of the logical functions
Maxterm canonical form:
y = f (x1 , x2 , x3 ) = (1 + D0 )(1 + D1 )(0 + D2 )
(0 + D3 )(1 + D4 )(1 + D5 )(1 + D6 )(1 + D7 )
(7)
After the removal of the components with a
value of 1
y = f (x1 , x2 , x3 ) = D2 · D3
Figure: Truth table of
the three-argument
function
(8)
This function can be represented in the
symbolic (numerical) form:
Y
y=
2, 3
(9)
The proper notation of the minterm canonical form of a given function is:
y = (x1 + x2 + x3 )(x1 + x2 + x3 )
Jakub Możaryn
Basics of Automation and Control
(10)
Synthesis of the logical functions
Canonical forms are an algebraic form of representing of the arbitrarily
complex logical functions.
They are created using only three logical operations: disjunction, conjunction, and negation.
Functionally complete set
A set of logical functions that enables the creation of algebraic notations
of any logical functions.
Basic functionally complete set
Set of logical functions: disjunction, conjunction, and negation.
Functionally complete sets are eg.
disjunction and negation,
conjunction and negation,
NOR function,
NAND function.
Jakub Możaryn
Basics of Automation and Control
Minimization of logical functions
In general, using the laws of the Boolean algebra, canonical forms can
be transformed to reduce the number of elementary logical operations
involved, which is called minimization of logical functions.
The basic activity when looking for the possibility of minimizing canonical forms is searching for pairs of minterms or maxterms over which the
so-called minimization based on the distributivity operation can be performed.
Distributivity operation - minterm canonical form
Distributivity operation (concatenation) while minimizing canonical
alternate form is a performing following operations
a · b + a · b = a · (b + b) = a · 1 = a
(11)
where: a - the equal part of both elements, b - the variable with a
negative sign.
Example:
x1 x2 x3 + x1 x2 x3 = x1 x2
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Basics of Automation and Control
(12)
Minimization of logical functions
Distributivity operation - maxterm canonical form
Distributivity operation (concatenation) while minimizing maxterm
canonical form is a performing following operations
(a + b)(a + b) = a + (bb) = a + 0 = a
(13)
gdzie: a represents the equal part of both elements, b - the variable with
a negative sign
Example:
(x1 + x2 + x3 )(x1 + x2 + x3 ) = x1 + x2
(14)
The minimization method consisting of successive transformations of the
original notation of a function in canonical form is called algebraic transformation method.
Jakub Możaryn
Basics of Automation and Control
Minimization of logical functions
Other minimization methods:
Quine–McCluskey method,
Karnaugh′ s table method (K-table),
only improve the searching procedure and performing the distributivity
operation.
The function form obtained by performing all possible simplifiaction based
on concatenation in the minterm canonical form is called normal minterm
form.
The function form obtained by performing all possible simplifiaction based
on concatenation in the minterm canonical form is called normal maxterm
form.
NOTE: Normal forms are not always a description that uses the
smallest possible number of the logical operations.
Jakub Możaryn
Basics of Automation and Control
Minimization of logical functions - factorization
It is possible to reduce the number of logical operations in normal minterm
form if from two or more elementary products it is possible to find a
common factor (distributivity rule), eg.
x1 x2 x3 + x1 x2 x3 = x1 (x2 x3 + x2 x3 )
(15)
It is possible to reduce the number of logical operations in normal maxterm form if from two or more elementary sums it is possible to find a
common factor (asociativity rule), eg.
(x1 + x2 + x3 )(x1 + x2 + x3 ) = x1 + (x2 + x3 )(x2 + x3 )
Such operations are called logical factorization.
Jakub Możaryn
Basics of Automation and Control
(16)
Minimization of logical functions - Example
Minterm canonical function
y = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 +
x1 x2 x3 + x1 x2 x3 + x1 x2 x3
(17)
There are following possibilities of minimizing
y = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 +
|
{z
} |
{z
}
Figure: Truth table of
the three-argument
function
+ x1 x2 x3 + x1 x2 x3 = x1 x2 + x1 x2 + x1 x2
|
{z
}
(18)
The result shows the possibility of further minimization - the middle component can be minimized based on the first and to the third component.
Jakub Możaryn
Basics of Automation and Control
Minimization of logical functions - Example
By using the Boolean algebra theorem x + x = x, the middle term can be
treated as if it occurred twice.
Therefore:
y = x1 x2 + x1 x2 + x1 x2 = x1 x2 + x1 x2 + x1 x2 + x1 x2 = x2 + x1
| {z } | {z }
(19)
The obtained form of the function y = x2 + x1 is a minimal form.
Maxterm canonical form of the given function is following:
y = (x1 + x2 + x3 )(x1 + x2 + x3 )
(20)
The maxterms appearing in it differ in the negation sign for the x3 variable,
so as a result of combining both factors, we get the following minimal form
y = x1 + x2 = x2 + x1
Jakub Możaryn
Basics of Automation and Control
(21)
Karnaugh table method
Karnaugh tables are a specific form of function value tables
Figure: Karnaugh table
Figure: Truth table for
3-argument function
Figure: Karnaugh table with the state numbers
Jakub Możaryn
Basics of Automation and Control
Karnaugh table method
In the Karnaugh tables, the values of the y dependent variable are typed
into the table fields that match the argument values listed on the borders
of the table.
A characteristic feature of Karnaugh tables is that adjacent argument
state values only differ by one position (argument values are consecutive
numbers in Gray code).
This allows the minterms or maxterms with numbers in the adjacent
fields to be concatenated.
Adjacent fields are e.g. fields 0 and 1, 0 and
2, 4 and 6, 0 and 4, etc.
Figure: Karnaugh table
with the state numbers
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Basics of Automation and Control
Karnaugh table method - minterm form
Figure: Karnaugh table
- concatenation of ones
The function takes the value 1 in argument
states 0 and 1, which means that minterm
canonical form of the function is the logical
sum of the minterms K0 and K1 that can be
concatenated:
y = K0 + K1 = x1 · x2 · x3 + x1 · x2 · x3 = x1 · x2
(22)
The ones in fields 0 and 1 are said to be concatenated ones.
Concatenation is determined directly by based on the same argument
values for both fields.
The fields 0 and 1 correspond to the values of x1 = 0 and x2 = 0; therefore
y = 00− = x1 · x2
Jakub Możaryn
Basics of Automation and Control
(23)
Karnaugh table method - maxterm form
Figure: Karnaugh table
- concatenation of zeros
The function takes the value 0 in argument
states 0 and 1, which means that maxterm
canonical form of the function is the logical
sum of the maxterms D0 and D1 that can be
concatenated:
y = D0 · D1 = (x1 + x2 + x3 ) · (x1 + x2 + x3 )
(24)
The zeros in fields 0 and 1 are said to be concatenated zeros.
Concatenation is determined directly by based on the same argument
values for both fields.
The fields 0 and 1 correspond to the values of x1 = 0 i x2 = 0; thus
y = 00− = x1 + x2
Jakub Możaryn
Basics of Automation and Control
(25)
Karnaugh table method
Figure: Karnaugh tables for two- and four-argument functions
The Karnaugha tables also allow the minimization of five- and sixargument functions.
Jakub Możaryn
Basics of Automation and Control
Minimization of not-fully defined logical functions
Not-fully defined logical functions are functions which for some argument states do not have specified values.
In tables of values of such functions in undefined states, a dash is entered
instead of the dependent variable value. In numerical notations of functions, the numbers of undefined states are given in parentheses, e.g.
X
y (x1 , x2 , x3 , x4 ) =
0, 1, 2, 3, 4, 9, 11(5, 7, 13, 15)
(26)
y (x1 , x2 , x3 , x4 ) =
Y
6, 8, 10, 12, 14(5, 7, 13, 15)
Jakub Możaryn
Basics of Automation and Control
(27)
Minimization of not-fully defined logical functions
Example:
By concatenating ones together, the result
is the normal minterm form of the function.
In this case, it is preferable to assume the
dependent variable with the value 1 in all
unknown states.
Figure: Karnaugh table
- not-fully defined
logical functions
y = x1 · x2 + x1 · x3 + x4
Jakub Możaryn
Basics of Automation and Control
(28)
Minimization of not-fully defined logical functions
Example:
By concatenating zeros together, the result
is the normal maxterm form of the function.
In this case, it is preferable to assume the
dependent variable with the value 0 in few
unknown states, and ones in the rest of the
unknown states.
Figure: Karnaugh table
- not-fully defined
logical functions
y = (x2 + x3 ) · (x1 + x4 )
(29)
Thus, the function obtained by concatenating zeros is different from
the function obtained by concatenating ones - although the desired
action is the same - the differences apply to undefined states.
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 9 - Synthesis and minimization of the logical functions
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 10 - Switching circuits.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Switching circuits
Design of the combinatorial logic circuits
Combinatorial logic circuits are realized using:
switch and relay technology,
from logical elements (gate networks),
with the use of systems of medium scale of integration (MSI) – using
function blocks (digital circuits, dozens of logic gates per chip),
using computer technology e.g. programmable logic controllers
(PLC’s).
Switch
A switch is an electrically operated element. It is a device having a contact
or several contacts, the status of which (closing or opening) depends on
the value of the input signal affecting the relay.
Jakub Możaryn
Basics of Automation and Control
Switching circuits
Normal state of the switch
The condition in which the switch is not affected by external signals.
Hence the names of the switches:
SPST-NO (Single-Pole Single-Throw, Normally-Open) (NO),
also called make contact, which is made up of two active elements
that do not normally touch each other.
SPST-NC (Single-Pole Single-Throw, Normally-Closed) (NC),
also called break contact, which is made up of two active elements
that in normal state do touch each other.
Transfer contact (Single-pole Double-Throw - SPDT) form
three contacts that act as NO and NC contacts.
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Basics of Automation and Control
Switching circuits
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Basics of Automation and Control
Switching circuits - Types
Due to the function to be performed in a switching system we can
distinguish between:
Input switches, allowing the system to react to external signals;
these are manually controlled switches (operator elements),
mechanically, magnetically, temperature and pressure transmitters,
etc.
Relays, are used for processing and amplification of the signals
provided by the input switches.
Output switches, also known as contactors, are
power-adapted to control actuators, e.g. motors, brakes, heaters,
etc. Contactors are equipped with contacts adapted to conduct
sufficiently large currents, necessary to supply various types of
devices.
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Basics of Automation and Control
Switching circuits - Input switches
Input switches
Input switches, allowing the system to react to external signals; these
are manually controlled switches (operator elements), mechanically,
magnetically, temperature and pressure transmitters, etc.
Figure 1: Input switches
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Basics of Automation and Control
Switching circuits - Sensors
Figure 2: Symbols of electronic proximity sensors
Inductive Sensor – an automation element that reacts when the metal is
near its active surface (sensor field).
Optical sensor – an automation element, it reacts to objects crossing the
light beam between the transmitter and receiver or to the beam reflected
from the object.
Capacitive sensor – an automation element that reacts to approaching
its active surface (sensor field) of any material (medium).
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Basics of Automation and Control
Switching circuits - Sensors
Figure 3: Capacitive Sensor
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Basics of Automation and Control
Switching circuits - Relays
Relays
Relays, are used for processing and amplification of the signals provided
by the input.
Figure 4: Relay
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Basics of Automation and Control
Switching circuits - Relays
Relays are used to:
obtain the required number of contacts corresponding to the same
input signal,
converting low-power signals to equivalent but higher power,
transmitting signals between circuits with different voltages or other
types of current (DC-AC),
implementation of feedback in switching sequential circuits.
Figure 5: Relay
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Basics of Automation and Control
Switching circuits - Output switches
Output switches (contactors)
output switches, also known as contactors, are power-adapted to control
actuators, e.g. motors, brakes, heaters, etc.
Contactors are equipped with contacts adapted to conduct sufficiently
high currents, necessary to supply various types of devices, and avoid electric arcs.
Figure 6: Contactor (contactor adapted to conduct sufficiently high currents)
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Basics of Automation and Control
Switching circuits - Notation
Switching circuits of elementary functions - scientific notation and assembly notation
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Basics of Automation and Control
Switching circuits - Notations
Switching circuits of binary functions - scientific notation and assembly
notation
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Basics of Automation and Control
Switching circuits - Example: logical function realisation
Purposefulness of using intermediary relays.
Example: Realize the function given in the Karnaugh array, using the
relays.
y = a · b + a · c + b · c · d = a · (b + c) + b · c · d
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Basics of Automation and Control
(1)
Switching circuits - Example: logical function realisation
Purposefulness of using intermediary relays.
Example: Realize the function given in the Karnaugh array, using the
relays.
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Basics of Automation and Control
Switching circuits - Example: logical function realisation
Purposefulness of using intermediary relays.
Example: Realize the function given in the Karnaugh array, using the
relays.
Figure 8: Solution 2 - Relays
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Basics of Automation and Control
Switching circuits - Example: Room temperature control
Example: Designing a ventilation control system - (example 1 from
lecture 1 - option 2)
Binary output signal y of the room ventilation control
y = 0, fan doesn’t work,
y = 1, fan works.
(3)
is generated from binary input signals x1 , x2 and x3 from the
temperature input switches (sensors) located in this room, with the same
switching threshold T :
xi = 0 if T < Ti ,
(4)
xi = 1 if T ≥ Ti
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Basics of Automation and Control
Switching circuits - Example: Room temperature control
Example: Designing a room temperature control system (example 1 from lecture 1 - option 2)
State no.
0
1
2
3
4
5
6
7
x1
0
0
0
0
1
1
1
1
Jakub Możaryn
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
y2
0
0
0
1
0
1
1
1
Basics of Automation and Control
Switching circuits - Example: Room temperature control
Example: Designing a room temperature control system (example 1 from lecture 1 - option 2)
Figure 9: Karnaugh table of y2
Solution:
y = x1 · x2 + x1 · x3 + x2 · x3 = x1 · (x2 + x3 ) + x2 · x3
Jakub Możaryn
Basics of Automation and Control
(5)
Switching circuits - Example: Room temperature control
Example: Designing a room temperature control system (example 1 from lecture 1 - option 2)
y = x1 · x2 + x1 · x3 + x2 · x3 = x1 · (x2 + x3 ) + x2 · x3
Figure 10: Relay circuit of y2 with intermediate relays
Jakub Możaryn
Basics of Automation and Control
(6)
Switching circuits - Example: electro-pneumatic system
control
Switching circuits are used as the control part of electropneumatic and
electro-hydraulic control systems.
They act on a pneumatic or hydraulic actuator via electrically operated
pneumatic or hydraulic valves.
Figure 11: Switching circuit diagram of the electropneumatic control system
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Basics of Automation and Control
Switching circuits - Bridges
Any complex logic can be realized by using serial or parallel connection of
the NO or NC switches. Such switching circuits are called series-parallel
circuits or circuits of the class Π .
Sometimes it is possible to simplify layout of the class Π by placing switches
between parallel branches. Such relay circuits are called bridge circuits or
class H circuits. This leads to a reduction in the number of switches.
An example of a bridge system is the so-called elementary bridge.
Figure 12: Elementary bridge and equivalent series-parallel system
y =a·c +b·d +a·e ·d +b·e ·c
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Basics of Automation and Control
(7)
Basics of Automation and Control
Lecture 10 - Switching circuits.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 11 - Gate circuits.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
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Basics of Automation and Control
Systems with logical elements
Logic gates
Logic elements (logic gates) are devices with a binary output signal and
binary input signals, the operation of which (the dependence of the output
signal value on the state of input signals) is described by a specific logic
function.
Logical elements are implemented in various techniques, eg electrical,
pneumatic, hydraulic elements, with different signal parameters corresponding to the values ”0” and ”1”.
The basic stage when designing circuits from logical elements is creating
the so-called structural diagrams, composed of symbols of logical elements informing only about the type of logical function performed
(and not about the technique of implementing the element).
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Basics of Automation and Control
Gate circuits
To implement arbitrarily complex logic circuits, a set of elements performing logical functions is necessary, creating a functionally complete
system.
An exemplary functionally complete system consists of the functions disjunction (OR), conjunction (AND), and negation (NOT), and is
called basic functionally complete system.
In practice, however, single-element systems are more useful. Arbitrarily
complex systems can be constructed using only elements that perform
the NOR function or using only elements that perform the NAND
function.
NOR
y =a+b
(1)
y =a·b
(2)
NAND
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Basics of Automation and Control
Gate circuits
1. Polish norm:
PN-78/M-42019
2. IEEE Std. 91 - 1973:
IEEE Standard Graphic
Symbols for Logic
Diagrams
3. Industry standard
(polish):
BN-71/3100-01
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Basics of Automation and Control
Gate circuits: AND, OR, NOT
Example: Realize minterm form of a function defined as the Karnaugh
array.
y = x1 · x3 + x1 · x2 + x2 · x4
Jakub Możaryn
Basics of Automation and Control
(3)
Gate circuits: AND, OR, NOT
Example: Realize the minterm form of the function defined as the Karnaugh array.
(4)
y = x1 · x3 + x1 · x2 + x2 · x4
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Basics of Automation and Control
Boolean algebra
Boolean algebra axioms
Conjunction
Disjunction
0=1
(5)
1=0
(10)
x ·0=0
(6)
x +0=x
(11)
x ·1=x
(7)
x +1=1
(12)
x ·x =x
(8)
x +x =x
(13)
x ·x =0
(9)
x +x =1
(14)
x1 · x2 = x2 · x1
Commutativity
(15)
x1 + x2 = x2 + x1
(16)
Associativity
x1 · (x2 · x3 ) = (x2 · x1 ) · x3
(17) x1 + (x2 + x3 ) = (x2 + x1 ) + x3 (18)
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Basics of Automation and Control
Boolean algebra
Distributivity of conjunction over disjunction
(x1 + x2 ) · x3 = x1 · x3 + x2 · x3
(19)
Distributivity of disjunction over conjunction
(x1 · x2 ) + x3 = (x1 + x3 ) · (x2 + x3 )
(20)
De Morgan laws
x1 · x2 = x1 + x2
(21)
x1 + x2 = x1 · x2
(22)
Double negation law (involution law)
x =x
(23)
On the basis of the above theorems, a number of other dependencies can
be created that are useful in transforming logical functions. The symbols
x0 , x1 , x2 , x3 in these statements can represent either a single argument
or any complex logical function.
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Basics of Automation and Control
Gate circuits: NOR, NAND
Construction of systems replacing the elements of (a) negation, (b)
disjunction, (c) conjunction with NOR or NAND elements.
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Basics of Automation and Control
Gate circuits: NOR, NAND
Example: Realize minterm form of function defined as the Karnaugh
array, using NOR elements.
Eliminate conjunctions or disjunctions by double negating and exploiting de Morgan laws.
y = x1 · x3 + x1 · x2 + x2 · x4 =
= x1 + x3 + x1 + x2 + x2 · x4 =
= x1 + x3 + x1 + x2 + x2 + x4 =
= x1 + x3 + x1 + x2 + x2 + x4 =
(24)
Jakub Możaryn
Basics of Automation and Control
Gate circuits: NOR, NAND
Example: Realize minterm form of function defined as the Karnaugh
array, using NOR elements.
y == x1 + x3 + x1 + x2 + x2 + x4 =
Jakub Możaryn
Basics of Automation and Control
(25)
Gate circuits: NOR, NAND
Example: Realize maxterm form of a function defined as the Karnaugh
array using NAND elements.
Eliminate conjunctions or disjunctions by double negating and exploiting de Morgan laws.
y = (x1 + x2 ) · (x1 + x4 ) · (x2 + x3 ) =
= x1 · x2 · (x1 + x4 ) · x2 · x3 =
= x1 · x2 · (x1 · x4 ) · x2 · x3
(26)
Jakub Możaryn
Basics of Automation and Control
Gate circuits: NOR, NAND
Example: Realize maxterm form of a function defined as the
Karnaugh array using NAND elements.
y = x1 · x2 · (x1 · x4 ) · x2 · x3
Jakub Możaryn
Basics of Automation and Control
(27)
Gate circuits: NOR, NAND
Replacing multi-input NOR, NAND elements with two-input elements
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Basics of Automation and Control
Logic static hazard
The following factors influence the course of transitional processes in a
combination system:
non-step character of changes in the values of signals occurring in
real systems,
delays caused by signal lines when sending signals through them,
delays caused by components in signal processing.
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Basics of Automation and Control
Logic hazard
Two types of transition states can be distinguished in combinational
circuits:
transition states (I) in which, according to the equation describing
the operation of the system, a change of one of the input signals
should not cause any change on the output,
transitions states (II) in which, according to the equation describing
the operation of the system, a change of one of the input signals
should change the output,
Logic static hazard
The phenomenon of the occurrence of short-term changes in the value of
the output signal during the duration of the first transition states (II) is
called static hazard.
Logic dynamic hazard
The phenomenon of the occurrence of additional changes in the value of
the output signal in the second kind of transients (II) is called dynamic
hazard.
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Basics of Automation and Control
Logic static hazard
Static hazard in zeros
Signals when x2 = x3 = 0 (sudden
value of y = 1, although according
to the array y = 0)
y = (x1 + x2 ) · (x1 + x3 )
(28)
System equation without hazard
y = (x1 +x2 )·(x1 +x3 )·(x2 +x3 ) (29)
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Basics of Automation and Control
Logic dynamic hazard
Figure 1: Illustration of the causes of logic dynamic hazard (when changing the
output y from 1 to 0 - several short changes in the output)
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Basics of Automation and Control
Basics of Automation and Control
Lecture 11 - Gate circuits.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 12 - Introduction to digital sequential systems.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
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Basics of Automation and Control
Process-dependent sequential systems
Basic concepts
Let us again use an example of a system controlling the operation of
actuators, according to the given work cycle..
After placing the sheet, the operator calls series of actuators movements
by pressing the appropriate ’START’ button. The course of these movements is presented in the so-called step diagram. The discrete control
system can be implemented as a process-dependent or time-dependent
system.
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Basics of Automation and Control
Process-dependent sequential systems
Internal states
The term internal state is introduced to describe the operation of sequential circuits.
In the considered example, in each
of the distinguished internal states, a
different operation of the work cycle
is performed, which requires the generation of an appropriate set of outputs.
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Basics of Automation and Control
Process-dependent sequential systems
Internal states
In internal state 0: yA = 0, yB = 0, yC = 0
In internal state 1: yA = 1, yB = 0, yC = 0
In internal state 2: yA = 1, yB = 1, yC = 0
To distinguish between internal states in the system, a set of
binary signals appropriate to the number of states is used.
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
The state of the process-dependent sequence system at the moment t
determine the values (state) of three groups of signals:
input signals x1t , x2t , .., xnt = X t (input state)
output signals y1t , y2t , .., ymt = Y t (output state)
signals representing the internal state Q1t , Q2t , .., Qkt = Q t
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Basics of Automation and Control
Process-dependent sequential systems
Due to the method of generating the output signals, there are two
types of sequential machines distinguished:
The distinguishing parts of the systems are:
a block (circuit) implementing the so-called a transition function,
a block (circuit) implementing the so-called an output function λ1 (Moore machine), λ2 (Mealy machine).
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
A transition function
The transition function determines what internal state the system will
take at the next moment (Q t+1 ) under the influence of input signals
present at the current moment (X t ), being in the current internal state
(Q t )
Q t+1 = δ(X t , Q t )
(1)
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Basics of Automation and Control
Process-dependent sequential systems
An output function
In Moore machine, the actual state of the outputs depends only on the
current internal state
Y t = λ1 (Q t )
(2)
In Mealy machine, the current state of outputs is a function of not only
the current internal state, but also the current state of the inputs
Y t = λ2 (Q t , Xt )
(3)
Functions λ1 and λ2 are called output functions of the Moore and Mealy
machines respectively.
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Basics of Automation and Control
Process-dependent sequential systems
Due to the order of changes in internal states, the following systems
are distinguished:
systems with linear programming (unbranched)
systems with branched programming
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Basics of Automation and Control
Process-dependent sequential systems
Due to the way the system reads the information about the state of
inputs, there are
asynchronous systems
synchronous systems.
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Basics of Automation and Control
Process-dependent sequential systems
From the appearance of the input state causing the change of the internal
state, according to the transition function, until the new internal state is
reached, the system is in transition (unstable) state.
In unstable state
Q t+1 ̸= Q t
(4)
The changes in the internal state are delayed in relation to the changes
in the state of the inputs by the duration of the transient states of
the system realizing the transition function. This delay is the time interval
between the present moment t and the next moment t + 1. Until the
next input state appears, causing the internal state to change, the system
remains in stable state during which:
Q t+1 = Q t
Jakub Możaryn
Basics of Automation and Control
(5)
Process-dependent sequential systems
Asynchronous systems
In the asynchronous system information about the inputs state is read by
the system in a continuously. This means that the change of the internal
state takes place immediately after the appearance of the corresponding
state of the inputs.
Synchronous systems
In the synchronous system changes in the internal state can only take
place at certain times, determined by the so-called clock signal (sequence
of rectangular pulses with a constant period), depending on the state of
the input signals at these times.
Information about the state of the inputs is intercepted discontinuously
by the synchronous system - at certain times, called samples (sampling
moments).
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
States of the synchronous system sychronized with the rising edge of
the clock signal.
An input state X n of the synchronous system is the state of the
input signals at the beginning of the internal state Q n .
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
States of the synchronous system sychronized with the the falling edge
of the clock signal.
An input state X n of the synchronous system is the state of the
input signals at the end of the internal state Q n .
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
Both in the case of Moore and Mealy machines, blocks that realize the
output function are combinational systems/circuits.
Blocks that perform the transition function, in the case of asynchronous
systems, can be built directly on the basis of the transition function as
combinational systems, with the feedback,
assembly: combination circuit - a block of typical memory elements,
the so-called flip-flops; they are said to be systems with a separate
flip-flop block.
Jakub Możaryn
Basics of Automation and Control
Process-dependent sequential systems
In the case of a system with a separate block of flip-flops, the task
of the combinatorial circuit is to generate the input signals of the flipflops (flip-flops excitation). This system realizes the so-called excitation
function
q t = δ1 (Q t , X t )
(6)
where q t is the current state of the input signals of the flip-flops. The
form of the excitation function depends on the transition functions of a
given system and the type of flip-flops used.
Synchronous systems can be implemented only with a dedicated flip-flop
block.
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 12 - Introduction to digital sequential systems.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 13 - Elementary sequential automata.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Asynchronous flip-flops
Flip-flops (latches)
Flip-flops are Moore’s sequential systems with two internal states, having
the so-called full transition system - the ability to go directly from any
internal state to any internal state and the so-called full output system
- each internal state should correspond to a different output state.
Jakub Możaryn
Basics of Automation and Control
Asynchronous flip-flops
Flip-flops (latches)
Flip-flops are Moore’s sequential systems with two internal states, having
the so-called full transition system - the ability to go directly from any
internal state to any internal state and the so-called full output system
- each internal state should correspond to a different output state.
It is assumed that the status of outputs in flip-flops is identical to
the internal status.
Jakub Możaryn
Basics of Automation and Control
Asynchronous flip-flops
The simplest sequential circuits are asynchronous Moore circuits with two
internal states, marked as 0 and 1, and two input signals: S (set) and R
(reset) and the output signal y , whose state coincides with the internal
state of Q.
Jakub Możaryn
Basics of Automation and Control
Asynchronous flip-flops
The simplest sequential circuits are asynchronous Moore circuits with two
internal states, marked as 0 and 1, and two input signals: S (set) and R
(reset) and the output signal y , whose state coincides with the internal
state of Q.
Such flip-flops are called asynchronous SR flip-flops.
Jakub Możaryn
Basics of Automation and Control
Asynchronous flip-flops
The simplest sequential circuits are asynchronous Moore circuits with two
internal states, marked as 0 and 1, and two input signals: S (set) and R
(reset) and the output signal y , whose state coincides with the internal
state of Q.
Such flip-flops are called asynchronous SR flip-flops.
For all types of flip-flops, their output function is assumed to be
y =Q
(1)
Therefore, a transition table is sufficient to describe the operation of flipflops.
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Basics of Automation and Control
Asynchronous flip-flops
There are three types of asynchronous flip-flops, with different transition
functions (tables)
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Basics of Automation and Control
Asynchronous flip-flops
Dominant Reset asynchronous flip-flop
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Basics of Automation and Control
Asynchronous flip-flops
Dominant Set asynchronous flip-flop
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Basics of Automation and Control
Asynchronous flip-flops
Dominant Memory asynchronous flip-flop
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Basics of Automation and Control
Asynchronous flip-flops
Dominant Memory asynchronous flip-flop
Transition function
Q t+1 = Q t · S + Q t · R + S · R = Q t · (S + R) + S · R
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Basics of Automation and Control
(2)
Asynchronous flip-flops
Multi-input flip-flop - gate circuits
S =a·b
(3)
R =c +d
(4)
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 13 - Elementary sequential automata.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 14 - Designing asynchronous systems with linear programs.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
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Basics of Automation and Control
Asynchronous systems with linear programs.
Example: Design process-dependent control system with two doubleacting pneumatic actuators A and B, equipped with sensors a, b, c, d, e
informing about the positions of each actuator piston, located as shown in
the figure.
Figure 1: Location of the actuators and servo-valves.
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Basics of Automation and Control
Asynchronous systems with linear programs
The control system should execute the cycle of movements consisting of 6
states:
1
1 – the piston of the actuator A extends fully,
2
2 – the piston of the actuator B extends partially out (to the switch d),
3
3 – the piston of the actuator B moves back,
4
4 – piston of the actuator B extends fully (to the switch e),
5
5 – the piston of the actuator B moves back,
6
6 – the piston of the actuator A moves back,
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Basics of Automation and Control
Asynchronous systems with linear programs
Figure 2: Work cycle diagram.
The work cycle is initiated by a pulse from the START button (x);
starting the work cycle is possible only when the piston rods of both
actuators are withdrawn (x · a · c).
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Basics of Automation and Control
Asynchronous systems with linear programs
The several variants of the control system are possible. Each of these
variants can be realized in a electric or pneumatic version.
1
Moore system - code with a fixed spacing - mono-stable valves,
2
Moore system - code with a fixed spacing - bi-stable valves,
3
Moore system - 1-of-n code - mono-stable valves,
4
Moore system - 1-of-n code - bi-stable valves,
5
Mealy system - code with a fixed spacing - mono-stable valves,
6
Mealy system - code with a fixed spacing - bi-stable valves,
7
Mealy system - 1-of-n code - mono-stable valves,
8
Mealy system - 1-of-n code - bi-stable valves.
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Basics of Automation and Control
Asynchronous systems with linear programs
Example: Moore system - code with a fixed spacing - mono-stable valves,
Figure 3: Diagram of the actuating system and block diagram of the designed
system
A different set of outputs (state of the outputs) is required for each movement in the work cycle. Thus, the Moore system must have 6 internal
states to realize 6 movements.
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Basics of Automation and Control
Asynchronous systems with linear programs
Figure 4: The graph of the Moore
system - assignment of the states
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Basics of Automation and Control
Asynchronous systems with linear programs
Figure 5: The graph of the Moore
system - assignment of the sensor
signals indicating transition between
states
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Basics of Automation and Control
Asynchronous systems with linear programs
Figure 6: The graph of the Moore
system - assignment of the output
signals in each state
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Basics of Automation and Control
Asynchronous systems with linear programs
Figure 7: The graph of the Moore system
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Basics of Automation and Control
Asynchronous systems with linear programs
Because there are 6 states, three binary signals are required to code them
while using fixed spacing code.
Figure 8: The graph of the Moore system - coding of the states
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Basics of Automation and Control
Asynchronous systems with linear programs
Graph with internal state codes (pseudo-ring code - constant space code)
Figure 9: The graph of the Moore system
The core of the discrete control system (state transition block) is thus a
set of flip-flops generating the signals Q1 , Q2 , and Q3 .
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Basics of Automation and Control
Asynchronous systems with linear programs
The dependence of the output signals y1 and y2 on the signals Q1 , Q2 and
Q3 is determined on the basis of the graph.
Output table
Q1 ↓ Q2 , Q3 →
0
1
00 01 11
0
1 1
1 − 1
10
−
1
y1
Q1 ↓ Q2 , Q3 →
0
1
00 01 11
0
0 1
0 − 0
10
−
1
y2
Q1 ↓ Q2 , Q3 → 00 01 11
0
00 10 11
1
10 −− 10
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Basics of Automation and Control
10
−−
11
y1 , y2
Asynchronous systems with linear programs
From the output table the following output functions are obtained:
Q1 ↓ Q2 , Q3 → 00
0
00
1
10
01 11 10
10 11 −−
−− 10 11
y1 , y2
y1 = Q1 + Q3
(1)
y2 = Q1 · Q2 + Q2 · Q3 = Q2 (Q1 + Q3 )
(2)
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Basics of Automation and Control
Asynchronous systems with linear programs
In order to determine the excitations of the flip-flops, there is created a
so-called simplified transition table
Q1t ↓ Q2t , Q3t → 00
0
001
1
000
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01
11
10
011
111 − − −
− − − 110
100
Q1t+1 , Q2t+1 , Q3t+1
Basics of Automation and Control
Asynchronous systems with linear programs
The simplified transition table is further rewritten into universal table
indicating the flip-flops output signals changes between transitions.
Q1t ↓ Q2t , Q3t → 00
0
001
1
000
01
11
10
011
111 − − −
− − − 110
100
Q1t+1 , Q2t+1 , Q3t+1
Q1t ↓ Q2t , Q3t → 00
0
001
1
000
01
11
10
011
111 − − −
− − − 110
100
Q1t+1 , Q2t+1 , Q3t+1
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Basics of Automation and Control
Asynchronous systems with linear programs
Based on the universal table, the excitations ensuring the correct order
of changes of internal states are determined.
NOTE: Not to be confused with concatenation in Karnaugh table.
NOTE The excitation of the flip-flops depends on the adopted code
and the number of code variables, and not on the task.
FLIP-FLOP Q1
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Basics of Automation and Control
S1 = Q2
(3)
R1 = Q2
(4)
Asynchronous systems with linear programs
FLIP-FLOP Q2
S2 = Q3
(5)
R2 = Q3
(6)
S3 = Q1
(7)
R3 = Q1
(8)
FLIP-FLOP Q3
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Basics of Automation and Control
Asynchronous systems with linear programs
The external signals are combined with the determined excitations,
conditioning the transition to the next internal states.
Jakub Możaryn
Basics of Automation and Control
Asynchronous systems with linear programs
Required excitations
of flip-flops
FLIP-FLOP Q1
S1 = Q2 · d
(9)
R1 = Q 2 · c
(10)
FLIP-FLOP Q2
S2 = Q3 · b
(11)
R2 = Q3 · e
(12)
FLIP-FLOP Q3
S3 = Q1 ·x ·a ·c (13)
R 3 = Q1 · c
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Basics of Automation and Control
(14)
Asynchronous systems with linear programs
Logical diagram of
the servo-valves
control system and its
mathematical
description:
STATES
S1 = Q2 · d
R1 = Q2 · c
S2 = Q3 · b
R2 = Q3 · e
S3 = Q1 · x · a · c
R3 = Q1 · c
(15)
OUTPUTS
y1 = Q1 + Q3
y2 = Q2 (Q1 + Q3 )
(16)
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 14 - Designing asynchronous systems with linear programs.
Jakub Możaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Możaryn
Basics of Automation and Control
Basics of Automation and Control
Lecture 15- Process identification
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Mozaryn
Basics of Automation and Control
Controlled process
Controlled process
Controlled process is a technological process that is under influence of
disturbances, where an external control algorithm performs the desired
action (control signal/control variable) and enforces desirable behaviour of
this process.
Mathematical description of the controlled process (simplified SISO - single
input single output)
y = f (u, z)
(1)
where: y - process variable, u - control variable, z - disturbance.
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Basics of Automation and Control
Controlled process
Gob (s) =
ym (s) PV (s)
=
u(s)
CV (s)
Process variables
Process variables are output variables (yi ;i =, . . . , n) that characterize
controlled process.
Process variables characterize the controlled process and their desired
course is defined in a control task.
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Basics of Automation and Control
(2)
Controlled process
Gob (s) =
ym (s) PV (s)
=
U(s)
CV (s)
(3)
Input variables
An amount of supplied energy or matter are an input variables
xi ;i = 1, ..., n of controlled process
To realize technological process there should be provided the relevant
streams of matter or streams of energy. The desired course of the process
variables depend on these streams and their parameters.
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Basics of Automation and Control
Controlled process
Gob (s) =
ym (s) PV (s)
=
U(s)
CV (s)
(4)
Disturbances
Disturbances (zi ; i = 1, . . . , n) are input signals which adversely affects
the course of the process variables.
Disturbances may directly affect the process or distort the streams of
energy or streams of matter, eg. in a temperature control in furnace such
interference are changes in the calorific value of the fuel.
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Basics of Automation and Control
Controlled process
Gob (s) =
ym (s) PV (s)
=
U(s)
CV (s)
(5)
Control variables
Control variables (ui ; i = 1, ..., n) are the input variables generated by the
controller.
Actuators, as a result of an influence of the control signals, shape
streams of matter or energy according to the control task.
Jakub Mozaryn
Basics of Automation and Control
Controlled process
Gob (s) =
ym (s) PV (s)
=
U(s)
CV (s)
Symbols:
u(s) = CV (s) CV - control variable,
ym (s) PV - process variable (from the sensor).
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Basics of Automation and Control
(6)
Selection of elements of control systems
Figure: Schematic diagram of the process with actuator (electromagnetic
valve) in a) normal model, b) reverse mode
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Basics of Automation and Control
Classification of controlled processes
Due to the type of equations:
linear,
nonlinear.
Due to the behavior in the steady state of step response:
static - having the ability to achieve equilibrium,
astatic - not having the ability to achieve equilibrium.
Due to the number of process variables:
one-dimensional,
multi-dimensional.
Due to the stability of parameters in time:
time invariant (stationary),
nonstationary.
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Basics of Automation and Control
Controlled process
Step response of the static processes: 1- first order lag element,
2, 3 – higher order lag elements,
4 – oscillatory, 5 - proportional.
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Step response of the astatic processes: 1- integral element, 2 integral element with first order
lag, 3 - integral element with first
order lag and delay.
Basics of Automation and Control
Experimental determination of the time characteristics of
controlled process
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Basics of Automation and Control
Models of the static process
The characteristic features of the step response of the higher order lag
elements are fixed time gains T1 and T2 defined by the tangent to the
step response at the point of inflection (as given in a picture).
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Basics of Automation and Control
Models of the static process
model 1 - first order lag with delay
kob
∆ym (s)
=
e −T0 s
∆u(s)
(Tz s + 1)
(7)
model 2 - Strejc model
G (s) =
G (s) =
∆ym (s)
kob
=
(8)
∆u(s)
(Tz s + 1)n
model 3 - Strejc model with delay
G (s) =
kob
∆ym (s)
=
e −T0 s
∆u(s)
(Tz s + 1)n
(9)
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Basics of Automation and Control
First order lag model with delay
Model 1 - Tangent method
T0 = T1 ;
Tz = T2
(10)
Model 1 - Secant method
Assumption: The step response of the model in 2 points corresponds
with the step response of the process.
P = 0, 5PV → t1 ; P = 0, 632PV → t2
(11)
Using the time equation of step response of the first order lag element:
t
y (t) = ust k(1 − e − T ),
(12)
the following equations are obtained:
T0 =
t1 − t2 ln 2
,
1 − ln 2
Tz = t2 − T0 =
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t2 − t1
.
1 − ln 2
Basics of Automation and Control
(13)
(14)
Higher order lag elements
Model 2 - Strejc model,
G (s) =
n
1
2
3
4
5
6
1
y (s)
=
u(s)
(Ts + 1)n
T1 /T
0
0,282
0,805
1,425
2,100
2,811
T2 /T
1
2,718
3,695
4,463
5,119
5,699
(15)
T1 /T2
0
0,104
0,218
0,319
0,410
0,493
Table: Parameters of the higher order
lag elements
Jakub Mozaryn
G (s) =
y (s)
1
=
u(s)
(Ts + 1)6
Basics of Automation and Control
(16)
Static processes models - example
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Basics of Automation and Control
Static processes models - example
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Basics of Automation and Control
Astatic process models - identification
Integral element with first order lag
Gob (s) =
1
Tz s(T0 s + 1)
Gob (s) =
Integral element with first order lag
and delay
(17) Gob (s) =
1 −T0 s
e
(18)
Tz s
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1
e −T0 s (19)
Tz s(T1 s + 1)
1 −(T0 +T1 )s
e
T
s
Basics of Automation andzControl
Gob (s) =
(20)
Basics of Automation and Control
Lecture 15- Process identification
Jakub Mozaryn
Institute of Automatic Control and Robotics, Department of Mechatronics, WUT
Jakub Mozaryn
Basics of Automation and Control