Feedback and Control Systems
SYSTEM MODELING
At the end of this chapter, the students shall be able to:
2.1 Discuss the two methods of modeling dynamic systems.
2.2 (a) Find the Laplace transform of a given function in time domain, and obtain the time domain
function given its Laplace transform; (b) Solve differential equations using Laplace transform
methods.
2.3 (a) Write a transfer function associated with a given differential equation and vice versa; (b) Solve
the output of a system with a given differential equation or transfer function for a given input.
2.4 Obtain transfer functions for single and multiple loops, passive and active (op-amp) electrical
networks.
2.5 Determine the transfer function of translational, rotational and translational-rotational mechanical
systems, including systems with gears.
2.6 Obtain transfer function relating the output displacement to the input voltage of electromechanical
systems.
2.7 Define terms associated with state-space modeling of systems and enumerate steps in obtaining
the state-space representation of a system.
2.8 Obtain the state-space model of electrical and mechanical systems.
2.9 Convert a transfer function into state-space representation and vice versa.
2.1 Introduction
Intended Learning Outcome: Discuss the two methods of modeling dynamic systems.
In the previous discussion, the analysis and design sequence that included obtaining the system’s
schematic and demonstrated this step for a position control system. The next step is to develop
mathematical models from schematics of physical systems. Two methods will be discussed: (1) transfer
functions in the frequency domain and (2) state equations in the time domain.
It should be remembered that in any case, the first step in developing a mathematical model is to apply the
fundamental physical laws of science and engineering. When electric networks are being modeled, Ohm’s
law and Kirchhoff’s laws are applied initially. For mechanical systems, Newton’s laws will be used.
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From previous courses, it is seen that a differential equation can describe the relationship between the
input and the output of the system. The form of the differential equation and its coefficients are a
formulation or description of the system. Although the differential equation relates the system to its input
and output, it is not a satisfying representation from a system perspective. It is much preferred that a
mathematical representation such as shown in figure 2.1(a) where the input, output and the system are
distinct and separate parts. Also, a representation where several interconnected subsystems like the
cascade connection of figure 2.1(b) can also be conveniently written as a single system like that of figure
2.1(a). It is then preferred that a single mathematical function, called the transfer function, will represent the
system.
Figure 2.1. Block diagram representation of a system, showing the relationship between the input r(t) and the output c(t)
Note in figure 2.1, rሺtሻ represents the reference input while cሺtሻ represents the controlled variable.
A major advantage of frequency-domain modeling is that they rapidly provide stability and transient
response information. Thus, we can immediately see the effects of varying system parameters until an
acceptable design is met.
The primary disadvantage of this approach, also called the classical approach is its limited applicability. It
can be applied only to linear, time-invariant systems or systems that can be approximated as such.
With an increasing design scope, the modern or time-domain approach of modeling systems is devised.
The state-space approach is a unified method for modeling, analyzing, and designing a wide range of
system. It can model nonlinear systems and systems with nonzero initial conditions. Time-varying systems
and systems with multiple inputs and multiple outputs can be compactly represented in state-space model
which is similar to the form of single input, single output systems.
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The time-domain approach can also be used for the same class of systems modeled by classical approach.
This alternate model gives the control systems designer another perspective from which to create the
design. The main drawback of the time-domain approach is its lack of intuition compared to the classical
approach. A few calculations must be made before the physical interpretation of the model becomes
apparent.
2.2 The Laplace Transform: A Review
Intended Learning Outcomes: (a) Find the Laplace transform of a given function in time domain, and obtain
the time domain function given its Laplace transform; (b) Solve differential equations using Laplace
transform methods.
The Laplace transform satisfies the requirements for the convenient mathematical representation of
systems using the transfer function as discussed previously. Furthermore, the Laplace transform makes the
relationships between systems algebraic.
The Laplace transform is defined as
ஶ
ℒ ሾfሺtሻሿ = Fሺsሻ = න fሺtሻeିୱ୲ dt
଴ష
(2.1)
where s = σ + jω, a complex variable. The function fሺtሻ is in t-domain (or time domain) and has a
Laplace transform Fሺsሻ in s-domain (or the complex frequency domain) if the integral of Equation 2.1
exists.
The inverse Laplace transform, is defined as
where
ℒ ିଵ ሾFሺsሻሿ = fሺtሻuሺtሻ =
u ሺ tሻ = ቄ
0,
1,
1 ஢ା୨ஶ
න
Fሺsሻeିୱ୲ ds
j2π ஢ି୨ஶ
t < 0
t>0
(2.2)
(2.3)
called the unit step function. Multiplication of a function with a unit step function yields a function that is zero
for negative values of t.
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Tables 2.1 and 2.2 summarize the Laplace transform of commonly used functions and the properties of the
Laplace transform, respectively.
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Find the Laplace transform of the function fሺtሻ = eିୟ୲ cosሺωt + θሻ
Example 2.1
ሺs + aሻ cos θ − ω sin θ
ሺs + aሻଶ + ωଶ
Answer:
Example 2.2
Find the inverse Laplace transform of Fሺsሻ = ୱሺୱାଶሻሺୱାଷሻమ.
Answer:
ଵ଴
fሺtሻ =
5
40
10
− 5eିଶ୲ + eିଷ୲ + teିଷ୲
9
9
3
Example 2.3
Solve the differential equation
dଶ y
dy
+ 12 + 32y = 32uሺtሻ
ଶ
dt
dt
using Laplace transform when all the initial conditions are zero.
Answer:
yሺtሻ = ሾ1 + eି଼୲ − 2eିସ୲ ሿuሺtሻ
1. Find the Laplace transform of the function fሺtሻ = 3teିଶ୲ sinሺ4t + 60° ሻ.
Drill Problem 2.1
a. fሺtሻ = sinhଶ at
2. Find the Laplace transform of the following functions
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b. fሺtሻ = t cos 5t
c. fሺtሻ = sinସ t
3. Determine the inverse Laplace transform of the following functions in s-domain.
a. Fሺsሻ = s2 −4s−12.
2s−56
b. Fሺsሻ = ሺୱାଷሻሺୱାସሻሺୱమ
൫ୱమ ାଷୱାଵ଴൯ሺୱାହሻ
ାଶୱାଵ଴଴ሻ
c. Fሺsሻ = ሺୱା଼ሻሺୱమ ା଼ୱାଷሻሺୱమାହୱା଻ሻ
ୱయ ାସୱమ ାଶୱା଺
4. What is the time domain function that has Fሺsሻ = s2൫s2 + ω2 ൯ as its Laplace transform?
1
a. y ᇱᇱ − y ᇱ − 6y = 0, with initial conditions yሺ0ሻ = 6 and y ᇱ ሺ0ሻ = 13.
5. Determine the solution of the following differential equations using Laplace transform.
b. y ᇱᇱ − 4y ᇱ + 4y = 0, with initial conditions yሺ0ሻ = 2.1 and y ᇱ ሺ0ሻ = 3.9.
c. y ᇱᇱ + ky ᇱ − 2k ଶ y = 0, with initial conditions yሺ0ሻ = 2 and y ᇱ ሺ0ሻ = 2k.
2.3. The Transfer Function
Intended Learning Outcomes: (a) Write a transfer function associated with a given differential equation and
vice versa; (b) Solve the output of a system with a given differential equation or transfer function for a given
input.
The Laplace transform, as stated before, can be used to establish algebraic input-system-output
relationship as depicted in figure 2.1(a). Furthermore, it will allow algebraic combination of mathematical
representations of subsystems to yield a total system representation.
For an nth-order, linear, time-invariant differential equation,
a୬
d୬ cሺtሻ
d୬ିଵ cሺtሻ
d ୫ r ሺ tሻ
d୫ିଵ rሺtሻ
ሺ
ሻ
+
a
+
⋯
+
a
c
t
=
b
+
b
+ ⋯ + b଴ rሺtሻ
୬ିଵ
଴
୫
୫ିଵ
dt ୬
dt ୬ିଵ
dt ୫
dt ୫ିଵ
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(2.4)
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Feedback and Control Systems
where cሺtሻ is the output, rሺtሻ is the input, and a୧ ’s, b୧ ’s and the form of the differential equation
େሺୱሻ
represents the system. Taking the Laplace transform of equation 2.4, and solving for the ratio of ୖሺୱሻ with
zero initial conditions yield
Cሺsሻ
b୫ s୫ + b୫ିଵ s୫ିଵ + ⋯ + b଴
ሺ
ሻ
=G s =
Rሺsሻ
a୬ s୬ + a୬ିଵ s୬ିଵ + ⋯ + a଴
(2.5)
with Gሺsሻ, called the system’s transfer function, evaluated at zero initial conditions. Thus, a system can
now be represented by a block diagram as seen in figure 2.2.
Figure 2.2. System representation using transfer function
From here, it can be seen that equation 2.5 can be used to determine the output when the system’s
transfer function and the input are known, as
Cሺsሻ = RሺsሻGሺsሻ
(2.6)
Example 2.4
dcሺtሻ
+ 2cሺtሻ = rሺtሻ
dt
Find the transfer function of the system represented by the differential equation
assuming zero initial conditions. Then find the system response to a step input rሺtሻ = uሺtሻ.
Answer:
The transfer function is
Gሺsሻ =
and the system response for the specified input is
cሺtሻ =
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1
s+2
1 1 ିଶ୲
− e
2 2
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Feedback and Control Systems
1. Find the transfer function Gሺsሻ = Cሺsሻ/Rሺsሻ corresponding to the differential equation
Drill Problems 2.2
dଶ c
dc
dଶ r
dr
dଷ c
+
3
+
7
+
5c
=
+ 4 + 3r
ଷ
ଶ
ଶ
dt
dt
dt
dt
dt
2. A system is described by the following differential equation
dଷ y
dଶ y
dy
dଷ x
dଶ x
dx
+
3
+
5
+
y
=
+
4
+ 6 + 8x
ଷ
ଶ
ଷ
ଶ
dt
dt
dt
dt
dt
dt
ଢ଼ሺୱሻ
Find the expression for the transfer function ଡ଼ሺୱሻ.
3. Find the differential equation corresponding to the transfer functions
a. Gሺsሻ = ୖሺୱሻ = ୱమ ା଺ୱାଶ
େሺୱሻ
ଶୱାଵ
b. Gሺsሻ = ୖሺୱሻ = ୱమ ାହୱାଵ଴
େሺୱሻ
଻
c. Gሺsሻ = ୖሺୱሻ = ሺୱାଵ଴ሻሺୱାଵଵሻ
େሺୱሻ
ଵହ
d. Gሺsሻ = ୖሺୱሻ = ୱయ ାଵଵୱమ ାଵଶୱାଵ଼
େሺୱሻ
ୱାଷ
4. For each of the systems described by the transfer functions in number 3, find the response of the
system to a step input.
5. Find the ramp response for a system whose transfer function is
s
Gሺsሻ =
ሺs + 4ሻሺs + 8ሻ
In general, a physical system than can be represented by a linear, time-invariant differential equation can
be modeled as a transfer function. The subsequent sections will demonstrate the use of the transfer
functions to model electrical networks, mechanical systems, and electro-mechanical systems.
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2.4 Electrical Network Transfer Functions
Intended Learning Outcome: Obtain transfer functions for single and multiple loops, passive and active (opamp) electrical networks.
Passive Networks. In this section, mathematical models of electric circuits including passive networks and
operational amplifiers will be obtained. Equivalent circuits for the electric networks will first consist of
resistors, inductors and capacitors. Table 2.3 summarizes the components and the relationships between
voltage and current and between voltage and charge under zero initial conditions.
As stated before, the first step in formulating the mathematical model of a system is to use fundamental
laws governing the system. Since the systems being modeled in this section are electrical networks, Ohm’s
Law and Kirchhoff’s Laws will be used.
Find the transfer function relating the capacitor voltage Vେ ሺsሻ to the input voltage Vሺsሻ for the circuit
Example 2.5
shown below.
Answer:
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Vେ ሺsሻ
୐େ
Gሺsሻ =
=
ୖ
ሺ
ሻ
ଶ
V s
s + s+
ଵ
୐
Example 2.6
Determine the transfer function Hଵ ሺsሻ =
୚ి ሺୱሻ
୚ሺୱሻ
୍మ ሺୱሻ
୚ሺୱሻ
ଵ
୐େ
using mesh analysis. Obtain the transfer function Hଶ ሺsሻ =
for the network shown below using nodal analysis.
Answers:
Hଵ ሺsሻ =
Iଶ ሺsሻ
LCsଶ
=
Vሺsሻ ሺRଵ + R ଶ ሻLCsଶ + ሺRଵ R ଶ C + Lሻs + Rଵ
భ మ
s
Vେ ሺsሻ
େ
Hଶ ሺsሻ =
=
ୋ ୋ ୐ାେ
ୋ
Vሺsሻ
ሺGଵ + Gଶ ሻs ଶ + భ మ
s+ మ
ୋ ୋ
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୐େ
୐େ
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From the foregoing examples, a technique in which mesh equations can be written can be developed. For
two loop electrical network as shown in example 2.6, the mesh equations can be written as
Sum of
Sum of impedances
ሺ
ሻ
൩
൥
I
s
=
applied
voltages൩
common to
ଶ
the two meshes
around mesh 1
Sum of
Sum of impedances
Sum
of
impedances
൩ Iଵ ሺsሻ + ቂ
ቃ Iଶ ሺsሻ = ൥applied voltages൩
−൥
common to
around mesh 2
the two meshes
around mesh 2
Sum of impedances
ቂ
ቃ Iଵ ሺsሻ − ൥
around mesh 1
(2.7a)
(2.7b)
This technique can be expanded to three-loop electrical network. The use of the technique is illustrated in
the following example.
Example 2.7
Write, but do not solve, the mesh equations for the network shown below
Answer:
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Find the transfer function Gሺsሻ = V୐ ሺsሻ/Vሺsሻ for the circuit shown below. Solve the problem in two ways
Example 2.8
– mesh analysis and nodal analysis.
Answer:
Gሺsሻ =
V୐ ሺsሻ sଶ + 2s + 1
=
Vሺsሻ sଶ + 5s + 2
Operational Amplifiers. An operational amplifier is an electronic amplifier used as a basic building block to
implement transfer functions. It has the following characteristics:
•
•
•
•
Differential input vଶ ሺtሻ − vଵ ሺtሻ
High input impedance, Z୧ = ∞ (ideal)
Low output impedance, Z୭ = 0 (ideal)
High constant gain amplification, A = ∞ (ideal)
The output of the operational amplifier v୭ ሺtሻ is
v୭ ሺtሻ = A൫vଶ ሺtሻ − vଵ ሺtሻ൯
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(2.8)
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Figure 2.3. Operational amplifier
Example 2.9
Determine the transfer function
୚౥ ሺୱሻ
୚౟ ሺୱሻ
for each of the op-amp configurations shown below. Assume that the
op-amp has ideal characteristics.
(a)
Answers:
a.
b.
୚౥ ሺୱሻ
୚౟ ሺୱሻ
୚౥ ሺୱሻ
୚౟ ሺୱሻ
(b)
= − ୞మ ሺୱሻ, an inverting amplifier
୞ ሺୱሻ
భ
= 1 + ୞మ ሺୱሻ, a non-inverting amplifier
୞ ሺୱሻ
భ
Example 2.10
Find the transfer function
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୚౥ ሺୱሻ
୚౟ ሺୱሻ
for the circuit given below.
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V୭ ሺsሻ
sଶ + 45.95s + 22.55
= −1.232
V୧ ሺsሻ
s
Answer:
Example 2.11
Find the transfer function
Answer:
Drill Problems 2.3
୚౥ ሺୱሻ
୚౟ ሺୱሻ
for the op-amp configuration shown below:
V୭ ሺsሻ
Rଵ R ଶ Cଵ Cଶ sଶ + ሺR ଶ Cଶ + R ଶ Cଵ + Rଵ Cଵ ሻs + 1
=
V୧ ሺsሻ
Rଵ R ଶ Cଵ Cଶ sଶ + ሺR ଶ Cଶ + Rଵ Cଵ ሻs + 1
1. Find the transfer function Gሺsሻ =
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୚౥ ሺୱሻ
୚౟ ሺୱሻ
for each of the networks shown below:
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2. Find the transfer function Gሺsሻ =
୚ై ሺୱሻ
୚ሺୱሻ
for each of the networks shown below:
3. Using mesh and nodal analysis, find the transfer function Gሺsሻ =
below:
୚౥ ሺୱሻ
୚ሺୱሻ
for each of the networks shown
4. For each of the operational amplifiers shown, find the transfer function Gሺsሻ =
5. Determine the transfer function Gሺsሻ =
System Modeling
୚౥ ሺୱሻ
୚౟ ሺୱሻ
୚౥ ሺୱሻ
୚౟ ሺୱሻ
.
for each of the networks shown below:
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2.5 Mechanical System Transfer Function
Intended Learning Outcome: Determine the transfer function of translational, rotational and translationalrotational mechanical systems, including systems with gears.
In the previous section, the transfer function for electrical networks was obtained. In this section, transfer
function for translational and rotational mechanical systems will be obtained. It will be evident later on that
the transfer functions of both systems are mathematically indistinguishable. Hence, an electrical network
can be interfaced to a mechanical system by cascading their transfer functions provided that one system is
not loaded by the other.
Translational Mechanical Systems. Table 2.4 below shows the components of mechanical systems and
relationships between force, velocity and displacement.
In mechanical systems there are three components:
•
•
•
Spring, with the spring constant K as the parameter
Viscous damper, with the coefficient of friction f୴ as the parameter
Mass, M as its parameter
These components can be compared to the passive electrical components:
•
Capacitor
•
Resistor
•
Inductor
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Comparing Tables 2.3 and 2.4, the following analogy between the mechanical and electrical systems can
be formulated.
•
The capacitor and inductor of the electrical system are energy-storage devices, whereas for
mechanical system, these are analogous to the spring and the mass.
•
Both the resistor of the electrical system and viscous damper of the mechanical system dissipate
energy.
•
The mechanical force is analogous to electrical voltage and the mechanical velocity is analogous to
electrical current.
•
Summing forces in terms of velocity is analogous to summing voltages written in terms of current
and the resulting mechanical differential equations are analogous to mesh equations.
Example 2.12
ଡ଼ሺୱሻ
Find the transfer function ୊ሺୱሻ for the system shown below.
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Answer:
Gሺsሻ =
Xሺsሻ
1
=
ଶ
Fሺsሻ Ms + f୴ s + K
Many mechanical systems are similar to multiple-loop and multiple-node electrical networks, where more
than one simultaneous differential equation is required to describe the system. In mechanical systems, the
number of equations of motion required is equal to the number of linearly independent motions. Linear
independence implies that a point of motion in a system can still move if all other points of motion are held
still; the number of these linearly independent motions is called the degrees of freedom of the system. For
mechanical systems with multiple degrees of freedom, use superposition: for each free-body diagram, hold
all other points of motion still and find the forces acting on the body only due to its own motion. Do this for
all of the bodies and the result will be a system of simultaneous equation of motion.
Example 2.13
Find the transfer function
Answer:
where
ଡ଼మ ሺୱሻ
୊ሺୱሻ
for the system shown below.
Gሺsሻ =
D=ቤ
Xଶ ሺsሻ f୴య ሺsሻ
=
Fሺsሻ
D
ൣMଵ s ଶ + ൫f୴భ + f୴య ൯s + ሺKଵ + K ଶ ሻ൧
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−൫f୴య s + K ଶ ൯
−൫f୴య s + K ଶ ൯
ቤ
ൣMଶ sଶ + ൫f୴మ + f୴య ൯s + ሺK ଶ + K ଷ ሻ൧
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The method by which the equation in each degree of freedom is obtained similar to the one done with
electrical networks, as
Sum of impedances
Sum of impedances
Sum of applied
൤
൨ Xଵ ሺsሻ − ൤
൨ Xଶ ሺsሻ = ൤
൨
connected to the motion at xଵ
between xଵ and xଶ
forces at xଵ
Sum of impedances
Sum of impedances
Sum of applied
൨ X ሺsሻ + ൤
൨ X ሺsሻ = ൤
൨
−൤
between xଵ and xଶ ଵ
connected to the motion at xଶ ଶ
forces at xଵ
(2.9a)
(2.9b)
This can also be expanded to systems with multiple degrees of freedom, as illustrated in the example
below.
Example 2.14
Write, but do not solve, the equations of motions for the mechanical network below.
Answer:
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Find the transfer function Gሺsሻ = Xଶ ሺsሻ/Fሺsሻ for the translational mechanical system shown below.
Example 2.15
Answer:
Drill Problems 2.4
Gሺsሻ =
1. Find the transfer function Gሺsሻ =
ଡ଼మሺୱሻ
୊ሺୱሻ
sሺsଷ
3s + 1
+ 7sଶ + 5s + 1ሻ
for the translational mechanical system shown below.
2. Find the transfer function Gሺsሻ = Xଶ ሺsሻ/Fሺsሻ for the translational mechanical network shown below.
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3. Find the transfer function Gሺsሻ = Xଷ ሺsሻ/Fሺsሻ for the translational mechanical system shown below.
4. Find the transfer function Gଵ ሺsሻ =
ଡ଼య ሺୱሻ
୊ሺୱሻ
for the translational mechanical system shown below.
5. Write, but do not solve, the equations of motion for the translational mechanical system shown below
Rotational Mechanical Systems. The rotational mechanical systems are handled the same way as
translational mechanical systems, except that the torque replaces force and angular displacement replaces
translational displacement. The mechanical components for rotational systems are the same as those for
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translational systems, except that the components undergo rotation instead of translation. Table 2.5 shows
the components along with the relationships between torque, angular velocity and angular displacement.
For rotational systems, the mass is replaced by inertia, and its parameter, called the moment of inertia, J.
The concept of degrees of freedom carries over to rotational systems, except that we test a point of motion
by rotating it while holding still all other points of motion. Writing the equations of motion for rotational
systems is similar to writing them for translational systems; the only difference is that the free-body diagram
consists of torques rather than forces.
Find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ for rotational systems shown.
Example 2.16
Answer:
where
System Modeling
Gሺsሻ =
θଶ ሺsሻ K
=
Tሺsሻ
D
ሺJ sଶ + Dଵ s + Kሻ
D= ฬ ଵ
−K
−K
ฬ
ሺJଵ sଶ + Dଵ s + Kሻ
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Feedback and Control Systems
Notice that the above system has that now well-known form
Sum of impedances
Sum of impedances
Sum of applied
൤
൨ θ ሺsሻ − ൤
൨ θ ሺsሻ = ൤
൨
connected to the motion at θଵ ଵ
between θଵ and θଶ ଶ
torques at θଵ
Sum of impedances
Sum of impedances
Sum of applied
൨ θ ሺsሻ + ൤
൨ θ ሺsሻ = ൤
൨
−൤
between θଵ and θଶ ଵ
connected to the motion at θଶ ଶ
torques at θଶ
(2.10a)
(2.10b)
Example 2.17
Write, but do not solve, the Laplace transform of the equations of motion for the system shown below.
Answer:
Find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ for the rotational mechanical system shown below.
Example 2.18
Answer:
System Modeling
Gሺsሻ =
1
2sଶ + s + 1
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Feedback and Control Systems
Systems with Gears. Gears provide mechanical advantage to rotational systems. Gears allow a system to
match the driving torque and the load – a trade-off between speed and torque. The linearized interaction
between gears is seen in figure 2.4.
Figure 2.4. A gear system
An input gear with radius rଵ and Nଵ teeth is rotated through angle θଵ ሺtሻ due to a torque Tଵ ሺtሻ. An output
gear with radius rଶ and Nଶ teeth responds by rotating through angle θଶ ሺtሻ and delivering a torque Tଶ ሺtሻ.
The distance traveled along each gear’s circumference is the same, thus
or
rଵ θଵ = rଶ θଶ
(2.11)
θଶ rଵ Nଵ
= =
θଵ rଶ Nଶ
(2.12)
since the ratio of the number of teeth along the circumference is in the same proportion as the ratio of the
radii. Thus, the ratio of the angular displacement of the gears is inversely proportional to the ratio of the
number of teeth. For the input and output torques, assuming that the gears are lossless (they neither
absorb nor store energy),
Tଶ θଵ Nଶ
=
=
Tଵ θଶ Nଵ
(2.13)
Figure 2.5. Transfer functions for angular displacement and torques in lossless gears.
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Feedback and Control Systems
Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical
impedance by the ratio
ଶ
Number of teeth of
gear on the ‫ ܖܗܑܜ܉ܖܑܜܛ܍܌‬shaft
൲
൮
Number of teeth of
gear on the ‫ ܍܋ܚܝܗܛ‬shaft
Example 2.19
Represent the rotational mechanical system shown to an equivalent system without gears.
Answer:
Find the transfer function Gሺsሻ = θଶ ሺsሻ/Tଵ ሺsሻ for the system shown below
Example 2.20
Answer:
System Modeling
Gሺsሻ =
θଶ ሺsሻ
Nଶ /Nଵ
= ଶ
Tଵ ሺsሻ Jୣ s + Dୣ s + K ୣ
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Feedback and Control Systems
where
Nଶ ଶ
Jୣ = Jଵ ൬ ൰ + Jଶ
Nଵ
Dୣ = Dଵ ൬
Nଶ ଶ
൰ + Dଶ
Nଵ
Kୣ = Kଶ
In order to eliminate gears with large radii, a gear train is used to implement large gear ratios by cascading
smaller gear ratios. Also, gears may exhibit inertia and damping. Handling of cases when the gears are
non-ideal is illustrated in the next example.
Find the transfer function Gሺsሻ = θଵ ሺsሻ/Tଵ ሺsሻ for the system shown below.
Example 2.21
Answer:
where
and
System Modeling
Gሺsሻ =
1
θଵ ሺsሻ
= ଶ
Tଵ ሺsሻ Jୣ s + Dୣ s
Jୣ = Jଵ + ሺJଶ + Jଷ ሻ ൬
Nଵ ଶ
Nଵ Nଷ ଶ
൰ + ሺJସ + Jହ ሻ ൬
൰
Nଶ
Nଶ Nସ
Nଵ ଶ
Dୣ = Dଵ + Dଶ ൬ ൰
Nଶ
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Feedback and Control Systems
Find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ for the rotational mechanical system with gears shown in
Example 2.22
the figure below.
Answer:
Gሺsሻ =
sଶ
1/2
+s+1
Example 2.23
Gሺsሻ = Xሺsሻ/Tሺsሻ
For the combined translational and rotational mechanical system shown below, find the transfer function
Answer:
System Modeling
Xሺsሻ
8
=
ଶ
Tሺsሻ 59s + 13s + 6
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Feedback and Control Systems
1. For each of the rotational mechanical systems shown below, determine the transfer function Gሺsሻ =
Drill Problems 2.5
θଵ ሺsሻ/Tሺsሻ.
2. For the rotational mechanical system shown below, find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ.
3. Find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ for the rotational mechanical system shown in the figure
below.
4. For the rotational mechanical system shown, find the transfer function Gሺsሻ = θଶ ሺsሻ/Tሺsሻ
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Feedback and Control Systems
function Gሺsሻ = Xሺsሻ/Tሺsሻ.
5. Given the combined translational and rotational system shown in the figure below, find the transfer
2.6 Electromechanical System Transfer Function
Intended Learning Outcome: Obtain transfer function relating the output displacement to the input voltage
of electromechanical systems.
A motor is an electromechanical component that yields a displacement output for a voltage input, that is, a
mechanical output generated by an electrical input. The transfer function for the armature-controlled dc
servomotor will be derived. Its schematic and block diagram is shown in the figure below.
Figure 2.6. The schematic and block diagram representation of a DC motor.
electromagnet called the fixed field. A rotating circuit, called the armature, through which the current iୟ ሺtሻ
In the above figure, a magnetic field is developed by stationary permanent magnets or a stationary
flows, passes through this magnetic field at right angles and feels a force F = Bℓiୟ ሺtሻ, where B is the
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Feedback and Control Systems
magnetic field strength and ℓ is the length of the conductor. The resulting torque turns the rotor, the rotating
member of the motor.
There is another phenomenon that occurs in the motor: A conductor moving at right angles to a magnetic
field generates a voltage at the terminals of the conductor, equal to e = Bℓv, where e is the voltage and v
is the velocity of the conductor normal to the magnetic field. Since the current-carrying armature is rotating
in a magnetic field, its voltage is proportional to speed. Thus,
vୠ ሺtሻ = K ୠ
dθ୫ ሺtሻ
dt
(2.13)
wherevୠ ሺtሻ is the back electromotive force, K ୠ is the back emf constant (in volt-second/radian) and
ୢ஘ౣ ሺ୲ሻ
ୢ୲
= ω୫ ሺtሻ is the angular velocity of the motor. Taking Laplace transform,
Vୠ ሺsሻ = K ୠ sθ୫ ሺsሻ
(2.14)
The relationship between the armature current and, iୟ ሺtሻ, the applied armature voltage, eୟ ሺtሻ and the
back emf vୠ ሺtሻ is found by writing a loop equation around the Laplace transformed armature circuit.
R ୟ Iୟ ሺsሻ + Lୟ sIୟ ሺsሻ + Vୠ ሺsሻ = Eୟ ሺsሻ
(2.15)
The torque developed by the motor is proportional to the armature current;
T୫ ሺsሻ = K ୲ Iୟ ሺsሻ
(2.16)
whereK ୲ is the motor torque constant (in Newton-meters/ampere), which depends on the motor and
magnetic field characteristics. Solving equation 2.16 in terms of the armature current and substituting this
and the back emf of Equation 2.14 to equation 2.15,
ሺR ୟ + Lୟ sሻT୫ ሺsሻ
+ K ୠ sθ୫ ሺsሻ = Eୟ ሺsሻ
K୲
System Modeling
(2.17)
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Feedback and Control Systems
To solve for the transfer function of the motor Gሺsሻ = θ୫ ሺsሻ/Eୟ ሺsሻ, an expression for T୫ in terms of
θ୫ ሺsሻ must be obtained. The equivalent mechanical loading on a motor is shown in a figure below.
Figure 2.7. Equivalent mechanical loading on a DC motor.
J୫ is the equivalent inertia at the armature and includes both the armature inertia and the load inertia
reflected to the armature. D୫ is the equivalent viscous damping at the armature and includes both the
armature viscous damping and the load viscous damping reflected to the armature. From figure 2.7,
T୫ ሺsሻ = ሺJ୫ sଶ + D୫ sሻθ୫ ሺsሻ
(2.18)
Substituting this to equation 2.17 and assuming the armature inductance Lୟ is small compared to the
armature resistance R ୟ which is very usual for a DC motor, the transfer function of the motor is solved as
Gሺsሻ =
θ୫ ሺsሻ
K ୲ /ሺR ୟ J୫ ሻ
=
Eୟ ሺsሻ s ቂs + ଵ ቀD + ୏౪୏ౘ ቁቃ
୎ౣ
Note that equation 2.19 is of the form
Gሺsሻ =
୫
K
s ሺs + αሻ
ୖ౗
(2.19)
(2.20)
To find the constants J୫ and D୫ , refer to figure 2.8, with a motor with inertia Jୟ and damping Dୟ at the
armature driving a load consisting of inertia J୐ and damping D୐ .
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Feedback and Control Systems
Figure 2.8. DC motor driving a rotational load
If all of the inertias and damping values are known, the load inertia and load damping can be reflected back
to the armature as some equivalent inertia and damping to be added to armature inertia and armature
damping. Thus,
Nଵ ଶ
J୫ = Jୟ + J୐ ൬ ൰
Nଶ
D୫ = Dୟ + D୐ ൬
Nଵ ଶ
൰
Nଶ
(2.21a)
(2.21b)
The electrical constants of the motor can be obtained through a dynamometer test of the motor, where a
dynamometer measures the torque and speed of a motor under the condition of a constant applied voltage.
It can be shown that the relationship between motor torque T୫ when a voltage eୟ is applied and the motor
speed ω୫ is
T୫ = −
Kୠ K୲
K୲
ω୫ + eୟ
Rୟ
Rୟ
(2.22)
which is linear. The y-intercept is called the stall torque and is the torque of the motor when the angular
velocity is zero. Thus,
Tୱ୲ୟ୪୪ =
K୲
e
Rୟ ୟ
(2.23)
The x-intercept is called the no-load speed and is the angular velocity of the motor at zero torque. Thus,
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Feedback and Control Systems
ω୬୭ି୪୭ୟୢ =
eୟ
Kୠ
(2.24)
Thus if Tୱ୲ୟ୪୪ and the ω୬୭ି୪୭ୟୢ is known, through the dynamometer test, the electrical constants can be
found as
K ୲ Tୱ୲ୟ୪୪
=
Rୟ
eୟ
and
Kୠ =
eୟ
ω୬୭ି୪୭ୟୢ
(2.25)
(2.26)
The figure bellow shows the plot as a result of dynamometer test.
Figure 2.9. Torque-speed curves with an armature voltage as a parameter
Given the system and torque-speed curve below respectively, find the transfer function Gሺsሻ =
Example 2.24
θ୐ ሺsሻ/Eୟ ሺsሻ.
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Feedback and Control Systems
Answer:
Gሺsሻ =
θ୐ ሺsሻ
0.0417
=
Eୟ ሺsሻ sሺs + 1.667ሻ
Find the transfer functionGሺsሻ = θ୐ ሺsሻ/Eୟ ሺsሻ for the motor and the load shown below. The torque-speed
Example 2.25
curve is given by T୫ = −8ω୫ + 200 when the input voltage is 100 volts.
Answer:
System Modeling
Gሺsሻ =
1/20
sሾs + ሺ15/2ሻሿ
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Feedback and Control Systems
1. For the motor, load, and torque-speed curve shown, find the transfer function Gሺsሻ = θ୐ ሺsሻ/Eୟ ሺsሻ.
Drill Problems 2.6
inertia. Find the transfer function Gሺsሻ = θଶ ሺsሻ/Eୟ ሺsሻ.
2. The motor whose torque-speed characteristics and the load drives are shown. Some of the gears have
3. A dc motor 55 N-m of torque at a speed of 600 rad/s when 12 volts are applied. It stalls out at this
s/rad, respectively, find the transfer function Gሺsሻ = θ୐ ሺsሻ/Eୟ ሺsሻ of this motor if it drives an inertia
voltage with 100 N-m of torque. If the inertia and damping of the armature are 7 kg-m2 and 3 N-m-
load of 105 kg-m2 through a gear train as shown below.
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Feedback and Control Systems
4. In this chapter, the transfer function of a dc motor relating the angular displacement output to armature
input voltage. Often, we want to control the output torque rather than the displacement. Derive the
transfer function of the motor that relates output torque to input armature voltage.
5. Find the transfer function Gሺsሻ = Xሺsሻ/Eୟ ሺsሻ for the system shown below.
2.7. The General State-Space Representation and Application
Intended Learning Outcome: Define terms associated with state-space modeling of systems and
enumerate steps in obtaining the state-space representation of a system.
In previous articles, the frequency-domain, or the classical technique of modeling physical systems was
discussed. This approach is based on converting a system’s differential equation to a transfer function, thus
generating a mathematical model of the system that algebraically relates the representation of the output to
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Feedback and Control Systems
a representation of the input. Replacing a differential equation with an algebraic equation not only simplifies
the representation of individual subsystems but also simplifies modeling interconnected subsystems.
This section presents a method by which an alternative to transfer functions in modeling dynamic systems
is presented; the method is called state-space modeling. It can be used for the same class of systems that
the transfer functions can model. The discussions presented here are on the introductory level; advanced
techniques for state-space modeling is beyond the scope of this course. Also, since state-space
representations of systems are expressed as matrices, the student may want to review matrices and linear
algebra.
The following are the steps in writing a state-space model for physical systems:
1. Select a particular subset of all possible system variables and call the variables in this subset state
variables.
2. For an nth-order system, write n simultaneous, first-order differential equations in terms of the
state variables. These systems of simultaneous differential equations are called state equations.
3. If the initial condition of all the state variables at t ଴ as well as the system input for t ≥ t ଴ is known,
the simultaneous differential equations for the state variables can be solved for t ≥ t ଴ .
4. The state variables with the system’s input will be algebraically combined and all of the other
system variables for t ≥ t ଴ will be found. The algebraic equations that will arise are called the
output equations.
5. The state equations and the output equations of the system are its state-space representation or
state-space model.
The application of such process is demonstrated using an RL and an RLC circuit.
Example 2.26
Find a possible state-space representation for the RL circuit shown below. Follow the steps enumerated
above.
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Feedback and Control Systems
Example 2.27
Find a possible state-space representation for the RLC circuit shown below.
If the system is linear, the state and output equations can be written in vector-matrix form. But before the
general state-space representation is presented, some definitions are in order.
•
A linear combination of n variables, x୧ , for i = 1 to n is given by the following sum, S,
S = k ୬ x୬ + k ୬ିଵ x୬ିଵ + ⋯ + Kଵ xଵ
where each K ୧ is a constant.
•
A set of variables is said to be linearly independent if none of the variables can be written as a
linear combination of the others.
•
A system variable is any variable that responds to an input or initial condition in a system.
•
The state variables are the smallest set of linearly independent system variables such that the
•
determine the value of all system variables for all t ≥ t ଴ .
A state vector is a vector whose elements are the state variables.
•
The state equations are a set of n simultaneous, first order differential equations with n variables
values of the members of the set at time t ଴ along with known forcing functions completely
•
The state space is the n-dimensional space whose axes are the state variables.
•
The output equation is the algebraic equation that expresses the output variables of a system as
where the n variables to be solved are the state variables.
linear combinations of the state variables and the inputs.
The general state-space representation of a linear system is given as
‫ܠ‬ሶ = ‫ ܠۯ‬+ ۰‫ܝ‬
System Modeling
(2.27a)
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Feedback and Control Systems
‫ = ܡ‬۱‫ ܠ‬+ ۲‫ܝ‬
(2.27b)
for t ≥ t ଴ and initial conditions, ‫ܠ‬ሺt ଴ ሻ, where:
‫ = ܠ‬state vector
‫ܠ‬ሶ =
‫ܠ܌‬
‫ܜ܌‬
= derivatives of the state vectors with respect to time
‫ =ܡ‬output vector
‫ = ܝ‬input or control vector
‫ = ۯ‬system matrix
۰ = input matrix
۱= output matrix
۲ = feedforward matrix
The state equation is given by equation 2.27a and the vector ‫ܠ‬, the state vector, contains the state
variables. As discussed previously, if the initial conditions of the state variables and the input vector ‫ ܝ‬are
known, this equation can be solved for the state variables. Equation 2.27b, the output equation, can be
used to solve for other system variables.
The choice of state variables for a given system is not unique. However, they must satisfy the following
requirements:
1. They must be linearly independent.
2. The minimum number of state variables to completely describe the system is equal to the order of
the differential equation representing the system.
Although the rule specifies the minimum number of state variables to be identified, the number can exceed
this minimum as long as all the chosen state variables are linearly independent. In some cases, this can
simplify the writing of state equations and output equations.
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Feedback and Control Systems
Example 2.28
Write the state-space representation of the systems in examples 2.26 and 2.27 in vector matrix form.
Answers:
For the RL network, the state equation is
di
R
= − i + v ሺ tሻ
dt
L
and an output equation is
vୖ ሺtሻ = Riሺtሻ
Thus, letting xሶ = ቂୢ୲ቃ , x = ሾiሿ , A = ቂ− ୐ ቃ , B = ሾ1ሿ , y = ሾvୖ ሿ , C = ሾRሿ , D = ሾ0ሿ , and u = vሺtሻ .
ୢ୧
ୖ
R
xሶ = ൤− ൨ x + vሺtሻ
L
Therefore
y = ሾRሿx
For the RLC network, the state equations are
dq
=i
dt
di
1
R
1
=− q− i+ v
dt
LC
L
L
1
v୐ ሺtሻ = − q − Ri + v
C
and an output equation is
Thus, letting ‫ܠ‬ሶ = ൤
dq/dt
q
0
1
0
൨, ‫ = ܠ‬ቂ ቃ, ‫ = ۯ‬൤
൨, ۰ = ൤
൨, y = v୐ , ۱ = ሾ−1/C −Rሿ,
t
−1/LC −R/L
1/L
di/dt
D = ሾ1ሿ, u = v. Therefore
System Modeling
൤
dq/dt
0
1
q
0
൨=൤
൨ቂ ቃ + ൤
൨v
−1/LC −R/L t
1/L
di/dt
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Feedback and Control Systems
q
v୐ = ሾ−1/C −Rሿ ቂ ቃ + v
t
2.8 State-Space Representation of Electrical and Mechanical Systems
Intended Learning Outcome: Obtain the state-space model of electrical and mechanical systems.
This section discusses the methods by which electrical and mechanical systems will be modeled through
state-space. In the previous section, it is stated that the choice of state variables are arbitrary. This section
describes a systematic technique by which these state variables can be selected. The approach is done as
follows:
1. Write a simple derivative equation for each energy-storage element and solve for each derivative
term as a linear combination of any of the system variables and input that are present in the
equation.
2. Select the differentiated variable as a state variable.
3. Express all other system variables in the equations in terms of the state variables and the input.
4. The output variables are written as linear combinations of the state variables and the input.
These steps will be demonstrated for electrical and mechanical systems through the following examples.
Example 2.29
Find a state-space representation of the network shown below if the output is the current through the
resistor.
Answer:
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Feedback and Control Systems
For mechanical systems, it is more convenient to obtain the state-equations directly from the equations of
motion rather than from the energy-storage elements. Thus, in mechanical systems, the position and
velocity of each point of linearly independent motions are selected as state variables. The procedure is
illustrated in the following example.
Example 2.30
Find the state and output equations of the translational mechanical system shown below
Answers:
xଵ
1
‫ = ܠ = ܡ‬ቂx ቃ = ቂ
0
ଶ
0
0
xଵ
0 0ቃ vଵ
቎ ቏
1 0 x૛
v૛
Find the state-space representation of the electrical network shown. The output is v୭ ሺtሻ.
Example 2.31
Answers:
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Feedback and Control Systems
Represent the translational mechanical system shown below in state-space, where xଷ ሺtሻ is the output.
Example 2.32
Answers:
1. Represent the electrical network shown in the figure below in state-space, where v୭ ሺtሻ is the output.
Drill Problems 2.7
2. Find the state-space representation of the network shown below if the output is v୭ ሺtሻ.
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Feedback and Control Systems
3. Represent the system shown below in state-space where the output is xଷ ሺtሻ.
4. Represent the rotational mechanical system shown below in state-space where θଵ ሺtሻ is the output.
5. Represent the system shown below in state-space where the output is θ୐ ሺtሻ.
2.9 The Transfer Function and the State-Space Relationship
Intended Learning Outcome: Convert a transfer function into state-space representation and vice versa.
To convert a transfer function into state-space representation, state variables must be chosen. One choice
would be that of phase variables, that is, each subsequent state variable is defined to be the derivative of
the previous state variable.
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Feedback and Control Systems
Consider the differential equation
d୬ y
d୬ିଵ y
dy
+
a
+ ⋯ + aଵ + a଴ y = b଴ u
୬ିଵ
୬
୬ିଵ
dt
dt
dt
(2.28)
A convenient way to choose state variables is to choose the output, yሺtሻ and its ሺn − 1ሻ derivatives.
Choosing the state variables, x୧ ,
xଵ = y
xଶ =
xଷ =
⋮
Differentiating both sides yields
dy
dt
dଶ y
dt ଶ
d୬ିଵ y
x୬ = ୬ିଵ
dt
xሶ ଵ =
dy
= xଶ
dx
xሶ ଷ =
dଷ y
= xସ
dx ଷ
dଶ y
xሶ ଶ = ଶ = xଷ
dx
xሶ ୬ିଵ =
⋮
d୬ିଵ y
= x୬
dx ୬ିଵ
xሶ ୬ = −a଴ xଵ − aଵ xଶ − ⋯ − a୬ିଵ x୬ + b଴ u
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In vector-matrix form,
(2.28)
and the output equation is
(2.29)
Thus, to convert a transfer function into state equations, convert the transfer function to a differential
equation by cross-multiplication and taking the inverse Laplace transform, assuming zero initial conditions.
Then the differential equation is represented in state-space in phase-variable form.
Example 2.33
Cሺsሻ
24
= ଷ
Rሺsሻ s + 9sଶ + 26s + 24
Find the state-space representation in phase-variable form for the transfer function
Gሺsሻ =
Answer:
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Generating an equivalent block diagram based on the phase variable form of the state-space
representation of the previous example will yield
which means that the output is generated by integrators and constant-gain blocks. The previous example
has a constant numerator. The next example illustrates the procedure when the numerator is a polynomial
of lower degree than the denominator.
Example 2.34
Find the state-space representation of the transfer function
Answer:
System Modeling
Gሺsሻ =
s ଶ + 7s + 2
sଷ + 9sଶ + 26s + 24
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In the previous example, it can be seen that the denominator yields the state equations, while the
numerator yields the output equation. A similar equivalent block diagram can also be obtained for this
example, illustrating how the state-space representation is implemented using block diagrams, as in
Example 2.35
Find the state equations and the output equation for the phase-variable representation of the transfer
function
Gሺsሻ =
sଶ
2s + 1
+ 7s + 9
Answer:
To convert a state-space representation into its equivalent transfer function, given the state and output
equations,
‫ܠ‬ሶ = ‫ ܠۯ‬+ ۰‫ܝ‬
‫ = ܡ‬۱‫ ܠ‬+ ۲‫ܝ‬
System Modeling
(2.30a)
(2.30b)
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Take the Laplace transform, assuming zero initial conditions,
s‫܆‬ሺsሻ = ‫܆ۯ‬ሺsሻ + ۰‫܃‬ሺsሻ
‫܇‬ሺsሻ = ۱‫܆‬ሺsሻ + ۲‫܃‬ሺsሻ
For the first equation, solving for ‫܆‬ሺsሻ,
Substituting this to the second equation
‫܆‬ሺsሻ = ሺs۷ − ‫ۯ‬ሻି૚ ۰‫܃‬ሺsሻ
‫܇‬ሺsሻ = ሾ۱ሺs۷ − ‫ۯ‬ሻି૚ ۰ + ۲ሿ‫܃‬ሺsሻ
(2.31)
in which the quantity ۱ሺs۷ − ‫ۯ‬ሻି૚ ۰ + ۲ is called the transfer function matrix since it relates the output
‫܇‬ሺsሻ to the input ‫܃‬ሺsሻ. However, if the input and the output are scalars, then
Gሺsሻ =
Y ሺsሻ
= ۱ሺs۷ − ‫ۯ‬ሻି૚ ۰ + ۲
Uሺsሻ
(2.32)
Note that this equation evaluates an inverse of a matrix. A review on how to evaluate this must be done by
the student. One method is by
ሺs۷ − ‫ۯ‬ሻି૚ =
The method is illustrated in the following examples.
adjሺs۷ − ‫ۯ‬ሻ
det ሺs۷ − ‫ۯ‬ሻ
Example 2.36
Given the system defined as
Find the transfer function Gሺsሻ = Yሺsሻ/Uሺsሻ.
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Answer:
Gሺsሻ =
10ሺsଶ + 3s + 2ሻ
s ଷ + 3sଶ + 2s + 1
Example 2.37
Convert the state and output equations to transfer function.
Answer:
Gሺsሻ =
3s + 5
sଶ + 4s + 6
Drill Problems 2.8
1. Find the state-space representation in phase-variable form for the system shown below.
2. Write the state equations and the output equation for the phase-variable representation.
3. Represent the transfer function in state-space. Give the answer in vector-matrix form.
Gሺsሻ =
s ଶ + 3s + 8
ሺs + 1ሻሺsଶ + 5s + 5ሻ
4. Find the transfer function Gሺsሻ = Yሺsሻ/Rሺsሻ for the following system represented in state-space.
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5. Determine the transfer function Gሺsሻ = Yሺsሻ/Rሺsሻ equivalent to the state and output equations
shown below.
References:
N. Nise. (2011). Control Systems Engineering 6th Edition. United States of America: John Wiley & Sons.
R. Dorf& R. Bishop. (2011). Modern Control Systems 12th Edition. New Jersey: Prentice Hall.
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