@1a introductory concepts of control systems 2

What is a control system?

Information
INPUT aircraft, missiles,
about the
Desired to the economic
Performance Difference systems, cars, etc system:
ERROR system
REFERENCE OUTPUT

Controller System to be controlled
+
–

Objective: To make the system OUTPUT and the desired REFERENCE as close
as possible, i.e., to make the ERROR as small as possible.

Key Issues: 1) How to describe the system to be controlled? (Modeling)

2) How to design the controller? (Control)
11
Copyrighted by Ben M. Chen

Back to systems – block diagram representation of a system

u(t) y(t)
System

u(t) is a signal or certain information injected into the system, which is called the
system input, whereas y(t) is a signal or certain information produced by the
system with respect to the input signal u(t). y(t) is called the system output. For
example,

R1
input: voltage source

+ output: voltage across R2
+
u(t)  R2 y(t)
─ R2
y (t )   u (t )
R1  R2

16

Linear systems

u(t) y(t)
System

Let y1(t) be the output produced by an input signal u1(t) and y2(t) be the output
produced by another input signal u2(t). Then, the system is said to be linear if

a) the input is  u1(t), the output is  y1(t), where  is a scalar; and

b) the input is u1(t) + u2(t), the output is y1(t) + y2(t).

Or equivalently, the input is  u1(t) +  u2(t), the output is  y1(t) +  y2(t). Such a
property is called superposition. For the circuit example on the previous page,

  u1 (t )   u 2 (t )  
R2 R2 R2
y (t )  u1 (t )   u 2 (t )   y1 (t )   y 2 (t )
R1  R2 R1  R2 R1  R2

It is a linear system! We will mainly focus on linear systems in this course.
17

Example for nonlinear systems

Example: Consider a system characterized by

y (t )  100u 2 (t )

Step One:
y1 (t )  100u12 (t ) & y 2 (t )  100  u 2 (t )
2

Step Two: Let u (t )  u1 (t )  u 2 (t ), we have

y (t )  100u 2 (t )  100[u1 (t )  u2 (t )]2  100[u12 (t )  u2 (t )  2u1 (t )u2 (t )]
2

 y1 (t )  y2 (t )  200u1 (t )u2 (t )  y1 (t )  y2 (t )
The system is nonlinear.

Exercise: Verify that the following system

y (t )  cos(u (t ))

is a nonlinear system. Give some examples in our daily life, which are nonlinear.
18

Time invariant systems

u(t) y(t)
System

A system is said to be time-invariant if for a shift input signal u( t t0), the output of
the system is y( t t0). To see if a system is time-invariant or not, we test

a) Find the output y1(t) that corresponds to the input u1(t).

b) Let u2(t) = u1(tt0) and then find the corresponding output y2(t).

c) If y2(t) = y1(tt0), then the system is time-invariant. Otherwise, it is not!

In common words, if a system is time-invariant, then for the same input signal, the
output produced by the system today will be exactly the same as that produced
by the system tomorrow or any other time.

19

Example for time invariant systems

Consider the same circuit, i.e.,

R2
y (t )   u (t )
R1  R2

Obviously, whenever you apply a same voltage to the circuit, its output will always
be the same. Let us verify this mathematically.

Step One:
R2 R2
y1 (t )   u1 (t )  y1 (t  t 0 )   u1 (t  t 0 )
R1  R2 R1  R2

Step Two: Let u 2 (t )  u1 (t  t 0 ), we have

R2 R2
y 2 (t )   u 2 (t )   u1 (t  t 0 )  y1 (t  t 0 )
R1  R2 R1  R2

By definition, it is time-invariant!
20

Example for time variant systems

Example 1: Consider a system characterized by

y (t )  cos(t )u (t )

Step One:

y1 (t )  cos(t )  u1 (t )  y1 (t  t 0 )  cos(t  t 0 )  u1 (t  t 0 )

Step Two: Let u 2 (t )  u1 (t  t 0 ) , we have

y 2 (t )  cos(t )  u 2 (t )  cos(t )  u1 (t  t 0 )  y1 (t  t 0 )

The system is not time-invariant. It is time-variant!

Example 2: Consider a financial system such as a stock market. Assume that you
invest $10,000 today in the market and make $2000. Is it guaranteed that you will
make exactly another $2000 tomorrow if you invest the same amount of money? Is
such a system time-invariant? You know the answer, don’t you?
21

Systems with memory and without memory

u(t) y(t)
System

A system is said to have memory if the value of y(t) at any particular time t1 depends
on the time from   to t1. For example,

+ dy (t ) 1
t

u(t) ~ C y(t) u (t )  C
dt
 y (t ) 
C  u(t )dt

─

On the other hand, a system is said to have no memory if the value of y(t) at any
particular time t1 depends only on the time t1. For example,

R1 + R2
+ y (t )   u (t )
u(t)  R2 y(t) R1  R2
─

22

Causal systems

u(t) y(t)
System

A causal system is a system where the output y(t) at a particular time t1 depends on
the input for t  t1. For example,

+ dy (t ) 1
t

u(t) ~ C y(t) u (t )  C
dt
 y (t ) 
C  u( )d

─

On the other hand, a system is said to be non-causal if the value of y(t) at a particular
time t1 depends on the input u(t) for some t > t1. For example,

y (t )  u (t  1)

in which the value of y(t) at t = 0 depends on the input at t = 1.

23

System stability

u(t) y(t)
System

The signal u(t) is said to be bounded if |u(t)| <  <  for all t, where  is real scalar.
A system is said to be BIBO (bounded-input bounded-output) stable if its output y(t)
produced by any bounded input is bounded.

A BIBO stable system:

y (t )  e u ( t )  | y (t ) |  e u ( t )  e   e   

A BIBO unstable system:

t t t

y (t )   u( )d

Let u (t )  1, which is bounded. Then, y (t )   u( )d   d  
 

24

Linear differential equations

General solution:

n th order linear d n xt  d n 1 xt 
differential equation  a n 1    a0 xt   u t 
dt n dt n 1

General solution xt   x ss t   xtr t 

Steady state response x ss t   particular integral obtained from assuming
with no arbitrary solution to have the same form as u t 
constant

Transient response with xtr t   general solution of homogeneou s equation
n arbitrary constants d n xtr t  d n 1 xtr t 
 a n 1    a0 xtr t   0
dt n dt n 1

33

General solution of homogeneous equation:

n th order linear d n xtr t  d n 1 xtr t 
homogeneous equation n
 a n 1 n 1
   a0 xtr t   0
dt dt

Roots of polynomial roots : z1 , ,z n
from homogeneous
equation given by z z1  z z n   z n  an 1 z n 1    a0

General solution xtr t   k1e z1t    k n e z n t
(distinct roots)

General solution  
xtr t   k1  k 2 t  k 3t 2 e13t  k 4  k 5t  e 22t  k 6 e 31t  k 7 e 41t
(non-distinct roots) if roots are 13, 13, 13, 22 , 22 , 31, 41

34

Particular integral:

x ss t  Any specific solution (with no arbitrary constant)
of
d n xt  d n 1 xt 
 an 1    a0 xt   u t 
dt n dt n 1
Method to determine Trial and error approach: assume x ss t  to have
x ss t  the same form as u t  and substitute into
differential equation

Example to find x ss t  for Try a solution of he 3t
dxt  dx t 
2 xt   e3t dt
 2 x t   e 3t 3he 3t 2he 3t e 3t  h0.2
dt
x ss t   0.2e 3t

Standard trial solutions u t  trial solution for xss t 
et het
t ht
tet h1h2t et
a cos t   b sin  t  h1 cos t   h2 sin  t 

35

Time-domain System Models
&
Dynamic Responses

36

RL circuit and governing differential equation

Consider determining i(t) in the following series RL circuit:

i (t)
t=0 5

3V 7H v(t)

where the switch is open for t < 0 and is closed for t  0.

Since i(t) and v(t) will not be equal to constants or sinusoids for all time, these
cannot be represented as constants or phasors. Instead, the basic general
voltage-current relationships for the resistor and inductor have to be used:

37

For t < 0 5 i (t)

i (t)
t =0 5
t<0
3 v(t) = 7
d i(t) 5 i (t)
7
dt
i (t) = 0
5

3 7 d i(t)
v(t) = 7
dt

3 5 i (t) = 0

i (t) = 0
voltage cross
5
over the switch
d i(t)
3 KVL 7 v(t) = 7 =0
dt
38

t 0
0 5 i (t)

i (t)
5

3 7 d i (t)
v(t) = 7
dt

Applying KVL: Mathematically, the above d.e. is often
written as
di t 
7  5 i t   3, t  0
di t 
 5 i t   u t , t  0
dt
7
dt
and i(t) can be found from determining the
general solution to this first order linear where the r.h.s. is u t   3, t  0
differential equation (d.e.) which governs and corresponds to the dc source or
the behavior of the circuit for t  0. excitation in this example.
39

Steady state response

iss t  
3
Since the r.h.s. of the governing d.e. , t 0
5
dit 
7  5it   u t   3, t  0
dt

Let us try a steady state solution of diss t  d 3 3
7 5iss t   7    5   3, t0
dt dt  5  5
iss t   k , t  0
and is a solution of the governing d.e.
which has the same form as u(t), as a
possible solution.
In mathematics, the above solution is
called the particular integral or solution
di t 
7 ss  5iss t   3
dt and is found from letting the answer to
 70   5k   3 have the same form as u(t). The word
3 "particular" is used as the solution is only
 k
5
one possible function that satisfy the d.e.
40

In circuit analysis, the derivation of iss(t) by letting the answer to have the same form
as u(t) can be shown to give the steady state response of the circuit as t  .

t 

i (t) = k Using KVL, the steady state
5
response is
3 7 d i(t)
v(t) = 7
dt

3  0  5k  0  5k
3
k
5 i (t) = 5 k
5
3
 it   , t  
i (t) = k 5
5

3 d i(t)
7 v(t) = 7 =0
dt This is the same as iss(t).

41

Transient response

To determine i(t) for all t, it is necessary to find the complete solution of the
governing d.e.
di t 
7  5i t   u t   3, t  0
dt

From mathematics, the complete solution can be obtained from summing a
particular solution, say, iss(t), with itr(t): i t   iss t   itr t , t  0

where itr(t) is the general solution of the homogeneous equation

di t 
5
 t
7  5i t   0, t  0 i tr t   k 1 e z1 t
 k1 e 7 , t0
dt

di t  where k1 is a constant (unknown now).
7 tr  5itr t 
dt ditr t  5
replaced by z  t
dt
itr t   k1e 7  0, t
 7 z  5z  7 z  5
1 0

Thus, it is called transient response.
5
z1   42
7 Copyrighted by Ben M. Chen

Complete response

To see that summing iss(t) and itr(t) gives the general solution of the governing ODE

di t 
7  5i t   3, t  0
dt
note that
3 d  3   5 3   3,
i ss t   , t0 satisfies 7     t0
5 dt 5 5
5
 t
d  7t   7t 
5 5
itr t   k1e 7 , t 0 satisfies 7  k1e   5 k1e   0, t  0
dt 








 t
5
3  t 
5
3  t 
5
i ss t   itr t   3  k 1 e 7 , t  0 satisfies 7
d   k1e 7   5  k1e 7   3
5 dt 5  5 
   

5
3  t
i t   iss t   itr t    k1e 7 , t  0 is the general solution of the ODE
5
43

3
5
i ss ( t )
0 steady state
response
t <0 t 0
Switch
close

k1
5t
i tr ( t ) = k 1e 7 , t0
k1 transient response
e
0

t =0 t = 7 (Time constant) k1 is to be
5
determined later
k1 + 3
5
i ss ( t ) + i tr ( t )

3 complete response
5

0 44
Complete response

Note that the time it takes for the transient or zero-input response itr(t) to decay to
1/e of its initial value is
7
Time taken for itr(t) to decay to 1/e of initial value 
5

and is called the time constant of the response or system. We can take the
transient response to have died out after a few time constants. For the RC circuit,

100 At the time equal to 3 time
90
constants, the magnitude is
about 5% of the peak.
80

70
Magnitude in Percentage

At the time equal to 4 time
60
50
about 1.83% of the peak.
40
x 100/e
30
At the time equal to 5 time
20
10 about 0.68% of the peak
0
0 1 2 3 4 5 6 7 8 9 10
Time (second) 45
7/5

A cruise control system

 acceleration
x
A cruise-control
x displacement
system friction
force b x mass force u
m

By the well-known Newton’s Law of motion: f = m a, where f is the total force applied
to an object with a mass m and a is the acceleration, we have
b u
u  bx  m
 x   
x x

m m
This a 2nd order Ordinary Differential Equation with respect to displacement x. It can

be written as a 1st order ODE with respect to speed v = x :

b u
v
 v  model of the cruise control system, u is input force, v is output.
m m
60

Assume a passenger car weights 1 ton, i.e., m = 1000 kg, and the friction coefficient
of a certain situation b = 100 N·s/m. Assume that the input force generated by the car
engine is u = 1000 N and the car is initially parked, i.e., x(0) = 0 and v(0) = 0. Find the
solutions for the car velocity v(t) and displacement x(t).

For the velocity model,
model
b u
v
 v  v  0.1v  1

m m
The steady state response: It is obvious that vss = 10 m/s = 36 km/h

The transient response: Characteristic polynomial z + 0.1 = 0, which gives z1 = 0.1.
10
0.1t 0.1t
vtr (t )  k1e  v(t )  vss  vtr (t )  10  k1e
8

v(0) = 0 implies that k1 = 10 and hence

Velocity (m/s)
6

4
 0.1t
v(t )  vss  vtr (t )  10  10e
2

61
What is the time constant for this system? 0
0 20 40 60 80 100
Time (second)

For the dynamic model in terms of displacement,

u  bx  m
 x    0.1x  1
x 

The steady state response: From the solution for the velocity, which is a constant, we
can conclude that the steady state solution for the displacement is xss = vsst = 10t.

The transient response: Characteristic polynomial z2 + 0.1z = 0, which gives z1 = 0.1
and z2 = 0. The transient solution is then given by

xtr (t )  k1e 0.1t  k 2 e 0t  k1e 0.1t  k 2

and hence the complete solution

x(t )  xss  xtr (t )  10t  k1e 0.1t  k 2  v(t )  x(t )  10  0.1k1e 0.1t


x(0) = 0 implies k1 + k2 = 0 and v(0) = 0 implies 10  0.1 k1 = 0. Thus, k1 = 100, k2 =  100.

x(t )  10t  100e 0.1t  100
62

Complete response for the car displacement

800

700

600

500
Displacement (m)

400

300

200

100

0
0 10 20 30 40 50 60 70 80
Time (second)

Exercise: Show that the car cruise control system is BIBO stable for its velocity model
and it BIBO unstable for its displacement model.
63

Behaviors of a general 2nd order system

Consider a general 2nd order system (an RLC circuit or a mechanical system or
whatever) governed by an ODE

a(t )  by (t )  cy (t )  u (t )
y 

Its transient response (or natural response) is fully characterized the properties of
its homogeneous equation or its characteristic polynomial, i.e.,

a(t )  by (t )  cy (t )  0  az 2  bz  c  0
y 

The latter has two roots at

two real distinct roots if b2  4ac > 0
 b  b 2  4ac
z1, 2  = two complex conjugate roots if b2  4ac < 0
2a
two identical roots if b2  4ac = 0

These different types of roots give different natures of responses. 64

Overdamped systems

Overdamped response is referred to the situation when the characteristic polynomial
has two distinct negative real roots, i.e., ab > 0 & b2  4ac > 0. For example,

(t )  6 y (t )  5 y (t )  0
y 
which has a characteristic polynomial,

z 2  6 z  5  0  z1, 2  1,  5  y (t )  k1e t  k 2 e 5t , y (t )  k1e t  5k 2 e 5t


Assume that y (0)  0, y (0)  4 , which implies
 1

0.8

y (0)  k1  k 2  0 0.6

k1  1, k 2  1 0.4

y (0)   k1  5k 2  4
 0.2

Magnitude
0
and thus, -0.2

y (t )  e t  e 5t -0.4

-0.6

-0.8
What is the dominating time constant? -1 65
0 1 2 3 4 5 6 7 8 9 10
Time (second)

Underdamped systems

Underdamped response is referred to the situation when the characteristic polynomial
has two complex conjugated roots negative real part, i.e., ab > 0 & b2  4ac < 0. For
example,
(t )  2 y (t )  101 y (t )  0
y 


z 2  2 z  101  0  z1, 2  1  j10  y (t )  k1e ( 1 j10 )t  k 2 e ( 1 j10 )t  e t (k1e j10t  k 2 e  j10t )
y (t )  e t [k1 (cos10t  j sin 10t )  k 2 (cos10t  j sin 10t )]  e t [(k1  k 2 ) cos10t  j (k1  k 2 ) sin 10t ]



y (0)  k1  k 2  0
j (k 1k 2 )  1  y (t )  e t sin 10t
y (0)  (k1  k 2 )  j10(k 1k 2 )  10


The time constant for such a system is determined by the exponential term.
66

Underdamped response

et
1

0.8

0.6

0.4

0.2
Magnitude

0

-0.2

-0.4

-0.6

-0.8

-1
0 1 2 3 4 5 6 7
 et
Time (second)

67

Critically damped systems

Critically damped response is corresponding to the situation when the characteristic
polynomial has two identical negative real roots, i.e., ab > 0 & b2  4ac = 0. For example,

(t )  2 y (t )  y (t )  0
y 

z 2  2 z  1  0  z1, 2  1  y (t )  k1e  t  k 2te t  e t (k1  k 2t )
y (t )  e t (k 2  k1  k 2t )


 1.4

1.2

y (0)  k1  1
k1  1, k 2  2
1

y (0)  k 2  k1  1
 0.8

Magnitude
0.6

and thus 0.4

y (t )  e t (1  2t ) 0.2

68
0
0 2 4 6 8 10
Time (second)

Never damped (unstable) systems

Never damped response is corresponding to the situation when the characteristic
polynomial has at least one root with a nonnegative real part. For example,

(t )  y (t )  0
y

z 2  1  0  z1, 2  1  y (t )  k1e  t  k 2 et  y (t )   k1e  t  k 2 et



450

400
y (0)  k1  k 2  2
k1  k 2  1 350

y (0)  k 2  k1  0
 300

Magnitude
250

and thus 200

y (t )  e t  et
150

100

50

It is an unstable system. We’ll study more on it. 0
69
0 1 2 3 4 5 6
Time (second)

Review of Laplace Transforms

70

Introduction

Let us first examine the following time-domain functions:

1 2

1.5

0.5 1

0.5

Magnitude
Magnitude

0 0

-0.5

-0.5 -1

-1.5

-1 -2
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
Time in Seconds Time in Seconds

A cosine function with a frequency f = 0.2 Hz. x (t )  cos0.4t   sin 0.8t  cos1.6t 
Note that it has a period T = 5 seconds. What are frequencies of this function?

Laplace transform is a tool to convert time-domain functions into a frequency-domain
ones in which information about frequencies of the function can be captured. It is
often much easier to solve problems in frequency-domain with the help of Laplace
transform. 71

Laplace Transform

Given a time-domain function f (t), the one-sided Laplace transform is defined as
follows:

F ( s )  L f (t )   f (t )e  st dt , s    j
0

where the lower limit of integration is set to 0 to include the origin (t = 0) and to capture
any discontinuities of the function at t = 0.

The integration for the Laplace transform might not convert to a finite solution for
arbitrary time-domain function. In order for the integration to convert to a final value,
which implies that the Laplace transform for the given function is existent, we need
  
 (  j ) t t  jt
| F ( s ) |   f (t )e dt   f (t )  e e dt   f (t )  e  t dt  
0 0 0

for some real scalar  = c . Clearly, the integration exists for all   c, which is called
the region of convergence (ROC). Laplace transform is undefined outside of ROC.
72

Example 1: Find the Laplace transform of a unit step function f(t) = 1(t) = 1, t  0.
  
1 1  1  1  1  1
F ( s )  1(t )e  st dt   e  st dt   e  st   e      e 0     0     1 
0 0 s 0 s  s  s  s  s

Clearly, the about result is valid for s =  + j with  > 0.

Example 2: Find the Laplace transform of an exponential function f (t) = e – a t, t  0.
   
1  s  a t 1
F ( s)   f (t )e  st dt   e at e  st dt   e  s  a t dt   e 
0 0 0
sa 0 sa
Again, the result is only valid for all s =  + j with  >  a.

imag axis imag axis

real axis real axis
0 a

ROC for Example 1 ROC for Example 2 73

Example 3: Find the Laplace transform of a unit impulse function  (t), which has the
following properties:

1.  (t) = 0 for t  0

2.  (0)  
 
3.   (t )dt  1,  f (t ) (t )dt  f (0) for any f (t)
 

4.  (t) is an even function, i.e.,  (t) =  (t)

Its Laplace transform is significant to many system and control problems. By definition,

L (t )    (t )e  st dt  e 0  1
0

Its ROC is the whole complex plane.

Obviously, impulse functions are non-existent in real life. We will learn from a tutorial
question on how to approximate such a function.
74

Inverse Laplace Transform

Given a frequency-domain function F(s), the inverse Laplace transform is to convert
it back to its original time-domain function f (t).
ROC
 1  j
f (t )  L1 F ( s ) 
1
 F ( s )e ds
st
1
2 j  1  j

This process is quite complex because it requires knowledge about complex analysis.

We use look-up table rather than evaluating these complex integrals.

The functions f(t) and F(s) are one-to-one pairing each other and are called Laplace
transform pair. Symbolically,

f (t )  F ( s )
75

Properties of Laplace transform:

1. Superposition or linearity:

La1 f1 (t )  a2 f 2 (t )  a1L f1 (t ) a2 L f 2 (t )  a1F1 ( s )  a2 F2 ( s )

Example: Find the Laplace transform of cos(t). By Euler’s formula, we have

cos(t ) 
2

1 j t
e  e  j t 
By the superposition property, we have

1
 
 1
  
Lcos(t )  L  e jt  e  jt   L e jt  e  jt
1

2  2 2

1  1   1  1 s  j  s  j s
       2
2  s  j   s  j  2 ( s  j )s  j  s   2
   

76

2. Scaling:

If F(s) is the Laplace transform of f(t), then

1 s
L f ( at )  F 
a a

Example: It was shown in the previous example that

Lcos(t ) 
s
s2   2
By the scaling property, we have

s  s 
1 
Lcos(2t ) 
1 2 2 s
  2  2
2 s 2
2s  4 2  s  4
2
 2
 
4  4 

which may also be obtained by replacing  by 2. 77

3. Time shift (delay): f(t)


f(t  a)
L f (t  a )  1(t  a)  e  as F ( s )
a

For a function delayed by ‘a’ in time-domain, the equivalence in s-domain is

multiplying its original Laplace transform of the function by eas.

Example:

Lcos t  
s s
 Lcos( (t  a ))  1(t  a )  e  as
s2   2 s2   2

78

4. Frequency shift:


Example: Given

cos(t ) 
s 
and sin(t ) 
s2   2 s2   2

Using the shift property, we obtain the Laplace transforms of the damped sine and
cosine functions as


 
L e  at cos(t ) 
sa
( s  a) 2   2
and  
L e  at sin(t ) 
( s  a) 2   2

79

5. Differentiation:

  Lf (t )  sL f (t )  f (0 )  sF ( s )  f (0 )
 df (t )    
L
 dt 
 d 2 f (t ) 
  L(t )  s L f (t )  sf (0 )  f (0 )  s F ( s )  sf (0 )  f (0 )
2   2  
L 2
f
 dt 

6. Integration:
t  1
L   f  d   L f (t )  F ( s )
1
0  s s

Example: The derivative of a unit step function 1(t) is a unit impulse function (t).

d
1(t )   (t )
dt

d  1 t  1
L (t )  L  1(t )  sL (t )  1(0  )  s  0  1
1
1 L (t )  L     d   L (t ) 
1
 dt  s 0   s s

80

7. Periodic functions:

If f (t) is a periodic function, then it can be represented as the sum of time-shifted
functions as follows:

f (t )  f1 (t )  f 2 (t )  f 3 (t )  
 f1 (t )  f1 (t  T )  f1 (t  2T )  

Applying time-shift property, we obtain

F ( s )  F1 ( s )  F1 ( s )e  sT  F1 ( s )e  s 2T 
T  st
F1 ( s )  f (t )e dt
0
 F1 ( s )(1  e  sT  e  s 2T )   sT

1 e 1  e  sT
81

8. Initial value theorem:

Examining the differentiation property of the Laplace transform, i.e.,

  Lf (t )  sL f (t )  f (0 )  sF ( s )  f (0 )
 df (t )    
L
 dt 
we have
  df (t )   df (t )  st 
sF ( s )  f (0 )  L    e dt   e  st df (t )  0, as s  
 dt  0  dt 0

Thus,
f (0  )  sF ( s ), as s   or f (0  )  lim [ sF ( s )]
s 

This is called the initial value theorem of Laplace transform. For example, recall that

sa
f (t )  e  at cos(t )  F (s) 
( s  a) 2   2

s( s  a) s 2  as
f (0)  e cos(0)  1
0
 lim [ sF ( s )]  lim  lim 2 1
s  ( s  a )   s   s  2 as  ( a   )
2 2 2 2
s 

82

9. Final value theorem:

Examining again the differentiation property of the Laplace transform, i.e.,

  Lf (t )  sL f (t )  f (0 )  sF ( s )  f (0 )
 df (t )    
L
 dt 
we have
 
  df (t )  df (t )  st df (t ) 0t 
lim [ sF ( s )  f (0 )]  lim L    lim  e dt   e dt  f (t ) 0   f ()  f (0  )
s 0 s  0  dt  s  0 0  dt 0  dt

Thus, The result is only valid for a function whose
f ()  lim [ sF ( s )] F(s) has all its poles in the open left-half
s 0 plane (a simple pole at s = 0 permitted)!

This is called the final value theorem of Laplace transform. For example,
sa
f (t )  e  at cos(t )  F (s) 
( s  a) 2   2
 s( s  a) s 2  as
f ( )  e cos()  0  lim [ sF ( s )]  lim  lim 2 0
s 0 ( s  a )   s  0 s  2 as  ( a   )
2 2 2 2
s 0

83

Summary of Laplace transform properties

Property f (t) F (s)

Linearity

Scaling

Time shift

Frequency shift

Time derivative
t
Time integration  f  d
0
F1 ( s )
Time periodicity
1  e  sT
Initial value f (0  )

Final value

Convolution
84
Courtesy of Dr Melissa Tao

Some commonly used Laplace transform pairs

f (t )  F ( s) f (t )  F ( s)

 (t )  1 
sin t 
s2   2
1
1(t )  s
s cos t 
s2   2
1 s sin    cos
t  sin(t   ) 
s2 s2   2

tn 
n! s cos    sin 
cos(t   ) 
s n 1 s2   2

1 
e  at  e  at sin t 
sa s  a 2   2

te  at 
1 sa
e  at cos t 
s  a 2 s  a 2   2
85

Why Laplace transform?

Allows us to work with algebraic
equations rather than differential
equations.

Provides an easy way
to solve system
Applicable to a Laplace transform is
problems involving
wider variety of significant for many
initial conditions.
inputs than phasor reasons

analysis

Capable of giving us the total response (natural
and forced) of the circuit in one single operation.

86
Courtesy of Dr Melissa Tao

Frequency-domain Descriptions
of Linear Systems

87

Frequency domain model of linear systems – transfer functions

A linear system expressed in terms of an ODE is called a model in the time domain. In
what follows, we will learn that the same system can be expressed in terms of a
rational function of s, the Laplace transform variable. Recall the cruise control system

u  bx  m
 x  m  bx  u
x 
Assume that the initial conditions are zero. Taking Laplace transform on its both sides,

(ms 2  bs) X ( s )  ms 2 X ( s )  bsX ( s )  Lm  bx  Lu  U ( s )
x 
we obtain a rational function of s, i.e.,

X (s) 1 1
H (s)   2 X ( s )  H ( s )U ( s )  U (s)
U ( s ) ms  bs ms  bs
2

H(s) is the ratio of the system output (displacement) and input (force) in frequency
domain. Such a function is called the transfer function of the system, which fully
characterizes the system properties. 96

More example: the series RLC circuit

The ODE for the general series RLC circuit was derived earlier as the following:

d 2 vC t  dv t 
LC  RC C  vC t   v(t )
dt 2 dt
Assume that all initial conditions are zero (note that for deriving transfer functions, we
always assume initial conditions are zero!).

 d 2 vC t  dvC t  
( LCs  RCs  1)VC ( s )  L  LC
2
 RC  vC t   Lv(t )  V ( s )
 dt 2 dt 
VC ( s ) 1 1
H (s)   VC ( s )  H ( s )V ( s )  V ( s)
V ( s ) LCs 2  RCs  1 LCs  RCs  1
2

H(s) is the ratio of the system output (capacitor voltage) and input (voltage source) in
frequency domain. The circuit (or the system) is fully characterized by the transfer
function. If the H(s) and V(s) are known, we can compute the system output. As such,
it is important to study the properties of H(s)! 97

System poles and zeros

As we have seen from the previous examples, a general linear time-invariant system
can be expressed in a frequency-domain model or transfer function:

U(s) Y(s)
H(s)

with

N ( s ) bm s m  bm 1s m 1    b1s  b0
H (s)   n n 1
, mn
D( s) s  an 1s    a1s  a0

n is called the order of the system. The roots of the numerator of H(s), i.e., N(s), are
called the system zeros (because the transfer function is equal to 0 at these points),
and the roots of the denominator of H(s), i.e., D(s), are called the system poles (the
transfer function is singular at these points). It turns out that the system properties are
fully captured by the locations of these poles and zeros…
98

Examples:
1
1 m
The cruise control system has a transfer function H ( s )  2 
ms  bs s ( s  b )
m
It has no zero at all and two poles at s  0, s   b m

1
The RLC circuit has a transfer function H ( s ) 
LCs 2  RCs  1
It has no zero and two poles at

 RC  ( RC ) 2  4 LC  RC  ( RC ) 2  4 LC
s1  , s2 
2 LC 2 LC
which are precisely the same as the roots of the characteristic polynomial of its ODE.

5s 2  10 s  5 5( s  1) 2
The system H ( s )  3  has two zeros (repeated)
s  6 s  11s  6 ( s  1)( s  2)( s  3)
2

at s = 1 and three poles at s  1, s  2, s  3 , respectively.
99

Response to sinusoidal inputs

Let us consider the series RL circuit whose current governed by the following ODE

di t  I (s) 0 .2
5  5 i t   v t   ( 5 s  5) I ( s )  V ( s )  H (s)  
dt V (s) s  1

Let the voltage source be a sinusoidal v(t) = cos (2t), which has a Laplace transform

s 0 .2 0.2 s A Bs  C
V (s)   I ( s)  V (s)   2   2
s 4
2
s 1 s 1 s  4 s 1 s  4

Using the coefficient matching method, we have to match

( A  B ) s 2  ( B  C ) s  ( 4 A  C )  0 .2 s 0.04 s  0.16 0.04 0.04 s 0.08  2 0.04
I (s)     
s2  4 s  1 s 2  22 s 2  22 s  1
 A B  0  A  0.04
 
  B  C  0.2   B  0.04
4 A  C  0

C  0.16
 i (t )  0.04[cos(2t )  2 sin( 2t )  e t ]

The steady state response is the given by

iss (t )  0.04[cos(2t )  2 sin(2t )]  0.0894 cos(2t  63.4349 ). 100

Let us solve the problem using another approach. Noting that v(t) = cos (2t), which
has an angular frequency of  = 2 rad/sec, and noting that the transfer function

I (s) 0 .2
H (s)  
V (s) s  1

is nothing more than the ratio or gain between the system input and output in the
frequency domain. Let us calculate the ‘gain’ of H(s) at the particular frequency
coinciding with the input signal, i.e., at s = j with  = 2.

0 .2 0 .2
H ( j 2 )  H ( j )   2    0.0894 e  j1.1071  0.0894   63 .4349 
j  1   2 j 2  1

It simply means that at  = 2 rad/sec, the input signal is amplified by 0.0894 and its
phase is shifted by  63.4349 degrees. It is obvious then the system output, i.e., the
current in the circuit at the steady state is given by

iss (t )  0.0894 cos(2t  63.4349 ).

which is identical to what we have obtained earlier. Actually, by letting s = j, the
above approach is the same as the phasor technique for AC circuits. 101

Frequency response – amplitude and phase responses

It can be seen from the previous example that for an AC circuits or for a system with a
sinusoidal input, the system steady state output can be easily evaluated once the
amplitude and phase of its transfer function are known. Thus, it is useful to ‘compute’
the amplitude and phase of the transfer function, i.e.,

H ( j  )  H ( s ) s  j  H ( j  )   H ( j  )

It is obvious that both magnitude and phase of H(j) are functions of . The plot of
|H(j)| is called the magnitude (amplitude) response of H(j) and the plot of H(j) is
called the phase response of H(j). Together they are called the frequency response
of the transfer function. In particularly, |H(0)| is called the DC gain of the system (DC is
equivalent to k cos(t) with  = 0).

The frequency response of a circuit or a system is an important concept in system
theory. It can be used to characterize the properties of the circuit or system, and used
to evaluated the system output response. 102

Example:

Let us reconsider the series RL circuit with the following transfer function:
I (s) 0 .2 0 .2 0 .2
H (s)    H ( j )   H ( j )   H ( j )    tan 1 
V (s) s  1 j  1 1 2
Thus, it is straightforward, but very tedious, to compute its amplitude and phase.
  0.01  H ( j )  0.2, H ( j )  0.57  0.2

0.15
  0.1  H ( j )  0.199,  H ( j )  5.71

Magnitude
0.1
  0.2  H ( j )  0.196,  H ( j )  11.31
0.05
  0.5  H ( j )  0.179,  H ( j )  26.57 

0
-2 -1 0 1 2 3
 1  H ( j )  0.141,  H ( j )  45 10 10 10 10 10 10
Frequency (rad/sec)
0
2  H ( j )  0.089,  H ( j )  63.43

 5  H ( j )  0.039,  H ( j )  78.69 Phase (degrees)
-50
  10  H ( j )  0.020,  H ( j )  84.29

  100  H ( j )  0.002,  H ( j )  89.43 -100
-2 -1 0 1 2 3
10 10 10 10 10 10
  1000  H ( j )  0.0002,  H ( j )  89.94 Frequency (rad/sec)
103

Note that in the plots of the magnitude and phase responses, we use a log scale
for the frequency axis. If we draw in a normal scale, the responses will look awful.

There is another way to draw the frequency response. i.e., directly draw both
magnitude and phase on a complex plane, which is called the polar plot. For the
example considered, its polar plot is given as follows:
0.1 0.2

0.08 0.15

Magnitude
0.06 0.1

0.04
0.05

0.02
0
imag axis

-2 -1 0 1 2 3
10 10 10 10 10 10
0
Frequency (rad/sec)
0
-0.02
-20

Phase (degrees)
-0.04
-40
-0.06
-60
-0.08 -80

-0.1 -100
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 10
-2 -1
10 10
0
10
1
10
2
10
3

read axis Frequency (rad/sec)

The red-line curves are more accurate plots using MATLAB. 104

What can we observe from the frequency response?

0.2

0.15

Magnitude
0.1

0.05

0
-2 -1 0 1 2 3
10 10 10 10 10 10
Frequency (rad/sec)
0

-20
Phase (degrees)

-40

-60

-80

-100
-2 -1 0 1 2 3
10 10 10 10 10 10
Frequency (rad/sec)

At low frequencies, magnitude
At low frequencies, magnitude For large frequencies, the magnitude
For large frequencies, the magnitude
response is relatively aalarge.
response is relatively large. response is small and thus signals
response is small and thus signals
Thus, the corresponding output
Thus, the corresponding output with large frequencies is attenuated
with large frequencies is attenuated
will be large.
will be large. or blocked.
or blocked.
105

Lowpass systems

0.2
1

0.8 0.15

0.6
0.1
0.4
0.05
0.2

-0.2
0
Lowpass System 0

-0.05

-0.4
-0.1
-0.6

  0.1 rad/sec
-0.15
-0.8

-1 -0.2
0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100
Time (second) Time (second)

0.04
1

0.8 0.03

0.6
0.02
0.4

0.2
0.01
0

-0.2
Lowpass System 0

-0.4
-0.01
-0.6

-0.8   10 rad/sec -0.02
0 1 2 3 4 5 6 7 8 9 10
-1
0 1 2 3 4 5 6 7 8 9 10 Time (second)
Time (second)

106

Bode plot

The plots of the magnitude and phase responses of a transfer function are called the
Bode plot. The easiest way to draw Bode plot is to use MATLAB (i.e., bode function).
However, there are some tricks that can help us to sketch Bode plots (approximation)
without computing detailed values. To do this, we need to introduce a scale called dB
(decibel). Given a positive scalar a, its decibel is defined as 20  log 10 ( a ) . For example,

a 1  20  log 10 ( a )  0  a  0 dB

a  10  20  log 10 ( a )  20  a  20 dB

a  100  20  log 10 ( a )  40  a  40 dB

a    20  log 10 (   )  20  log 10 ( )  20  log 10 (  )  a   in dB   in dB

a  20  log 10 (
 
 )  20  log 10 ( )  20  log 10 (  )  a   in dB   in dB

In the dB scale, the product of two scalars becomes an addition and the division of
two scalars becomes a subtraction. 107

Bode plot – an integrator

We start with finding the Bode plot asymptotes for a simple system characterized by
1 1 1
H (s)   H ( j )   H ( j )  H ( j )    90 
s j 
Examining the amplitude in dB scale, i.e.,
1
20  log 10 H ( j )  20  log 10  20  log 10  dB

it is simple to see that

  1  20  log 10 H ( j1)  20  log 10 1  0 dB

  10  20  log 10 H ( j10 )  20  log 10 10  20 dB

   2  101  20  log 10 H ( j 2 )  20  log 10  2  20  log 10 101
 20  log 10 10  20  log 10 1  20  20  log 10 1 dB

Thus, the above expressions clearly indicate that the magnitude is reduced by  20 dB
when the frequency is increased by 10 times. It is equivalent to say that the magnitude
times
108
is rolling off 20 dB per decade.

The phase response of an integrator is  90 degrees, a constant. The Bode plot of an
integrator is given by
Bode Diagram
20
rolling off
0
20 dB per
Magnitude (dB)

-20 decade

-40

-60
-89

-89.5
constant
Phase (deg)

-90
phase
-90.5 response
-91
-1 0 1 2 3
10 10 10 10 10
Frequency (rad/sec)
109

@1a introductory concepts of control systems 2

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