Chapter 4
Continuous Time Signals
Time Response
Poles and Zeros
Poles:
(1) the values of the Laplace transform variable, s, that cause
the transfer function to become infinite
(2) Any roots of the denominator of the transfer function that
are common to roots of the numerator.
Zeros:
(1) the values of the Laplace transform variable, s, that cause
the transfer function to become zero
(2) Any roots of the numerator of the transfer function that
are common to roots of the denominator.
Figure 4.1
a. System showing
input and output;
b. pole-zero plot
of the system;
c. evolution of a
system response.
Follow blue arrows
to see the evolution
of the response
component generated
by the pole or zero.
Poles and Zeros
1)
2)
3)
4)
A pole of the input function generates the form
of the forced response ( steady state response)
A pole of the transfer function generates the form
of the natural response
A pole on the real axis generates an exponential
response of the form e-at
The zeros and poles generate the amplitudes for
both the forced and the natural responses.
Figure 4.2
Effect of a real-axis
pole upon transient
response
Ex: Evaluating response using poles
Problem:
Given the system, write the output, c(t), in general terms. Specify the
natural and forced parts of the solution.
R (s )
1
s

(s  3)
(s  2)(s  4)(s  5)
C (s )


By inspection, each system pole generates an exponential as part of the
natural response, and the input’s pole generates the s s response, thus
k1
C (s ) 
s

k3
k2
k4


(s  2) (s  4) (s  5)
c (t )  k 1

k 2e 2t  k 3e 4t  k 4e 5t
Forced
Response
Natural
Response
Figure 4.4
a. First-order system;
b. pole plot
Steady State Step Response
The steady state response (provided it exists) for a
unit step is given by
where G(s) is the transfer function of the system.
First Order Systems
We define the following indicators:
Steady state value, y: the final value of the step response
(this is meaningless if the system has poles in the RHP).
Time Constant, 1/a: The time it takes the step response to
rise to 63% of its final value. Or the time for e-at to decay
37% of its initial value.
First Order Systems
Rise time, Tr: The time elapsed for the waveform to go from
0.1 to 0.9 of its final value
Settling time, Ts: the time elapsed until the step response
enters (without leaving it afterwards) a specified deviation
band, ±, around the final value. This deviation , is
usually defined as a percentage of y, say 2% to 5%.
Overshoot, Mp: The maximum instantaneous amount by
which the step response exceeds its final value. It is
usually expressed as a percentage of y
Figure 4.3: Step response indicators
Figure 4.5
First-order system
response to a unit
step
Figure 4.6
Laboratory
results
of a system step
response test
Figure 4.7
Second-order
systems, pole
plots,
and step
responses
Figure 4.8
Second-order
step response
components
generated by
complex poles
Figure 4.9
System for
Example 4.2
Factoring the denominator we get s1 = -5+j13.23 & s2 = -5-j13.23
The response is damped sinusoid
Figure 4.10
Step responses
for second-order
system
damping cases
The General Second Order system
Natural Frequency
n
Is the frequency of oscillation of the system without damping
Damping Ratio

Exponential decay Frequency
1 Natural period (seconds)

Natural Frequency
2 Exponentail timec onstant
General System
n 2
b
G (s )  2
 2
s  as  b s  2n s  n 2
n  b
a  2n
Figure 4.11
Second-order
response as a
function of
damping ratio
Relating  and n to the pole locations we have
s1,2  n  n
 2 1
Figure 4.12 Systems for Example 4.4
n  b
Since
then
a  2n

a
2 b
Using values of a & b from each system we
find
 1.155
for (a)
overdamped
for (b)
critically damped
 1
underdamped
  0.894 for (c)
Figure 4.13
Second-order
underdamped
responses for
damping ratio
values
Figure 4.14
Second-order
underdamped
response
specifications
Figure 4.15
Percent
overshoot vs.
damping ratio
Figure 4.16
Normalized rise
time vs. damping
ratio for a
second-order
underdamped
response
Figure 4.17
Pole plot for an
underdamped
second-order
system
Figure 4.18
Lines of constant
peak time, Tp , settling
time,Ts , and percent
overshoot, %OS
Note: Ts < Ts ;
1
Tp < Tp 2; %OS
1<
2
1
%OS
2
Figure 4.19
Step responses
of second-order
underdamped
systems
as poles move:
a. with constant
real part;
b. with constant
imaginary part;
c. with constant
damping ratio
Figure 4.20 Pole plot for Example 4.6
Find  , n ,T p , % OS and T s
Solution:
  cos   cos[tan 1 (7 / 3)]  0.394
n  7 2  32  7.616
 
Tp 
  0.449 sec ond
d 7
% OS = e
 ( / (1 2 ) ) X 100 
26%
and approximate settling time is
4
4
Ts 
  1.333 seconds
d 3
Figure 4.22
The Cybermotion
SR3 security robot
on patrol. The
robot navigates by
ultrasound and path
programs transmitted
from a computer,
eliminating the need
for guide strips on
the floor. It has video
capabilities as well as
temperature, humidity,
fire, intrusion, and gas
sensors.
Courtesy of Cybermotion, Inc.
Figure 4.23
Component responses of
a three-pole system:
a. pole plot;
b. component
responses: nondominant
pole is near
dominant second-order
pair (Case I), far from the
pair (Case II), and
at infinity (Case III)
Figure 4.24 Step responses with additional poles
24.542
s 2  4s  24.542
245.42
T 2 (s ) 
(s  10)(s 2  4s  24.542)
73.626
T 3 (s ) 
(s  3)(s 2  4s  24.54)
T1 (s ) 
c1 (t )  1  1.09e 2t cos(4.532t  23.8 )
c 2 (t )  1  0.29e 10t  1.189e 2t cos(4.532t  53.34 )
c 2 (t )  1  1.14e 3t  0.707e 2t cos(4.532t  78.63 )
Note that c2(t) is better
approximate of c1(t) than
c3(t) since the 3rd pole is
farthest from dominant poles
Figure 4.25 Effect of adding a zero to a two-pole system
The system has 2
poles at -1+j2.828 and
-1-j2.828
Note: the farther the
location of the zero
from the dominant
poles the lesser its
effect on the response
Transfer Functions for Continuous
Time State Space Models
Taking Laplace transform in the state space model
equations yields
and hence
G(s) is the system transfer function.
Laplace transform solution: Example
Problem: Given the system represented in state space by the following equations: a)
solve the state equation and find output b) find eigenvalues and poles
1
0
 0
0
X   0
0
1  X  0  e t
 24 26 9 
1 
y  1 1 0 X
1 
X (0)  0 
 2 
Solution:
First we find (sI-A)-1 =
(s 2  9s  26)
(s  9)
1


2

24
(
s

9
s
)
s


2


24
s

(26
s

24)
s


s 3  9s 2  26s  24
Laplace transform solution: Example Cont.
Using U(s) = 1/(s+1), laplace transform of e-t, and B and x(0) from given equations, we
sustitue in X(s) = (sI-A)-1[x(0) + BU(s)], we get
(s 3  10s 2  37s  29)
X1 
(s  1)(s  2)(s  3)(s  4)
(2s 2  21s  24)
X2 
(s  1)(s  2)(s  3)(s  4)
s (2s 2  21s  24)
X3 
(s  1)(s  2)(s  3)(s  4)
The output is
 X 1 (s ) 
Y (s )  1 1 0  X 2 (s )   X 1 (s )  X 2 (s )
 X 3 (s ) 
(s 3  12s 2  16s  5)
Y (s ) 
(s  1)(s  2)(s  3)(s  4)
6.5
19
11.5
=


s 2 s 3 s 4
Laplace transform solution: Example Cont.
Taking Laplace transform of Y(s) we get
y (t )  6.5e 2t  19e 3t  11.5e 4t
b)
We set det(sI-A) = 0 to find poles and eigenvalues which are -2, -3, and -4
Solution of Continuous Time State Space Models
A key quantity in determining solutions to state
equations is the matrix exponential defined as
The explicit solution to the linear state equation is
then given by
Time domain solution, Example:
Problem: for the state equation and initial state vector shown,
where u(t) is a unit step. Find the state-transition matrix and then
solve for x(t).
0 1
0
x 
x (t )    u (t )

  8 6 
1 
Solution:
1 
x (0)   
0 
Find eigenvalues using det(sI-A) = 0. Hense s2+6s+8 = 0 and s1= -2 s2
= -4. Now we assume state transition matrix as
 (K 1e 2t  K 2e 4t ) (K 3e 2t  K 4e 4t ) 
(t )  
2t
4t
2t
4t 
(K 5e  K 6e ) (K 7e  K 8e ) 
Solve for the constants using (0)  I
and  (0)=A
 (2e 2t  e 4t ) (1 / 2e 2t  1 / 2e 4t ) 
(t )  

2t
4t
2t
4t
(1e  2e ) 
(4e  4e )
Time domain solution, Example: Cont.
Also,
(1 / 2e 2(t  )  1 / 2e 4(t  ) ) 
(t   )B   2(t  )

4(t  )
 2e
) 
(e
and
(2e 2t  e 4t ) 
(t )x (0)  
4t 
(4e  4e ) 
Then,
t
t


2t
2
4t
4
1 / 2e  e d   1 / 2e  e d  ) 
t
0
0



(
t


)
Bu
(

)
d


0
t
t


2t
2
4t
4
 e  e d   2e  e d  )



0
0
1 / 8  1 / 4e 2t  1 / 8e 4t 
=
2t
4t 
1 / 2e  1 / 2e 

Time domain solution, Example Cont.
Final result is,
t
x (t )  (t )x (0)   (t   )Bu ( )d 
0
1 / 8  7 / 4e 2t  7 / 8e 4t 

2t
4t 
-7/2e  7 / 2e 

State Space via Laplace transform Example:
Problem: Find the state transition matrix using laplace for the following
system
0 1
0
x 
x (t )    u (t )

  8 6 
1 
1 
x (0)   
0 
Solution: we find
(t ) as inverse Laplace transform of (sI - A )1
First find
1 
s
(sI  A )  

8
(
s

6)


For which
s  6 1  
 8 s   2
 s
(sI  A ) 1   2

s  6s  8 
 s 2
s 6
 6s  8
8
 6s  8
1

s 2  6s  8 

s

s 2  6s  8 
State Space via Laplace transform Example Cont.
Expanding each term in the matrix using partial fractions
 2
(


s

2
s
(sI  A ) 1  
( 4 
 s  2 s
1
1/ 2 1/ 2 
) (

)
4
s 2 s 4 

4
1
2 
) (

)
4
s  2 s  4 
Finally, taking the inverse Laplace transform, we obtain
 (2e 2t  e 4t ) (1 / 2e 2t  1 / 2e 4t ) 
(t )  

2t
4t
2t
4t
(e  2e ) 
(4e  4e )
Step Response for Canonical Second Order Transfer Function
On applying the inverse Laplace transform we
finally obtain
Figure 4.5: Pole location and unit step response of a
canonical second order system.
Summary

There are two key approaches to linear dynamic models:
 the, so-called, time domain, and
 the so-called, frequency domain

Although these two approaches are largely equivalent, they
each have their own particular advantages and it is
therefore important to have a good grasp of each.
Time domain

In the time domain,
 systems are modeled by differential equations
 systems are characterized by the evolution of their
variables (output etc.) in time
 the evolution of variables in time is computed by
solving differential equations
Frequency domain

In the frequency domain,
 modeling exploits the key linear system property that
the steady state response to a sinusoid is again a
sinusoid of the same frequency; the system only
changes amplitude and phase of the input in a fashion
uniquely determined by the system at that frequency,
 systems are modeled by transfer functions, which
capture this impact as a function of frequency.