NAME 467: Control Engineering

Chapter 4: Time Response

Professor Dr. Zobair Ibn Awal

Bangladesh University of Engineering and Technology (BUET)

Dhaka 1000, Bangladesh

February 12, 2024

1/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Chapter 4: Time Response
Learning outcomes
1 Use poles and zeros of transfer functions to determine the
time response of a control system (Sections 4.1–4.2).
2 Describe quantitatively the transient response of first-order
systems (Section 4.3).
3 Write the general response of second-order systems given the
pole location (Section 4.4).
4 Find the damping ratio and natural frequency of a
second-order system (Section 4.5).
5 Find the settling time, peak time, percent overshoot, and rise
time for an underdamped second-order system (Section 4.6).

2/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Define poles and zeros of a transfer function and discuss their role
in system response.

3/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Definition of Pole and Zero

Definition of Pole Definition of Zero

The poles of a transfer function The zeros of a transfer function
are (1) the values of the Laplace are (1) the values of the Laplace
transform variable, s, that cause transform variable, s, that cause
the transfer function to become the transfer function to become
infinite or (2) any roots of the zero, or (2) any roots of the
denominator of the transfer numerator of the transfer
function that are common to function that are common to
roots of the numerator. roots of the denominator.

4/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Role in System Response
The output response of a system is the sum of two responses: the
forced response and the natural response1 . Although many
techniques, such as solving a differential equation or taking the
inverse Laplace transform, enable us to evaluate this output
response, these techniques are laborious and time-consuming.
Productivity is aided by analysis and design techniques that yield
results in a minimum of time. If the technique is so rapid that we
feel we derive the desired result by inspection, we sometimes use
the attribute qualitative to describe the method. The use of poles
and zeros and their relationship to the time response of a system is
such a technique. Learning this relationship gives us a qualitative
‘handle’ on problems. The concept of poles and zeros,
fundamental to the analysis and design of control systems,
simplifies the evaluation of a system’s response.
1
The forced response is also called the steady-state response or particular
solution. The natural response is also called the homogeneous solution.
5/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Analyze the roles of poles (×) and zeros (⃝) on the response of
(s+2)
the system G (s) = (s+5) to a unit step input.

6/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The system is shown in the following figure.

Given the transfer function G (s), a pole (×) exists at s = −5,

and a zero (⃝) exists at s = −2. These values are plotted on
the complex s-plane in the following figure using an × for poles
and a ⃝ for the zero.

7/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The response of the system is obtained by multiplying the transfer
function with the input function.

(s + 2) 1 A B
C (s) = G (s)R(s) = = +
(s + 5) s s (s + 5)
...............................................................

To find A To find B
A 1 (s + 2) (s + 2) B 1

s +2

s +2
= ⇒A= = ⇒B=
s s (s + 5) (s + 5) s +5 s s +5 s
(s + 2) (s + 2) (−5 + 2)
∴A= ∴B= =
(s + 5) s −5
s→0 s→−5
(0 + 2) 2 −3 3
= = = =
(0 + 5) 5 −5 5

...............................................................
8/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Thus,
2 3
A B
C (s) = + = 5 + 5
s (s + 5) s (s + 5)
Taking Laplace inverse
" #
2 3
−1 −1
L [C (s)] = L 5
+ 5
s (s + 5)
" # " #
2 3
−1 −1
=L 5
+L 5
s (s + 5)
   
2 −1 1 3 −1 1
= ×L + ×L
5 s 5 (s + 5)
2 3 −5t
∴ c(t) = + e
5 5

9/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The evolution of a system response can be seen in the following
figure. The arrows show the evolution of the response component
generated by the pole or zero.

10/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 From the development summarized in the figure above, we draw
the following conclusions:
1 A pole (×) of the input function generates the form of the
forced response (that is, the pole at the origin generated a
step function at the output).
2 A pole (×) of the transfer function generates the form of the
natural response (that is, the pole at −5 generated e −5t ).
3 A pole (×) on the real axis generates an exponential
response of the form e −αt , where −α is the pole location on
the real axis. Thus, the farther to the left a pole is on the
negative real axis, the faster the exponential transient
response will decay to zero (again, the pole at −5 generated
e −5t ).
4 The zeros (⃝) and poles (×) generate the amplitudes for
both the forced and natural responses (this can be seen
from the calculation of A and B).

11/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
(s + 3)
For the system G (s) = , write the output
(s + 2)(s + 4)(s + 5)
c(t) by inspection for a unit step input, in general terms. Specify
the forced and natural part of the solution.

12/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:
By inspection we can observe
the following:
The response of the system
C (s) due to a unit step
1
input R(s) = is
s
C (s) =R(s)G (s)
(s + 3)
=
s(s + 2)(s + 4)(s + 5)

The poles are at: 0 (due to

input) and -2,-4 and- 5
(due to system) and the
zeros are at: -3

13/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:
The input’s pole generates
K1
a forced response, say
s
The system’s pole
generates a natural
response in exponential
K2 K3
form, say , and
s +2 s +4
K4
s +5
The system zero influences
the amplitudes K1 , K2 , K3
and K4

14/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:
Hence the response in
s-domain 
K1
C (s) = +
 s Forced 
K2 K3 K4
+ +
s + 2 s + 4 s + 5 Natural
Response in time domain,

C (t) =[K1 ]Forced

+[K2 e −2t
+K3 e −4t
+K4 e −5t ]Natural

15/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
A system has a transfer function,
10(s + 4)(s + 6)
G (s) = . Write by inspection, the
(s + 1)(s + 7)(s + 8)(s + 10)
output, c(t), in general terms if the input is a unit step.

16/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

17/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 By inspection we can observe the following:
The response of the system C (s) due to a unit step input
1 10(s+4)(s+6)
R(s) = is C (s) = R(s)G (s) = s(s+1)(s+7)(s+8)(s+10)
s
The poles are at: 0 (due to input) and -1,-7,-8 and -10 (due
to system) and the zeros are at: -4, and -6
K1
The input’s pole generates a forced response, say
s
The system’s pole generates a natural response in exponential
K2 K3 K4 K5
form, say , , and
s +1 s +7 s +8 s + 10
The system zero influences the amplitudes K1 , K2 , K3 , K4 and
K5
Hence the response in s-domain
   
K1 K2 K3 K4 K5
C (s) = + + + +
s Forced s + 1 s + 7 s + 8 s + 10 Natural
Response in time domain,
C (t) = [K1 ]Forced + K2 e −t + K3 e −7t + K4 e −8t + K5 e −10t Natural
 

18/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 First Order System

19/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Discuss the performance specifications of a first order system
without zeroes for a step input.

20/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Let us discuss first-order systems without zeros to define a
performance specification. A first-order system without zeros can
be described by the transfer function given below.

C (s) a
G (s) = =
R(s) (s + a)

The block diagram and pole-zero plot of the system are as follows:

21/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 If the input is a unit step, where R(s) = 1/s, the Laplace
transform of the step response is C (s), where

a A B
C (s) = G (s)R(s) = = +
s(s + a) s (s + a)
...............................................................

To find A To find B
A a a B a a
= ⇒A= = ⇒B=
s s(s + a) (s + a) s +a s(s + a) s
a a a a
∴A= = ∴B= =
(s + a) (0 + a) s −a
s→0 s→−a
=1 = −1

...............................................................

22/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Thus,

A B 1 1
C (s) = + = −
s (s + a) s (s + a)
Taking the inverse transform, the step response is obtained by
     
−1 −1 1 1 −1 1 −1 1
L [C (s)] = L − =L −L
s (s + a) s (s + a)

∴ c(t) = 1 − e −at = cf (t) + cn (t)

Where, the input pole at the origin generated a forced response
cf (t) = 1, and the system pole at −a generated the natural
response cn (t) = −e −at .

23/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Let us now examine the significance of parameter a, the only
parameter needed to describe the transient response. When
t = 1/a,
e −at = e −a×1/a = e −1 = 0.37
t=1/a
or
c(t) = 1 − e −at = 1 − 0.37 = 0.63
t=1/a t=1/a

We now use the above description (three equations) to define three

transient response performance specification:
1 Time Constant
2 Rise Time, Tr
3 Settling Time, Ts

24/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Time Constant
The term 1/a is called the time constant. Time constant can
be described as the time for e −at to decay to 37 percent of its
initial value or it is the time it takes for the step response to rise to
63 percent of its final value.

25/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Time Constant

The reciprocal of the time constant has the units (1/seconds),

or frequency. Thus, we can call the parameter a the
exponential frequency. Since the derivative of e −at is −a
when t = 0, a is the initial rate of change of the exponential
at t = 0. Thus, the time constant can be considered a
transient response specification for a first order system, since
it is related to the speed at which the system responds to a
step input.
The time constant can also be evaluated from the pole plot.
Since the pole of the transfer function is at −a, we can say
the pole is located at the reciprocal of the time constant, and
the farther the pole from the imaginary axis, the faster
the transient response.

26/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Rise Time, Tr
Rise time is defined as the time for the waveform to go from 0.1 to
0.9 of its final value. Rise time is obtained by solving
c(t) = 1 − e −at for the difference in time at c(t) = 0.9 and
c(t) = 0.1.

27/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Rise Time, Tr

1 − e −at1 = 0.1 1 − e −at2 = 0.9

⇒ − e −at1 = 0.1 − 1 ⇒ − e −at2 = 0.9 − 1
⇒ − e −at1 = −0.9 ⇒e −at2 = −0.1
⇒ loge e −at1 = loge 0.9 ⇒ loge e −at2 = loge 0.1
⇒ − at1 = −0.11 ⇒ − at2 = −2.31
0.11 2.31
∴t1 = ∴t2 =
a a

2.31 0.11 2.2

Therefore, rise time is Tr = t2 − t1 = − =
a a a

28/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Settling Time, Ts
Settling time is defined as the time for the response to reach, and
stay within, 2 percent of its final value. Strictly speaking, this is
the definition of the 2 percent settling time. Other percentages, for
example 5 percent, also can be used. Letting c(t) = 0.98 and
solving for time t, we find the settling time to be as following.

29/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Settling Time, Ts

1 − e −at2 = 0.98
−at1
1−e =0 ⇒ − e −at2 = 0.98 − 1 = −0.02
−at1
⇒−e = −1 ⇒ loge e −at2 = loge 0.02
−at1
⇒ loge e = loge 1 ⇒ − at2 = −4
⇒ − at1 = 0.0 4
∴t2 =
∴t1 = 0 a

4 4
Therefore, settling time is Ts = t2 − t1 = −0=
a a

30/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
50
A system has a transfer function, G (s) = . Find the time
(s + 50)
constant (Tc ), rise time (Tr ) and settling time (Ts ).

31/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Comparing the expressions
K 50
G (s) = =
(s + a) (s + 50)

∴ a = K = 50
Time constant in seconds
1 1
= = 0.02
a 50
Rise time in seconds
2.2 2.2
Tr = = = 0.04
a 50
Settling time (98 percent) in seconds

4 4
Ts = = = 0.08
a 50

32/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Describe how first order transfer function can be constructed via
testing.

33/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
First order TF via experiment

Often it is not possible or practical to obtain a system’s transfer

function analytically. Perhaps the system is closed, and the
component parts are not easily identifiable. Since the transfer
function is a representation of the system from input to output,
the system’s step response can lead to a representation even
though the inner construction is not known. With a step input, we
can measure the time constant and the steady-state value, from
which the transfer function can be calculated.

34/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
First order TF via experiment
K
Consider a simple first-order system, G (s) = , whose step
(s + a)
response is
K A B
C (S) = G (s)R(s) = = +
s(s + a) s (s + a)
...............................................................

To find A To find B
A K B K
= =
s s(s + a) s +a s(s + a)
K K
⇒A = ⇒B =
(s + a) s
K K
⇒A = ⇒B =
(s + a) s
s→0 s→−a
K K K K
∴A = = ∴B = =−
(0 + a) a −a a
35/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Therefore,
K K /a K /a
C (S) = = −
s(s + a) s (s + a)
If we can identify K and a from laboratory testing, we can obtain
the transfer function of the system. However, the system has to
demonstrate first-order characteristics:
1 Non-zero initial slope.
2 No overshoot.
The following two steps will determine the values of K and a:
1 From the response, we measure the time constant, that is, the
time for the amplitude to reach 63 percent of its final value.
2 To find K , we realize that the forced response reaches a
steady state value of K /a. The value of K /a can be obtained
from the experiment (the steady state response). Substituting
the value of a, we will find K .

36/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Determine the first-order transfer function of the system from the
following experimental result.

37/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Laboratory results of a system step response test.

38/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

The system shows first

order characteristics. That
is, (1) Non-zero initial slope
and (2) No overshoot.
The final value is about
0.72.
The time constant is the
time to reach 63 percent of
the final value. That is the
time to reach the value of
Laboratory results of a system 0.63 × 0.72 = 0.45. It
step response test. takes about 0.13 seconds
to reach the value of 0.45.

39/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

Thus from time constant a

1
can be obtained, = 0.13
a
1
or a = = 7.7 per
0.13
second.
The steady state value is
K
= 0.72. Thus,
a
K = 0.72 × a =
0.72 × 7.7 = 5.54.
Laboratory results of a system Thus the transfer function
step response test. is,
5.54
G (s) =
(s + 7.7)

40/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Find the output response, c(t), for each of the systems shown in
the following figure. Also find the time constant, rise time, and
settling time for each case.

41/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution: (a)
The response due to a step input is
5 A B
C (s) = = +
s(s + 5) s (s + 5)

where, A = 1 and B = −1. Hence the response in time

domain is c(t) = 1 − e −5t
The given system is a 1st order system
a 5
Comparing the expressions: G (s) = = gives
(s + a) (s + 5)
a=5
1 1
Time constant (in seconds), Tc = = = 0.2
a 5
2.2 2.2
Rise time (in seconds), Tr = = = 0.44
a 5
4 4
Settling time (98 percent), in seconds, Ts = = = 0.8
a 5
42/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution: (b)
The response due to a step input is
20 A B
C (s) = = +
s(s + 20) s (s + 20)

where, A = 1 and B = −1. Hence the response in time

domain is c(t) = 1 − e −20t
The given system is a 1st order system
a 20
Comparing the expressions: G (s) = = gives
(s + a) (s + 20)
a = 20
1 1
Time constant (in seconds), Tc = = = 0.05
a 20
2.2 2.2
Rise time (in seconds), Tr = = = 0.11
a 20
4 4
Settling time (98 percent), in seconds, Ts = = = 0.2
a 20
43/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
44/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
For the system shown in the following figure, (1) find an equation
that relates settling time of the velocity of the mass to M; (2) find
an equation that relates rise time of the velocity of the mass to M.
Plot the step response using M = 1 and M = 2. Use the plots to
find the time constant, rise time, and settling time.

45/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:
The equation of motion can be
written as
dv (t)
M + 6v (t) = f (t)
dt
where, v (t) = ẋ(t).
Considering all initial conditions as zero, the Laplace transform
results,

MsV (s) + 6V (s) = F (s)

⇒V (s) (Ms + 6) = F (s)
1
V (s) 1 M K
⇒G (s) = = = =
F (s) Ms + 6 6 (s + a)
s+
M

46/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 k 1 6
G (s) = , where K = and a = . This is a first order
(s + a) M M
system whose properties can be determined as follows:
1 Settling time (98% of steady-state response),
4 M 2
Ts = = 4 × = M
a 6 3
2.2 M 1.1
2 Rise time, Tr = = 2.2 × = M
a 6 3

47/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
48/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
49/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Second Order System

50/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Discuss the four basic responses of second-order systems
attributable to pole locations.

51/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 1. Over-damped responses
Poles: Two real at σ1 , σ2
Natural response: Two
exponentials with time constants
equal to the reciprocal of the
pole locations, or

c(t) = K1 e −σ1 t + K2 e −σ2 t

52/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 2. Under-damped responses
Poles: Two complex at σd ± jωd
Natural response: Damped
sinusoid with an exponential
envelope whose time constant is
equal to the reciprocal of the
pole’s real part. The radian
frequency of the sinusoid, the
damped frequency of oscillation,
is equal to the imaginary part of
the poles, or

c(t) = Ae −σd t cos(ωd t − ϕ)

c(t) = K1 e −σd t cos ωd t

+K2 e −σd t sin ωd t

53/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 3. Undamped responses
Poles: Two imaginary at ±jω1
Natural response: Undamped
sinusoid with radian frequency
equal to the imaginary part of
the poles, or

c(t) = A cos(ω1 t − ϕ)

c(t) = K1 cos ω1 t + K2 sin ω1 t

54/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 4. Critically damped responses
Poles: Two real at −σ1
Natural response: One term is
an exponential whose time
constant is equal to the
reciprocal of the pole location.
Another term is the product of
time, t, and an exponential with
time constant equal to the
reciprocal of the pole location, or

c(t) = K1 e −σ1 t + K2 te −σ1 t

55/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
For each of the following transfer functions, write, by inspection,
400
the general form of the step responses: (1) G (s) = s 2 +12s+400 , (2)
900 225 625
G (s) = s 2 +90s+900 , (3) G (s) = s 2 +30s+225 and (4) G (s) = s 2 +625 .

56/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 1 √
−12 ± 122 − 4 × 1 × 400
s= = −6 ± j19.08
2
c(t) = K1 + K2 e −6t cos(19.08t + ϕ)
2 √
−90 ± 902 − 4 × 1 × 900
s= = −78.54, −11.46
2
c(t) = K1 + K2 e −78.54t + K3 e −11.46t
3 √
−30 ± 302 − 4 × 1 × 225
s= = −15, −15
2
c(t) = K1 + K2 e −15t + K3 te −15t
4
√
s = ± −625 = ±j25
c(t) = K1 + K2 cos(25t + ϕ)

57/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Exercise
For each of the transfer functions shown below, find the locations
of poles and zeros, plot them on the s-plane, and then write an
expression for the general form of the step response without
solving for the inverse Laplace transform. State the nature of each
2
response (over-damped, under-damped, and so on): T (s) = ,
s +2
5 10(s + 7)
T (s) = , T (s) = ,
(s + 3)(s + 6) (s + 10)(s + 20)
20 (s + 2) (s + 5)
T (s) = 2 , T (s) = 2 , T (s) =
s + 6s + 144 s +9 (s + 10)2

58/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

1 Pole: −2; c(t) = A + Be −2t ; First order response.

2 Poles: −3, −6; c(t) = A + Be −3t + Ce −6t ; Over-damped
response.
3 Poles: −10, −20; Zero: −7; c(t) = A + Be −10t + Ce −20t ;
Over-damped response.
√ √
4 Poles: (−3 + j3 15), (−3√ − j3 15);
c(t) = A + Be −3t cos (3 15t + ϕ); under-damped response.
5 Poles: j3, −j3; Zero: −2; c(t) = A + B cos (3t + ϕ);
Un-damped response.
6 Poles: −10, −10; Zero: −5; c(t) = A + Be −10t + Cte −10t ;
Critically damped response.

59/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Define natural frequency and damping ratio for a 2nd order system.
Discuss the time scale problem with regard to damping ratio.

60/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Note

Now that we have become familiar with second-order systems

and their responses, we generalize the discussion and establish
quantitative specifications defined in such a way that the
response of a second-order system can be described to a designer
without the need for sketching the response.

We define two physically meaningful specifications for

second-order systems. These quantities can be used to describe the
characteristics of the second-order transient response just as time
constants describe the first-order system response. The two
quantities are called natural frequency and damping ratio.

61/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Definition

The two important parameters for a second order system transient

responses are:
1 Natural frequency, ωn : The natural frequency of a
second-order system is the frequency of oscillation of the
system without damping.
2 Damping ratio, ζ: The damping ratio is the ratio of
exponential decay frequency of the envelope (σ) to the
natural frequency.

62/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Note

We have already seen that a second-order system’s

underdamped step response is characterized by damped
oscillations. Our definition is derived from the need to
quantitatively describe this damped oscillation regardless of
the time scale. Thus, a system whose transient response goes
through three cycles in a millisecond before reaching the steady
state would have the same measure as a system that went through
three cycles in a millennium before reaching the steady state.

63/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Note
For example, the underdamped curve in Figure 4.10 has an
associated measure that defines its shape. This measure remains
the same even if we change the time base from seconds to
microseconds or to millennia. A viable definition for this quantity is
one that compares the exponential decay frequency of the
envelope to the natural frequency. This ratio is constant
regardless of the time scale of the response. Also, the
reciprocal, which is proportional to the ratio of the natural period
to the exponential time constant, remains the same regardless of
the time base.

64/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Deduce the general form of 2nd order transfer function in terms of
natural frequency and damping ratio. Discuss the 2nd order system
transient responses as a function of damping ratio.

65/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The general second-order system shown in Figure 4.7(a) can be
transformed to show the quantities ζ and ωn . Consider the general
system

b
G (s) =
s 2 + as + b

Without damping, the poles would be on the jω-axis, and the

response would be an undamped sinusoid. For the poles to be
purely imaginary, a = 0. Hence,
b
G (s) =
s2 + b
By definition, the natural frequency, ωn , is the frequency of
oscillation of √
this system. Since the poles of this system are on the
jω-axis at ±j b, √
ωn = b
∴ b = ωn2
66/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Now to determine the term a, let us assume an under-damped
system, the complex poles have a real part, σ equal to −a/2. The
a
magnitude of this value (| − |) is then the exponential decay
2
frequency described in Section 4.4. Hence,
a
Exponential decay frequency |σ| |− |
ζ= = = 2 = a
Natural frequency (rad/second) ωn ωn 2ωn

from which
a = 2ζωn
Our general second-order transfer function finally looks like this:

ωn2
G (s) =
s 2 + 2ζωn s + ωn2

The above equation is also called the standard form of second

order system.

67/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 ωn2
G (s) =
s 2 + 2ζωn s + ωn2
It is now possible to deduce the root location in terms of damping
ratio and natural frequency. The roots of the characteristic
polynomial can be expressed as:
p
−2ζωn ± (2ζωn )2 − 4ωn2
s1,2 =
p2
−2ζωn ± 4ζ 2 ωn2 − 4ωn2
s1,2 =
2p
−2ζωn ± 2ωn ζ 2 − 1
s1,2 =
2p
s1,2 = −ζωn ± ωn ζ 2 − 1

68/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The following figure depicts the relation along with time response:

69/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 This section defined two specifications, or parameters, of
second-order systems: natural frequency, ωn , and damping ratio, ζ.
We saw that the nature of the response obtained was related to
the value of ζ. Variations of damping ratio alone yield the
complete range of overdamped, critically damped, under-damped,
and un-damped responses.

70/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
For each of the following systems find out ζ and ωn and comment
12 16
on the response: (1) G (s) = s 2 +8s+12 , (2) G (s) = s 2 +8s+16 , (3)
20 400
G (s) = s 2 +8s+20 , (4) G (s) = 2 , (5)
s + 12s + 400
900 225
G (s) = 2 , (6) G (s) = 2 , (7)
s + 90s + 900 s + 30s + 225
625
G (s) = 2 .
s + 625

71/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Deduce the time domain expression for a second-order
under-damped responses for various damping ratio values.

72/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 ωn2
G (s) =
s2 + 2ζωn s + ωn2
1
R(s) =
s
C (s) =?
c(t) =?
Now,
C (s) = R(s)G (s)
1 ωn2
⇒ C (s) = × 2
s s + 2ζωn s + ωn2
ωn2
⇒ C (s) =
s(s 2 + 2ζωn s + ωn2 )

73/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Let us assume the following form,

ωn2 K1 K2 s + K3
= + 2
s(s 2 + 2ζωn s + ωn2 ) s s + 2ζωn s + ωn2

⇒ ωn2 = K1 (s 2 + 2ζωn s + ωn2 ) + (K2 s + K3 )s ... (1)

Now, for s → 0 we obtain the following from equation (1),

ωn2 = K1 (02 + 2ζωn × 0 + ωn2 ) + (K2 × 0 + K3 ) × 0

⇒ ωn2 = K1 × ωn2
∴ K1 = 1

74/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Now, to find K2 and K3 let us put the value of K1 in equation (1)
and equate,

ωn2 = K1 (s 2 + 2ζωn s + ωn2 ) + (K2 s + K3 )s

⇒ ωn2 = 1 × (s 2 + 2ζωn s + ωn2 ) + (K2 s + K3 )s
⇒ ωn2 = s 2 + 2ζωn s + ωn2 + K2 s 2 + K3 s
⇒ ωn2 = s 2 (K2 + 1) + s(K3 + 2ζωn ) + ωn2

∴ s 2 (K2 + 1) + s(K3 + 2ζωn ) = 0 ... (2)

Now, balancing the coefficients in equation (2) gives, (K2 + 1) = 0
and (K3 + 2ζωn ) = 0. Thus,

K2 = −1

K3 = −2ζωn
Therefore, putting the values of K1 = 1, K2 = −1, and
K3 = −2ζωn in assumed form,
75/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 K1 K2 s + K3
C (s) = + 2
s s + 2ζωn s + ωn2
1 (−1)s + (−2ζωn )
⇒ C (s) = + 2
s s + 2ζωn s + ωn2
1 s + 2ζωn
⇒ C (s) = − 2
s s + 2ζωn s + ωn2
1 (s + ζωn ) + ζωn
⇒ C (s) = − 2
s (s + 2ζωn s + ζ 2 ωn2 ) − ζ 2 ωn2 + ωn2
1 (s + ζωn ) + ζωn
⇒ C (s) = −
s (s + ζωn )2 + ωn2 − ζ 2 ωn2
1 (s + ζωn ) + ζωn
⇒ C (s) = −
s (s + ζωn )2 + ωn2 (1 − ζ 2 )
ζ p
(s + ζωn ) + p (ωn 1 − ζ 2 )
1 1 − ζ2
⇒ C (s) = − p
s (s + ζωn )2 + (ωn 1 − ζ 2 )2

76/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Now we can write the above in the following form,
1 A(s + a) + Bω
C (s) = −
s (s + a)2 + ω 2
ζ p
Where, A = 1, a = ζωn , B = p , ω = ωn 1 − ζ2
1 − ζ2
Therefore,
1 A(s + a) + Bω
C (s) = −
s (s + a)2 + ω 2
1 A(s + a) Bω
⇒ C (s) = − 2 2
−
s (s + a) + ω (s + a)2 + ω 2
From the properties of Laplace transformation we know that
A(s + a)
L Ae −at cos ωt =
 
(s + a)2 + ω 2
Bω
L Be −at sin ωt =
 
(s + a)2 + ω 2
77/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Hence,
c(t) = L −1 [C (s)]
 
−1 1 A(s + a) Bω
=L − 2 2
−
s (s + a) + ω (s + a)2 + ω 2
= 1 − Ae −at cos ωt − Be −at sin ωt
= 1 − e −at (A cos ωt + B sin ωt)
!
p ζ p
∴ c(t) = 1−e −ζωn t cos ωn 1 − ζ 2t + p sin ωn 1 − ζ 2 t
1 − ζ2
A plot of this response appears in the figure below for various
values of ζ, plotted along a time axis normalized to the natural
frequency. We now see the relationship between the value of ζ and
the type of response obtained: The lower the value of ζ, the more
oscillatory the response. The natural frequency is a time-axis scale
factor and does not affect the nature of the response other than to
scale it in time.

78/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
79/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Define the following performance specifications for a second order
underdamped system: Rise Time, Peak Time, Percent Overshoot,
Settling Time and Delay Time.

80/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Rise time, Tr . The time required for the waveform to go from 0.1
of the final value to 0.9 of the final value.

81/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Peak time, Tp . The time required to reach the first, or maximum,
peak.

82/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Percent overshoot, %OS. The amount that the waveform
overshoots the steady state (or final) value, at the peak time,
expressed as a percentage of the steady-state value.

83/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Settling time, Ts . The time required for the transient’s damped
oscillations to reach and stay within ±2% of the steady-state value.

84/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Delay time, td . The delay time is the time required for the
response to reach half the final value the very first time.

85/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
86/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Notice that the definitions for settling time (Ts ) and rise time
(Tr ) are basically the same as the definitions for the
first-order response.
All definitions are also valid for systems of order higher than
two, although analytic expressions for these parameters cannot
be found unless the response of the higher-order system can
be approximated as a second-order system.
Rise time (Tr ), peak time (TP ), and settling time (Ts ) yield
information about the speed of the transient response. This
information can help a designer determine if the speed and the
nature of the response do or do not degrade the performance
of the system. For example, the speed of an entire computer
system depends on the time it takes for a hard drive head to
reach steady state and read data; passenger comfort depends
in part on the suspension system of a car and the number of
oscillations it goes through after hitting a bump.

87/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The transient response of the system may be described in terms
of two factors:
1 The swiftness of response, as represented by the rise time
and the peak time.
2 The closeness of the response to the desired response, as
represented by the overshoot and settling time.

88/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Deduce the expression of Peak Time for a second order
under-damped system.

89/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp

1 ωn2
R(s) = → G (s) = 2 → C (s)
s s + 2ζωn s + ωn2

Tp =?
Here,
C (s) = R(s)G (s)
1 ωn2
⇒ C (s) = × 2
s s + 2ζωn s + ωn2
ωn2
⇒ C (s) =
s(s 2 + 2ζωn s + ωn2 )

90/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp

Tp is found by differentiating time response, c(t) and finding the

first zero crossing after t = 0. This task is simplified by
‘differentiating’ in the frequency domain by using the following
Laplace transform property (Item 7 of Table 2.2):
 
df
L = sF (s) − f (0)
dt

Assuming zero initial conditions and using the equation c(t) we

get,  
dc(t)
L = sC (s) = 0
dt

91/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp

However,

ωn2
sC (s) = s ×
s(s 2 + 2ζωn s + ωn2 )
ωn2
=
s 2 + 2ζωn s + ωn2
ωn2
=
s 2 + 2ζωn s + ζ 2 ωn2 − ζ 2 ωn2 + ωn2
ωn2
=
(s + ζωn )2 − ζ 2 ωn2 + ωn2
ωn2
=
(s + ζωn )2 + ωn2 (1 − ζ 2 )

92/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp

ωn2
=  p 2
(s + ζωn )2 + ωn (1 − ζ 2 )
p
ω n (1 − ζ 2 )
ωn2 × p
ωn (1 − ζ 2 )
=  p 2
(s + ζωn )2 + ωn (1 − ζ 2 )
ω
p n
p
ωn (1 − ζ 2 )
(1 − ζ 2 )
=  p 2
(s + ζωn )2 + ωn (1 − ζ 2 )
Aωd
=
(s + a)2 + ωd2
ωn p
Where, A = p , ωd = ωn (1 − ζ 2 ) and a = ζωn .
(1 − ζ 2 )
93/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp
Hence,
 
dc(t) Aωd
L =
dt (s + a)2 + ωd2
 
dc(t) −1 Aωd
⇒ =L
dt (s + a)2 + ωd2
dc(t)
⇒ = Ae −at sin ωd t
dt
dc(t) ωn
q
⇒ =p e −ζωn t sin ωn (1 − ζ 2 )t
dt (1 − ζ 2 )

dc(t)
The condition = 0 to satisfy the above relation is therefore,
dt
p nπ
ωn 1 − ζ 2 t = nπ or, t = p
ωn 1 − ζ 2
94/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Peak Time, Tp

Each value of n yields the time the peak time, Tp , is found by

for local maxima or minima. letting n = 1 in the above
Letting n = 0 yields t = 0, the equation:
first point on the curve in π
Figure 4.14 that has zero slope. Tp = p
The first peak, which occurs at ωn 1 − ζ 2

95/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Deduce the expression of % Over Shoot for a second order
under-damped system.

96/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Percent Over Shoot, %OS

1 ωn2
R(s) = → G (s) = 2 → C (s)
s s + 2ζωn s + ωn2

%OS =?
We will use the general expression of time response, c(t), of a
second order under-damped system
!
−ζωn t
p ζ p
c(t) = 1 − e cos ωn 1 − ζ 2 t + p sin ωn 1 − ζ 2 t
1 − ζ2

97/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Percent Over Shoot, %OS
From the definition of percent over shoot we have the following,
cmax − cfinal
%OS = × 100
cfinal

The term cmax is found by evaluating c(t) at the peak time, c(Tp )
π
where Tp = p .
ωn 1 − ζ 2
!
p ζ p
c(t) = 1 − e −ζωn t × cos ωn 1 − ζ 2t + p sin ωn 1 − ζ 2t
1 − ζ2
⇒ c(Tp ) = 1 − e −ζωn Tp
!
p ζ p
× cos ωn 1 − ζ 2 Tp + p sin ωn 1 − ζ 2 Tp
1 − ζ2

98/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Percent Over Shoot, %OS

 
ζπ
− p  !
1 − ζ2 ζ
⇒ c(Tp ) = 1 − e cos π + p sin π
1 − ζ2
 
ζπ
− p 
⇒ c(Tp ) = 1 + e 1 − ζ2
 
ζπ
− p 
⇒ cmax = 1 + e 1 − ζ2

99/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Percent Over Shoot, %OS

For unit step, cfinal = 1 Hence,

cmax − cfinal
%OS = × 100
cfinal
 
− √ ζπ
1−ζ 2
1+e −1
⇒ %OS = × 100

1
− √ ζπ
1−ζ 2
∴ %OS = e × 100

100/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Percent Over Shoot, %OS
Notice that the percent overshoot is a function only of the
damping ratio, ζ. The equation is plotted in the following figure.

101/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Deduce the expression of Settling Time for a second order
under-damped system.

102/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Settling Time, Ts

1 ωn2
R(s) = → G (s) = 2 → C (s)
s s + 2ζωn s + ωn2

Ts =?
We will use the general expression of time response, c(t), of a
second order under-damped system
!
p ζ p
c(t) = 1 − e −ζωn t cos ωn 1 − ζ 2 t + p sin ωn 1 − ζ 2 t
1 − ζ2
ωd
−ζ p t
!
1−ζ 2 ζ
=1−e cos ωd t + p sin ωd t
1 − ζ2

103/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Settling Time, Ts
Let us assume,
!
ζ
A cos(ωd t − ϕ) = cos ωd t + p sin ωd t
1 − ζ2
Again,
A cos(ωd t − ϕ) = A (cos ωd t cos ϕ + sin ωd t sin ϕ)
ζ
Comparing these yields, A cos ϕ = 1 and A sin ϕ = p .
1 − ζ2

...............................................................
q
A = A2 cos2 ϕ + A2 sin2 ϕ
−1 A sin ϕ
ϕ = tan s
A cos ϕ ζ2 1
ζ = 1+ 2
=p
−1
= tan p 1 − ζ 1 − ζ2
1 − ζ2

104/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Settling Time, Ts
Hence,
ωd
−ζ p t
1 1 − ζ 2 cos(ω t − ϕ)
c(t) = 1 − p e d
1 − ζ2
In order to find the settling time, Ts we must find the time for
which the response c(t) in Eq. (4.28) reaches and stays within
±2% of the steady-state value, cfinal . Using our definition, the
settling time is the time it takes for the amplitude of the decaying
sinusoid in Eq. (4.28) to reach 0.02. A conservative approach,
cos(ωd t − ϕ) = 1, thus yields the following:
 p 
⇒ ζωn Ts = − ln 0.02 1 − ζ 2
1
e −ζωn Ts = 0.02
 p 
p − ln 0.02 1 − ζ 2
1 − ζ2 ⇒ Ts =
−ζωn Ts
p
2 ζωn
⇒e = 0.02 1 − ζ

105/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Settling Time, Ts

 p 
− ln 0.02 1 − ζ 2
Ts =
ζωn
From the above expression it can be verified that numerator varies
from 3.91 to 4.74 as ζ varies from 0 to 0.9. Let us agree on an
approximation for the settling time that will be used for all values
of ζ; let it be
4
Ts =
ζωn

106/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
How can we evaluate Rise Time for a second order under-damped
system?

107/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Rise Time, Tr

A precise analytic relationship between rise time, Tr and damping

ratio, ζ, cannot be found. However, the following approach can be
used to determine the rise time, Tr :
1 First designate ωn t as the normalized time variable and select
a value for ζ.
2 Solve for the values of ωn t that yield c(t) = 0.9 and
c(t) = 0.1.
3 Subtracting the two values of ωn t yields the normalized rise
time, ωn Tr , for that value of ζ.
4 Continuing this with other values of ζ, we obtain the results
plotted in Figure 4.16.

108/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Evaluation of Rise Time, Tr

109/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Given the transfer function
100
G (s) =
s2 + 15s + 100
find the Tp , %OS, Ts and Tr .

110/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

Comparing,

100 ωn2
G (s) = =
s 2 + 15s + 100 s 2 + 2ζωn s + ωn2

2ζωn = 15
ωn2 = 100 15
√ ⇒ζ=
⇒ ωn = 100 = 10 2 × ωn
15
⇒ζ= = 0.75
2 × 10

Therefore, ωn = 10 and ζ = 0.75 .

111/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution:

π π
1 Tp = p = √ = 0.475 sec.
ωn 1 − ζ 2 10 × 1 − 0.752
 
ζπ 0.75π
!
− p  −√
2 %OS = e 1 − ζ 2 × 100 = e 1 − 0.752 × 100 =
2.838  p   √ 
− ln 0.02 1 − ζ 2 − ln 0.02 1 − 0.752
3 Ts = = = 0.577
ζωn 0.75 × 10
sec.
1.76ζ 3 − 0.417ζ 2 + 1.039ζ + 1
4 Tr = = 0.229 sec.
ωn

112/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Discuss the significance of pole locations for an under-damped
second order system with respect to damping ratio, natural
frequency, peak time, settling time and percent overshoot.

113/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 The pole plot for a general, under-damped second-order system,
previously shown in Figure 4.11, is reproduced and expanded in
Figure 4.17 for focus.

p
s1,2 = −ζωn ± ωn ζ2 − 1
p
= −ζωn ± jωn 1 − ζ 2
= −σd ± jωd

Here, ωd is the imaginary part

of the pole and is called the
damped frequency of oscillation,
and σd is the magnitude of the
real part of the pole and is the
exponential damping frequency.

114/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 We see from the Pythagorean theorem that the radial distance
from the origin to the pole is the natural frequency, ωn , and the
cos θ = ζ.

s1,2 = −σd ± jωd

q
Hyp = (−σd )2 + (ωd )2
q
= ζ 2 ωn2 + ωn2 (1 − ζ 2 )
q
= ζ 2 ωn2 + ωn2 − ζ 2 ωn2 )
= ωn
ζωn
cos θ = =ζ
ωn

115/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 4 π
Ts = =
ζωn σd

Settling time (Ts ) is inversely

proportional to the real part of
the pole. Since vertical lines on
the s-plane are lines of constant
real value, they are also lines of
constant settling time.

116/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 π π
Tp = p =
ωn 1 − ζ 2 ωd

Peak Time (Tp ) is inversely

proportional to the imaginary
part of the pole. Since
horizontal lines on the s-plane
are lines of constant imaginary
value, they are also lines of
constant peak time.

117/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 ζ = cos θ
Finally, radial lines are lines of
constant ζ. Since percent
overshoot is only a function of
ζ, radial lines are thus lines of
constant percent overshoot,
%OS.

118/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Examples
See examples 4.6 and 4.7.

119/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Concept
System with additional poles and zeros.

120/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Pole

Under certain conditions, a system with more than two poles

or with zeros can be approximated as a second-order system
that has just two complex dominant poles.
Once the approximation is justified, then the formula for
percent overshoot, settling time, and peak time can be
applied to these higher-order systems by using the location of
the dominant poles.
Let us investigate the effect of an additional pole on the
second-order response. Here, let us assume that in a
three-pole system with two dominant
p second-order
2
complex poles (−ζωn ± jωn 1 − ζ ) and one real pole
(−αr ) has no zeros.

121/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
122/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Pole
How much farther from the dominant poles does the third
pole have to be for its effect on the second-order response to
be negligible? The answer of course depends on the accuracy
for which we are looking. It can be assumed that the
exponential decay is negligible after five time constants. Thus,
if the real pole is five times farther to the left than the
dominant poles, we assume that the system is
represented by its dominant second-order pair of poles.
What about the magnitude of the exponential decay? Can it
be so large that its contribution at the peak time is not
negligible? We can show, through a partial-fraction expansion,
that the residue of the third pole, in a three-pole system with
dominant second-order poles and no zeros, will actually
decrease in magnitude as the third pole is moved farther
into the left half-plane.

123/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Exercise
Determine the validity of a second-order approximation for each of
these two transfer functions:
700
1 G (s) =
(s + 15)(s 2 + 4s + 100)
360
2 G (s) =
(s + 4)(s 2 + 2s + 90)

124/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Zero

We have seen that the zeros of a response affect the residue,

or amplitude, of a response component but do not affect the
nature of the response — exponential, damped sinusoidal, and
so on.
Let us now discuss the contribution of zero under three
different cases:
Add a zero in the left half plane
Add a zero in the right half plane
Pole-zero cancellation

125/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Zero
In a general approach, it can be component consisting of the
seen using C (s) as the response derivative of the original
to a system T (s) with unity response.
numerator that if we add a zero, We can see here that as the zero
yielding (s + a)T (s), the moves away from the dominant
response would be poles, the response approaches
that of the two-pole system.
(s + a)C (s) = sC (s) + aC (s)

Zero in the left

If a, the negative of the zero, is
very large, the Laplace
transform of the response is
approximately aC (s), or a
scaled version of the original
response. If a is not very large,
the response has an additional
126/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Zero
In a general approach, it can be opposite direction from the
seen using C (s) as the response scaled response. This
to a system T (s) with unity phenomenon is known as a
numerator that if we add a zero, nonminimum-phase system. If a
yielding (s + a)T (s), the motorcycle or airplane was a
response would be nonminimum-phase system, it
would initially veer left when
(s + a)C (s) = sC (s) + aC (s) commanded to steer right.
Zero in the right
An interesting phenomenon
occurs if a is negative, placing
the zero in the right half-plane.
If the derivative term, sC (s), is
larger than the scaled response,
aC (s), the response will initially
follow the derivative in the
127/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
System with Additional Zero

Pole-zero cancellation
Assume a three-pole system with a zero. If the pole term, (s + p3 ),
and the zero term, (s + z), cancel out, we are left with

K (s +z)

T (s) =
(s 
 +p3 )(s 2 + as + b)


as a second-order transfer function. From another perspective, if

the zero at −z is very close to the pole at −p3 , then a
partial-fraction expansion will show that the residue of the
exponential decay is much smaller than the amplitude of the
second-order response.

128/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Exercise

129/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Consider the pitch control model of Unmanned Free-Swimming
Submersible (UFSS) vehicle.

130/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
The transfer function relating pitch angle, θ(s), to elevator surface
angle, δe (s), is as follows:

θ(s) −0.125(s + 0.435)

=
δe (s) (s + 1.23)(s 2 + 0.226s + 0.0169)

131/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
i. Using only the second-order poles shown in the transfer
function, predict percent overshoot, rise time, peak time, and
settling time.
ii. Using Laplace transforms, find the analytical expression for
the response of the pitch angle to a negative step input in
elevator surface deflection.
iii. Evaluate the effect of the additional pole and zero on the
validity of the second-order approximation.
iv. Plot the step response of the vehicle dynamics and verify your
conclusions found in (iii).

132/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution (i)

Let us apply second order approximation to

−0.125(s + 0.435) ωn2

G (s) = ≈
(s + 1.23)(s 2 + 0.226s + 0.0169) s 2 + 2ζωn s + ωn2

Taking the characteristic polynomial s 2 + 2ζωn s + ωn2 we obtain

the following:

ωn2 =0.0169 2ζωn =0.226

√
∴ ωn = 0.0169 = 0.13 rad/sec 0.226 0.226
∴ζ= = = 0.869
2ωn 2 × 0.13

133/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution (i)
√
2
%OS = e −ζπ/ 1−ζ × 100
√
1−0.8692
= e −0.869π/ × 100 = 0.402%
1.76ζ 3 − 0.417ζ 2 + 1.039ζ + 1
Tr =
ωn
1.76 × 0.869 − 0.417 × 0.8692 + 1.039 × 0.869 + 1
3
= = 21.1s
0.13
π
Tp = p
ωn 1 − ζ 2
π
= √ = 48.8s
0.13 1 − 0.8692
4
Ts =
ζωn
4
= = 35.4s
0.869 × 0.13
134/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution (ii)

135/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution (iii)

θ(t) = 2.616 + 0.0645e −1.23t − 4.39e −0.113t cos(0.0643t + 52.38)

Looking at the relative amplitudes between the coefficient of the

e −1.23t term and the cosine term, we see that there is pole-zero
cancellation between the pole at −1.23 and the zero at −0.435.
Further, the pole at −1.23 is more than five times farther from the
jω axis than the second-order dominant poles. We conclude that
the response will be close to that predicted.

136/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
Solution (iv)

137/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Ships at sea undergo motion about their
roll axis, as shown in the following figure.
Fins called stabilizers are used to reduce
this rolling motion. The stabilizers can be
positioned by a closed-loop roll control
system that consists of components, such
as fin actuators and sensors, as well as the
ship’s roll dynamics.

Assume the roll dynamics, which relates the roll-angle output,

θ(s), to a disturbance-torque input, TD (s), is

θ(s) 2.25
= 2
TD (s) s + 0.5s + 2.25

138/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Question
Do the following:
Find the natural frequency, damping ratio, peak time, settling
time, rise time, and percent overshoot.
Find the analytical expression for the output response to a
unit step input.
Plot the response.

139/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Homework/Assignment
Problems: 1, 2, 6, 8, 14 to 20, 23 to 25, 29, 30, 32, 33, 69 to 71.

140/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering
 Thank You!

141/141 Professor Dr. Zobair Ibn Awal NAME 467: Control Engineering