Chapter 4 time domain analysis

Chapter 4: Time Domain Analysis
Outline
 Introduction
 Test Inputs used in control system Analysis &
Design
 System Poles and Zeros
• Pole-zero plot
 Standard form for 1st and 2nd order systems
 First order system
 Second order system
• Effect of poles on system response
• Effect of zeros on system response**
 Performance of FB systems(in the classical sense)
• Transient performance
• Steady-state accuracy
 Routh’s Stability criterion
1

Introduction
• The first step in analyzing a control system is to
derive a mathematical model of the system (
Chapter 3). Once such model is obtained, the
different analysis techniques will be used to
study system performance.
• In analyzing and designing control systems, we
must have a basis of comparison of
performance of various control systems. This
basis may be setup by specifying particular test
input signals and by comparing the responses
of various systems to these input signals.
2

Common Input signals used in Control System
Two classes of inputs commonly used to characterize
the
performance of feedback control systems are:
(a) The “Singularity” Functions: These functions have a
discontinuity @ t=0
(i) The Dirac Delta (Impulse) Function
- used to characterize the response of a system to brief,
intense inputs 3
(a) Unit pulses of different
extents
(b) The impulse
function

Common Input signals used in Control System…
(ii) The Unit-step Function us(t)
- used to characterize a system’s transient
response to a sudden change.
(iii) The Unit-Ramp Function r(t)
4
Us(t) = 0 for t < 0
= 1 for t > 0
r(t) = 0 for t < 0
= t for t > 0

(iii) The Unit-Ramp Function r(t)
- Used to characterize a system’s ability to follow a
time-varying input, and the transient behavior
around a discontinuity in the slope of an input
function.
Remark:
- The singularity functions are related to each other
by differentiation & integration , as shown below.
5

(b) Sinusoidal Inputs
- The response of linear systems to sinusoidal
inputs of the form
Is of fundamental importance to control
engineering and system dynamics ( Studied in
Chapter 7)
6

Example 4.1
In practice a machine, described a second-order
transfer function G(s), will be subjected to inputs
that change suddenly. Use the unit-step response
to determine how long it will take the machine’s
response to settle to a new steady-state value
after a change.
Solution: for

7

Example 4.1…
Where
Taking inverse LT yields
8

Example 4.1…
Solution:
The response is shown above, and indicates that it
takes this system 2.5 - 3 seconds to respond to the
step. 9
0 0.5 1 1.5 2 2.5 3 3.5
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
Step Response
Time (sec)
Response

System Poles and Zeros
Consider a system with TF
If we factor the numerator & denominator
polynomial and write
Where
-- are the roots of the characteristics
polynomial D(s), and know as the system poles.
-- are the roots of the numerator
polynomial N(s), and know as the system zeros.
10

Remark:
• Note that because of the coefficients of N(s) &
D(s) are real ( they come from the modeling
parameters), the system poles & zeros must be
either
(a) purely real or
(b) appear as complex conjugates
And In general we write
or
Example 4.2
Find the poles and zeros of the system
11

Example 4.2…
Solution:
So that we have
(a) a pair of real zeros at and
(b) three poles at and
Remark
The system poles and zeros completely
characterize the transfer function (and therefore
the system itself ) except for an overall gain
constant K:
12

Pole-zero plot
• The pole-zero plot for example 4.2 is shown below;
where the pole positions are denoted by an X, and the
zeros as drawn as an O.
The poles-zero plot is used extensively throughout
control theory and system dynamics to provide a
qualitative indication of the dynamic behavior of the
system 13

Pole-zero plot
Aside: In MATLAB a system may be specified by its
poles and zeros using the function
zpk(zeros, poles, gain), for example
sys=zpk([0 -2],[-3 , -1+i*2 , -1-i*2],5)
step(sys)
Will plot the system response of the system in
Example4.2
•The characteristics equation of a system is
Complex poles & Zeros
• We have noted that in general and that poles
and zeros may appear as complex conjugate pairs:
14

Pole-zero plot
Complex poles & Zeros…
For example the pole-zero plot
Corresponds to the transfer function
15

Pole-zero plot
• The homogeneous response we will have a pair of
complex exponential terms associated with each
pair of conjugate pair of poles, such as
But Ci and Ci+1 are also complex (say a+jb), so we
can write
16

Pole-zero plot
• so that the contribution of the complex conjugate
pole pair to the homogeneous response may be
written
Where
and
Which is shown below ( for )
17

Pole-zero plot
Example 4.3
Find and plot the poles and zeros of
and then determine the modal response
components of this system
Solution:
18

Pole-zero plot
Example 4.3…
• The pole-zero plot is
-And the modal components are (1) Ce− t
(corresponding to the pole at s =−1), and (2) Ae−t
sin(2t + φ) (corresponding to the complex conjugate
pole pair at s =−1± j2), and where the constants C,A,
and φ are determined from the initial conditions. 19

Pole-zero plot
Note: A pair of purely imaginary poles ( on the
imaginary axis of the s-plane) implies and
there will be no decay. The system will act as a
pure oscillator.
• The effect of pole locations in the s-plane on the
modal response component is summarized in the
figure below.
20

Pole-zero plot
Observe:
(a) Poles in the left-half of the s-plane ( the LHP),
that is
generate components that decay with time.
(b) conversely, poles in the right-half s-plane ( the
RHP), that is generate components that
grow with time.
(c) Poles that lie on the imaginary axis ( )
generate components that are purely oscillatory,
and neither grow nor decay with time.
(d) A pole at the origin of the s-plane ( ),
generates a component that is a constant.
In addition we note that oscillator frequency and 21
00

Pole-zero plot
Observe:
(e) The rate of decay/growth is determined by the real
part of
the pole , and poles deep in the LHP
generate
rapidly decaying components.
(e) For complex conjugate pole pairs, the oscillatory
frequency is
determined by the imaginary part of the pole pair
22

Standard form of 1st and 2nd order systems
These are
(a) All pole system ( with no zeros) , and
(b) Have unity gain ( )
First Order system
We define the first-order standard form as
Where the single parameter is the time constant.
As a differential equation
And the system has a single pole at .
The step response is
23

First Order system…
24

First Order system…
Common Step Response Descriptors:
(a) Settling Time: the time taken for the response
to reach 98% of its final value. Since
and , we take
as the definition of Ts
(a) Rise Time: commonly taken as the time taken
for the step response to rise from 10% to 90%
of the steady state response to a step input.
And it is found as follows:
25

Second Order System
•The standard unity gain second-order system has
a transfer function
With two parameters
(i) - the undamped natural frequency, and
(ii) - the damping ratio ( ).
•The system poles are the roots of
,that is
leading to four cases
26

Second Order System…
i) - the poles are real and distinct
ii) - the poles are real and coincident
iii) - the poles are complex conjugates
iv) -the poles are purely imaginary
27

Pole positions for an underdamped( )
second order system
when plotted on the s-plane
28

We note that
(a) The poles lie at a distance ωn from the origin, and
(b) The poles lie on radial lines at an angle
θ = cos−1(ζ )
as shown above
• The inﬂuence of ζ and ωn on the pole locations
may therefore be summarized:
29Lines of
increasin
g
Lines of

Second Order System Step response
(a) The over damped case( )
where the constants C1 and C2 are determined
from p1 and p2 .
30

(b) The critically damped case( )
The step response takes a special form
31

(c) The under damped case( ) ,with a pair
of complex conjugate poles
the step response becomes

where
32

(c) The under damped case( )…
Referring the LT pairs ( chapter 2), i.e.,

33

Remarks:
• As , the system becomes sluggish whereas the
overshoot decrease.
• As , the system becomes too oscillatory
34
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
t (sec)
Response
 = 0
0.2
0.4
0.6
0.8
1.0
Overshoot
Damped oscillatory
response
1
0 t

• Hence , in most of the practical situations, we
impose a requirement on the control system to have
Specific Application where is used ( or very
close to 1)
 welding robot ( no oscillation is acceptable at all)
 pick and place robot
(d) The undamped case ( )
35

Performance of Feedback systems
Quantitative performance specifications of FB control
system, in the classical sense, are
Transient performance (tr, tp , ts and Mp)
Steady-state accuracy
Transient Performance
Consider the closed loop TF ( as example)
(4.1)
Where
•The response of (4.1) to any input will contain the
following functions of time.
 A constant term resulting from the s-factor
 An term resulting from the ( ) factor
 &
36

• Additional terms determined by the specific input
forcing will also appear in the response, but the
intrinsic dynamic features of the system, the so called
response mode or natural modes, are determined by
the characteristic polynomial itself.
Therefore, transient performance of a control system
can be studies effectively by considering the response
of the system for step input.
37

Step Response Based Second-Order System
Specifications
(a) Rise Time (TR): Applies to over- and underdamped
systems. As in the case of first-order systems, the usual
deﬁnition is the time taken for the step response to rise
from 10% to 90% , 0% to 100% of the ﬁnal value:
38

Specifications
(a) Rise Time (TR)…
• For a second-order system there is no simple
(general) expression for TR
• However, for underdamped second order systems
the 0% -100% rise time is normally used.
TR 

 ( first time occuring)
or
39

Specifications
(b) Peak Time (Tp): Applies only to underdamped
systems, and is deﬁned as the time to reach the ﬁrst
peak of the oscillatory step response.
40

Specifications
(b) Peak Time (Tp)…
• Tp is found by diﬀerentiating the step response
ystep(t), and equating to zero.

• The first peak occur @ the point corresponding to π
41

 Performance of Feedback systems
Specifications
(c) Maximum Overshoot(Mp)
After simple manipulation , we get
• The height of the first peak of the response, expressed
as a percentage of the steady-state response.
• Note that the percent overshoot depends only on
• Conversely we can find ζ to give a specific %
overshoot from the above:
42

Specifications
(d) Settling Time (Ts)
•The most common deﬁnition for the settling time Ts is
the time for the step response ystep(t) to reach and
stay within 2% of the steady-state value yss.
• A conservative estimate can be found from the decay
envelope, that is by ﬁnding the time for the envelope
to decay to less than 2% of its initial value as
illustrated below.
giving
43

Constant Ts, Tp and zeta lines (Nise Norm pp 184)
44

 Effect of zeros on system response
Consider a system:
Which can be represented as two cascade blocks
Then if the response of the a system 1/D(s) is v(t), then
and as the zero (at s =−b) moves deeper into the l.h.s –
plane, the relative contribution of the derivative term
decreases and the system response tends toward a
scaled version of the all pole response v(t).
• A zero near a pole will tend to reduce its effect on the
system response ( and result in a smaller RESIDUE).
45
U(s)
u(t)
V(s)
v(t)
Y(s)
y(t)

In general, the presence of the derivative terms in the
response means that:
• The response is faster(shorter peak-time TP and rise-
time TR ).
• Greater overshoot in the response (if any). A zero
may cause overshoot in the response of an over-
damped second-order system.
Example 4.4
The following MATLAB step response compares the
responses for the under-damped system
With similar unity-gain systems with zeros at s =−1,−2,
−3
, , 46

Example 4.4
Note the increase in the overshoot, and the decrease in
TP as the zero approaches the origin.
47

Example 4.5
The following MATLAB step response compares the
resp onse for the unity-gain overdamped system
with two real poles at s =−3 and s =−4 with the similar
system with a zeros at s =−1:
48

Steady State Accuracy
• As opposed to transient performance , steady-state
accuracy depends on the system characteristics
equation as well as the specific reference signal used.
• Hence, we can’t take a single input to specify the
stead-state performance of a control system.
• Steady-state performance specification is given by
Consider a polynomial input signal ( see why poly!)
As k , the input is becoming faster and faster.
49

• Consider a unity-feedback control system shown
below( the analysis can be extended for different
configuration)
&
(i) Unit-step input, R(s)=1/s
where 50
+
-Reference
input
error
Controller Plant
Feedback
path

(ii) Unit-ramp input, R(s)=1/s2
where
(iii) Unit-parabola input, R(s)=1/s3
• proceeding similarly we find
where
51

Type-Number
Let consider
,
Where N-is number of free-integrator (a.k.a Type-
number)
52
Step Ramp Parabola
Type 0 (N=0)
Type 1 (N=1)
0
Type 2 ( N=2)
0 0

 Stability & Routh’s Test
• The concept of stability, which is qualitative property of
control system, is very important because every control
system must be stable.
• If a control system is not stable, it will usually burn out
or disintegrate.
Three types of stability exist
- BIBO**
- Marginal stability ( stability in the sense of Lyapunov)
- Asymptotic stability
Definition
A system is stable if every bounded input excites a
bounded output. Otherwise the system is said to be
unstable.
Note: It is difficult to deduce stability from the definition
because there are infinitely many bounded inputs to be53

Theorem 4.1
A system with proper rational Transfer function G(s) is
stable iff every pole of G(s) has a negative real part or,
equivalently, lies inside the open left half s-plane.
Routh-Hurwitz Test
Consider a system with transfer function
G(s)=N(s)/D(s). It is assumed that N(s) and D(s) have
no common factor. To determine the stability of G(s) by
using Theorem 4.1, we must first compute the poles of
G(s) or, equivalently, the roots of D(s).
If the degree of D(s) is three or higher, hand
computation of the roots is complicated. Therefore, it is
desirable to have a method of determining stability
without solving for the roots. And Routh test or the
Routh-Hurwitz test , explained in the next slide , is such54

Routh-Hurwitz Test
Consider characteristic equation of an nth order system
given by
a0sn+a1sn-1+a2sn-2+…+an-1s+an=0
55
Sn a0 a2 a4 a6
. . .
Sn-1 a1 a3 a5 a7
. . .
Sn-2 b1 b2 b3 b4
. . .
Sn-3 c1 c2 c3 c4
. . .
Sn-4 d1 d2 d3 d4
. . .
. . . . .
. . . . .
. . . . .
S2 e1 e2 0 0
S1 f1 0 0 0
0

Routh-Hurwitz Test…
• The coefficients b1, b2, b3 etc, are evaluated, in terms
of the elements from the first two rows, as indicated
below:
b1= (a1a2 - a0a3)/a1
b2= (a1a4 - a0a5)/a1
b3= (a1a6 - a0a7)/a1
and so on.
• The computation of the elements in the third row is
continued until all the elements become zero.
• The same procedure is followed in evaluating the
entries designed by c’s , d’s , e’s , and so on interms of
the elements of the previous two rows. i.e.,
c1= (b1a3 - a1b2)/b1 , c2= (b1a5 - a1b3)/b1 , c3= (b1a7 -
a1b4)/b1
d = (c b - b c )/c , d = (c b - b c )/c , d = (c b -56

Theorem 4.2 (Routh Test)
A polynomial with a positive leading coefficient is a
Hurwitz polynomial if and only if every entry in the
Routh table is positive or, equivalently, iff every entry in
the first column of the table ( namely, a0 , a1 , b1 , c1 , d1 ,
… , e1 , f1 , g1 ) is positive.
Remark:
• This theorem implies that if a zero or a negative
number appears in the table, then the polynomial is not
Hurwitz. In this case, it is unnecessary to complete the
table.
• The Routh test can be used to determine the number
of roots of D(s) lying in the open right half s-plane. To be
more specific, if none of the entries in the first column of57

Chapter 4 time domain analysis

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