Chapter 5 – The Performance of Feedback Control Systems
The ability to adjust the transient and steady-state response of a feedback control
system is a beneficial outcome of the design of control systems.
One of the first steps in the design process is to specify the measures of
performance.
In this chapter we introduce the common time-domain specifications such as
percent overshoot, settling time, time to peak, time to rise, and steady-state tracking
error.
We will use selected input signals such as the step and ramp to test the response of
the control system.
The correlation between the system performance and the location of the system
transfer function poles and zeros in the s-plane is discussed.
We will develop valuable relationships between the performance specifications and
the natural frequency and damping ratio for second-order systems.
Relying on the notion of dominant poles, we can extrapolate the ideas associated
with second-order systems to those of higher order.
Illustrations
Introduction
Steady-State: exists a long time following any input signal initiation
Transient Response: disappears with time
Design Specifications: normally include several time-response indices
for a specified input command as well as a desired steady-state
accuracy.
Illustrations
Test Input Signals
A unit impulse function is also useful for test
signal purposes. It’s characteristics are
shown to the right.
Illustrations
Performance of a Second-Order System
Y( s )
K
 R( s )
2
s  p s  K
n
Y( s )
2
2
s  2  n  s  n
2

with a unity step input
Y( s )
y( t)
Illustrations
n
2

cos
 s 2  2    s   2   s
n
n 

1
1

e
    n t

 sin n    t  
2
1

1
 
Performance of a Second-Order System
Illustrations
Performance of a Second-Order System
Illustrations
Performance of a Second-Order System
Rise Time, Tr
Peak Time, To
Percentage Overshoot, P.O.
Settling Time, Ts
Normalized Rise Time Tr1
Illustrations
Performance of a Second-Order System
Standard performance measures are usually defined in terms of
the step response of a system. The transient response of a system
may be described using two factors, the swiftness and the
closeness of the response to the desired response.
The swiftness of the response is measured by the rise time (Tr)
and the peak time (Tp).
Underdamped systems: 0-100% rise time is used
Overdamped systems: 10-90% rise time is used
The closeness is measured by the overshoot and settling time.
Using these measurements the percent overshoot (P.O.) can be
calculated.
Illustrations
Performance of a Second-Order System
PO
Ts
M pv  fv
fv
 100
4
 n

Tp
n 
2
1  

 
M pv
1  e
 
PO
Illustrations
1 
100 e

1 
2
2
Performance of a Second-Order System
Naturally these two
performance
measures are in
opposition and a
compromise must be
made.
Illustrations
Performance of a Second-Order System
Illustrations
Performance of a Second-Order System
Illustrations
Performance of a Second-Order System
Illustrations
Effects of a Third Pole and Zero on the Second-Order System
T( s )
Illustrations
1
s 2  2  s  1  s  1
Effects of a Third Pole and Zero on the Second-Order System
Illustrations
Effects of a Third Pole and Zero on the Second-Order System
Example 5.1 - Parameter Selection
Select the gain K and the parameter p so that
the percent overshoot is less than 5% and the
settling time (within 2% of the final value)
should be less than 4 seconds.
Illustrations
Effects of a Third Pole and Zero on the Second-Order System
Example 5.1 - Parameter Selection
Ts
4
 n
 4sec
 n  1
When the closed-loop roots are chosen as:
r1 1  j 1
r2 1  j 1
We have Ts
4sec and an overshoot of 4.3%.
Therefore, 
1
and n
2
T( s )
K
Illustrations
n
1
2

2
G( s )
K
n
1  G( s )
s  p s  K
s  2  n s  n
2
2
2
P
2  n
2
2
2
Effects of a Third Pole and Zero on the Second-Order System
Example 5.2 – Dominant Poles of T(s)
2
n
Y( s )
R( s )
For  n
T( s )
Illustrations
s
T( s )
3, 
a
2
( s  a)
0.16 , and a
62.5( s  2.5)
s 2  6 s  25(s  6.25)
  1   s 
2
 2  n s  n
2.5 :
Effects of a Third Pole and Zero on the Second-Order System
62.5( s  2.5)
Example 5.2 – Dominant
2
 Poles of T(s)
T( s )
s  6 s  25 ( s  6.25)
As a first approximation, we neglect the real pole and obtain:
T( s )
10( s  2.5)
s 2  6s  25
We now have 
0.6 and n
accompanying zero for which
5 for dominant poles with one
a
0.833
 n
Using the previously mentioned charts (Figure 5.13a), we find that the
percent overshoot is 55%. We expect the setting time to within 2% of the
final value to be:
T( s )
4
4
 n
0.6 5
1.33sec
Using computer simulations the actual percent overshoot is equal to 38%
and the settling time is 1.6 seconds.
Thus, the effect of the damping of the third pole of T(s) is to dampen the
overshoot and increase the settling time (hence the real pole cannot be
neglected.
Illustrations
The s-Plane Root Location and The Transient Response
Illustrations
Steady-State Error of Feedback Control Systems
For
Step Input - Position Error Constant
Ramp Input - Velocity Error Constant
Acceleration Input - Acceleration Error Constant
Illustrations
The Steady-State Error of Nonunity Feedback Systems
For a system in
which the feedback
is not unity (Fig
5.21), the units of the
output are usually
different from the
output of the sensor.
In Fig. 5.22, K1 and
K2 convert from
rad/s to volts.
Illustrations
The Steady-State Error of Nonunity Feedback Systems
T( s )
E( s )
E( s )
Illustrations
K1 G( s )
1  K1 G( s )
R( s )  Y( s )
1
1  K1 G( s )
( 1  T( s ) )  R( s )
 R( s )
ess
lim s  E( s )
s 0
1
1  K1 G( 0)
Performance Indices
A performance index is a quantitative measure of the
performance of a system and is chosen so that emphasis
is given to the important system specifications.
A system is considered an optimum control system
when the system parameters are adjusted so that the
index reaches an extremum value, commonly a
minimum value.
Illustrations
Performance Indices
There are several performance indices:
(1) Integral of the square of the error, ISE
T
ISE
 2
 e ( t) dt
0
(2) Integral of the absolute magnitude of the error, IAE
T
IAE


0
e( t ) d t
(3) Integral of time multiplied by absolute error, ITAE
T
ITAE

 t  e( t ) d t
0
(4) Integral of time multiplied by the squared error, ITSE
T
ITSE
Illustrations

2
 t e ( t) dt
0
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
System Performance Using MATLAB and Simulink
Illustrations
Exercises and Problems
Chapter 5 –
Illustrations
E5.5, E5.16, DP5.4 – Select 3 more problems of
your choice. Submit One Set of Multiple Choices,
and Matching Concepts