1. CONTROL SYSTEMS
THEORY
Forced Response Errors
CHAPTER 6
STB 35103

2. Objectives
 To
find the steady-state error for a unity
feedback system
 To specify a system’s steady-state error
performance
 To design system parameters to meet
steady-state error performance
specifications

3. Introduction

In chapter 1, we learnt about 3
requirements needed when designing a
control system




Transient response
Stability
Steady-state errors (SSE)
Up until now we only covered until
transient response and stability

4. Review on transient response

We learned in Chapter 4, there are 4 types
of transient response for a second-order
system.




Overdamped
Underdamped
Undamped
Critically damped

5. Review on transient response

An example of elevator response

The transient response for elevator can be
considered as overdamped. The system is
stable but has steady-state error

6. Introduction

What is steady-state error?

Steady-state error is the difference between
the input and output for a certain test input as
t →∞

Test input used for steady-state error analysis
and design are
Step
 Ramp
 Parabola


7. Introduction

Test waveforms

8. Introduction

Example of systems tested using the test
signal.

Targeting system:

Targeting a static target. (e.g. a stopping car). We
test the system using step input because the position
of the car is in constant position.

Targeting a car moving with constant velocity. We
test the system using ramp input because the car is
moving in constant velocity.

Targeting an accelerating car. We test the system
using parabola input because the car is accelerating.

9. Introduction

We are only concerned with the difference
between the input and the output of a
feedback control system after the steady
state has been reached, our discussion is
limited to stable systems where the
natural response approaches zero when
(time) t approaches infinity.

10. SSE for unity feedback system

Unity feedback system can be represented
as

Steady state error can be calculated from
a system’s closed-loop transfer function,
T(s), or the open-loop transfer function,
G(s), for unity feedback systems.

11. SSE for unity feedback system

Closed loop transfer function, T(s) is calculated
by solving the unity feedback system using the
block diagram reduction method for feedback
system.
1
T ( s) =
G ( s)
1 + G ( s ) × (1)

12. SSE for unity feedback system

Open-loop transfer function for a unity
feedback system is the value of G(s) multiply
1.
1
1

13. SSE for unity feedback system

Steady state error in terms of T(s).

To find E(s), the error between the input, R(s) and
output, C(s), we write
E ( s) = R(s) − C ( s)
= R ( s ) − R ( s )T ( s )
= R ( s ) 1 − T ( s ) 



)
We can find final value of the error, e(∞ in terms of
T(s) using
e ( ∞ ) = lim sR( s ) 1 − T ( s ) 


s →0

We can only use this equation if T(s) is stable, E(s) has no
poles in the right-half plane or poles on the imaginary axis
other than the origin

14. SSE for unity feedback system

Example 7.1

Find the steady state-error for a unity feedback
system that has T(s) = 5/(s2+7s+10) and the
input is a unit step.

Solution:
R(s) =unit step = 1/s
T(s) = 5/(s2+7s+10), we must check the
stability of T(s) using Routh table or poles.

15. SSE for unity feedback system

Example 7.1 (cont.)

We know from the unity feedback system
E ( s) = R(s) − C (s )

C ( s) = R( s )T ( s )
So, E(s) can be calculated using both equation
E ( s) = R(s) − C (s)
= R ( s ) − R ( s )T ( s )
= R ( s ) 1 − T ( s ) 



16. SSE for unity feedback system

Example 7.1 (cont.)

E(s) in example 7.1 is
1
5

E ( s ) = 1 − 2
s  s + 7 s + 10 


1  s 2 + 7 s + 10
5
=  2
− 2

s  s + 7 s + 10 s + 7 s + 10 
2
 s2 + 7s + 5 
1
s + 7s + 5
=  2
=
s  s + 7 s + 10  s s 2 + 7 s + 10
(
)

17. SSE for unity feedback system

Example 7.1 (cont.)

Before calculating the final value of the error
we must check the position of E(s) poles
s2 + 7s + 5
s 2 + 7s + 5
E ( s) =
=
2
s ( s + 2 ) ( s + 5)
s s + 7 s + 10
(

)
The poles for E(s) are at (0,0), (-2,0) and
(-5,0). Since all the poles are not on the right
half plane or the imaginary axis we can use the
equation to calculate final error value in terms
of T(s).

18. SSE for unity feedback system

Example 7.1 (cont.)
e ( ∞ ) = lim sR ( s ) 1 − T ( s ) 


s →0
5
1

= lim s  ÷1 − 2

s →0
s   s + 7 s + 10 

5 5 1

= 1 −  =
=
 10  10 2

19. SSE for unity feedback system

Steady state error in terms of G(s)
 We can find final value of the error, e(∞
)
in
terms of G(s) using
sR ( s )
e ( ∞ ) = lim
s →0 1 + G ( s )

We are going to use three types of input R(s);
step, ramp and parabola. So the final value of
the error for this types of input can be
described as

20. SSE for unity feedback system

Step input e(∞
)
e ( ∞ ) = estep ( ∞ ) =

1
sR(s)
1 + lim G ( s )
s →0
)
Ramp input e(∞
1
sR(s)
e ( ∞ ) = eramp ( ∞ ) =
lim sG ( s )
s →0

)
Parabola input e(∞
1
sR(s)
e ( ∞ ) = e parabola ( ∞ ) =
lim s 2G ( s )
s →0

21. SSE for unity feedback system
Steady state error with no integration
 Example 7.2


Find the steady-state errors for inputs of 5u(t),
5tu(t), and 5t2u(t) to the system below.
No integration

Solution hint
5u(t) = unit step = 5(1/s)
 5tu(t) = ramp = 5(1/s2)
 5t2u(t) = parabola = 5(2/s3) = 10(1/s3)


22. SSE for unity feedback system

23. SSE for unity feedback system

Example 7.2 (cont)
5
5
5
e ( ∞ ) = estep ( ∞ ) =
=
=
1 + lim G ( s ) 1 + 20 21
s →0
5
5
e ( ∞ ) = eramp ( ∞ ) =
= =∞
lim sG ( s ) 0
s →0
10
5
e ( ∞ ) = e parabola ( ∞ ) =
= =∞
2
lim s G ( s ) 0
s →0

24. SSE for unity feedback system

Try to solve steady state errors for
systems with one integration in Example
7.3.

25. SSE for unity feedback system

From the previous slides, the final error
value for three kinds of input; step, ramp
and parabola, are as follows
1
e ( ∞ ) = estep ( ∞ ) =
1 + lim G ( s )
position constant, K p
1
e ( ∞ ) = eramp ( ∞ ) =
lim sG ( s )
velocity constant, K v
s →0
s →0
1
e ( ∞ ) = eramp ( ∞ ) =
lim s 2G ( s )
s →0
acceleration constant, K a

26. SSE for unity feedback system

Steady state error via static error
constants

Example 7.4 (Figure 7.7 (a) )

27. SSE for unity feedback system

Solution

First step is to calculate the static error constants.
500( s + 2)( s + 5)( s + 6) 500(0 + 2)(0 + 5)(0 + 6)
K p = lim G ( s ) = lim
=
= 5.208
s →0
s →0 ( s + 8)( s + 10)( s + 12)
s (0 + 8)(0 + 10)(0 + 12)
s (500)( s + 2)( s + 5)( s + 6)
K v = lim sG ( s ) = lim
=0
s →0
s →0
( s + 8)( s + 10)( s + 12)
s 2 (500)( s + 2)( s + 5)( s + 6)
K a = lim s G ( s ) = lim
=0
s →0
s →0
( s + 8)( s + 10)( s + 12)
2

28. SSE for unity feedback system

Next step is to calculate the final error value.
1
Step input, e(∞) =
= 0.161
1+ K p
1
Ramp input, e(∞) =
=∞
Kv
1
Parabola input,e(∞) =
=∞
Ka

Try to solve the remaining problems in Figure
7.7 (a) and (c).

29. SSE for unity feedback system

System Type

We are still focusing on unity negative
feedback system.

Since steady-state errors are dependent upon
the number of integrations in the forward path,
we give a name to this system attribute.

30. SSE for unity feedback system

Below is a feedback control system for
defining system type.




We define the system type to be the value of n
in the denominator.
Type 0 when n = 0
Type 1 when n = 1
Type 2 when n = 2

31. SSE for unity feedback system

Relationship between input, system type,
static error constant, and steady-state
errors can be summarized as

32. SSE for unity feedback system

Steady-state error specifications.

We can use the static error constants to
represent the steady-state error characteristic
of our system.

Conclusion that we can made based on static
error constants.

Problem: What information is contained in the
specification Kv = 1000.

33. SSE for unity feedback system

Kv = 1000
Solution:
1. The system is stable.
2. The system is of Type 1, since only Type 1
have Kv that are finite constant


34. SSE for unity feedback system
3.
A ramp input is the test signal. Refer to table.
4.
The steady-state error between the input
ramp and the output ramp is 1/Kv per unit of
slope.