Errors in Automatic control system

1. 1
Steady State Error Analysis

2. 2
Test Waveform for evaluating steady-state
error

3. 3
G(s)
H(s)
R(s)
+
-
C(s)
G(s)
R(s)
+
-
C(s)
Unity feedback
H(s)=1
Non-unity feedback
H(s)≠1
E(s)
E(s)
Steady-state error analysis

4. 4
Steady-state error analysis
For unity feedback system:
)
(
)
(
)
( s
C
s
R
s
E 
 System error
For a non-unity feedback system:
)
(
)
(
)
(
)
( s
C
s
H
s
R
s
E 
 Actuating error

5. 5
Steady-state error analysis
Consider a unity feedback system, if the inputs are step response, ramp &
parabolic (no sinusoidal input). We want to find the steady-state error
)
(
lim t
e
e
t
ss



Where, )
(
)
(
)
( t
c
t
r
t
e 

By Final Value Theorem:
)
(
lim
)
(
lim
0
s
sE
t
e
e
s
t
ss






6. 6
Steady-state error analysis
Consider Unity Feedback System
)
(
)
(
)
( s
C
s
R
s
E 
 (1)
)
(
1
)
(
)
(
)
(
s
G
s
G
s
R
s
C

 (2)
Substitute (2) into (1)
)
(
)
(
1
1
)
(
)
(
1
)
(
)
(
)
( s
R
s
G
s
R
s
G
s
G
s
R
s
E





 (3)

7. 7
Steady-state error analysis
Based on equation (3), it can be seen that E(s) depends on:
(a) Input signal, R(s)
(b) G(s), open loop transfer function
Now, assume:
 
 
j
j
N
i
M
i
p
s
S
z
s
K
s
G








1
1
)
(
type N
Cases to be considered:
3
2
1
)
(
)
(
1
)
(
)
(
1
)
(
)
(
s
s
R
C
s
s
R
B
s
s
R
A




8. 8
Case (A): Input is a unit step R(s)=1/s
)
(
1
1
)
(
)
(
1
1
)
(
s
G
s
s
R
s
G
s
E




)
(
lim
_
0
s
sE
Error
State
Steady
e
s
ss























 )
(
1
1
lim
)
(
1
1
lim
0
0 s
G
s
G
s
s
e
s
s
ss





















p
s
K
s
G 1
1
)
(
lim
1
1
0
where )
(
lim
0
s
G
K
s
p

 
“Static Position
Error Constant”

9. 9
If N = 0, Kp = constant finite
K
e
p
ss 


1
1
If N ≥ 1, Kp = infinite 0
1
1
1
1






p
ss
K
e
For unit step response, as the type of system increases (N ≥ 1), the steady
state error goes to zero

10. 10
Case (B): Input is a unit ramp R(s)=1/s2
)
(
1
1
)
(
)
(
1
1
)
(
2
s
G
s
s
R
s
G
s
E




)
(
lim
_
0
s
sE
Error
State
Steady
e
s
ss























 )
(
1
lim
)
(
1
1
lim
0
2
0 s
sG
s
s
G
s
s
e
s
s
ss
V
s
s
K
s
sG
s
sG
1
)
(
lim
1
)
(
lim
0
1
0
0






















where )
(
lim
0
s
sG
K
s
v

 
“Static Velocity
Error Constant”

11. 11
If N = 0, 


v
ss
K
e
1
If N =1, Kv = finite finite
K
e
v
ss 

1
,
0
)
(
)
(




j
i
v
p
s
z
s
s
K


If N ≥2, Kv = infinite 0
1
1




v
ss
K
e
For unit ramp response, the steady state error in infinite for system of type
zero, finite steady state error for system of type 1, and zero steady state error
for systems with type greater or equal to 2.

12. 12
Case (C): Input is a parabolic, R(s)=1/s3
)
(
1
1
)
(
)
(
1
1
)
(
3
s
G
s
s
R
s
G
s
E




)
(
lim
_
0
s
sE
Error
State
Steady
e
s
ss























 )
(
1
lim
)
(
1
1
lim 2
2
0
3
0 s
G
s
s
s
G
s
s
e
s
s
ss
a
s
s
K
s
G
s
s
G
s
1
)
(
lim
1
)
(
lim
0
1
2
0
2
0






















where
)
(
lim 2
0
s
G
s
K
s
a

 
“Static Acceleration
Error Constant”

13. 13

14. 14
If N = 0, 


a
ss
K
e
1
If N =1, Ka = 0 


a
ss
K
e
1
,
0
)
(
)
(
2




j
i
a
p
s
z
s
s
K


If N = 2, Ka = constant finite
K
e
a
ss 

1
 Increasing system type (N) will accommodate more different inputs.
If N ≥3 , Ka = infinite 0
1
1




a
ss
K
e

15. 15
Example 3
R(s)
+
-
C(s)
)
2
(
)
1
(
3


s
s
s
If r(t) = (2+3t)u(t), find the steady state error (ess) for the
given system.
Solution:




)
(
lim
0
s
G
K
s
p
2
3
)
(
lim
0



s
sG
K
s
v
2
2
3
3
1
2
3
1
2








v
p
ss
K
K
e