ME3015, System Dynamics and Control
Steady-State Error Analysis
ME3017
Dr. Nader Sadegh
Steady-State Gain

Steady-State Gain
U
G
Y
u=cosntantu=Acost, =0
 yss=AG(0)cos(t+0)=AG(0) yss/u=G(0) if G(s) is asy.
stable

ME3015, System Dynamics and Control
R

Feedback Loop
E
G
Y
H

The objective is to find the steady-state value of
the error e(t) without solving for it!
1
Feedback Steady-State Error
Error Transfer Function:

Y (s )
G( s )
E (s )
1



R(s ) 1  G(s )H (s )
R(s ) 1  G(s )H (s )
Assume that all the closed-loop poles (i.e.,
the roots of 1+G(s)H(s)=0) are in the left
half of the complex plane.
 By the FRP the steady state error to
r=constant:
ME3015, System Dynamics and Control

ess 
r
1  G ( 0) H ( 0)
Example

Cruise Controller
E
Y
R
-
k/(Ts+1)
C(s)
ME3015, System Dynamics and Control
speed
H(s)=1

G(s)H(s)=kC(s)/(Ts+1)G(0)H(0)=kC(0)
If C(s)=Kp, then ess=1/(1+kKp)
 If C(s)=Kp+Ki/s=(Kps+Ki)/sess=0

2
Input Disturbance
Consider Single Feedback Loop

Output Error:

D
E
R
G
C
Y
H
ME3015, System Dynamics and Control
R (s )
GHD(s )
E (s ) 

1  CGH (s ) 1  CGH (s )
 Assuming constant r(t)=r and d(t)=d
e( ) 
r
GH (0)d

1  CGH (0) 1  CGH (0)
If C(s) conatins a pure integrator (1/s) then
e()=0

Cruise Controller Revisited

Consider the vehicle going uphill:
v
bv
u
mg
ME3015, System Dynamics and Control


Eq. of motion:


mdv/dt=u-bv-mgsin
Transfer Function:
V (s ) 
k
U (s )  D(s )
Ts  1
k
1
m
, T  , d  mg sin 
b
b
3
PI Cruise Control
G=k/(Ts+1)
 C=(Kps+Ki)/s
ME3015, System Dynamics and Control


R
D
E
V
G
C
Conclusion: The integral controller
eliminates the steady state error to
both a constant reference and input
disturbance
Steady-State Error to Sinusoidal
Reference Input

Consider a single feedback loop with a
sinusoidal r(t)=Asint or cost
R
ME3015, System Dynamics and Control

E
E (s )
1

R(s ) 1  G(s )H (s )
G
Then ess(t)=Bsin(t+) where
H
Y
B
1

A 1  G ( j) H ( j)

For small steady-state error |GH(j)|>>1
4
Example

Feedback loop in Figure 10.7.1:
Proportional control of 1/[s(Is+c)].
Determine the steady-state error to a
r(t)=Rsint for small .
 Plot E/R vs. =/n for different closed-loop
damping ratio.
ME3015, System Dynamics and Control


Solution: The error transfer (Gerr=E/R)
function is given by Gerr=1/(1+GH) or
Gerr=1/(1+Kp/(Is2+cs))=(Is2+cs)/(Is2+cs+Kp)
2
2
2
 Gerr=(s +2ns)/(s +2ns+n )
2
 Where n =Kp/I and =c/(2In)
ME3015, System Dynamics and Control


  2  2  j
G err ( j ) 
1   2  2  j
E
 G err ( j ) 
R
 2  4 2
1     2  
2 2
2
5
Result
6
z=0.1
z=0.3
z=0.5
z=0.7
z=1
5
ME3015, System Dynamics and Control
E/R
4
3
2
1
0
0
0.5
1
1.5
2
2.5
w/wn
3
3.5
4
4.5
5
6