Chapter 7
Steady State Error
Test waveforms for evaluating steady-state errors of position control systems
Steady-state error is the difference between the input
and the output for a prescribed test input as t →∞
Test inputs for steady-state error analysis and design vary with target type
Steady-state error:
a. step input;
b. ramp input
Closed-loop control system error:
a. general representation;
b. representation for unity feedback systems
System with:
a. finite steady-state error for a step input;
b. zero steady-state error for step input
For pure gain K as in (a), Css= K ess or ess =1/K Css there will always be error
But if the pure gain is replaced by an integrator as in (b) the steady state error
will be zero.
Feedback control system for Example 7.1
Problem Find the steady-state error for the system in(a) if T(s) = 5/(s2+7s+10)
and the input is a unit step.
Solution:
s 2  7s  5
Using E(s) = R(s)[1-T(s)]
E ( s) 
s( s 2  7 s  10)
We can apply the final value theorem e()  lim sE ( s )
s0
To yield e(∞) =1/2
Steady-state error in terms of G(s)
E(s) =R(s) – C(s),
C(s) =E(s)G(s) so
E(s) = R(s)/1+G(s)
Using final value theorem
R( s )
e()  lim sE ( s)  lim s
s 0
s 0 1  G ( s )
Steady-state error in terms of G(s)
For step input, the steady state error will be zero if there is at least one pure
integration in the forward path.
s(1 / s)
1

s 0 1  G ( s )
1  lim G( s)
e()  lim sE ( s)  lim
s 0
s 0
For ramp input, the steady state error will be zero if there is at least 2 pure
integration in the forward path.
s (1 / s 2 )
1
e()  lim sE ( s )  lim

s 0
s 0 1  G ( s )
lim sG ( s )
s 0
For parabolic input, the steady state error will be zero if there is at least 3
pure integration in the forward path.
s (1 / s 3 )
1
e()  lim sE ( s )  lim

s 0
s 0 1  G ( s )
lim s 2G ( s)
s 0
Steady state error for systems with no integrations
Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the
system. u(t) is a step function.
Solution First verify the closed loop system is stable.
5
5
5


For input 5u(t) e() 
1  lim G( s) 1  20 21
s 0
5
5
 
lim sG ( s) 0
s 0
10
10


For input 5t2u(t) e() 
2
lim s G( s) 0
For input 5tu(t) e() 
s 0
Steady state error for systems with one integration
Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the
system. u(t) is a step function.
Solution First verify the closed loop system is stable.
For input 5u(t)
e ( ) 
5
5
 0
1  limG (s ) 
s 0
For input 5tu(t) e () 
5
5
1


lim sG (s ) 100 20
s 0
For input
5t2u(t)
e() 
10
10


2
lim s G( s) 0
s 0
Static error constants
Position constant
Velocity constant
K p  lim G (s )
s 0
K v  lim sG (s )
s 0
step input
ramp input
s 2G (s ) parabolic input
Acceleration constant K a  lim
s 0
Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the
expected error for standard step, ramp, and parabolic inputs..
Solution First verify the closed
loop system is stable.
K p  limG (s ) 
s 0
500X 2X 5
 5.208
8X 10X 2
K v  lim sG (s )  0
s 0
K a  lim s 2G (s )  0
s 0
Thus for step input,
e() 
1
 0.161
1 K p
For ramp input e () 
1

Kv
For parabolic input
1
e ( ) 

Ka
Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the
expected error for standard step, ramp, and parabolic inputs.
Solution First verify the closed
loop system is stable.
K p  lim G (s )  
s 0
K v  lim sG (s ) 
s 0
500X 2X 5X 6
 31.25
8X 10X 12
K a  lim s 2G (s )  0
s 0
Thus for step input,
For ramp input
For parabolic input
e() 
1
0
1 K p
1
1

 0.032
K v 31.25
1
e ( ) 

Ka
e ( ) 
Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the
expected error for standard step, ramp, and parabolic inputs.
Solution First verify the
closed loop system is stable.
K p  lim G (s )  
s 0
K v  lim sG (s )  
s 0
K a  lim s 2G (s ) 
s 0
Thus for step input,
For ramp input
For parabolic input
500X2X4X5X6X7
 875
8X10X12
e() 
1
0
1 K p
e ( ) 
1
0
Kv
e ( ) 
1
1

 1.14X 103
Ka 975
Feedback control system for defining system type
System type is the value of n in the denominator, or
the number of pure integrations in the forward path.
Relationships between input, system type, static error
constants, and steady-state errors
Feedback control system for Example 7.6
Problem Find the value of K so that there is 10% error in the steady state..
Solution:
Since the system is Type 1, the error must apply to a ramp input, thus
1
e ( ) 
 0.1
Kv
and
K X5
K v  lim sG (s ) 
 10
s 0
6X7X8
and
K = 672
Applying Routh-Hurwitz we see that the system is stable at this gain.
Feedback control system showing disturbance
C (s )  E (s )G 1 (s )G 2 (s )  D (s )G 2 (s )
But C (s )  R (s ) - E (s )
So
E (s ) 
G 2 (s )
1
R (s ) 
D (s )
1  G1 (s )G 2 (s )
1  G1 (s )G 2 (s )
Applying final value theorem we obtain
sG 2 (s )
s
R (s )  lim
D (s )
s 0 1  G (s )G (s )
s 0 1  G (s )G (s )
1
2
1
2
e ()  lim sE (s )  lim
s 0
=
e R ( )

e D ( )
Previous system rearranged to show disturbance as input and error as output, with
R(s) = 0
Assume step disturbance D(s) = 1/s
We get
e D ( )  
1
1
 limG1 (s )
s 0 G (s )
s 0
2
lim
This shows that the steady-state error
produced by a step disturbance can be
reduced by increasing the dc gain of
G1(s) or decreasing the dc gain of G2(s)
Feedback control system for Example 7.7
Problem Find the steady state error due to a step disturbance for the shown
system..
Solution: The system is stable, we find
e D ( )  
1
1
 limG1 (s )
s 0 G (s )
s 0
2
lim

1
1

0  1000
1000
Forming an equivalent unity feedback system from a general non-unity
feedback system
G(s) = G1(s)G2(s)
H(s) = H1(s)/G1(s)
Non-unity feedback control system for Example 7.8
Problem Find the system type, the appropriate error constant, and the steady state
error for a unit step input..
Solution: The system is stable, we convert the system into an equivalent unity
feedback system
G (s )
100(s  5)
Ge (s ) 
 3
1  G (s )H (s )  G (s ) s  15s 2  50s  400
The system is Type 0, and Kp= -5/4. The steady-state error is
1
1
e () 

 4
1 K p 1 5 / 4
Steady-State error for systems in State Space
The Laplace transform of the error is
E(s) = R(s) –Y(s), and since
Y(s) = R(s)T(s).
Then
E(s) = R(s)[1-T(s)]
Using T(s) =Y(s)/U(s) = C(sI-A)-1B + D
We have
E(s) = R(s)[1-C(sI-A)-1B]
Applying final value theorem, we have
lim sE (s )  lim sR (s )[1  C (sI  A )1 B ]
s 0
s 0
Steady-State error for systems in State Space
Example: Evaluate the ss error for the system described by
 5 1 0
A   0 2 1  ;
 20 10 1 
0 
B  0  ;
1 
C   1 1 0
Solution:
Substitute A,B, and C in lim sE (s )  lim sR (s )[1  C (sI  A )1 B ]
s 0
s 0
s 4


e ()  lim sR (s ) 1  3

2
s 0
 s  6s  13s  20 
 s 3  6s 2  12s  16 
= lim sR (s )  3

2
s 0
s

6
s

13
s

20


For unit step, R(s)=1/s and e(∞)=4/5, for unit ramp, R(s) = 1/s2 and e(∞)= ∞