1. No category

Steady-State Error Analysis
How to evaluate steady-state errors for
unity feedback systems
How to evaluate steady-state errors for
non-unity feedback systems
How to evaluate steady-state errors for
disturbance inputs
How to specify a system’s steady-state
error performance
Static Error constants and System Type
Definition and Test Inputs
Steady-state error is the difference
between the input and output for
a defined or prescribed test input to a
system as t.
There may be different types of test inputs.
Test
waveforms
for
evaluating
steady-state
errors of
control
systems
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Application




The analysis is aimed at systems that
can reach a steady state value and
hence stable systems.
Here the natural response approached
zero as t.
Since unstable systems indicate a loss
of control, steady state analysis is not
applicable.
Stability should be tested first.
Steady-state error:
a. step input;
b. ramp input
Closed-loop control
system error:
a. general representation;
b. representation for unity feedback systems
Show practical
(hand)
2
System with:
a. finite steady-state error for a step input;
b. zero steady-state error for step input
System with:
a. finite steady-state
error for a step input;
b. zero steady-state
error for step input
(a)
C (s)
K

R( s) 1  K
E (s)
1

R( s) 1  K
1
ess 
due to a step input
1 K
(b)
C (s)
K

R( s) K  s
E (s)
s

R( s) K  s
ess  0
due to a step input
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Unity Feedback
R( s)
1  G (s)
ess  e()  lim e(t )
E (s) 
t 
sR ( s )
s 0 1  G ( s )
 lim
(a) Step Input
s (1/ s )
1  G (s)
1
1
 lim

s  0 1  lim G ( s )
1 Kp
e( )
 lim
s 0
s 0
The term lim G ( s ) is the dc gain of the forward TF G ( s )
s 0
or the position error constant K p
For e() to be zero,
lim G ( s )  K p  
s 0
(b) Ramp input
s (1/ s 2 )
s 0 1  G ( s )
1
1
 lim
 lim
s  0 s  sG ( s )
s  0 s  sG ( s )
1
 lim
s  0 sG ( s )
The term lim sG ( s ) is the velocity error constant K v
e( )
 lim
s 0
For e() to be zero,
lim sG ( s )  K v  
s 0
(c) Parabolic input
e( )
s (1/ s 3 )
1  G ( s)
1
 lim 2
s 0 s  s 2 G ( s )
1
 lim 2
s 0 s G ( s )
 lim
s 0
The term lim s 2 G ( s ) is the accleratrion error constant K a
s 0
For e() to be zero,
lim s 2 G ( s )  K a  
s 0
4
Example
e()  lim
s 0
sR ( s )
1  G (s)
5
s
5
5
5
e()  lim


s  0 1  lim G ( s )
1  20 21
(a) Due to a step input 5u (t ), LT is
s 0
sR ( s )
s 0 1  G ( s )
e()  lim
(b) Due to a ramp input 5tu (t ), LT is
e()  lim
s 0
5
s2
5
5
 
lim sG ( s ) 0
s 0
5
e()  lim
s 0
sR ( s )
1  G (s)
(c) Due to a parabolic input 5t 2 u (t ), LT is
10
s3
10
10


2
s  0 lim s G ( s )
0
e()  lim
s 0
Example (with
integrator)
e()  lim
s 0
sR ( s )
1  G ( s)
(a) Due to a step input 5u (t ), LT is
e()  lim
s 0
5
s
5
5
 0
1  lim G ( s ) 
s 0
5
(previous eg.) which is finite is now zero.
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The plant has an integrator or is now a Type 1. Without
the integrator, the plant is type 0.
Note the value
6
e()  lim
s 0
sR ( s )
1  G (s)
(b) Due to a ramp input 5tu (t ), LT is
5
s2
5
5
1


s  0 lim sG ( s )
100 20
e()  lim
s 0
With Type 0, the value is , now the value is
finite since the plant Type is 1.
sR ( s )
s 0 1  G ( s )
e()  lim
(c) Due to a parabolic input 5t 2 u (t ), LT is
e()  lim
s 0
10
s3
10
10


2
lim s G ( s ) 0
s 0
No improvement over the Type 0 plant
Feedback control system for defining
system type
7
Relationships between input, system type, static
error constants, and steady-state errors
Error Constants
Example
Fig. 7.10
8
A Routh Hurwitz test should be conducted to ensure stability. Do this!!
Example
9
Example
10
Example
Feedback control system showing disturbance
We need to evaluate E(s)/R(s). Two
inputs exist R(s) and D(s).
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C ( s)  R( s)  E (s)
Also,
C ( s )  E ( s )G1 ( s )G2 ( s )  D ( s )G2 ( s )
Solving for E ( s ) from these two Eqs.
E (s) 
G2 ( s )
1
R( s) 
D( s)
1  G1 ( s )G2 ( s )
1  G1 ( s )G2 ( s )
Now, the steady state value of error can be solved as
e()  lim sE ( s )  lim
s 0
s 0
sG2 ( s )
s
R( s )  lim
D( s)
s  0 1  G ( s )G ( s )
1  G1 ( s )G2 ( s )
1
2
e()  eR ()  eD ()
Error due to input R( s)
s
eR ()  lim
R( s)
s  0 1  G ( s )G ( s )
1
2
Error due to disturbance D( s)
sG2 ( s )
eD ()   lim
D( s)
s  0 1  G ( s )G ( s )
1
2
Assuming the disturbance D( s) 
eD ()  
1
s
1
1
lim
 lim G1 ( s )
s 0 G ( s )
s 0
2
The steady state error due to D( s ) can be reduced by
(a) increasing the dc gain of G1 ( s )
(b) decreasing the the dc gain of G2 ( s )
It is worth noting that the denominators of the two
TFs are the same
E (s)
E (s)
and
R( s)
D( s)
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Steady State Errors
Disturbances - examples
Non unity feedback (without disturbances)
Non unity feedback (with disturbances)
Examples
Example
eD ()  

1
1
lim
 lim G1 ( s )
s 0 G ( s )
s 0
2
1
1

0  1000
1000
Note that the dc gain of G2 ( s ) =K p 2 is
infinite since the system is of Type 1.
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Example
G1 ( s )  1000; G2 ( s ) 
eD ()  

s2
s4
1
1
lim
 lim G1 ( s )
s 0 G ( s )
s 0
2
1
 9.98e  4
2  1000
Steady State Error – Nonunity
Feedback
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Forming an equivalent unity feedback system
from a general nonunity feedback system
Ge ( s )
Example – non unity feedback
G (s) 
100
1
; H 2 (s) 
s ( s  10)
s5
G (s)
1  G (s) H (s)  G (s)
100( s  5)
 3
s  15s 2  50 s  400
The equivalent TF Ge ( s ) =
The system is Type 0, the static error constant is
K p  lim Ge ( s ) 
s 0
e( ) 
100 * 5
 1.25
400
1
1

 4
1  K p 1  1.25
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Example – non unity feedback
Evaluate e() for a unit step input. Repeat for a unit ramp.
OR, the equivalent TF can also be solved using
Ge ( s ) =
subst. G ( s) 
Ge ( s ) 

G ( s)
1  G ( s) H ( s)  G ( s)
100
1
and H ( s) 
in the above
s4
s 1
100
s4
 100   1   100 
1 



 s  4  s 1  s  4 
100( s  1)
s 2  95s  4
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Nonunity feedback control
system with disturbance
E ( s)  R( s)  C ( s) H ( s)
C ( s )  E ( s )G1 ( s )G2 ( s )  D( s )G2 ( s )




G1 ( s )G2 ( s )
G2 ( s )
E ( s )  1 
 R( s)  
 D( s)
 1  G1 ( s )G2 ( s ) H ( s ) 
1  G1 ( s )G2 ( s ) H ( s ) 
Now, the steady state value of error can be solved as
 



G1 ( s )G2 ( s )
G2 ( s )

e()  lim sE ( s )  lim s  1 
 R( s)  
 D( s) 
s 0
s 0
1  G1 ( s )G2 ( s ) H ( s ) 

  1  G1 ( s )G2 ( s ) H ( s ) 
For both R ( s )  D( s ) 
1
s

 
 
lim G1 ( s )G2 ( s )
lim G2 ( s )

s 0
s 0


e()  lim sE ( s )  lim  1 
s 0
s 0
lim
1

G
(
s
)
G
(
s
)
H
(
s
)
lim
1

G
(
s
)
G
(
s
)
H
(
s
)








1
2
1
2
  s 0
  s 0
For zero e()
lim G1 ( s )G2 ( s )
s 0
lim 1  G1 ( s )G2 ( s ) H ( s ) 
s 0
 1 and
lim G2 ( s )
s 0
lim 1  G1 ( s )G2 ( s ) H ( s ) 
0
s 0
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Sensitivity Analysis
How will changes in the system
parameters affect the dynamic behavior of
the system?
The degree to which changes in the
system parameters is called Sensitivity.
The higher the sensitivity, the less
desirable the effect of this parameter
change on the system or transfer function
Feedback can provide reduces sensitivity
to parameter changes
Assume F  K / ( K  a)
If K  10 and a  100, evaluating F  0.091
If a  300 (increasing by 3 times), F  0.032
The change in the parameter a is
 300-100 

 *100%  a 200% change
 100 
causes a change in the parameter F is
 0.032-0.091 

 *100%  a -65% change
 0.091 
Definition
Sensitivity is the ratio of fractional change in the
function F to the factional change in a parameter
P as the fractional change in the parameters
approaches zero.
This is expressed as
S F , P  lim
P  0
Fractional change in the function F
Fractional change in the parameter P
 lim
P  0
F/ F
F/ P P  F
 lim

P/ P P 0 P/ F F  P
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Example – Sensitivity of a CLTF
Given the block diagram, calculate the sensitivity
to changes in the parameter a.
Explain how to reduce the sensitivity.
T (s) 
SF ,P 
K
s  as  K
P F
F P
2
Using the sensitivity equation (the function is T and the parameter is a )
ST ,a 

a T
T (s)  a

a
 Ks

K
 s 2  as  K 2


2
s  as  K
ST ,a 




 as
s 2  as  K
The sensitivity of the CLTF to changes in the parameter a
can be reduced by increasing K .
Example – Sensitivity of a steady state error with ramp input
Find the sensitivity of e() due to changes in K and a with a ramp input
First, determine the function F
1
1
a
F  e( ) 


K v lim sG ( s ) K
SF ,P 
P F
F P
s 0
The sensitivity of the function e() to changes in
the parameter a (of course P  a ) is
Se,a 
a e a  1 

1
e  a a  K 
K
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SF ,P 
P F
F P
The sensitivity of the function e() to changes in
the parameter K is (now P  K )
Se, K 
K e
K  a 

 1
a  K 2 
e K
K
Changes in either parameter a or K on e()
results in no reduction or increase in sensitivity
of the TF. The negative sign indicates a decrease
in e() as K increases.
Example – Sensitivity of a steady state error with a step input
Find the sensitivity of e() due to changes in K and a with a step input
The steady state error for this Type 0 is
e() =
1

1+K p
1
K
1
ab

ab
ab  K
The sensitivity of e() due to changes in a is
Se,a 
a e

e a
 ab  K  b  ab
a
K
*

2
ab
ab  K
 ab  K 
ab  K
2
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The sensitivity of e() due to changes in K is
Se, K 
K e

e K
K
ab
K
*

2
ab
 ab  K  ab  K
ab  K
The equations show that the sensitivity due to
changes in K and a is less that unity for positive a and b.
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