Steady-state Error

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Outline
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Steady-State Error
M. Sami Fadali
Professor of Electrical Engineering
University of Nevada
Steady-state error.
Nonunity feedback.
Unity feedback: error constants.
Error due to disturbance.
1
Error
2
Error Block Diagram
e(t )  r (t )  c(t )
e()  _|Å e(t )
R(s)
T(s)
t 
 _|Å sE ( s )
C(s)

+
E(s)
E ( s )  R ( s )  C ( s )  1  T ( s )R( s )
s 0
Final Value Theorem
• Assume stability for the limit to exist.
• Steady state practically reached after (3-5)
•  = largest time constant of the system
e()  _|Å sE ( s )
s 0
 _|Å 1  T ( s )sR ( s )
s 0
3
4
Example
Error and Stability
Find the steady-state error due to a unit step.
_|Å
• Stable range:
• System must remain stable
→
_|Å
→
• Steady-state error cannot be reduced below
5
Unity Feedback
th
order, type
G (s) 
K   j s  1
s 0
i 1
G( s) 
R ( s ) s l 1 K   i s  1
i 1
n l
s
l
  s  1   
m
i
i 1
j 1
j 1
n l
s l   i s  1
i 1
n l
s 0
K   j s  1
m
s R( s)
e()  _|Å sE ( s )  _|Å
s 0
s 0 1  G ( s )
 _|Å
s 0
s (1 s )
1

1  G ( s ) 1  _|Å G ( s )
s 0
R( s)
E (s) 
1  G( s)
s l   i s  1
Zero for type > 0
e()  _|Å sE ( s )  _|Å

j 1
n l
Error depends
on input & type
number of
system.
Step: Position Error Constant
R(s) E(s)
C(s)
+
G(s)
m
6
j
s  1
7
 K , Type 0

K p  _|Å G ( s )  , Type 1
s 0
, Type 2

1 K p  1, Type 0

e   0,
Type 1
0,
Type 2

8
Parabolic: Acceleration Error Constant
Ramp: Velocity Error Constant
s (1 s 2 )
1

e()  _|Å sE ( s )  _|Å
s 0
s 0 1  G ( s )
_|Å sG ( s)
s 0
K   j s  1
m
G( s) 
j 1
n l
K v  _|Å
s l   i s  1
s 0
i 1
0, Type 0

sG ( s )   K , Type 1
, Type 2

Type 0
,

e   1 K v , Type 1
0,
Type 2

Step 1(t)
Type 0
1/(Kp+1)
Type 1
0
K   j s  1
s 0
m
G( s) 
j 1
n l
s l   i s  1
i 1
0, Type 0

K a  _|Å s 2G ( s )  0, Type 1
s 0
 K , Type 2

Type 0
,

e   ,
Type 1
1 K , Type 2
 a
10
9
Example
Unity Feedback Error
Input
s (1 s 3 )
1

e()  _|Å sE ( s )  _|Å
s 0
s 0 1  G ( s )
_|Å s 2G ( s)
For unity FB, find the steady-state error due to a
(i) unit step, (ii) unit ramp, (iii) unit parabolic.
Type 2
0
Type 0:
Ramp t

1/Kv
0
Parabolic t2/2


1/Ka
11
_|Å
→
Type 0: Infinite steady-state error due to a unit
ramp or parabolic.
12
Example
Example
For unity FB, find the steady-state error due to a
(i) unit step, (ii) unit ramp, (iii) unit parabolic.
For unity FB, find the steady-state error due to a
(i) unit step, (ii) unit ramp, (iii) unit parabolic.
Type 1:
_|Å
_|Å
Type 2:
→
Type 1: Infinite steady-state error due to a unit
parabolic and zero error due to a unit step.
13
→
Type 2: Zero steady-state error due to a unit step
and due to a unit ramp.
14
Percentage Error
e A ()  _|Å sE A ( s )  Ae()
Input Amplitude Scaling
E ( s )  1  T ( s )R ( s )
E A ( s )  1  T ( s )AR( s )  AE ( s )
e A ()  _|Å sE A ( s )  Ae()
s 0
• Error is scaled if input is scaled.
• Table for unity feedback assumes unit input.
• For amplitude A 1, multiply all error expressions
by the input amplitude A.
15
s 0
e()% 
e A ()% 
e( )
100%
1
Ae()
100%  e()%
A
• Expressed as a percentage of the input amplitude.
• Unaffected by input scaling.
16
Error Due to Disturbance
Unity vs. Nonunity Feedback
• Disturbance: “nuisance” input non included in model.
Nonunity feedback: get the steady-state error
using the closed-loop transfer function
.
Unity feedback: use system type and error
constants.
– Type 0, use position error constant
• Output due to disturbance is ideally zero.
• All output due to disturbance is error.
• Total error = tracking error + error due to disturbance.
• For linear systems, use superposition.
(step)
– Type 1, use velocity error constant (ramp)
– Type 2, use acceleration error constant
(parabolic)
17
18
Steady-State Error
_|Å
Includes
Example
sign
→
• Find the steady-state error due to a unit step input
and a unit step disturbance.
_|Å
→
+
R(s)
_|Å
→
+
G2

+
D(s)
C(s)
G1
• For unity feedback systems, we can use
.
error constants to calculate
Use superposition (linear system)
19
20
Example
Effect of Disturbance
D(s)
Find the error due to (a) a unit step disturbance,
(b) a unit step input, (c) the total error.
C(s)
+
G1

_|Å
→
G2
G1
TD ( s ) 
1  G1G2
 sG1 D ( s )
eD ()  _|Å
s 0 1  G G
1 2
21
(b) Unit step error (c) Total error
b) Unit step tracking error: Unity feedback, Type 0
_|Å
→
_|Å
→
c) Total error
23
_|Å
→
22
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