B_lecture9 Examples of Steady-state Error Automatic control System
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Transcript of B_lecture9 Examples of Steady-state Error Automatic control System
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Lecture 9 examples of steady-state error
1
Examples of steady-state error
Example 1
The forward transfer functions of the negative unity feedback system are as
follows:
2( )
( 1)(0.5 1)G s
s S s
Determine the steady-state error for the following inputs, respectively.
Check the stability of the system before applying the final-value theorem.
2 21 11. ( ) 1( ) 2. ( ) 3 3. ( ) 4. ( ) 1( ) 32 2
r t t r t t r t t r t t t t
Answer:
1. Check the stability of the system
3 2 3 2
( ) 2 4( )
1 ( ) 0.5 1.5 2 3 2 4
G ss
G s s s s s s s
The characteristic equation 3 23 2 4 0s s s
3
2
1
0
1 2
3 4
2
3
4
s
s
s
s
The system is stable.
2. Manner 1.
2 2
3 2
( 3 2)( ) ( ) ( ) ( ) ( ) ( ) ( )
3 2 4r
s s sE s R s C s R s s R s R s
s s s
12
1 3 20 0
( 3 2) 1lim ( ) lim 0
3 2 4ssr r
s s
s s se s E s s
s s s s
22
2 3 2 20
( 3 2) 3lim 1.5
3 2 4ssr
s
s s se s
s s s s
32
3 3 2 30
( 3 2) 1lim
3 2 4ssr
s
s s se s
s s s s
4Superposition 4 1 2 3 0 1.5ssr ssr ssr ssre e e e
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Lecture 9 examples of steady-state error
2
Manner 2
Because the system is a unity feedback system
The error definition between ( ) ( ) ( )e t r t c t and ( ) ( ) ( )e t r t b t are same.
Open loop transfer function 2
( )( 1)(0.5 1)
G ss s s
1 typesystem. 2K
Find error constants pK vK aK
0 0
2lim ( ) limps s
K G ss
0lim ( ) 2vs
K sG s
2
0lim ( ) 0as
K s G s
1
10
1 1ssr
p
Re
K
2
31.5
2ssr
v
Ve
K
3
1
0ssr
a
Ae
K
Superposition 4 1 2 3 0 1.5ssr ssr ssr ssre e e e
Example 2
The control system is showed in Fig. example 2.
Find the steady-state error of the system.
5.0
s
1
)12.0(
10
ss
1.0n
tr 1 c
_
e
Fig. example 2
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Lecture 9 examples of steady-state error
3
Answer:
1. Check the stability of the system
Open loop transfer function 2
1 10 10(0.5 1)( ) ( 0.5)
(0.2 1) (0.2 1)
sG s
s s s s s
Close loop transfer function 3 2
( ) 5 10( )
1 ( ) 0.2 5 10
G s ss
G s s s s
The characteristic equation 3 20.2 5 10 0s s s
3
2
1
0
0.2 5
1 10
3
10
s
s
s
s
The system is stable.
2. Find the steady-state error of the system.
Manner 1.
3 2
3 2
( ) 1 1 0.2
1 10( ) 1 ( ) 0.2 5 101 ( 0.5)
(0.2 1)
RE s s s
R s G s s s s
s s s
3 2
3 2 20 0
0.2 1lim ( ) lim 0
0.2 5 10ssr R
s s
s s se sE s s
s s s s
Manner 2.
2 type system
( ) 1r t t 0ssre
When ( ) 0r t 3 2
10
( ) 10(0.2 1)
1 10( ) 0.2 5 101 ( 0.5)(0.2 1)
nE s ss s
N s s s s
s s s
3 20 0
10 0.1lim ( ) lim 0
0.2 5 10ssn n
s s
se sE s s
s s s s
Superposition 0 0 0ss ssr ssne e e