Do Now Please answer questions below in your binder Take out your homework
Homework 6 Answer Automatic control System
4
Homework 6 Answer 1 Homework 6 Answer Problem 1 The forward transfer functions of the negative unity feedback system are as follows: 1. 10 () (0.1 1)(0.5 1) Gs s s s 2. 2 7( 1) () ( 4)( 2 2) s Gs ss s s Determine the steady-state error for a unit-step input, a unit-ramp input, and a parabolic input, respectively. Check the stability of the system before applying the final-value theorem. . Answer 1. Check the stability of the system The characteristic equation 3 2 0.05 0.6 10 0 s s s 3 2 1 0 0.05 1 0.6 10 1 6 10 s s s s The system is stable. Manner 1. 1 () () () 1 () s ER E sRs Rs Gs When () 1( ) rt t ， 0 0 1 1 1 () , lim lim 0 10 1 (0.1 1)(0.5 1) ss s s s Rs e sE s s s s s s When () 1( ) rt t t ， 2 2 0 0 1 1 1 () , lim lim 0.1 10 1 (0.1 1)(0.5 1) ss s s s Rs e sE s s s s s s When 2 1 () 1( ) 2 rt t t ， 3 3 0 0 1 1 1 () , lim lim 10 1 (0.1 1)(0.5 1) ss s s s Rs e sE s s s s s s
-
Upload
abaziz-mousa-outlawzz -
Category
Documents
-
view
213 -
download
1
description
Automatic control System
Transcript of Homework 6 Answer Automatic control System
-
Homework 6 Answer
1
Homework 6 Answer
Problem 1
The forward transfer functions of the negative unity feedback system are as
follows:
1. 10
( )(0.1 1)(0.5 1)
G ss s s
2. 2
7( 1)( )
( 4)( 2 2)
sG s
s s s s
Determine the steady-state error for a unit-step input, a unit-ramp input, and a
parabolic input, respectively.
Check the stability of the system before applying the final-value theorem.
.
Answer
1.
Check the stability of the system
The characteristic equation 3 20.05 0.6 10 0s s s
3
2
1
0
0.05 1
0.6 10
1
6
10
s
s
s
s
The system is stable.
Manner 1.
1( ) ( ) ( )
1 ( )s ERE s R s R s
G s
When ( ) 1( )r t t 0 0
1 1 1( ) , lim lim 0
101
(0.1 1)(0.5 1)
ss ss s
R s e sE ss s
s s s
When ( ) 1( )r t t t 2 20 0
1 1 1( ) , lim lim 0.1
101
(0.1 1)(0.5 1)
ss ss s
R s e sE ss s
s s s
When 21
( ) 1( )2
r t t t 3 30 0
1 1 1( ) , lim lim
101
(0.1 1)(0.5 1)
ss ss s
R s e sE ss s
s s s
-
Homework 6 Answer
2
Manner 2.
1 typesystem 10K
Find error constants pK vK aK
0lim ( )ps
K G s
0lim ( ) 10vs
K sG s
2
0lim ( ) 0as
K s G s
When ( ) 1( )r t t 1
01 1
ssr
p
Re
K
When ( ) 1( )r t t t 1
0.110
ssr
v
Ve
K
When 21
( ) 1( )2
r t t t 1
0ssr
a
Ae
K
2.
Check the stability of the system
The characteristic equation 4 3 26 10 15 7 0s s s s
4
3
2
1
0
1 10 7
6 15 0
7.5 7
9.4
7
s
s
s
s
s
The system is stable.
Manner 1.
1( ) ( ) ( )
1 ( )s ERE s R s R s
G s
When ( ) 1( )r t t 0 0
2
1 1 1( ) , lim lim 0
7( 1)1
( 4)( 2 2)
ss ss s
R s e sE sss s
s s s s
When ( ) 1( )r t t t 2 20 0
2
1 1 1( ) , lim lim 8 / 7 1.14
7( 1)1
( 4)( 2 2)
ss ss s
R s e sE sss s
s s s s
-
Homework 6 Answer
3
When 21
( ) 1( )2
r t t t 3 30 0
2
1 1 1( ) , lim lim
7( 1)1
( 4)( 2 2)
ss ss s
R s e sE sss s
s s s s
Manner 2.
1 Typesystem 7
8K
Find error constants pK vK aK
0lim ( )ps
K G s
0
7lim ( )
8v
sK sG s
2
0lim ( ) 0as
K s G s
When ( ) 1( )r t t 1
01 1
ssr
p
Re
K
When ( ) 1( )r t t t 1 8
1.147 7
8
ssr
v
Ve
K
When 21
( ) 1( )2
r t t t 1
0ssr
a
Ae
K
Problem 2
The control system is showed in Fig. problem 2
The error definition is ( ) ( ) ( )e t r t b t , Known ( ) 1( ) , ( ) 1( ) r t t n t t .
1. When 40K , Find the steady-state error of the system.
2. When 20K , Find the steady-state error of the system.
2.5
5
1
s105.0 s
K)(sR
_
)(sE )(sC
)(sN
Fig. problem 2
-
Homework 6 Answer
4
Answer
When 1 1
( ) , ( )R s N ss s
( ) 1
2.5( )1
(0.05 1)( 5)
2.5
( ) 52.5( )
1(0.05 1)( 5)
(0.05 1)( 5) 2.5(0.05 1) 1( )
(0.05 1)( 5) 2.5
ER
EN
E s
KR s
s s
E s sKN s
s s
s s sE s
s s K s
1 K=40 The system is a second-order system. Its characteristic equation
20.05 1.25 105 0s s All coefficient are positive, so the system is
stable.
0 0
(0.05 1)( 5) 2.5(0.05 1) 1 2.5lim ( ) lim 0.0238
(0.05 1)( 5) 2.5 5 2.5ss
s s
s s se sE s s
s s K s K
2K=20, the system is stable.
0
2.5lim ( ) 0.0455
5 2.5ss
se sE s
K
Compare them: greater Ksmaller sse
