88324 control system mid term2008
TRANSCRIPT
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8/9/2019 88324 Control System Mid Term2008
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Solution to Mid-term
1)
1)
3 65 1 1 5( ) ( ) ( ) ( )
3 3 6 3
9 3, 3 2, 1.06
2*3 2 2 2
t t
n
F s f t e es s
overdamped
= =
+ +
= = = =
2)
2 2 2 2
1( ) (sin 5 sin )
2
1 5( ) ( )
2 25
f t t t
F s
s s
=
=
+ +
Two under-damped components
3)
6 1144 12, 0.25
2*12 4n
underdamped = = = = =
Note that
2 2
3
20 20 135( )
( 3) 135 ( 3) 135135
20( ) sin 135
135
t
F ss s
f t e t
= =+ + + +
=
4)
2 2
2( )
9 9
2( ) cos3 sin 3
3
sF s under damped
s s
f t t t
= + + +
= +
5)
2
4
2 2
( ) 10( ) 5( ) 8 25
8 80.8
2*5 10
10 3 10( ) ( ) sin 3
3 ( 4) 3 3
n
t
X sF sU s s s
underdamped
F s f t e t s
= = =+ +
= = =
= =+ +
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2) Note that
50b t t
a a
K K KT gives
R R= +
0 100T= = (See figure), so
100 50
0 150, : 150 50 0
t
a
b t t
a a
K
R
K K KT so
R R
=
= = + =
we have :
13
1
6
t b
a
K K
R
= =
=
1) Since
'
1 1( ) ( ) ( )
6 3
1 50 1( ),
3 150 3
1(18* ( ) 36 ( )) 2*3 ( ) 2*3 ( )3
a a
a
f f L L L L
E s I s s
T I s N
T T T NT T s s s s s s
= +
= = =
= = + = +
So
(6 6) ( ) 6 ( ) 12 ( )
(12 18) ( ) (12 18) ( )
L L L
L L
T s s s s s
s s s s s
= + + +
= + = +
and
( ) 3 (12 18) ( )
1 1 1( ) 3 (12 18) ( ) 3 ( ) (12 18) ( )
6 3 2( ) 1
( ) (6 10)
a L
a L L L
L
a
I s s s s
E s s s s s s s s s s
s
E s s s
= +
= + + = + +
=+
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2) For
5
3
( ) ( ) 1
1 1 6( )
(6 10) 10 10(6 10)
( ) 0.1(1( )) 0.1
a a
L
t
L
e t E s
ss s s s
t t e
= =
= = + +
=
3)
1) Note for 0ref = , we have :
1( ) 0.5 1
10 1( ) (0.5 1)( ) 101
0.5 1
ms s Js b
kp D s s Js b kp
s Js b
++= =
+ + ++
+ +
when1
( ) , ( ( ) 1( )) D s D t t s
= =
1 0.5 1( )
(0.5 1)( ) 10m
ss
s s Js b kp
+ =
+ + +
so0
1 1( ) lim ( )
10 1 10m m
ss s
b kp kp = = =
+ +
2) For D(t)=0, then
( ) 1
10( )1
(0.5 1)( )ref
E s
kps
s Js b
=+
+ +
1 1( )
101
(0.5 1)( )
E skps
s Js b
=
++ +
for1
( ) , ( ) 1( )ref ref s t ts
= =
0
1( ) lim ( )
1 10se sE s
kp = =
+
Obviously, the disturbance will incur a comparable error in the tracking error at the steady state.