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(ELEC271)[2010](s)final~1833^_72156.pdf
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ELEC271 -Automatic Control Systems
Reference Solution to Final, Spring 2010
Easy to see that
J =4 (.y1(t).22 + .y2(t).22)|w1(t)=(t) +(.y1(t).22 + .y2(t).22)| w1(t)=0 . w2(t)=0 w2(t)=(t)
Hence it su.ces to .nd the controller with unweighted optimal transient according to Algorithm 9.1 in Section 9.1.

Step I (Spectral factorization): First we .nd d(s)= s2 +6s + 2 which satis.es a(.s)a(s)+ b(.s)b(s)= d(.s)d(s).
q(s)
Step II (Pole placement): Assume the controller to be C(s)= p(s) which is 2nd order and strictly proper. Then by a(s)p(s)+b(s)q(s)= d2(s) we can .nd the controller

(4 6.6 2)s+2
to be C(s)= .
s2+(2 6. 2)s+12.43
Step III (2-norm calculation): Now we can compute the optimal transient cost.
..
d(s) . a(s).
2 .
b(s) .
2 .
d(s) . p(s).
2 .
q(s) .
2. J . =4 +++
d(s) d(s) d(s) d(s)
22 22

..
2
.
(6 . 2)s +2 .
2 .
2
=4 +
. s2 +6s +2 s2 +6s +2 2
2

.
2 .
2.
.
(2 . 6)s +4 3 . 10 .
(4 6 . 6 2)s +2
+ + s2 +6s +2 s2 +6s +2 .
22

20 . 83 4 164 . 883 340 . 192 3
= + + +
666 6

= 886 . 144 2

11.9083
1
2
According to Example 4.25 on Page 180 in the textbook,
.e(t). =1
.11
.e(t).2 = (+ )
4n

1
where = ,n = K, i.e.
K
11
.e(t).2 =+ .
2 4 K
Since u(t)= Ke(t), we have
.u(t). = K
K2 .u(t).2 =+ K
2 4
(a)
11 1 . 19
J = .e(t).22 + .u(t). =+ + K +2 K =
4 K 4 K 4
9
when K = 1 the equality holds. Thus,Jopt = when C(s) = 1.
4
(b)
J = (2+ 2).e(t).22 + .u(t).2
2

K2 2+ 2 12
=+ K ++ K ++
4 K 24
To .nd the optimal K, let

dJ K 2+ 2
= +1 . =0,
dK 2 K2


2
By observation we have K = 2. Thus,Jopt =2+ 9 when C(s)= 2.
4
2
From Example 8.11 on Page 346, we have

1
. 1
2
.G(s). = 11
0
2 1.2 2
Let g(t)= L.1[G(s)]. Then 1
g(t)= L.1[]
s2 +2s +11
= L.1[]
(s + )2 +1 . 2 1 .1 . 2
= L.1[]
.1 . 2 (s + )2 +1 . 2
=1 e .t sin .1 . 2t
.1 . 2 To .nd the maximum value of g(t), consider
1 .t
g .(t)= (.e.t sin .1 . 2t + .1 . 2e cos .1 . 2t)
.1 . 2
.t
e
= . sin(.1 . 2t . cos.1 )
.1 . 2 The peak value of g(t) must be achieved at the smallest positive t when g.(t) = 0.
Hence cos.1
t = .
.1 . 2
Thus, cos.1
.L.1[G(s)]. = g()
.1 . 2
.1
. cos
1.2
= e
(a)
11
When = 0, G(s) = Thus the step response of G(s) is Y (s) = and the
(s+1)2 . s(s+1)2
time-domain response is y(t)=1 . e.t(1 + t) which has no overshoot. 3
(b)
3s+1 3s+1
When = 3, G(s) = Thus the step response of G(s) is Y (s) = and
(s+1)2 . s(s+1)2 3
the time-domain response is y(t)=1 . e.t(1 . 2t). Since when By(t) = 0, t = 2 and y(3 ) > 1, the response has overshoot.
2
(c)
By G(s) we have s +1
Y (s)= ,
s(s + 1)2 and
.t
y(t)=1 . [1 + (1 . )t]e.
.t
Assume .1. y(tc)=[ + (1 . )t]e= 0, then t