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(Elec271)[2011](s)midterm~682^_39663.pdf
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ELEC271 -Automatic Control Systems
Reference Solution to Midterm
April 4, 2011

1

Since the input is U(s)= 1 , The output is
s
Y (s)= G(s)U(s) 3s3 +5s2 +6s +4 1
=
s(s2 +2s + 4) s 11 2s +2
= ++
ss2 s2 +2s +4 11 2(s + 1)
= ++

ss2 (s + 1)2 +3 Hence
y(t)= L.1[Y (s)] L.1 .1 .. 1 .. (s + 1) .
=+ L.1 +2L.1
ss2 (s + 1)2 +3

.t
= [1+ t +2e cos 3t](t)
2

There are more than one way to solve the problem. The .rst method de.nes intermediate signals .rst. De.ne the input of B(s) to be v and its output to be w, as shown in Figure 1. Thus we have
v = A(s)(r . E(s)w) w = B(s)v y = C(s)w + D(s)v

Eliminate v and w and we have the solution yA(s)[B(s)C(s)+ D(s)] r = 1+ A(s)B(s)E(s) .
1


y
r
-
-
Figure 1: The original control system.

The other way is to simplify the diagram by equivalent substitutions. The original diagram as shown in Figure 1 is equivalent to the one in Figure 2.


y
r
-
-
Figure 2: The equivalent control system. Thus the transfer function is y .D(s) . A(s)B(s)
= C(s)+
rB(s) 1+ A(s)B(s)E(s) A(s)[B(s)C(s)+ D(s)]
=
1+ A(s)B(s)E(s)

For this problem, the order of P (s) is n = 2, and the number of poles n + m = 3, so m =1= n . 1. Hence to get a unique controller, we can .nd among proper controllers.
Assume that the 1st order proper controller is of the form q0s + q1
C(s)= .
p0s + p1
2

Then the closed loop characteristic polynomial is c(s)= a(s)p(s)+ b(s)q(s)=(s 2 . 1)(p0s + p1)+ s(q0s + q1). On the other hand from the desired closed loop poles, we can also know that c(s)=(s + 1)(s + 3)2 = s 3 +7s 2 + 15s +9. Comparing the coe.cients gives q0 = 16,q1 = 16,p0 =1,p1 = .9. Hence the controller is
16s + 16
C(s)= .
s . 9

One can also use Diophatine equations. In this case S(a(s),b(s), 2) = [T (a(s), 2) T (b(s), 2)]
. .
. . .
.
1 0 0 0
0 1 1 0
.1 0 0 1
0 .1 0 0

=
Solve

. .
. . .
.
=

. .
.
1
7
15
9

. .
.
, p0
p1
q0
S(a(s),b(s), 2)
q1

and you should conclude with the same controller.
Note that since all the coe.cients are not mutually independent, Kharitonov Theorem does not apply here. One had better use Routh table.
1 4 + K 2 + 3K
2 + K 4 + 2K
2 + K 2 + 3K
2 . K
2 + 3K


Since K [0, 1], 2+ K, 2 . K, 2+3K are positive for all K. Hence the polynomial is stable for all K [0, 1].
One can also use routh.m published with HW3 to verify this. The MATLAB command line runs like
3

>> syms K; %define K first, you need symbolic toolbox installed to make this work. >> routh([1 2+K 4+K 4+2*K 2+3*K]);
================================= s^4|1 K+4 3*K+2 s^3|K+2 2*K+4 0 s^2|K+2 3*K+2 0 s^1|2-K 0 0 s^0|3*K+2 0 0
4