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(elec271)[2009](s)solmid~PPSpider^_10313.pdf
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ELEC271 -Automatic Control Systems, Spring 2008-09
Reference Solution to Mid-term
March 25, 2009
1 Solution
d (cos t)(t) = (cos t).(t) + (cos t).(t)
dt
= .( sin t)(t) + (cos t)(t) = .( sin t)(t)+ (t)
where (t) is the unit step function and (t) is the unit impulse function.
2 Solution
One of the suitable state space models looks like:
..
..
a1
. 1 0
a0
..
..
b11 b21
..
..
.. .. .
.
. ...
... .
A = . an.1 0 1 , [ b1 b2 ]=
,
..
a0
..
..
b1n b2n
..
..
an
1
. 00 .
0 0
,
[ d1 d2 ] ==
c
.
a0
Note that the system cannot be written in a controller form like
..
a1 an.1 an
. . .
a0 a0 a0
..
..
1 2
a0 a0
..
A =1 00 , [ b1 b2 ]= 00 , . ..
...
. .. .
..
..
..
.... ..
0 10 00
c = . c1 cn . , [ d1 d2 ]= . 00 . .
since the corresponding transfer function is
G(s) = [ G1(s) G2(s) ]
= [ c(sI . A).1b1 c(sI . A).1b2 ]
= [ 1(c1sn.1 + + cn) 2(c1sn.1 + + cn) ]
a0sn + + an =[ 1 2 ]G.(s)
where
n.1 +
c1s + cn
.
G(s)= .
a0sn + + an But the original DISO transfer function has di.erent numerators for di.erent inputs. Hence such state space models are not the realizations for the original DISO system.
In fact if one insists to express the state space model in a controller form, the knowledge of Kalman decomposition is needed, which is not covered in ELEC271 but in ELEC560 (Linear system theory).
For other related knowledge, refer to Page 39&42, Chapter 2 for realization of systems in form of transfer functions, and Page 46, Chapter 2 for realization of DISO systems.
Solution
Given the closed-loop poles, the characteristic polynomial is obvious:
c(s)=(s + 1)(s + 2)(s + 3)(s + 4)(s + 5) = s 5 + 15s 4 + 85s 3 + 225s 2 + 274s + 120
Denote that P (s) is of n-th order and the desired controller is of m-th order, then we know that n = 3 and n + m = 5, hence m =2= n . 1. By the knowledge on Page 97, Chapter 3, it is preferred to seek for a proper controller since the solution is unique in this case. Denote the
2
q0s +q1s+q2
controller to be C(s) = and the corresponding Diophantine equation is also easy
p0s2+p1s+p2
to be .gured out:
p0
p1
100 000
1
15
85
010
|
001 000
|
p2
=
000 100 q0 225
|
000 010
.
.
274
120
.
||
.
..
q1
000 001
q2
Hence the desired controller is exactly
225s2 + 274s + 120
C(s)= .
s2 + 15s + 85
4 Solution
4. (a)
The Routh table corresponding to a(s) is:
s3 1 2
s2 K K
s1 1
s0 K
When K [1, 4], K is always positive, so the elements of Routh table are all positive and every polynomial a(s) is stable.
4. (b)
We denote the elements of the Routh table of a(s) and aK1,K2 (s) respectively by rij and rij .
We have rij = K1rij for i = 0 and rij = K2rij for i = 1. In the following we show th