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(ELEC271)[2010](s)midterm~1833^_45467.pdf
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ELEC271 -Automatic Control Systems
Reference Solution to Midterm, Spring 2010
Y (s) A(s) C(s)
1.C(s)D(s)
=
R(s) C(s) B(s)
1+ A(s)
1.C(s)D(s) C(s)
A(s)C(s)= 1+ A(s)B(s) . C(s)D(s).
(a)
By Kharitonov Theorem, it su.ces to verify whether the following polynomials are stable
p1(s)= s 3 +2s 2 +5s +5 p2(s)= s 3 +2s 2 +5s +1 p3(s)= s 3 +2s 2 + s +1 p4(s)= s 3 +2s 2 + s +5
By Routh table one can conclude that since a4(s) is unstable, the polynomial s3 + 2s2 + a2s + a3 is not stable for all possible a2,a3.
(b)
In this case we can check the Routh table of the polynomial.
1
1 K
2 K
K
2
K

Easy to see that s3 +2s2 + Ks + K is stable for all possible K.
According to Theorem 3.29, we will check the characteristic polynomial of the closed-loop system when doing stabilizations. Moreover, we need also verify whether the controller will let the degree of the closed-loop characteristic polynomial drop.
(a)
Its not possible.
In this case the closed-loop characteristic polynomial is
c(s)= s 2 + KP .
When KP 0, the closed-loop poles are all on the imaginary axis, i.e. the closed-loop system is not internally stable. When KP < 0, there is always one positive real pole, so the system is not stable, either. To conclude, its not possible to stabilize P (s) with C(s)= KP .
(b)
Its not possible.
In this case the closed-loop characteristic polynomial is
c(s)= s 3 + KP s + KI .
By Routh table, we can see that its not possible for such c(s) to have all stable poles. At least one of them would have a non-negative real part.
2
(c)
Its possible.
A 1st order controller is of the form q1s + q0
C(s)= ,
p1s + p0 where all the coe.cients are real numbers, and p1 .
= 0. Then the closed-loop charac-teristic polynomial becomes
c(s)= p1s 3 + p0s 2 + q1s + q0.
Now this becomes a pole placement problem, and we have the full power to assign the closed-loop poles to anywhere as we wish. Since p1 .
= 0, we dont have to worry about the degree drop in c(s).
For instance, a 1st order controller 3s +1
C(s)=
s +3 will stabilize P (s), since in this case, c(s)= s 3 +3s 2 +3s +1=(s + 1)3 . Its stability can be easily seen, and its degree does not drop.
(d)
As stated in the previous parts.
In the frequency domain,
Y (s) = P (s)U(s)
= P (s) C(s) . R(s) .

.Y (s)
= P (s)[C1(s)R(s) . C2(s)Y (s)], 3
where C(s) = [C1(s) C2(s)]. So
Y (s) = P (s)C1(s)
R(s) 1 + P (s)C2(s)
s . 1
=

s2 + 2s . 3
1
=
s + 3.

The closed-loop characteristic polynomial is given by c(s)=(s + 1)(s . 3) + 4s = s 2 +2s . 3, which is obviously unstable. Hence the closed-loop system is not internally stable.
4