ELEN90055 Control Systems Worksheet 3 – Solutions to Starred Problems
Semester 2
Y G1 G4 R=G7+ 1+GG G3 1+GG G6
2. (a) and (b): The transfer function is
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Y=s(s+1)=Ae .
R 1 + Ae−sT s2 + s + Ae−sT
s , the transfer function can be written as
A ( 1 − 2 s ) .
(s2 +s)(1+ T2 s)+A(1− T2 s) ( s 2 + s ) ( 1 + T2 s ) + A ( 1 − T2 s ) = 0
⇒ T2 s 3 + ( 1 + T2 ) s 2 + ( 1 − A T2 ) s + A = 0 Routh’s Hurwitz array:
s3 T2 1−AT2 ⇒ T2>0 ⇒T>0
s2 1+T2 A ⇒T>−2
s1 (1+T2)(1−AT2)−AT2 0 ⇒ 1−AT2 +T(1−A)>0 1+T2 42
Using the approximation e−sT ≈ 1− T2 1 + T2
A 1 − T2 s Y = 1 + T2 s
R s2 +s+A1−T2 s 1 + T2 s
So the characteristic equation is:
s0 A 0 ⇒A>0 SotheconditionsonAandT isthatT >0,A>0and1−AT2 +T(1−A)>0(thinkwhy
42 we did not assume the first column of the Routh’s array is negative).
3. We first find the transfer function of the system:
H(s)= = Js+kK s = R(s) 1+ K 1
Js2 +kKs+K ⇒ Y(s)= K 1
Using the initial value theorem we have
Js2+kKs+K s limy(t) = lim sY(s) = 0.
We now use the final value theorem to compute the steady state value of the output, but before that we need to check the stability of the system. It can be easily verified that the system is stable as J, kK and K are all positive. So we can apply the final value theorem:
lim y(t ) = lim sY (s) = 1 t→∞ s→0
4. In this system, the equation for the error is
Routh’s Hurwitz array:
E(s) = R(s)−Y(s)
= 1 − G(s) R(s)
1+G(s) = 1 R(s)
= s2(s + 2) R(s)
s3 +2s2 +4s+4
We now check the stability of the system to see if we can use the final value theorem.
s2 2 4 s1 2×4−1×4=20
So all poles of the system are on the left-half plane and therefore the system is stable. (a) R(s) = 1s :
lim e(t) = lim sE(s) = lim s2(s + 2) = 0
s→0 s3 +2s2 +4s+4
(b) R(s) = 1 : s2
(c) R(s) = 2 : s3
lime(t)=limsE(s)=lims s2(s+2) 1 =0 t→∞ s→0 s→0 s3+2s2+4s+4s2
lime(t)=limsE(s)=lims s2(s+2) 2 =1
s→0 s3+2s2+4s+4s3
H1(s) = Y(s) = R(s)
H2(s)= Y(s) = W(s)
D(s) s2+1 1+ D(s)
1 s2+1 1+ D(s)
D(s) = s2 +1+D(s)
1 = s2 +1+D(s)
(b) The characteristic polynomial is s2 + s + k + 1 = 0. For the second order system, we know all poles are on the laft-half plane if and only if all coefficients of the polyno- mial are positive (or all negative). So k + 1 > 0, that is k > −1.
(c) Definee(t)=r(t)−y(t).Then
E(s) = R(s)−Y(s) = 1− s2 +s+k+1 R(s) = lime(t)=limsE(s)= 1
We now calculate the output from the disturbance w(t). 111
Y(s) = s2 +s+k+1 W(s) = s2 +s+k+1 s limy(t)=limsY(s)= 1
t→∞ s→0 k+1
t→∞ s→0 k+1
From (1) and (2), we observe that we can make limt→∞ e(t) and limt→∞ y(t) arbitrarily small by letting k to be large. Note that k should be greater than 0 to make the system stable (see part (b)).
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