ELEN90055 Control Systems Worksheet 2 – Solutions to Starred Problems
1. (a∗) f(t)=1+2t+t2+δ(t) F(s)=L{f(t)}=L{1}+L{2t}+L{t2}+L{δ(t)}
= 1 + 2 + 2! + 1 s s2 s3
(b∗) f(t)=(t+3)2
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f(t)=(t+3)2=t2+6t+9 ⇒ F(s)=2+6+9.
(c∗) f(t)=tsint
We use the multiplication by time property of Laplace transform:
L{tg(t)}=−dG(s). ds
We start with g(t) = sint. From the Laplace transform table we have L{sint}= 1 .
Then we use the above multiplication by time property and obtain
L{tsint}=−d 1 ds s2+1
= 2s (s2 + 1)2
2.(a∗)F(s)= 1 s(s+1)
We first perform partial fraction expansion. There are two ways to do it: Method 1. Cross multiplication:
F(s)= 1 = A+ B = A(s+1)+Bs = (A+B)s+A s(s+1) s s+1 s(s+1) s(s+1)
⇒ A=1, B=−1 1
Method 2. Direct formulas:
A = lim s 1 = 1
s→0 s(s + 1)
B= lim(s+1) 1 =−1
s→−1 s(s + 1) Then using the Laplace transform table we have
(b∗) F(s)= s2 s2 +1
f(t)=L−1{F(s)}=1(t)−e−t
Note that the transfer function is not strictly proper. So we first write F(s) as s2 s2 + 1 − 1 1
(c∗) F(s)= 3s+2 = s2 +4s+20
F(s)= s2+1 = s2+1 =1−s2+1 ⇒ f(t)=δ(t)−sint
3s+2 (s+2)2 +16
We will use L {sint}, L {cost} and the shift in frequency property of the Laplace transforms:
F(s) = 3s+2 = 3(s+2)−4/3 = 3 (s+2)2 +16 (s+2)2 +42
⇒ f(t)=3e−2tcos4t−e−2tsin4t (d∗) F(s)=s2+2s+3
s+2 (s+2)2 +42
− 4 (s+2)2 +42
We first do partial fraction expansion using the following two methods.
Method 1. Cross multiplication:
F(s)=s2+2s+3= B1 + B2 + B3
(s+1)3 s+1 (s+1)2 (s+1)3 = B1(s+1)2 +B2(s+1)+B3
⇒ s2 +2s+3 = B1(s2 +2s+1)+B2(s+1)+B3 = B1s2 +(2B1 +B2)s+B1 +B2 +B3
2B1+B2 =2 ⇒ B2 =0
B1+B2+B3 =3 ⇒ B3 =2
Method 2. Direct formulas:
1 3 Bkl B11 B12 B13
F(s) = ∑ ∑ (s+1)l = s+1 + (s+1)2 + (s+1)3 k=1 l=1
(3−l)! s→−1
d3−l 3s2+2s+3 3−l (s+1) 3
1 d3−l2
(s+1) lim s +2s+3
(3 − l)! s→−1 ds3−l
So we obtain that
B11= lim s+2s+3 =1
1d22 2 s→−1 ds2
d2 B12 = lim s + 2s + 3
B13 = lim s2 + 2s + 3 = 2
s→−1 and F(s) can be written as
= lim 2s + 2 = 0 s→−1
F(s)=s2+2s+3= 1 + 2 . (s+1)3 s+1 (s+1)3
We now find the inverse Laplace transform of this function. −1 1 −t
−1 d 1 −1 1 −t
L −dss+1 =L (s+1)2 =te −1d1 −122−t
L −ds(s+1)2 =L (s+1)3 =t e f(t)=e−t +t2e−t.
3. (a∗) y ̈(t)−2y ̇(t)+4y(t)=0; y(0)=1, y ̇(0)=2 s2Y(s)−sy(0)−y ̇(0)−2sY(s)−y(0)+4Y(s) = 0
⇒ Y(s)s2 −2s+4 = sy(0)+y ̇(0)−2y(0) = s
⇒ Y(s)= s = s = s−1 +√1 3
s2 −2s+4 (s−1)2 +3 (s−1)2 +3 3 (s−1)2 +3
t√1t√ ⇒ y(t)=ecos 3t+√esin 3t
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