Laplace Transform 2

(a)
(i)
1/[(s+1)(s^2+4)]= (1/5)[ 1/(s+1)+(-s+1)/(s^2+4)]
so, the inverse Laplace transform= 0.2 e^(-t)- 0.2 cos(2t)+ 0.1 sin(2t)
(ii)
f(t)= e^(-t)u(t-π)= e^(-π) e^(π-t) u(t-π)
F(s)= e^(-π) e^(-πs) /( s+1)

y"+4y=f(t), y(0)=1, y'(0)=0
取Laplace得: s^2Y-s-0+4Y=F(s)= e^(-π) e^(-πs) /s
Y= s/(s^2+4) + e^(-π) e^(-πs)/[(s+1)(s^2+4)]
= s/(s^2+4) + e^(-π) e^(-πs) (1/5) [ 1/(s+1) + (-s+1)/(s^2+4) ]
y= cos(2t) + (1/10) e^(-π) u(t-π) [2e^(π-t) -2cos(2t)+ sin(2t) ]

(b)
(i)
case 1: t > 3
f*g=∫[0~t] f(t-x) g(x) dx
=∫[0~2] 2f(t-x) dx = ∫[t-2~t] 2f(y) dy = 0 (因 f(y)=0)
case 2: 2 < t < 3
f*g=∫[0~t] f(t-x) g(x) dx
=∫[0~2] 2f(t-x) dx=∫[t-2~t] 2f(y) dy
=∫[t-2~1] 2y dy= 1-(t-2)^2
case 3: 1 < t < 2
f*g=∫[0~t] f(t-x) g(x) dx
=∫[0~t] 2f(t-x) dx =∫[0~t] 2f(y)dy = ∫[0~1] 2y dy= 1
case4: 0 < t < 1
f*g=∫[0~t] f(t-x) g(x) dx
=∫[0~t] 2f(t-x) dx = ∫[0~t] 2f(y) dy=∫[0~t] 2y dy= t^2
so,
f*g=t^2[1-u(t-1)]+[u(t-1)-u(t-2)]+[1-(t-2)^2][u(t-2)-u(t-3)]
=t^2+ (1-t^2) u(t-1)-(t-2)^2 u(t-2)+ (t^2-4t+3) u(t-3)
註: 用inverse Laplace of [ F(s)G(s) ] 比較快
以上4個 case, t之範圍均可含等號

(ii)
取Laplace得
2/(s^3) (sY-0)+ Y/s = 2/s^3
s(s+2)Y=2
Y= 1/s - 1/(s+2)
y= 1- e^(-2t)

2011-07-30 01:48:07 補充：
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