Laplace Transform

a) Resolving the given fraction into partial fraction:

1/[s(s + 2)2] = 1/(4s) - 1/[4(s + 2)] - 1/[2(s + 2)2]

Taking inverse Laplace transform the function is:

u(t)/4 - e-2t/4 - te-2t/2

b i) Using the definition of Laplace transform:

F(s) = ∫ (t = 0 → ∞) e-st f(t) dt

= ∫ (t = 0 → 1) 2e-st dt + ∫ (t = 1 → ∞) e-st e-2(t-1) dt

= [-2e-st/s] (t = 0 → 1) + e2 ∫ (t = 1 → ∞) e-(s+2)t dt

= (2/s - 2e-s/s) + e2 [-e-(s+2)t/(s + 2)] (t = 1 → ∞)

= (2/s - 2e-s/s) + e2 [e-(s+2)/(s + 2)]

= 2/s - 2e-s/s + e-s/(s + 2)

= 2/s + e-s[1/(s + 2) - 2/s]

ii) Taking Laplace transform on the diff. eqn given:

s2Y(s) - sy(0) - y'(0) + 4sY(s) - 4y(0) + 4Y(s) = 2/s + e-s[1/(s + 2) - 2/s]

s2Y(s) + 4sY(s) + 4Y(s) = 2/s + e-s[1/(s + 2) - 2/s]

(s2 + 4s + 4)Y(s) = 2/s + e-s[1/(s + 2) - 2/s]

(s + 2)2 Y(s) = 2/s + e-s[1/(s + 2) - 2/s]

Y(s) = 2/[s(s + 2)2] + e-s{1/(s + 2)3 - 2/[s(s + 2)2]}

= 2/[s(s + 2)2] + e-s/(s + 2)3 - 2e-s/[s(s + 2)2]

For inverse Laplace transform of 2/[s(s + 2)2], it is u(t)/2 - e-2t/2 - te-2t by (a)

Hence then inv. Lap. transform of 2e-s/[s(s + 2)2] is u(t - 1)/2 - e-2(t-1)/2 - te-2(t-1).

Inv. Lap. transform of 1/(s + 2)3 is t2e-2t/2, therefore inv. Lap. transform of e-s/(s + 2)3 is (t - 1)2e-2(t - 1)/2

Hence:

y(t) = u(t)/2 - e-2t/2 - te-2t + u(t - 1)/2 - e-2(t-1)/2 - te-2(t-1) + (t - 1)2e-2(t - 1)/2

= [u(t) + u(t - 1)]/2 - [e-2t + e-2(t-1)](t + 1/2) + (t - 1)2e-2(t - 1)/2

2011-06-27 23:33:42 補充：
u(t)