Solve using convolution theorem
(choose e^(-2t) as f or g)
note: we cannot use partial fractions to get to solution (if part of the process)
(choose e^(-2t) as f or g)
note: we cannot use partial fractions to get to solution (if part of the process)
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Calling L{y(t)} by Y(s)
(s² + 2s + 5) Y(s) = 1/(s + 2) ==> Y(s) = 1/[(s + 2)(s² + 2s + 5)].
It is useful to note that s² + 2s + 5 = (s + 1)² + 4.
So you want to use the convolution theorem. The inverse transform of the two terms independently are
L^(-1){1/(s + 2)} = e^(-2t) and L^(-1){1/[(s + 1)² + 4]} = ½ e^(-t) sin(2t).
It really seems like partial fractions would be the easy thing to do, but okay. It doesn't really make any difference who gets called f and who gets called g. The convolution is
t
∫ e^(-2(t - τ)) ½ e^(-τ) sin(2τ) dτ =
0
t
∫ ½e^(-2t) e^(τ) sin(2τ) dτ = ½ e^(-2t) [2/5 - e^t/5 (2cos(2t) - sin(2t)) ==>
0
y(t) = e^(-2t)/5 - e^(-t)cos(2t)/5 + e^(-t)sin(2t)/10.
(s² + 2s + 5) Y(s) = 1/(s + 2) ==> Y(s) = 1/[(s + 2)(s² + 2s + 5)].
It is useful to note that s² + 2s + 5 = (s + 1)² + 4.
So you want to use the convolution theorem. The inverse transform of the two terms independently are
L^(-1){1/(s + 2)} = e^(-2t) and L^(-1){1/[(s + 1)² + 4]} = ½ e^(-t) sin(2t).
It really seems like partial fractions would be the easy thing to do, but okay. It doesn't really make any difference who gets called f and who gets called g. The convolution is
t
∫ e^(-2(t - τ)) ½ e^(-τ) sin(2τ) dτ =
0
t
∫ ½e^(-2t) e^(τ) sin(2τ) dτ = ½ e^(-2t) [2/5 - e^t/5 (2cos(2t) - sin(2t)) ==>
0
y(t) = e^(-2t)/5 - e^(-t)cos(2t)/5 + e^(-t)sin(2t)/10.