# Thread: Laplace Transform - Convolution Theorem

1. ## Laplace Transform - Convolution Theorem

Q)The output y(t) of a linear system, with impulse response of h(t), subjected to an input function x(t) is given by y(t) = h(t) * x(t)

If x(t) = u(t) and H(s) = 4/s^2 +2s + 5 ,

Solve for y(t) by using the Convolution Theorem.

My approach:
I've tried to apply the partial fraction to find the coefficient of
(4/s^2 + 2s +5) X1/s = (As + B/s^2 +2s + 5) + (C/s)

A=-4/5; B=-8/5; C=4/5

Then apply,
inverse L{Y(s)} = y(t)
= inverse L{[(-4/5)s + (-8/5)/s^2 + 2s +5]+ 4/5s}
.....

but can't get the final answer (4/5e^-t cos2t - 2/5e^-t sin2t + 4/5) u(t)

2. Originally Posted by Octopus
Q)The output y(t) of a linear system, with impulse response of h(t), subjected to an input function x(t) is given by y(t) = h(t) * x(t)

If x(t) = u(t) and H(s) = 4/s^2 +2s + 5 ,

Solve for y(t) by using the Convolution Theorem.

My approach:
I've tried to apply the partial fraction to find the coefficient of
(4/s^2 + 2s +5) X1/s = (As + B/s^2 +2s + 5) + (C/s)

A=-4/5; B=-8/5; C=4/5

Then apply,
inverse L{Y(s)} = y(t)
= inverse L{[(-4/5)s + (-8/5)/s^2 + 2s +5]+ 4/5s}
.....

but can't get the final answer (4/5e^-t cos2t - 2/5e^-t sin2t + 4/5) u(t)

$(\mathcal{L}u)(s)=1/s$

So:

$
\mathcal{L}y=\mathcal{L}(h*u)=H(s)(\mathcal{L}u)(s )=\frac{4}{s^2 + 2s +5} \times \frac{1}{s}=\frac{4}{s^3 + 2s^2 +5s}
$

Now use partial fractions ...

CB