Finding the Convolution of Two Functions Using the Laplace Transform

[redundant6939]

Homework Statement

x(t) = cos(3πt)
h(t) = e^-2tu(t)

Find y(t) = x(t) * h(t)(ie convolution)

Homework Equations

Y(s) = X(s)H(s) and then take inverse laplace tranform of Y(s)

The Attempt at a Solution

L(x(t)) = [πδ(ω - 3π) + πδ(ω + 3π)]
L(h(t)) = \frac{1}{s+2}

Laplace Transform inverse :
y(t) = \frac{1}{2} \int \frac{e^{st}}{s+2} * δ(w-3π) + \frac{1}{2} \int \frac{e^{st}}{s+2} * δ(w+3π)dt = \frac{1}{2} \frac {e^{3sπ} + e^{-3sπ}}{s+2}

Im not sure about direc delta integration..

 

[vela]

The Laplace transform of $\cos 2\pi t$ isn't what you said it is. (It should be a function of $s$, not $\omega$.) You're thinking of the Fourier transform.

 

[redundant6939]

The Laplace transform of $\cos 2\pi t$ isn't what you said it is. (It should be a function of $s$, not $\omega$.) You're thinking of the Fourier transform.

Is it \frac {s}{s^2 + (3π)^2} ?

 

[vela]

Yes.

 

[redundant6939]

But isn't that the LT of cos(3pi*t)u(t)?

 

[vela]

Yes. Usually, it's assumed you're working with the unilateral Laplace transform, and you essentially ignore what's going on before t=0. Are you supposed to be using the bilateral Laplace transform?

 

[Ray Vickson]

But isn't that the LT of cos(3pi*t)u(t)?

The "bilateral" transform of $\cos(3 \pi t)$ does not exist, because the integral
\int_{-\infty}^{\infty} e^{-st} \cos(3 \pi t) \, dt
is not convergent. However, the "unilateral" LT of $u(t) \cos(3 \pi t)$ certainly does exist.