Finding the Convolution of Two Functions Using the Laplace Transform

[redundant6939]

Homework Statement

[itex]x(t) = cos(3πt)[/itex]
[itex]h(t) = e^-2tu(t)[/itex]

Find [itex]y(t) = x(t) * h(t) [/itex](ie convolution)

Homework Equations

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

The Attempt at a Solution

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

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

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 [itex] \frac {s}{s^2 + (3π)^2}[/itex] ?

 

[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
[tex] \int_{-\infty}^{\infty} e^{-st} \cos(3 \pi t) \, dt [/tex]
is not convergent. However, the "unilateral" LT of ##u(t) \cos(3 \pi t)## certainly does exist.