- #1

Jncik

- 103

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## Homework Statement

Hello, I'm revising this summer for signals and systems and I came across this convolution

[tex]cos(t)*u(t)[/tex]

## Homework Equations

having two signals x(t) and h(t), where x(t) is the input signal and h(t) the impulse response

the output y(t) is given by [tex]y(t) = x(t)*h(t) = \int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau[/tex]

## The Attempt at a Solution

[tex] cos(t)*u(t) = \int_{-\infty}^{+\infty}cos(\tau)u(t-\tau)d\tau[/tex]

**(1)**

u(t-τ) is 1 for t-τ>0 => τ<t

hence

(1) is

[tex] \int_{-\infty}^{t}cos(\tau)d\tau = sin(\tau)|_{\tau->t, \tau->-\infty} = sin(t) - sin(-\infty) = sin(t) - undefined in [-1, 1]

[/tex]

now, I had 2 other integrals in the same exercise where the result was just sin(t), and I thought this was supposed to be sin(t) as well but I get this undefined number and I think it's wrong because I've never calculated an integral that gave me an undefined number, at least in signals and system course..

after searching in my book it had a similar integration

[tex]\int_{-\infty}^{+\infty}u_{0}(t)cos(t) dt [/tex]

[tex]u_{0}(t)[/tex] is just u(t)...

now again using the same method I would have

[tex]\int_{-\infty}^{+\infty}u_{0}(t)cos(t) dt

= \int_{0}^{+\infty}cos(t) dt = sin(t)|_{t->0,t->+\infty} = sin(0) - sin(+\infty) = -sin(+\infty)

[/tex]

which is undefined again

but he finds this result

[tex] \int_{-\infty}^{+\infty}u_{0}(t)cos(t) dt = \int_{-\infty}^{+\infty} \delta(t) dt = 1[/tex]

I'm wondering, what am I doing wrong and why does he find such result?

thanks in advance