# Thread: Convolution: sin(t) * u(t)

1. ## Convolution: sin(t) * u(t)

A computer gives the answer as $\displaystyle -\cos t$, but by hand, I seem to be getting a non-converging integral. What am I doing wrong?

$\displaystyle \sin t * u(t) = \int_{-\infty}^\infty \sin \tau \cdot u(t - \tau) d\tau$

Since $\displaystyle t > \tau$ for result to be non-zero, we have:

$\displaystyle = \int_{-\infty}^t \sin \tau d\tau$

$\displaystyle = (- \cos \tau)|_{-\infty}^t$

Which does not converge.

2. ## Re: Convolution: sin(t) * u(t)

Originally Posted by VinceW
A computer gives the answer as $\displaystyle -\cos t$, but by hand, I seem to be getting a non-converging integral. What am I doing wrong?

$\displaystyle \sin t * u(t) = \int_{-\infty}^\infty \sin \tau \cdot u(t - \tau) d\tau$

Since $\displaystyle t > \tau$ for result to be non-zero, we have:

$\displaystyle = \int_{-\infty}^t \sin \tau d\tau$

$\displaystyle = (- \cos \tau)|_{-\infty}^t$

Which does not converge.
You're meant to assume sin(t) = 0 for t < 0. In which case the answer is 1 - cos(t).

3. ## Re: Convolution: sin(t) * u(t)

Why? I don't see any validity behind that assumption at all.

4. ## Re: Convolution: sin(t) * u(t)

Originally Posted by VinceW
Why? I don't see any validity behind that assumption at all.
Well, I guess it will depend on whether t < 0 makes sense in the context from which the question has come, won't it.

5. ## Re: Convolution: sin(t) * u(t)

Originally Posted by mr fantastic
Well, I guess it will depend on whether t < 0 makes sense in the context from which the question has come, won't it.
You are not helping with this problem. This is a pure math problem so of course t < 0 makes sense.

I'm wondering whether that convolution genuinely doesn't exist for $\displaystyle \sin t * u(t)$ and a converging convolution only exists for $\displaystyle [(\sin t) \cdot u(t)] * u(t)$