Convolution

**4Math** · March 21st 2011, 03:28 AM

Hi,

I have the following problem:

$h(t)=(sin t+cost)\theta(t)$
$w(t)=\theta(t)-\theta(t-2\pi)$

How do I compute the following convolution::
$h(t)\ast w(t)$

I know that the definition for convolution is as follow:

(1)

But how do I apply (1) to the given problem?

Do I distribute the parentheses

$w(t)=\theta(t)-\theta(t-2\pi)=\theta(t)-\theta(t)-\theta2\pi$

Any help would be greatly appreciate it.
Thank you

**Opalg** · March 21st 2011, 05:29 AM

If you apply the definition of convolution then you get $\displaystyle(h*w)(t) = \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau.$ You cannot simplify that any further unless you know something about the nature of the function $\theta$ .

**4Math** · March 21st 2011, 07:31 AM

Originally Posted by Opalg

If you apply the definition of convolution then you get $\displaystyle(h*w)(t) = \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau.$ You cannot simplify that any further unless you know something about the nature of the function $\theta$ .

How do I integrate this integral?

Kind regards

**4Math** · March 21st 2011, 12:49 PM

I guess that integration is solved by partialintegration.

$\int udv=uv-\int vdu$

My question is this:

what is $u$ and $dv$ in this integral: $\displaystyle \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau.$

$u=...$
$dv=....$

**zzzoak** · March 21st 2011, 02:02 PM

May be $\theta(t)$ is the Heaviside step function ?

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