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Convolution of a dirac delta function

by pka
Tags: convolution
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Mar19-08, 12:44 AM
P: 11 I've got a question about the convolution of a dirac delta function (or unit step). So, I know what my final answer is supposed to be but I cannot understand how to solve the last portion of it which involves the convolution of a dirac/unit step function. It looks like this:

10 * Inverse laplace of [ H(s) * e ^ (-5s) ]

where H(s) = (1/20) * (1 - e ^ -20t).

This is what I've done to lead me to the dirac/unit step. Btw, I'm calling it the dirac/unit step function because I get the dirac delta function in my answer whereas the answer has a unit step function. So, just for clarity's sake I will call the unit step function u(t - a) and the dirac delta function d(t - a).

Now, let's continue.

Saying L(s) = e ^ -5s. So that its inverse laplace, l(t) = d(t - 5).

Let's also say that M(s) = H(s) * L(s).

Convolution time!!! And I get m(t) = (1/20) * (1 - e ^ -20t) * Integral from 0 to t of d(tau - 5) d(tau).

I'm sorry about my notations, I don't know how to put in an integral sign or...any other fancies. =/

This is where my trouble starts. I thought the integral of a dirac delta function would be just h(t). But that's not right. In order for my answer to make any sense then the integral of d(tau - 5) should be just d(t - 5) to get a fairly simple answer of d(t - 5) * h(t).

Any help in this matter would be greatly appreciated. Links too! If I've posed a really simple question in too much writing then feel free to let me know or if I'm thinking about this way too hard then please...also let me know. But many thanks to any advice or help anyone can offer me.
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Mar19-08, 01:06 AM
P: 11
Actually...I think I've solved my problem!!!! In integral of the dirac delta should be just the unit step me what I need. And so...the convolution turns out to be m(t) = h(t) * u(t - a). So...then it's just 10 * m(t).

:D Can anyone tell me if my answer is correct in its thought and all that? :D

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