# Is there a process for deriving special cases of convolution?

For my signals and systems homework, I was met with the following problem:

Consider a CT system with impulse response given by

$$h(t) = e^{-|t|}$$

and a step input $$x(t) = u(t)$$. Using the deﬁnition of convolution, determine an expression for the output $$y(t) = x(t)∗h(t)$$.

I approached this problem using the standard method taught in my class. I set up the following integral:

$$\int_{-\infty}^{+\infty}h(\tau)x(t-\tau)d\tau$$

While evaluating the integral, I got stuck at the end as the integral approached infinity. I decided to rearrange $$h(t)$$ by expressing it in a different form. The form I then got it in was:

$$h(t) = e^{-t}u(t) + e^tu(-t)$$

Which then allowed me to match it to the following convolution on a table given with my homework.

From doing this problem it seems to me that there are special cases for convolutions. Ones that which the standard method of evaluation doesn't work.

Is there a process to derive these special cases? If so can an example be given? And why wouldn't the standard process work for this example?

• All convolutions can be solved in two ways: algebraically or graphically. If you get stuck in one of them, try the other :) – GKH Feb 8 '20 at 0:02

Well, indeed there are special cases for convolutions but yours is quite straightforward. You have to consider two cases: $$t<0$$ and $$t>0$$.