Convolution of step function with exponentials

Question

I have a problem in signals and systems to solve which is basically math

So suppose we have two functions x(t) = H(t) which is the Heaviside unit step function as input and an impulse response

$$h(t)=e^{-\left | t \right |} + e^{-\left | 2t \right |}$$

which is $$h(t)=(e^t+e^{2t})H(-t)+(e^{-t}+e^{-2t})H()t)$$

We want to calculate the convolution between them to find the step response.

At a point i have reached this integral

$$\int_{-\infty }^{\infty }(e^\tau + e^{2\tau })H(-\tau )H(t-\tau )d\tau $$

I think it is zero but I cannot explain why this is true

Any help is welcome

Answer 1

You got yourself into trouble by making things more complicated than they need be. First of all, note that

$$(h\ast u)(t)=[(h_1+h_2)\ast u](t)=(h_1\ast u)(t)+(h_2\ast u)(t)$$

where $u(t)$ denotes the step function, and $h_1(t)$ and $h_2(t)$ are the two (two-sided) exponentials. So you can compute the two convolutions separately and add them afterwards.

I'll show you how I would compute the first of these convolutions (and the other one is completely analogous):

$$(h_1\ast u)(t)=\int_{-\infty}^{\infty}e^{-|\tau|}u(t-\tau)d\tau=\int_{-\infty}^{t}e^{-|\tau|}d\tau\tag{1}$$

Now you need to distinguish the two cases $t<0$ and $t>0$. For $t<0$, the integral in $(1)$ becomes

$$\int_{-\infty}^{t}e^{-|\tau|}d\tau=\int_{-\infty}^{t}e^{\tau}d\tau,\qquad t<0$$

For $t>0$ we have

$$\int_{-\infty}^{t}e^{-|\tau|}d\tau=\int_{-\infty}^{0}e^{\tau}d\tau+\int_{0}^{t}e^{-\tau}d\tau,\qquad t>0$$

I trust you can take it from here.