# Convolution of x(t) and x(-t)

Consider the signal $$x(t)=e^{-t}u(t)$$
where $$u(t)=\mathbb{1}(t\geq0)$$, i.e. the Heaviside function.

Find the signal $$y(t)=x(t)*x(-t)$$

My attempt:

$$y(t)=x(t)*x(-t)$$

$$=e^{-t}u(t)*e^{t}u(-t)$$

$$=e^{-t}u(t)*e^{t}u(-t)$$

$$= \int_{-\infty}^{\infty}[e^{-\tau}u(\tau)][e^{t-\tau}u(-t+\tau)]d\tau$$

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

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

$$= e^{t}\int_{0}^{t}e^{-2\tau}d\tau$$ (I think)

$$=e^{t}[(-\frac{1}{2}e^{-2\tau}]_0^t$$

$$=-\frac{1}{2}e^{t}[e^{-2t}-1]$$

$$=\frac{1}{2}[e^{t}-e^{-t}]$$

The solution and Wolfram Alpha give $$y(t)=\frac{1}{2}e^{-|t|}$$. Did I evaulate the step functions wrong? Guidance is appreciated.

• what if $t$ is a negative value? For that case, the integration region wouldn't be like from $0$ to $t$ May 15, 2020 at 18:42
• I don't think there is a close for expression unless $u$ has some special form. May 15, 2020 at 18:43
• @Oliver Diaz In this context $u$ means the unit step function, also known as the Heaviside function. May 15, 2020 at 19:06
• Do you mean $u(t)=\mathbb{1}_{(0,\infty)}(t)$? if so, the problem is trivial. May 15, 2020 at 19:19
• Yes, that's right. May 15, 2020 at 19:55

You need $$u(\tau)$$ and $$u(-t+\tau)$$ to both be $$1$$, so $$\tau > 0$$ and $$\tau > t$$. So, if $$t > 0$$, then this becomes $$\tau > t$$, and your integral (in the line where you say "I think") should actually have limits $$t$$ to $$\infty$$. This gives $$y(t) = \frac{1}{2}e^{-t}$$ for $$t > 0$$.
On the other hand, if $$t < 0$$, this becomes $$\tau > 0$$, so your integral should have limits $$0$$ to $$\infty$$. This gives $$y(t) = \frac{1}{2}e^{t}$$ for $$t < 0$$. Putting this together with the solution in the $$t > 0$$ case, you can see that $$y(t) = \frac{1}{2}e^{-\left\lvert t\right\rvert}$$ covers both cases in a single expression.
If $$u(t)=\mathbb{1}_{(0,\infty)}(t)$$, then $$u(\tau) u(-t+\tau)=\mathbb{1}_{(t,\infty)}(\tau), \qquad t>0$$ $$u(\tau)u(-t+\tau)=\mathbb{1}_{(0,\infty)}(\tau), \qquad t\leq0$$ Thus \begin{aligned} e^t\int_{\mathbb{R}}e^{-2\tau}u(\tau)u(-t+\tau)\,d\tau&= e^t\int^\infty_{t}e^{-2\tau}\,d\tau = \frac{1}{2}e^{-t} \end{aligned} for $$t>0$$, and \begin{aligned} e^t\int_{\mathbb{R}}e^{-2\tau}u(\tau)u(-t+\tau)\,d\tau&= e^t\int^\infty_0e^{-2\tau}\,d\tau = \frac{1}{2}e^t \end{aligned} for $$t\leq0$$. In other words, the solution is $$G(t):=\frac{1}{2}e^{-|t|}$$.