0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ what if $t$ is a negative value? For that case, the integration region wouldn't be like from $0$ to $t$ $\endgroup$
    – Cardinal
    May 15, 2020 at 18:42
  • $\begingroup$ I don't think there is a close for expression unless $u$ has some special form. $\endgroup$ May 15, 2020 at 18:43
  • $\begingroup$ @Oliver Diaz In this context $u$ means the unit step function, also known as the Heaviside function. $\endgroup$
    – Prasiortle
    May 15, 2020 at 19:06
  • $\begingroup$ Do you mean $u(t)=\mathbb{1}_{(0,\infty)}(t)$? if so, the problem is trivial. $\endgroup$ May 15, 2020 at 19:19
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Prasiortle
    May 15, 2020 at 19:55

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $G(t):=\frac{1}{2}e^{-|t|}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.