1
$\begingroup$

Question

The impulse response of a continuous time system is $h(t) = e^{-t}u(t)$. When $x(t) = u(t - 1)$ is input to this system what is the value of the output signal $y(t)$ at $t = 2.34$.

My Attempt

For convolution $y(t) = h(t) \star x(t)$ \begin{align} x(2.27) &= u(2.34-1) = 1\\ h(2.27) &= e^{-2.34}\cdot u(2.34) = 0.09632763823\\ \Longrightarrow y(t) &= 1 \cdot 0.09632 = 0.096 \end{align}

But the answer is wrong. Not too sure where my mistake is. Could someone correct my error?

Second Attempt \begin{align} y(t) &= ∫x(\tau)h(t-\tau)d\tau \quad\text{with limits negative and positive infinity}\\ &= ∫u(\tau-1)e^{-\tau}u(t-\tau)d\tau \\ &= ∫e^{-\tau}u(\tau-1)u(t-\tau)d\tau \\ &= ∫e^{-\tau}d\tau \quad \text{with limits 1 to 2.34}\\ &= 0.2715518029 \end{align}

Though the answer still appears to be wrong.

$\endgroup$
  • 1
    $\begingroup$ You need to do the convolution integral and then substitute $t=2.34$ into it, not before. $\endgroup$ – Peter K. May 8 '16 at 16:27
  • $\begingroup$ @PeterK. That's the "hard way" of doing it and likely to end in disaster. All the OP needs to do is calculate $$y(2.34) = \int_{-\infty}^\infty h(t)x(2.34-t)\,\mathrm dt = \int_0^\infty e^{-t}x(2.34-t)\,\mathrm dt = \int_0^? e^{-t}\,\mathrm dt,$$ which is a lot easier if the OP can figure out why the lower limit got changed to $0$ and what that $?$ in the upper limit should be in the integral on the right. $\endgroup$ – Dilip Sarwate May 8 '16 at 17:58
  • $\begingroup$ @Dilip Sarwate I have edited my post after attempting the method you described but not to clear as to why your using 0 over 1. Also if the ? is supposed to be 2.34 i have calculated using both 0 and 1 and have gotten the wrong answer $\endgroup$ – sky knight May 8 '16 at 18:09
  • $\begingroup$ Sorry, I had typos in what I wrote. Please replace $u(2.34-t)$ by $x(2.34-t)$ in two places in the displayed equation. $\endgroup$ – Dilip Sarwate May 8 '16 at 18:17
  • $\begingroup$ @DilipSarwate I've edited your previous comment with that correction. $\endgroup$ – Peter K. May 8 '16 at 18:40
1
$\begingroup$

Given a continuous time LTI system with impulse response $h(t) = e^{-t}\text{u}(t)$ the output $y(t)$ for any input $x(t)$ is found via the convolution-integral: $$y(t) = x(t) \star h(t) = \int_{-\infty}^{\infty} {x(\tau)h(t-\tau)} d\tau$$

Specifically when $x(t) = \text{u}(t-1)$, we can proceed with the most general approach by first taking the integral and then inserting t=2.34, or else as @DilipSarwate suggests, you can first substitude t=2.34 into the integrand and then take the integral. I prefer the first here: \begin{align} y(t) &= \int_{-\infty}^{\infty} {\text{u}(\tau-1)e^{-(t-\tau)} \text{u}(t-\tau)d\tau}\\ y(t) &= \int_{\tau=1}^{\tau=t} {e^{-(t-\tau)}d\tau}\\ y(t) &= 0 &\scriptstyle{\text{for t less than 1}}\\ y(t) &= e^{-(t-\tau)} \big\vert_{\tau=1}^{\tau=t} = 1 - e^{-(t-1)} &\scriptstyle{\text{ for t greater than 1, hence:}}\\ y(t) &= [1 - e^{-(t-1)}] \text{u}(t-1) &\scriptstyle{\text{, for all t}}\\ \end{align}

I hope it's clear how the integral limits are modified by the arguments of the $\text{u}(t)$ step function.

And inserting $t=2.34$ yields:

$$y(2.34) = [1 - e^{-(2.34-1)}] \text{u}(2.34-1) = [1 -e^{-1.34}]\text{u}(1.34) = 0.7382 $$

| improve this answer | |
$\endgroup$
2
$\begingroup$

\begin{align} y(2.34) &= h \star x\,\big\vert_{t=2.34}\\ &= \int_{-\infty}^\infty h(\tau)x(2.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{from the definition of the convolution integral}} \\ &= \int_0^\infty e^{-\tau}x(2.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{because } h(\tau)=0\text{ for }\tau < 0}\\ &= \int_0^\infty e^{-\tau}u(1.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{because }x(2.34-\tau) = u((2.34-\tau)-1)}\\ &= \int_0^{1.34} e^{-\tau}\,\mathrm d\tau &\scriptstyle{\text{because }u(1.34-\tau) = 0 \text{ for }\tau < 1.34}\\ &= 1-e^{-1.34} \end{align}

| improve this answer | |
$\endgroup$

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.