0
$\begingroup$

So, I have to find $H\{ x(t)\})$ (which is an LTI system), where $$x(t) = \sum_{k=0}^{\infty} a_ke^{ \ jw_kt}$$ and where the impulse response of the system is given by: $$h(t) = \frac{\delta(t+\tau)-\delta(t)}{\tau}$$

Here's is my attempt. I know that $$H\{x(t)\} = \int_{-\infty}^{+\infty}x(t-u)h(u)\ du \\ =\frac{1}{\tau}\left(\int_{-\infty}^{+\infty}x(t-u)\delta(u+\tau)\ du\ \ - \ \ \int_{-\infty}^{+\infty}x(t-u)\delta(u)\ du\right)$$ but I don't know where to go from there. If you could help, that would be really nice, thank you.

$\endgroup$
4
  • $\begingroup$ Try to calculate it in the spectral domain. $\endgroup$
    – Max
    Mar 23, 2018 at 7:02
  • $\begingroup$ I don't know what that is. We haven't seen this in class yet $\endgroup$
    – Skyris
    Mar 23, 2018 at 8:06
  • $\begingroup$ Recall that $\delta(s)$ is nonzero only at $s=0$. So the first part of $h(t)$ will sift out $x(-\tau)$. The second part of $h(t)$ is only relevant if $\tau = 0$, but then you have an indeterminate form as the denominator will also be 0. $\endgroup$
    – Andy Walls
    Mar 23, 2018 at 10:32
  • $\begingroup$ I managed to do it on my own finally (have a look at my solution). $\endgroup$
    – Skyris
    Mar 23, 2018 at 10:39

1 Answer 1

2
$\begingroup$

Okay... I managed to do it on my own. Here's the solution if you're interested.

We know that $$H\{x(t)\} = \sum_{k=0}^{\infty}a_k \ H\{e^{jw_kt}\}$$ since $H$ is a linear system.

We also know that $H\{e^{jw_kt}\} = F(w_k) \ e^{jw_kt}$, where $F(w_k)$ is the frequency response of the system.

By definition, we have that $$F(w_k) = \int_{-\infty}^{+\infty}h(t) \ e^{-jw_kt} \ dt \\ = \frac{1}{\tau} \left(\int_{-\infty}^{+\infty}\delta(t+\tau)e^{-jw_kt} dt \ - \ \int_{-\infty}^{+\infty}\delta(t) e^{-jw_kt} dt\right)\\ =\frac{1}{\tau} \left( \int_{-\infty}^{+\infty} \delta(u) e^{\ jw_k(\tau-u)} \ du \ - \ \int_{-\infty}^{+\infty} \delta(t)e^{\ jw_k(0-t)}\ dt \right) \\ = \frac{e^{ \ jw_k\tau} - e^0}{\tau}$$

Therefore we can simply write $$H\{x(t)\} = \sum_{k=0}^{\infty} \frac{e^{\ jw_k\tau}-1}{\tau} \ a_ke^{\ jw_kt}$$

$\endgroup$
3
  • $\begingroup$ You replaced $\delta(\tau)$ with $\delta(t)$, which is quite different. Was your original problem statement in error? $\endgroup$
    – Andy Walls
    Mar 23, 2018 at 12:32
  • $\begingroup$ I noticed the same switch from t to tau. Please edit your question to clarify $\endgroup$
    – user28715
    Mar 23, 2018 at 21:45
  • $\begingroup$ Yes, I made a mistake when copying the statement. Thank you. I am going to change this. $\endgroup$
    – Skyris
    Mar 24, 2018 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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