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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.

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  • $\begingroup$ Try to calculate it in the spectral domain. $\endgroup$ – Max Mar 23 '18 at 7:02
  • $\begingroup$ I don't know what that is. We haven't seen this in class yet $\endgroup$ – Skyris Mar 23 '18 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 '18 at 10:32
  • $\begingroup$ I managed to do it on my own finally (have a look at my solution). $\endgroup$ – Skyris Mar 23 '18 at 10:39
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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 $$H\{x(t)\} = \sum_{k=0}^{\infty} \frac{e^{\ jw_k\tau}-1}{\tau} \ a_ke^{\ jw_kt}$$

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  • $\begingroup$ You replaced $\delta(\tau)$ with $\delta(t)$, which is quite different. Was your original problem statement in error? $\endgroup$ – Andy Walls Mar 23 '18 at 12:32
  • $\begingroup$ I noticed the same switch from t to tau. Please edit your question to clarify $\endgroup$ – Stanley Pawlukiewicz Mar 23 '18 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 '18 at 16:59

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