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It's been a while since I've worked these types of problems, but I've been working out the following problem for a while now and can't figure out a few details:

enter image description here

And here is the solution:
enter image description here

Most of the solution makes sense, but I just have a few things I'm not able to clarify, namely:

  1. How were they able to determine the h(t) is a low pass filter? And which fourier coefficients will be passed through? Seems like n=2=-2 should be a valid coefficients too

  2. How did they get h(t) into the frequency domain?

*Update:
For the aspect of getting h(t) into the frequency domain, I've attempted the following but it doesn't line up with the what's given in the solution:
enter image description here

Any help is appreciated

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How were they able to determine the h(t) is a low pass filter?

Specifically, they noted that the time-domain response of the filter is of the general form \$\frac{\sin(x)}{x}\$, with some scaling factors for both time and amplitude.

"Everyone knows" (but you may have forgotten) that the Fourier transform of a sinc function is a single rectangular pulse centered at the origin, which means that the frequency-domain response of the filter is an "ideal" or "brick-wall" low-pass filter. Such a filter passes all frequencies below its cutoff frequency, but none at all above that frequency.

They then proceed to identify exactly what its cutoff frequency is.

Note that this filter is "non-causal" — its time-domain response extends infinitely far into the past, so it isn't physically realizable.

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  • \$\begingroup\$ Hi Dave Tweed, thank you for the additional info. I've dug out my old Signals&Systems book and things are started to come back to me. However, In attempting to get h(t) into the frequency domain, I've arrived at a solution that differs from the correct solution. I've added an update to my question and was hoping you could provide some further insight. Thanks again \$\endgroup\$ – InterestingGuy Sep 2 '19 at 19:00
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I'll answer your questions in reverse order, because the answers will make more sense that way.

How did they get h(t) into the frequency domain?

By looking the function up in a table of Fourier transforms, most likely. The same way you'd normally do a Fourier transform:
enter image description here

How were they able to determine the h(t) is a low pass filter?

By noticing that there exists some frequency \$\omega_x\$ for which \$H(\omega) = 0\$ for all \$\omega>\omega_x\$.

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  • \$\begingroup\$ Hi The Photon, Thank you for the additional insight. In following some Fourier transforms as you suggested, I've arrived at a solution that doesn't match the given solution. I've added an update to my problem to clarify. Would you have any guess as to what might be wrong with my attempt? Thanks again \$\endgroup\$ – InterestingGuy Sep 2 '19 at 19:05
  • \$\begingroup\$ @InterestingGuy, there are different definitions of the Fourier transform that differ by a factor of sqrt(2 pi). You might just be using a different one from the other source. \$\endgroup\$ – The Photon Sep 2 '19 at 19:21
  • \$\begingroup\$ Thank you very much for your help. Just one last clarification on your second answer, in my case, where H(n*fundamental_frequency) is not equal to 0 tells me which Fourier coefficients (Cn) make it through the filter, correct? \$\endgroup\$ – InterestingGuy Sep 2 '19 at 21:27

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