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I have wandered into the realm of signal processing and am currently focusing on learning more about filtering signals. I am currently trying to understand how to think about and solve a practice problem from my theory book. Given the signal

original signal

which one of the following signals corresponds to the high-pass version of the above signal?

alternatives

I am not sure at all how to reason about this. If the signals were plotted in the frequency domain, the high-pass version of the signal would be obvious to identify by looking at it's spectra. We would see heavily attenuated lower frequencies while the higher ones remain as before. In this case the signals are plotted against the sample index which I am not used to looking at and as such I don't know how to go about this problem. How can I build intuition for this?

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  • $\begingroup$ Which one looks to you like it contains lower frequencies? Which one looks like it contains higher frequencies? $\endgroup$
    – Jdip
    Commented Sep 29, 2022 at 18:09
  • $\begingroup$ @Jdip I do not know how to determine that based on the above plots. $\endgroup$
    – KSI
    Commented Sep 29, 2022 at 18:26
  • $\begingroup$ Sample index is the same as time (measured in different units) $\endgroup$
    – user20574
    Commented Sep 30, 2022 at 17:17

2 Answers 2

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  • Low-frequency signals are smooth. Low-pass filters tend to average a signal over time.
  • High-frequency content is jagged.
  • A high-pass filter blocks DC -- meaning that a filter that has been high-pass filtered has a zero average in the long term.

So eyeball the signals, and pick out the one whose jagged edges are intact, but that looks like it's varying around zero.

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    $\begingroup$ "A high-pass filter blocks DC" Interesting, how does this work? $\endgroup$
    – KSI
    Commented Sep 29, 2022 at 19:01
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    $\begingroup$ In the frequency domain, DC is the 0 frequency. High-passing removes it. $\endgroup$
    – Jdip
    Commented Sep 29, 2022 at 19:10
  • $\begingroup$ @Jdip That makes a lot of sense, thank you! I feel comfortable with reasoning about this problem now. $\endgroup$
    – KSI
    Commented Sep 29, 2022 at 19:17
  • $\begingroup$ The fourier transform (any flavor) with $f = 0$ is just the average of the signal. So if the zero-frequency response is zero, then the signal average is zero. $\endgroup$
    – TimWescott
    Commented Sep 29, 2022 at 20:56
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It is quite difficult to perform detailed frequency analysis by eye inspection. However, you still can infer some behavior by looking at time series.

  1. Hint 1: what are the signals of lowest frequency content (in the time domain)?
  2. Hint 2: what kind of smoothness/sharpness in shape would you expect from low/high frequency signals?

At the intersection, you may guess which one is high-pass filtered.

I do suspect that the exercice has been done the other way around. Take a low-frequency trend (like B) with non-zero mean. Add a high-pass signal with zero mean (like C), and get A.

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  • $\begingroup$ Regarding hint 1: Hmm, not sure I understand the question. Regarding hint 2: Perhaps I would expect a "sharper shape" as the high-pass filter is emphasizing the finer details while a low-pass filter contributes to "smooth" a shape. If this is correct then for sure alternative B is not correct. $\endgroup$
    – KSI
    Commented Sep 29, 2022 at 18:25
  • $\begingroup$ That's correct! B looks "smoother", so you can infer it contains the lower frequencies. $\endgroup$
    – Jdip
    Commented Sep 29, 2022 at 18:34
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    $\begingroup$ @Jdip Yeah I just saw it, the answer is now obviously D. I did not know that high-pass filters block DC offsets. $\endgroup$
    – KSI
    Commented Sep 29, 2022 at 19:16
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    $\begingroup$ I’d say it’s C… $\endgroup$
    – Jdip
    Commented Sep 29, 2022 at 20:09
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    $\begingroup$ @KSI, It looks like the scaling of D is identical to the original signal, meaning D was not filtered at all. C has values similar to if you took the original and subtracted out the low pass filter (i.e. B), which is what a high pass filter in some ways. $\endgroup$
    – Ash
    Commented Sep 29, 2022 at 21:15

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