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It is one of Fourier transform properties Derivative

Integration

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closed as unclear what you're asking by Stanley Pawlukiewicz, lennon310, A_A, AlexTP, Dilip Sarwate Apr 14 '18 at 16:46

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  • $\begingroup$ try what? j.w. exp^(jwt) $\endgroup$ – Muhammad Aldakkak Mar 31 '18 at 15:14
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    $\begingroup$ I'm voting to close this question as off-topic because it appears to be a homework problem with no effort shown. $\endgroup$ – Stanley Pawlukiewicz Mar 31 '18 at 16:49
  • $\begingroup$ No it is not I was watching course about signal processing and the instructor says that derivative in time domain is highpass filter in frequency domain, but i didn't understand why so I ask here.. video in the link below, and the attached images is captured from it youtube.com/watch?v=A5Mo1_to7fk $\endgroup$ – Muhammad Aldakkak Mar 31 '18 at 19:22
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    $\begingroup$ Do you understand that multiplication in the frequency domain is convolution in the time domain? $\endgroup$ – Stanley Pawlukiewicz Mar 31 '18 at 19:32
  • $\begingroup$ yes, I do. you mean that jw will have large value so it will pass high frequencies? $\endgroup$ – Muhammad Aldakkak Mar 31 '18 at 19:50
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First, it's actually NOT a high or lowpass filter since it doesn't have a pass band and there isn't any "cut off" frequency.

It's basically just a constant slope and this can be interpreted as

  • differentiation amplifies high frequencies more than low frequencies
  • integration amplifies low frequencies more than high frequencies

That's pretty intuitive: high frequency means that the signal is changing quickly, hence the "rate of change" or "derivative" is big as well. The math around this is also simple and straight forward.

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Basically, you start from the signal spectrum $X(\omega)$. In a relative way (leaving details and theory aside), the spectrum changes if it is multiplied by a weighting function, here $j\omega$ or $\frac{1}{j\omega}$. The first one is proportional to $|\omega|$ (in magnitude spectrum), an increasing function, and so it tends to give more weight to high $|\omega|$'s, and zero weight to the low frequencies. Hence the notion of high-pass.

In a converse way, $\frac{1}{|\omega|}$ is decreasing, and put higher weight on low frequencies. Hence the low-pass notion.

One should be more careful with the theory, and the transfer to the discrete vision of signals, but I hope you will get the intuition.

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  • $\begingroup$ Good, I was pondering about putting some graphs. Are they still needed? $\endgroup$ – Laurent Duval Apr 2 '18 at 14:33

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