It is one of Fourier transform properties Derivative


  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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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