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This question already has an answer here:

If I have a message signal $m(t)$ and it has a bandwidth $B$. I know that the bandwidth of $m^N(t)$ is $NB$. But what is the bandwidth of $\frac{d m(t)}{dt}$? Thanks!

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marked as duplicate by lennon310, Stanley Pawlukiewicz, Peter K. Dec 4 '18 at 15:33

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    $\begingroup$ @MBaz: This is about the bandwidth, not about the magnitude. $\endgroup$ – Matt L. Nov 30 '18 at 19:42
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    $\begingroup$ @MattL. Thanks for pointing it out -- I misread the question. $\endgroup$ – MBaz Nov 30 '18 at 22:44
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Hint:

Checking a table of Fourier transform properties you'll find that the Fourier transform of the derivative is given by

$$\mathcal{F}\left\{\frac{dm(t)}{dt}\right\}=j\omega M(\omega)\tag{1}$$

where $M(\omega)$ is the Fourier transform of $m(t)$. Now if $M(\omega)$ is band-limited, what does this tell you about the Fourier transform of $dm(t)/dt$?

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  • $\begingroup$ Would it be the same? Since the $jw$ would only affect the amplitude of the signal? $\endgroup$ – Neilerino Nov 30 '18 at 19:55
  • $\begingroup$ @Neilerino: Well, if $M(\omega)=0$ for a certain frequency range, then $j\omega M(\omega)$ must be zero too, right? $\endgroup$ – Matt L. Nov 30 '18 at 20:02
  • $\begingroup$ Yes! Therefore the bandwidth would be the same as $m(t)$. $\endgroup$ – Neilerino Nov 30 '18 at 20:07

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