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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2@MBaz: This is about the bandwidth, not about the magnitude. – Matt L. Nov 30 '18 at 19:42
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1@MattL. Thanks for pointing it out -- I misread the question. – 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 $dm(t)/dt$?
Matt L.
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Would it be the same? Since the $jw$ would only affect the amplitude of the signal? – Neilerino Nov 30 '18 at 19:55
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@Neilerino: Well, if $M(\omega)=0$ for a certain frequency range, then $j\omega M(\omega)$ must be zero too, right? – Matt L. Nov 30 '18 at 20:02
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