# Bandwidth of the Derivative of a Message [duplicate]

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!

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

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

$$\mathcal{F}\left\{\frac{dm(t)}{dt}\right\}=j\omega M(\omega)\tag{1}$$
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 $$dm(t)/dt$$?
• Would it be the same? Since the $jw$ would only affect the amplitude of the signal? – Neilerino Nov 30 '18 at 19:55
• @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
• Yes! Therefore the bandwidth would be the same as $m(t)$. – Neilerino Nov 30 '18 at 20:07