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Recently I came across an statement which says, If signal m(t) has bandwidth B then pow(m(t),n) will have bandwith nB. To prove it mathematically I startted with,

Let n=2, y(t)= m(t)xm(t)

y(t)=Convolution(M(f),M(f)) :: where M(f) is the fourier transform of M(t)

After that the book states that y(t) has bandwidth 2B. Can someone point out to what I am missing here?

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    \$\begingroup\$ \$sin^2(\omega t)=\frac{1}{2}-\frac{sin(2\omega t)}{2}\$ \$\endgroup\$
    – Chu
    Commented Nov 8, 2015 at 13:50

1 Answer 1

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Raising m(t) to the power 2 is of course squaring the signal and if the signal consisted of two sine waves spaced at f1 and f2 , f1 would transform to 2f1 and f2 would transform to 2f2. So if f1 were 100 Hz and f2 were 1000 Hz your original bandwidth would be 900 Hz and your bandwidth (after squaring) would be: -

2000 Hz - 200 Hz = 1800 Hz i.e. your bandwidth has doubled after squaring.

There will be other artefacts as well - sum and difference frequencies will be present and these are well known in modulation schemes because plenty of modulation schemes involve multiplication. For instance amplitude modulation: -

enter image description here

Here the audio signal occupying the baseband is multiplied by a "carrier frequency" and the final signal transforms from the baseband to be twice the spectral width of the baseband centred about the frequency of the carrier - maybe this is what your book means when it says "that y(t) has bandwidth 2B"?

So, it works for squaring and if you looked at some trig identities for sines raised to other powers you would find that it works for those other powers.

Because this isn't math stack exchange I feel no compulsion to prove for n>2 but wiki to the rescue: -

enter image description here

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List item

As you can see, whatever the power raised to results in a sinewave frequency of that power times \$\theta\$ somewhere in the answer.

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  • \$\begingroup\$ I agree with your example, but if Y(t)=m(t) x Cos(2*Pift). This signal has bandwidth 2B where B is m(t) bandwidth. This one shouls have (B+f) not 2B. \$\endgroup\$
    – Virange
    Commented Nov 8, 2015 at 13:49
  • \$\begingroup\$ I think I've already added that to my answer. \$\endgroup\$
    – Andy aka
    Commented Nov 8, 2015 at 13:50

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