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Bandwidth in frequency spectrum

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Member level 4
Dec 1, 2009
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Why in general, QAM uses twice the bandwidth as PAM?? is it because of its dimension happens to be twice as PAM?

Also, does cos(2*pi*f*t) look the same as sin(2*pi*f*t) in the frequency spectrum? assuming f is the same in both. is f the bandwidth?

A modulated signal will have a bandwidth approximately equal to the bandwidth of the baseband signal being carried. Some modulation formats will be wider because they are designed for spectral spreading.

By inspection, your examples of cos(2*pi*f*t) and sin(2*pi*f*t) will simply create a pure sinusoid at frequency f, if you sweep the time variable, t. If you took an FFT of cos(...) and sin(...), you'd get a single tone at frequency = f Hz.

Thanks for the reply.

I asked the second question is because, I think the reason of QAM uses twice the bandwidth as PAM is due to the fact that is has two time waveforms, and one is 90 degree shift of the other, so cos and sin. Therefore I am trying to find a way to visualize this fact on the frequency spectrum and hence explain why it uses twice the bandwidth

It's more complex than that. Sine and cosine are 90 degrees out of phase with one another, so they are not (fully) acting at the same time. If you sweep them together in the time domain, you simply draw a circle of constant radius (unless you are modulating either component). In effect, you can put two baseband signals on one carrier, which is the concept of baseband I & Q signals (Inphase and Quadrature, i.e. 90 degrees out of phase with the "in-phase" signal). You can put two completely independent signals onto one carrier, and easily demodulate them at the receiver.

I'm not sure how to put the bandwidth concept for QAM vs PAM into words... perhaps a signals guru can jump in and give some more insight?

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