Following figure is drawn as the block diagram of the digital modulator:

enter image description here

The output of the filter block (with the impulse response p(t)) is the convolution of the weighted impulse train with the impulse response of the filter, which in the time domain is equal to:

$\sum_{k} {a_k} p(t-kT)$

Considering that the convolution in the time domain is equivalent to the multiplication in the frequency domain, and since the Fourier transform of the impulse train is itself a impulse train, it appears that the output of the filter block in the frequency domain is equal to the product of the impulse train in the Fourier transform of p (t) which, since p(t) is a baseband signal (and assume it is bandlimited), so multiplying Fourier transform of impulse train by Fourier transform of p(t) results on a single impulse at the origin!!!! what's wrong in my understanding of what's going on in the frequency domain? can anyone give me a correct interpretation of the filter output in the frequency domain? thanks.


1 Answer 1


since the Fourier transform of the impulse train is itself a impulse train

Here is the rub. That's only the case if the impulse train is uniform, i.e. all impulses have the same amplitude.

We can interpret the process as "upsampling", i.e. you start with the original sequence $a_k$ and then insert $K-1$ zeros between samples. That leads to periodic repetition of the spectrum in the frequency domain.

If your bits are randomly distributed, then the spectrum of $a_k$ will be mostly white and the spectrum of the modulated impulse train will be a periodic repetition of that white spectrum, i.e. also be more or less white.

The idea of the low-pass filter is to remove the mirror spectra, so you are only left with a single base-band copy of it. This copy is then modulated with the carrier.


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