Given the spectrum of a signal x(t) , What is the minimum sample rate that would allow for x[n] to be recoverable?

So the specturm of the signal x(t) looks like the following , x axis is frequency:

Here I got this questions:

What is the minimum sample rate that would allow for x[n] to be recoverable?

Recoverable here means the shape of the spectrum is maintained but its placement on the

x-asix will vary, i.e. the spectrum will be centered on 0. It might be helpful to draw both

0 → 2π and 2π → 4π to answer the question.

My thought is that f_nyquist > f_max = 1250, so fs = 2*f_nyquist = 2500

However, when I draw specturm when fs = 2000 and fs=1000, it seems that the shape of original specturm is maintained:

So I wonder what is the minimum sample rate that would allow for x[n] to be recoverable? Is it because the shape of the two rectangle is the same and are symmetric so we can have lower minimum sample rate?

Answer 1

[Walter Roberson]

This is a square wave. The fourier transform of a square wave consists of an infinite series of odd harmonics; https://dsp.stackexchange.com/questions/34844/why-fourier-series-and-transform-of-a-square-wave-are-different

Therefore the minimum bandwidth to reproduce the square wave is infinite.

Answer 2

[Paul]

Hi Zhen,

Your question isn't really about Matlab. You might get more traction on a signal processing forum, like dsp.stackexchange.com

But we can use Matlab to help draw the diagrams.

First, define the Fourier transform of x(t)

X = @(f) double(abs(f) <= 1250 & abs(f) >= 750);

To draw the spectrum of the sampled signal x[n] = x(n*T) we need to add to X(f) copies of X(f) shifted at integer multiples of Fs.

Let Fs = 1500, for example.

Fs = 1600;

The we can plot a portion of the spectrum (using only the first four copies) of the sampled signal as

f = -5000:5000;
plot(f,X(f) + X(f-Fs) + X(f+Fs) + X(f-2*Fs) + X(f+2*Fs)),grid
ylim('padded')

Maybe you can try a similar approach to compare to your hand sketches.