What is the minimum sampling rate for the following signal?

\(x(t)=Sinc^{2}(100 \pi t)\)

 

The fundamental frequency of the unsquared signal is 50 Hz. Squaring it makes the fundamental frequency 100 Hz. The sampling theorem requires the minimum frequency to be at least twice that or 200 Hz.

FINAL ANSWER : 200 Hz.

 

Thanks papabravo. I confirmed your answer by doing the Fourier Transform of x(t) and noticed it's bandwidth.

I have a similar question for you that I'm not sure about. I posted this as a new thread since it's might go un-noticed.

The signal x(t) = cos (14*pi*t) is sampled at a sampling interval of T = 0.1 seconds. Can we recover the signal from its samples (why or why not).

I attached my attempt at a solution.

What I did was this: I did a fourier transform on the signal, X(w), which resulted in two delta functions: one located at -14pi, the other at 14pi. Now, the fourier transform of the sampled signal, Xs(w) equals X(w) plus X(w) shifted by integer multiples of n*2*pi/Ts which equals 20*pi.

From my plot of Xs(w), there are delta functions interspersed to the left and right of the original X(w). I don't know if the original signal can be recovered with an anti-aliasing filter.

 

Sampling rate must be 14 Hz. See the details in the other thread.