Question

find the Nyquist sampling frequency of a signal given by v(t) = 6*10^4sinc^3(400t)*10^4sinc^3(100t)

Answer 1

If you know one Fourier Transform pair and a few properties of Fourier Transform and Nyquist theorem, then you can easily answer this question. Lets come to all these three things one by one to find the answer.

Fourier Transform Pair: sinc(t) —→ rect(f), i.e. the Fourier Transform of sinc(t) is rect(f), Where sinc(t) = sin(pi t) / pi t by definition and rect(f) = 1 if -0.5<f<0.5, 0 otherwise.

Properties of FT: Let x(t) —→ X(f), x1(t) —-> X1(f) and x2(t) —-> X2(f)

Scaling: x(at) —-> X(f/a)/|a|

Convolution in TD: x1(t) * x2(t) —-> X1(f)X2(f)

Multiplication in TD: x1(t) x2(t) —-> X1(f) * X2(f)

Nyquist Theorem: For a real low-pass signal, sampling rate should be atleast twice the highest frequency present in the signal.

A function spreads on convolution with other function.

Knowing these three we can solve the above problem.

The given signal is the convolution of a sinc^2 and sinc^3 term, Lets go to each term individually.

First forget about the amplitude, as it does not contribute to the highest frequency,

First term is Sinc^2(400t) = sinc(400t) x sinc(400t), this multiplication in time domain will result in convolution in frequency domain. Lets find FT of sinc(400t).

Now we know sinc(t) —-> rect(f)

Hence, sinc(at) —-> rect(f/a) / |a|.

Hence sinc(400t) —-> (1 / 400) x rect(f / 400) ,

and sinc(100t) —-> (1/100) rect(f / 100).

one should understand that the rect(f/a) is 1 , -0.5a < f < 0.5a; 0 otherwise.

Hence rect(f / 400) = 1 if -200 < f < 200; 0 otherwise, hence highest frequency of sinc(400t) is 200 Hz.

and rect(f / 100) = 1 if -50 < f < 50, 0 otherwise, highest freq of sinc(100t) = 50 Hz.

Now, Highest Frequency of sinc^2(400t) = 200 + 200 = 400 Hz, addition as in frequency domain spectrum will spread.

Highest Frequency of sinc^3(100 t) = 50 + 50 + 50 = 150 Hz.

and highest frequency of sinc^2(400t) * sinc^3(100 t) = 150 Hz, because of the convolution in time domain multiplication will happen in frequency domain, one spectrum is zero beyond 150 Hz, multiplying by zero will result in zero. hence the convolved signal’s highest frequency = 150 Hz.

Hence Nyquist Rate = 150 * 2 = 300 Hz.

Note: Do not go for short tricks, short tricks are not helpful in every situation, you may get negative marks because of it. Fundamentals/ concepts works in every situation.

Thanks