Question

Let x(t) = 10sinc(50t) and y(t) = 2sinc²(20r). Find the Nyquist rate (minimum sampling rate) for each of the following signals.

a) x(t),
b) y(t),
c) x(t) * y(t),
d) x(t) ∙ y(t).

Answer 1

The Nyquist rate varies for different signals based on their highest frequency components; it's 50 samples per second for x(t), 20 samples per second for y(t), and will be higher for the convolution and product of x(t) and y(t) based on combined frequency contents.

option b is the correct

Step-by-step explanation:

The Nyquist rate is the minimum sampling rate required to avoid aliasing and is twice the highest frequency component in the signal.

For the signal x(t) = 10sinc(50t), the highest frequency component is 25 Hz, hence the Nyquist rate is 50 samples per second.

The signal y(t) = 2sinc²(20t) has its highest frequency component at 10 Hz, which gives a Nyquist rate of 20 samples per second. When we consider the convolution x(t) * y(t), the bandwidth will be the sum of the bandwidths of x(t) and y(t), requiring a Nyquist rate that considers the highest frequency components from both signals added together. For the product x(t) · y(t), the Nyquist rate will also need to consider the highest frequency, but taking into the account that frequency components will multiply, potentially leading to higher frequency contents that dictate a higher Nyquist rate.