Question

Find the Nyquist rate and Nyquist interval for the signal f(t) = \(\Big[\frac{sin⁡500πt}{πt}\Big]^2\).

(a) 1000 Hz, 1 msec

(b) 1 Hz, 1000 sec

(c) 1000 Hz, 1 sec

(d) 1000 Hz, 1000 sec

This question was addressed to me by my school teacher while I was bunking the class.

The above asked question is from Sampling in section Sampling Theorem of Signals and Systems

Answer 1

Right option is (a) 1000 Hz, 1 msec

Best explanation: Given f(t) = \(\Big[\frac{sin⁡500πt}{πt}\Big]^2 = \frac{1-cos⁡1000πt}{(πt)^2}\)

Frequency, ωm = 1000π

2πfm = 1000π

2fm = 1000 Hz

Nyquist rate, Fs = 2fm = 1000 Hz

Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{1000}\) = 1 msec.