Question

Find the Nyquist rate and Nyquist interval of sin(2πt).

(a) 2 Hz, \(\frac{1}{2}\) sec

(b) \(\frac{1}{2}\) Hz, \(\frac{1}{2}\) sec

(c) \(\frac{1}{2}\) Hz, 2 sec

(d) 2 Hz, 2 sec

I have been asked this question by my college director while I was bunking the class.

Enquiry is from Sampling in chapter Sampling Theorem of Signals and Systems

Answer 1

Correct choice is (a) 2 Hz, \(\frac{1}{2}\) sec

For explanation: We know that sin⁡ ω0 t ↔ jπ[δ(ω+ω0) – δ(ω-ω0)]

sin⁡ 2πt ↔ jπ[δ(ω+2π)-δ(ω-2π)]

Here ωm = 2π

But ωm = 2πfm

∴   fm = 1 Hz

Nyquist rate, Fs = 2fm = 2 Hz

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