Question

The sampling frequency of a signal is Fs = 5000 samples per second. Find its Nyquist interval.​

Answer 1

Answer:

Answer: a

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 = 12fm=12sec.

Explanation:

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