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Determine the Nyquist rate of the signal x(t) = 1 + cos⁡ 2000πt + sin⁡ 4000πt.

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in Signals and Systems by (1.2m points)
Determine the Nyquist rate of the signal x(t) = 1 + cos⁡ 2000πt + sin⁡ 4000πt.

(a) 2000 Hz

(b) 4000 Hz

(c) 1 Hz

(d) 6000 Hz

I got this question in class test.

The origin of the question is Sampling topic in chapter Sampling Theorem of Signals and Systems

1 Answer

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The correct answer is (b) 4000 Hz

The best explanation: Given x(t) = 1 + cos 2000πt + sin⁡ 4000πt

Highest frequency component in 1 is zero

Highest frequency component in cos⁡2000πt is ωm1 = 2000π

Highest frequency component in sin⁡4000πt is ωm2 = 4000π

So the maximum frequency component in x(t) is ωm = 4000π [highest of 0, 2000π, 4000π]

∴  2πfm = 4000π

2fm = 4000

Nyquist rate, Fs = 2fm = 4000 Hz.

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