Question

The Nyquist rate for the signal \(x\left( t \right) = \frac{1}{{2\pi }}\cos \left( {4000\;\pi t} \right)\cos \left( {1000\;\pi t} \right)\) will be
1. 5 kHz
2. 10 kHz
3. 15 kHz
4. 20 kHz

Answer 1

Correct Answer - Option 1 : 5 kHz

Concept:

Nyquist rate is the minimum rate at which a signal can be sampled without introducing errors, which is twice the highest frequency present in the signal.

Nyquist rate = 2 × highest frequency present in the signal

fs = 2fm

Calculation:

\(x\left( t \right) = \frac{1}{{2\pi }}\cos \left( {4000\;\pi t} \right)\cos \left( {1000\;\pi t} \right)\)

\( = \frac{1}{{4\pi }}\left[ {2\cos \left( {4000\;\pi t} \right)\cos \left( {1000\;\pi t} \right)} \right]\)

\( = \frac{1}{{4\pi }}\left[ {\cos \left( {3000\;\pi t} \right) + \cos \left( {5000\;\pi t} \right)} \right]\)

ω₁ = 3000 π ⇒ f₁ = 1500 Hz

ω₂ = 5000 π ⇒ f₂ = 2500 Hz

f_m = max (f₁, f₂) = 2500 Hz

Nyquist rate, f_s = 2f_m = 2 × 2500 = 5 kHz