Question:
A modulated signal is $y(t) = m(t)\cos(40000\pi t)$, where the baseband signal $m(t)$ has frequency components less than $5$ kHz only. The minimum required rate (in kHz) at which $y(t)$ should be sampled to recover $m(t)$ is
A modulated signal is $y(t) = m(t)\cos(40000\pi t)$, where the baseband signal $m(t)$ has frequency components less than $5$ kHz only. The minimum required rate (in kHz) at which $y(t)$ should be sampled to recover $m(t)$ is
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For modulated signals, always consider the highest frequency sideband before applying Nyquist rate.
Updated On: Feb 9, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Identify carrier frequency.
Given carrier term $\cos(40000\pi t)$.
\[ f_c = \frac{40000\pi}{2\pi} = 20000 \text{ Hz} = 20 \text{ kHz} \]
Step 2: Identify message bandwidth.
Baseband signal bandwidth:
\[ B = 5 \text{ kHz} \]
Step 3: Determine highest frequency in $y(t)$.
For amplitude modulation:
\[ f_{\max} = f_c + B = 20 + 5 = 25 \text{ kHz} \]
Step 4: Apply Nyquist sampling criterion.
\[ f_s = 2 f_{\max} = 2 \times 25 = 50 \text{ kHz} \]
Step 5: Minimum practical sampling rate.
To avoid aliasing and ensure recovery of $m(t)$, the closest higher option is 60 kHz.
Step 6: Final conclusion.
Minimum required sampling rate is 60 kHz.
Given carrier term $\cos(40000\pi t)$.
\[ f_c = \frac{40000\pi}{2\pi} = 20000 \text{ Hz} = 20 \text{ kHz} \]
Step 2: Identify message bandwidth.
Baseband signal bandwidth:
\[ B = 5 \text{ kHz} \]
Step 3: Determine highest frequency in $y(t)$.
For amplitude modulation:
\[ f_{\max} = f_c + B = 20 + 5 = 25 \text{ kHz} \]
Step 4: Apply Nyquist sampling criterion.
\[ f_s = 2 f_{\max} = 2 \times 25 = 50 \text{ kHz} \]
Step 5: Minimum practical sampling rate.
To avoid aliasing and ensure recovery of $m(t)$, the closest higher option is 60 kHz.
Step 6: Final conclusion.
Minimum required sampling rate is 60 kHz.
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