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. 10
• B. 50
• C. 35
• D. 60

Hint:

For modulated signals, always consider the highest frequency sideband before applying Nyquist rate.

Answer 1

The Correct Option is D

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.

Answer 2

This question asks for the sampling rate needed on the modulated signal \( y(t) = m(t)\cos(40000\pi t) \) so that the baseband message \( m(t) \), whose content is limited to below \( 5 \) kHz, can later be recovered. Instead of computing the highest frequency and doubling it directly, look at the full frequency occupancy of \( y(t) \) and test each option against it.

1. Option 10 kHz: The carrier is at \( f_c = \dfrac{40000\pi}{2\pi} = 20 \) kHz, so \( y(t) \) has energy sitting between \( 15 \) kHz and \( 25 \) kHz (and the mirror band below zero). A rate of \( 10 \) kHz is not even enough to represent a single tone near \( 20 \) kHz without severe aliasing, so this option leaves the signal completely unresolved.
2. Option 50 kHz: The topmost frequency present in \( y(t) \) is \( f_c + B = 20 + 5 = 25 \) kHz, and the classical Nyquist bound says \( f_s \geq 2f_{\max} \), which numerically gives exactly \( 50 \) kHz. This is the theoretical borderline value. Sampling exactly at this critical rate is a knife-edge case, since it depends on capturing samples at precisely the zero-crossing alignment of the highest-frequency component; any realizable system needs to clear this bound with room to spare, so \( 50 \) kHz cannot be treated as a safely working rate.
3. Option 35 kHz: This is less than twice \( 25 \) kHz, so components of \( y(t) \) near the upper edge of its band would fold back into the recovered spectrum and corrupt \( m(t) \) with aliased images. It fails the sampling requirement outright.
4. Option 60 kHz: This clears the critical \( 50 \) kHz bound with a comfortable margin, guaranteeing that even with a realizable (non-ideal, non-brick-wall) reconstruction filter, the replicated spectra after sampling stay separated and \( m(t) \) can be cleanly extracted.

Only the rate that safely exceeds the critical Nyquist value, rather than sitting exactly on it, is dependable in practice.

So the correct answer is 60 kHz.