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

Question

A continuous time periodic signal $ x(t) $ is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If $ T $ is the period of $ x(t) $, then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]

Answer 1

Step 1: The carrier frequency is $ f_c = \dfrac{40000\pi}{2\pi} = 20 $ kHz, and the message occupies frequencies up to $ B = 5 $ kHz.
Step 2: The highest frequency actually present in $ y(t) $ is $ f_c + B = 25 $ kHz, so the strict Nyquist requirement is $ f_s > 2 \times 25 = 50 $ kHz (a rate exactly equal to $ 50 $ kHz is only a marginal, non-robust bound).
Step 3: Checking the given choices, $ 10 $ kHz and $ 35 $ kHz both fall short of even the $ 50 $ kHz bound, and $ 50 $ kHz itself is only the borderline value, not a rate that clears it.
Step 4: The only option that comfortably exceeds $ 50 $ kHz and gives a workable margin for recovering $ m(t) $ is \[ \boxed{60 \text{ kHz}} \]

Answer 2

Think of this in terms of how much guard margin a real receiver needs beyond the textbook Nyquist number when pulling $ m(t) $ back out of $ y(t) $.

Option 10 kHz: Far too slow, this rate cannot even track a $ 20 $ kHz carrier component, let alone the sidebands extending to $ 25 $ kHz.
Option 50 kHz: Equals exactly $ 2(f_c+B) = 2 \times 25 $, the ideal-case Nyquist value. Real anti-aliasing and reconstruction filters are never perfectly sharp, so a design that samples at exactly this rate leaves zero room for the filter's transition band, risking overlap between spectral replicas.
Option 35 kHz: Well under the $ 50 $ kHz bound, this guarantees overlap of shifted copies of the spectrum and irrecoverable distortion of $ m(t) $.
Option 60 kHz: Provides roughly a $ 10 $ kHz cushion above the ideal bound, which is enough headroom for a realizable filter to separate the baseband image from its neighbours cleanly.

Weighing the margin each option leaves for practical filtering, only one choice sits safely clear of the critical value.

So the correct answer is 60 kHz.

Answer 3

Think of this in terms of how much guard margin a real receiver needs beyond the textbook Nyquist number when pulling $ m(t) $ back out of $ y(t) $.

Option 10 kHz: Far too slow, this rate cannot even track a $ 20 $ kHz carrier component, let alone the sidebands extending to $ 25 $ kHz.
Option 50 kHz: Equals exactly $ 2(f_c+B) = 2 \times 25 $, the ideal-case Nyquist value. Real anti-aliasing and reconstruction filters are never perfectly sharp, so a design that samples at exactly this rate leaves zero room for the filter's transition band, risking overlap between spectral replicas.
Option 35 kHz: Well under the $ 50 $ kHz bound, this guarantees overlap of shifted copies of the spectrum and irrecoverable distortion of $ m(t) $.
Option 60 kHz: Provides roughly a $ 10 $ kHz cushion above the ideal bound, which is enough headroom for a realizable filter to separate the baseband image from its neighbours cleanly.

Weighing the margin each option leaves for practical filtering, only one choice sits safely clear of the critical value.

So the correct answer is 60 kHz.

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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The Correct Option is D

Approach Solution - 1

Approach Solution -2

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