Given below are two statements: Statement š¼: The laws of nature put twoā¦
2020
Given below are two statements:
StatementĀ š¼: The laws of nature put two fundamental limits on data rate of a channel. The H.Nyquist limit restricts the number of independent samples per second to twice the band-width in a Noiseless channel
StatementĀ š¼š¼: Shannonās major result about noised channel is that maximum data rate of a channel whose band width isĀ \(š»\)Ā Hz, and whose signal-to-noise ratio isĀ \(š/š\)Ā is given by: Maximum number of bits/sec =Ā \(=H \log _2 ( 1 + \dfrac{S}{N})\)
InĀ the light of the above statements, choose the correct answer from the options given below
Answer: A. Both Statement š¼ and Statement š¼š¼ are true ā Answer: Both statements are true. Statement I (Nyquist limit): For a band-limited noiseless channel of bandwidth H Hz, the Nyquist sampling principle impliesā¦
- A.
Both StatementĀ š¼Ā and StatementĀ š¼š¼Ā are true
- B.
Both StatementĀ š¼Ā and StatementĀ š¼š¼Ā are false
- C.
StatementĀ š¼Ā is correct but StatementĀ š¼š¼Ā is false
- D.
StatementĀ š¼Ā is incorrect but StatementĀ š¼š¼Ā is true
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Correct answer: A
Answer: Both statements are true.
Statement I (Nyquist limit): For a band-limited noiseless channel of bandwidth H Hz, the Nyquist sampling principle implies up to 2H independent signal samples per second (the Nyquist rate). If each sample conveys one of M discrete signal levels, the maximum data rate in a noiseless channel is 2H Ā· log2(M) bits per second.
Statement II (Shannon capacity): For a channel with additive white Gaussian noise, bandwidth H Hz, and signal-to-noise ratio S/N, the maximum achievable reliable data rate is given by Shannon's formula: C = H Ā· log2(1 + S/N) bits per second. This is the fundamental limit when noise is present and optimal coding is used.
Relation and assumptions: Nyquist applies to ideal noiseless channels and describes how bandwidth limits discrete signaling; Shannon accounts for noise and gives the ultimate capacity under Gaussian noise assumptions. Both statements, as stated, are correct within their usual assumptions.
Nyquist (noiseless): maximum samples per second = 2H; maximum bits/s with M levels = 2H Ā· log2(M).
Shannon (noisy): channel capacity C = H Ā· log2(1 + S/N) bits/s.
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