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Given below are two statements:Statement I: 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.Statement II: Shannon's major result about noised channel is that maximum data rate of a channel whose band width is H Hz, and whose signal-to-noise ratio is S/N is given by:Maximum number of bits/sec \(= H \;log_2 \left(1+\frac{S}{N}\right)\) is given by:In the light of the above statements. choose the correct answer from the options given below
The correct answer is option 1.Noiseless channel: Nyquist Bit RateFor a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rateBitRate=2 X bandwidth x10g2LIn this formula, bandwidth is the bandwidth of the channel, L is the number of signals used to represent data, and BitRate is the bit rate in bits per second.Noisy channel: Shannon CapacityIn the reality, we cannot have a noiseless channel; the channel is always noisy. In 1844, Claude Shannon introduced a formula, called the Shannon capacity, to determine the theoretical highest data rate for a noisy channel:Capacity= bandwidth x log2(1+SNR)In this formula, bandwidth is the bandwidth of the channel, SNR is the signal-to-noise ratio, and capacity is the capacity of the channel in bits per second. Note that in the Shannon formula there is no indication of the signal level, which means that no matter how many levels we have, we cannot achieve a data rate higher than the capacity of the channel. In other words, the formula defines the characteristics of the channel, not the method of transmission. ? Hence both Statement I and Statement II are true the correct answer.
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