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Telecommunications Engineering DISCUSSION
My textbook gives two formulas for the maximum data rate of a channel. Nyquist says C = 2B × log2(M), which seems to allow any rate if I use enough signal levels. Shannon says C = B × log2(1 + S/N), which depends on noise and has no M in it.
For a 3.1 kHz telephone channel with 30 dB signal-to-noise ratio, which one gives the real limit? And how can a line with only 3.1 kHz of bandwidth carry tens of kilobits per second?
They answer different questions. Nyquist's result is about symbols: a noiseless baseband channel of bandwidth B can carry at most 2B independent symbols per second without intersymbol interference. With M distinguishable levels per symbol you get 2B × log2(M) bit/s. It says nothing about how large M can be.
Shannon's result sets that limit: noise decides how many levels can be told apart reliably. So both apply together. Nyquist caps the symbol rate, Shannon caps the information rate whatever the modulation.
The numbers for your channel: 30 dB is a power ratio of 1000, so C = 3100 × log2(1001) = 3100 × 9.97 = about 30.9 kbit/s. No scheme can do better on that channel with an arbitrarily low error rate.
On the Nyquist side, M = 4 would give 2 × 3100 × 2 = 12.4 kbit/s. To reach 30.9 kbit/s you would need log2(M) = 4.98, about 32 levels, and 30 dB of SNR is just what makes that number of levels distinguishable. Pushing M higher than the noise allows only adds errors, not capacity.
On Hz against bit/s: bandwidth limits how fast the signal can change, which is the symbol rate in baud, not the bit rate. Each symbol can carry several bits if it has many possible states. A modem sending 2400 symbols per second with 64 states per symbol delivers 6 × 2400 = 14.4 kbit/s through a voice channel.
So bit rate = symbol rate × bits per symbol, and the two are equal only for two-state signalling. That is why baud and bit/s are not interchangeable terms.
A rule of thumb from the Shannon formula at high SNR: capacity is roughly B × SNR(dB) / 3, because every extra 3 dB doubles the power ratio and adds one bit per second per hertz. 3100 × 30 / 3 = 31 kbit/s, close to the exact figure. So doubling the bandwidth doubles the capacity, while doubling the power adds only one bit/s per Hz.
The formula also does not go to zero at low SNR. At 0 dB, C = B, and useful communication is possible with the signal below the noise, as in spread-spectrum systems, at the price of a low rate for the bandwidth used.
Do not confuse either of these with the Nyquist sampling theorem, which is a third statement: a signal with no content above f_max is fully described by samples taken faster than 2 × f_max. It is about digitising an analog signal, not about channel capacity, although the same factor of two appears for the same underlying reason.
Also remember that Shannon's figure is a bound, approached only with ideal coding. Practical systems with good error-correcting codes get close to it, while a fixed modulation with no coding sits well below.
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