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I am confused with the shannon channel capacity formula Versus nyquist bit rate.

nyquist formula contains signal levels while Shannon formula doen't, how to refer to signal Levels in shannon's formula ?

I would appreciate an example for channel capacity Using both formulas

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The formulas are not calculating the same thing. The Nyquist bit rate formula is for a noiseless channel and calculates the maximum bit rate for a given channel bandwidth and number of signaling levels. Note that the formula uses the number of signal levels, not the actual levels. It simply combines the bandwidth, which determines how fast symbols can be sent, and the number of signal levels for each symbol which determines how many bits can be sent during one signaling interval, to the maximum number of bits/second that can be sent. The Shannon formula is for a channel with noise and combines the channel bandwidth and the signal-to-noise ratio to determine the maximum number of bits/second that can be sent over that channel. It does use signal level in the form of signal-to-noise ratio. The Nyquist formula, as already noted, does not use signal level because it is immaterial as it assumes there is no noise.

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  • \$\begingroup\$ Notice that the Nyquist rate (at least the way it's described on Wikipedia) doesn't relate to bit rate at all. It gives the maximum number of pulses (or symbols) that can be transmitted, so Nyquist is giving a baud rate instead of a bit rate. \$\endgroup\$ – The Photon Jul 29 '14 at 23:40
  • \$\begingroup\$ If you google 'Nyquist bit rate formula", you will several hits which describe the formula as calculating the bit rate. However, the main point of the question was how the formulas compare and the main answer is that Nyquist assumes no noise (and hence signal level is not important) and Shannon includes noise (in the form of SNR). \$\endgroup\$ – Barry Jul 30 '14 at 1:48
  • \$\begingroup\$ it looks like that version of the formula just adds an extra term of log_2 (L), where L is the number of levels per symbol. Naturally this translates baud to bits. It does seem to be a little unclear what is the Nyquist formula. \$\endgroup\$ – The Photon Jul 30 '14 at 3:34

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