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Example: Let us consider an example that relates the Nyquist and Shannon formulations

Example: Let us consider an example that relates the Nyquist and Shannon formulations Suppose that the spectrum of a channel is between 3 MHz and 4 MHz and SNR = 24 dB. Then B = 4 MHz – 3 MHz = 1 MHz SNR = 24 dB  in linear SNR = 251 Shannon: C = 10 6 x log 2 (1+251) = 8 Mbps

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Example: Let us consider an example that relates the Nyquist and Shannon formulations

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  1. Example: Let us consider an example that relates the Nyquist and Shannon formulations Suppose that the spectrum of a channel is between 3 MHz and 4 MHz and SNR = 24 dB. Then B = 4 MHz – 3 MHz = 1 MHz SNR = 24 dB  in linear SNR = 251 Shannon: C = 10 6 x log 2 (1+251) = 8 Mbps Based on Nyquist’s formula, how many signaling levels are required ? We have C = 2B log2 M 8 x 106 = 2x 106 x log2 M 4 = log2 M  M = 16

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