Example 1
Q: Given the signal-to-noise ratio (SNR) of 20 dB, and the bandwidth of 4kHz (using phone line), what is the maximum data rate according to Shannon¡¯s theorem?
Ans:
4 * log2(1 + 100) = 4 * log2 (101) = 26.63 kbps.
Note that the value of S/N = 100 is equivalent to the SNR of 20 dB
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Example 2
Q: If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio is 20 dB, what is the maximum achievable data rate?
Ans:
SNR of 20 dB = S/N = 100.
The Shannon limit is: 3* log2(101) ¡Ö 19.975 kbps
The Nyquist limit is:
2B log2V = 2 x 3 x log22 = 6 kbps.
The bottleneck is therefore the Nyquist limit, giving a maximum channel capacity of 6 kbps
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