ENC3250 Solutions pg 7

# ENC3250 Solutions pg 7 - C = 300 log2(1 100.3 = 300...

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-7- C = 300 log 2 (1 + 10 0.3 ) = 300 log 2 (2.995) = 474 bps 2.10 Using Nyquist's equation: C = 2B log 2 M We have C = 9600 bps a. log 2 M = 4, because a signal element encodes a 4-bit word Therefore, C = 9600 = 2B × 4, and B = 1200 Hz b. 9600 = 2B × 8, and B = 600 Hz 2.11 Nyquist analyzed the theoretical capacity of a noiseless channel; therefore, in that case, the signaling rate is limited solely by channel bandwidth. Shannon addressed the question of what signaling rate can be achieved over a channel with a given bandwidth, a given signal power, and in the presence of noise. 2.12 a. Using Shannon’s formula: C = 3000 log 2 (1+400000) = 56 Kbps b. Due to the fact there is a distortion level (as well as other potentially detrimental impacts to the rated capacity, the actual maximum will be somewhat degraded from the theoretical maximum. A discussion of these
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