ENC3250 Solutions pg 7

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

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Unformatted text preview: C = 300 log2 (1 + 100.3) = 300 log2 (2.995) = 474 bps 2.10 Using Nyquist's equation: C = 2B log2M We have C = 9600 bps a. log2 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 log2 (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 relevant impacts should be included and a qualitative value discussed. 2.13 C = B log2 (1 + SNR) 20 × 10 6 = 3 × 106 × log 2 (1 + SNR) log2(1 + SNR) = 6.67 1 + SNR = 102 SNR = 101 2.14 From Equation 2.1, we have LdB = 20 log (4πd/λ ) = 20 log (4πdf/v), where λf = v (see Question 2.4). If we double either d or f, we add a term 20 log(2), which is approximately 6 dB. 2.15 Decibels Losses Gains 1 0.8 1.25 2 0.63 1.6 3 0.5 2 4 0.4 2.5 5 0.32 3.2 6 0.25 4.0 7 0.2 5.0 2.16 For a voltage ratio, we have NdB = 30 = 20 log(V2 /V1 ) V2 /V1 = 1030/20 = 101.5 = 31.6 2.17 Power (dBW) = 10 log (Power/1W) = 10 log20 = 13 dBW -7- 8 0.16 6.3 9 0.125 8.0 10 0.1 10 ...
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.

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