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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 4bit 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.
 Fall '11
 Dr.TimothyCerner

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