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528<br />

Performance of Communication Systems Corrupted by Noise Chap. 7<br />

the uniform distribution is (-d/2, d/2), where d is the step size (d = 2 in Fig. 3–8c). Thus,<br />

with M = 2 n = 2V/d (from Fig. 7–16),<br />

q<br />

d/2<br />

e 2 2<br />

2<br />

q = e q f(e q ) de q = e 1 q (7–76)<br />

L - q<br />

L -d/ 2 d de q = d2<br />

12 = V2<br />

3M 2<br />

The noise power due to bit errors is evaluated by the use of Eq. (7–74), yielding<br />

e 2 b = [y k - Q(x k )] 2<br />

(7–77)<br />

where Q (x k ) is given by Eq. (7–71). The recovered analog sample y k is reconstructed from the<br />

received PCM code word using the same algorithm as that of Eq. (7–71). Assuming that the<br />

received PCM word for the kth sample is (b k1 , b k2 , .. . , b kn ), we see that<br />

y k = V a<br />

n<br />

j=1<br />

b kj A 1 2B j<br />

(7–78)<br />

The b’s will be different from the a’s whenever there is a bit error in the recovered digital<br />

(PCM) waveform. By using Eqs. (7–78) and (7–71), Eq. (7–77) becomes<br />

e 2 b = V 2 n<br />

c a (b kj - a kj )A 1 2B j 2<br />

d<br />

j=1<br />

= V 2 n n<br />

a a ab kj b k/ - a kj b k/ - b kj a k/ + a kj a k/ b2 -j-/ (7–79)<br />

j=1 /=1<br />

where b kj and b k are two bits in the received PCM word that occur at different bit positions<br />

when j Z . Similarly, a kj and b k are the transmitted (Tx) and received (Rx) bits in two different<br />

bit positions, where j Z (for the PCM word corresponding to the k th sample). The<br />

encoding process produces bits that are independent if j Z . Furthermore, the bits have a<br />

zero-mean value. Thus, b kj b k/ = b kj b k/ = 0 for j Z . Similar results are obtained for the<br />

other averages on the right-hand side of Eq. (7–79). Equation (7–79) becomes<br />

e 2 b = V 2 n<br />

2<br />

a ab kj - 2a kj b kj + a 2 kj b2 -2j<br />

(7–80)<br />

j=1<br />

Evaluating the averages in this equation, we obtain †<br />

b 2 kj = (+1) 2 P(+1Rx) + (-1) 2 P(-1Rx) = 1<br />

2<br />

a kj = (+1) 2 P(+1Tx) + (-1) 2 P(-1Tx) = 1<br />

a kj b kj = (+1) (+1)P(+1Tx, +1Rx) + (-1)(-1)P(-1Tx, -1Rx)<br />

+ (-1)(+1)P(-1Tx, +1Rx) + (+1)(-1)P(+1Tx, -1Rx)<br />

= [P(+1Tx, +1Rx) + P(-1Tx, -1Rx)]<br />

- [P(-1Tx, +1Rx) + P(+1Tx, -1Rx)]<br />

= [1 - P e ] - [P e ] = 1 - 2P e<br />

† The notation +1Tx denotes a binary 1 transmitted, -1Tx denotes a binary 0 transmitted, +1Rx denotes a<br />

binary 1 received, and -1Rx denotes a binary 0 received.

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