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Solutions to Chapter 4 - Communication Networks

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<strong>Communication</strong> <strong>Networks</strong> (2 nd Edition)<br />

<strong>Chapter</strong> 4 <strong>Solutions</strong><br />

location, on average, a non-alternating pattern will be discovered on the third observation, at<br />

which time the synchronizer will move one bit forward <strong>to</strong> the next bit.<br />

Thus, the average number of bits elapsed until the synchronizer observes the framing bit is:<br />

Average number of bits elapsed until frame bit found<br />

= 96.5 x [(2 x 193) + 1] = 37345 bits<br />

At this point, the synchronizer will begin observing the framing bit. It can’t be sure that it has<br />

located the framing bit until an alternating pattern has been observed for a sufficient number of<br />

bits, say n. The synchronizer can never be sure that it has locked on<strong>to</strong> the frame bit, so n must<br />

be chosen sufficiently large so that the synchronizer is reasonably confident that it has indeed<br />

found the frame bit.<br />

9. The CEPT-1 carrier system uses a framing byte at the beginning of a frame.<br />

a. Suppose that all frames begin with the same byte pattern. What is the probability that this<br />

pattern occurs elsewhere in the frame Assume that each information bit takes a value of 0 or<br />

1 independently and with equal probability.<br />

The probability that random information matches the framing pattern is quite small:<br />

p = (½) 8 =1/256<br />

b. Consider an arbitrary information bit position in the frame. Calculate the average number of<br />

times that the byte beginning in this bit position needs <strong>to</strong> be observed before it is found <strong>to</strong> not<br />

be the framing byte.<br />

The length N of a CEPT frame is 32 bytes, so N=256. The average number of frames M<br />

observed before a location mismatches the framing pattern is:<br />

M = 0(1 − p)<br />

+ 1p(1<br />

− p)<br />

+ 2 p<br />

1<br />

= p(1<br />

− p)<br />

(1 − p)<br />

2<br />

=<br />

1−<br />

p<br />

p<br />

2<br />

(1 − p)<br />

+ ... =<br />

8<br />

(1/ 2)<br />

=<br />

1−<br />

(1/ 2)<br />

8<br />

p(1<br />

− p){1<br />

+ 2 p + 3p<br />

2<br />

+ ...<br />

The first observation usually indicates that this is not the right position.<br />

c. Now suppose that the frame synchronizer begins at a random bit position in the frame.<br />

Suppose the synchronizer observes the byte beginning in the given bit position until it<br />

observes a violation of the alternating pattern. Calculate the average number of bits that<br />

elapse until the frame synchronizer locks on<strong>to</strong> the framing byte.<br />

The average number of bits observed until the beginning of frame is found is then:<br />

8 8<br />

N 2 1/ 2<br />

( MN + 1) = (<br />

2<br />

2 1−1/<br />

2<br />

8<br />

8<br />

2 2 −1/<br />

2<br />

N + 1) = (<br />

2 1−1/<br />

2<br />

8<br />

8<br />

) ≈ 256bits<br />

10. Suppose a multiplexer has two input streams, each at a nominal rate of 1 Mbps. To accommodate<br />

deviations from the nominal rate, the multiplexer transmits at a rate of 2.2 Mbps as follows. Each<br />

group of 22 bits in the output of the multiplexer contains 18 positions that always carry information<br />

bits, nine from each input. The remaining four positions consist of two flag bits and two data bits.<br />

Each flag bit indicates whether the corresponding data bit carries user information or a stuff bit<br />

because user information was not available at the input.<br />

Leon-Garcia/Widjaja 5

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