Solutions to Chapter 4 - Communication Networks
Solutions to Chapter 4 - Communication Networks
Solutions to Chapter 4 - Communication Networks
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
<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