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Unformatted text preview: CS 168 Computer Networks Jannotti Homework 1 Due: 20 February 2010, 4pm Problem 1 Sec. 2, prob. 2 A noiseless channel can carry an arbitrarily large amount of information, no matter how often it is sampled. Just send a lot of data per sample. For the 4kHz channel, make 8000 samples/sec. If each sample is 16 bits, the channel can send 128 kbps. If each sample is 1024 bits, the channel can send 8.2 Mbps. The key word here is “noiseless”. With a normal 4kHz channel, the Shannon limit would not allow this. Sec. 2, prob. 4 20dB = S/N ratio of 100. log 2 (101) is approximately 6.658. So Shannon’s limit is approximately 19.975 kbps. But the Nyquist limit is 2*3 kHz = 6kHz, giving 6 Kb/s. With 1 bit per sample, the bottleneck is the Nyquist limit, giving a maximum channel capacity of 6 Kb/s. Sec. 2, prob. 29 A sample of 125 μ sec corresponds to 8000 samples/sec. According to Nyquist theorem, this is the sampling frequency needed to capture all the information in a 4kHz channel, such as a telephone channel.in a 4kHz channel, such as a telephone channel....
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This note was uploaded on 12/03/2010 for the course CS 168 taught by Professor Jj during the Spring '10 term at Brown.
 Spring '10
 JJ
 Computer Networks

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