HW 4 Sol - same access code there would be no way to tell...

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ECE 445: Computer Networks and Telecommunications Fall 2007 HW Solution 4 — Due: October 25 Professor: Stephen Wicker TA: Prapun Suksompong 4.30 [5 pt] See p 306 of the textbook. It depends how far away the subscriber is. If the subscriber is close in, QAM-64 is used for 120 Mbps. For medium distances, QAM-16 is used for 80 Mbps. For distant stations, QPSK is used for 40 Mbps. 4.31 [5 pt] Uncompressed video has a constant bit rate. Each frame has the same number of pixels as the previous frame. Thus, it is possible to compute very accurately how much bandwidth will be needed and when. Consequently, constant bit rate service is the best choice. 4.33 [5 pt] A Bluetooth device can never be a master of more than one piconet. Since the piconet is defined by synchronization to the masters Bluetooth clock it is impossible to be the master of two or more piconets. The access code at the start of the frame is derived from the masters identity. This is how slaves tell which message belongs to which piconet. If two overlapping piconets used the
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Unformatted text preview: same access code, there would be no way to tell which frame belonged to which piconet. In effect, the two piconets would be merged into one big piconet instead of two separate ones. 4.35 [5 pt] An ACL channel is asynchronous, with frames arriving irregularly as data are produced. An SCO channel is synchronous, with frames arriving periodically at a well-defined rate. 4.39 [5 pt] The simplest choice is to do nothing special. Every incoming frame is put onto the backplane and sent to the destination card, which might be the source card. In this case, intracard traffic goes over the switch backplane. The other choice is to recognize this case and treat it specially, sending the frame out directly and not going over the backplane. See also p. 282 of the textbook. 4.40 [5 pt] The worst case is an endless stream of 64-byte (512-bit) frames. If the backplane can handle 10 × 10 9 = 10 10 bps, the number of frames it can handle is 10 10 / 512. This is 1 . 953 × 10 7 frames/sec. 4-1...
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This homework help was uploaded on 02/20/2008 for the course ECE 4450 taught by Professor Wicker during the Fall '07 term at Cornell.

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