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HW1_solution_edited

# HW1_solution_edited - TELCOM2000 Homework 1 Solution...

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TELCOM2000 Homework 1 Solution Stallings 2.1 One CD-quality channel requires 44,100 smp/s * 16 bits/smp = 705,600 bps The number of channels on a 10-Mbps Ethernet is 10000000/705600 = 14.17 or 14 channels. If you assumed stereo quality, the answer is 7 channels. Stallings 2.3 (a) Required data rate = 8,000 smp/s * 8 bits/smp * 24 channels = 1,536,000 bps or 1.536 Mbps (b) Each 3-minute (or 180-second) message requires 8,000 smp/s * 8 bits/smp * 180 s = 11,520,000 bits, or 11,520,000/8 = 1,440,000 bytes or 1.44 megabytes. Stalling 2.4 How many bits will it take to represent the following sets of outcomes? a. The uppercase alphabet A, B, …, Z 2 5 = 32; 32 > 26 (number of letters) _ 5 bits b. The digits 0, 1, …, 9 2 4 = 16; 16 > 10 (number of digits) _ 4 bits c. The seconds in a 24-hour day 2 17 = 131072; 131072 > 86400 (number of seconds) _ 17 bits d. The people in the United States (about 300,000,000 of them) 2 29 = 536,870,912 _ 29 bits e. Population of the world (about 6 billion) 2 33 = 8,589,934,592 _ 33 bits Stalling 2.11 a. #dots along the width = 300 d/i x 8.5i = 2550 dots. #dots along the height = 300 d/i x 11i = 3300 dots. (Notice that dots/inch times inches = dots) b. Total dots = 2550 x 3300 = 8,415,000 d/page. At 1 Byte/dot, the memory required = 8.415 MB/page, or 8 b/d x 8.415M d/page = 67.32 Mb/page.

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