Homework 4 Solutions

Homework 4 Solutions - CS421/S03 Chapter 14 (7/e) 1....

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CS421/S03 Solutions to Homework Four Chapter 14 (7/e) 1. Problem 14.1 a. The number of clusters M = 16. The total available bandwidth in each direction is 20 MHz. Therefore, the total number of channels in each way is 20 MHz/30 kHz = 666, and each cell can have 666/N channels. Because each system is duplicated 16 times, the total number of channels will be 16 × 666 = 10656. So, the number of simultaneous communications that can be supported by each system is 10656. b. Total number of channels available is K = 666. For a frequency reuse factor N , each cell can use k CE = K / N = 666/ N channels. For N = 4, k CE = 166 channels. For N = 7, k CE = 95 channels. For N = 12, k CE = 55 channels. For N = 19, k CE = 35 channels. c. The number of cells covered is 16 × N . For N = 4, area = 64 cells; For N = 7, area =112 cells; For N = 12, area = 192 cells; For N = 19, area = 304 cells. d. From part b, we know the number of channels that can be carried per cell for each system. The total number of channels available is just 100 times that number, for a result of 16600, 9500, 5500, 3500, respectively. 2 . Problem 14.4 (12.5 × 10 6 – 2(10 × 10 3 ))/(30 × 10 3 ) = 416 Chapter 15 . 3.
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Homework 4 Solutions - CS421/S03 Chapter 14 (7/e) 1....

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