exam1-s10-solutions

exam1-s10-solutions - ECE 5325/6325 Fall 2009: Exam 1...

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Unformatted text preview: ECE 5325/6325 Fall 2009: Exam 1 Solutions 1. (25 pts total) Question 1 was different between the 6325 and 5325 exams (but all other questions were the same). For the other half who havent seen the questions, they are also included on this problem. First, the 5325-only question 1. 5325 Question : You are called in to design a cellular system for Planarton, Utah. Unfortunately, your path loss measurements indicate that path loss follows the log-distance model with path loss exponent n = 2 . 95. You find a cellular standard that needs an S/I ratio of only 15 dB. (a) (9 pts) What is the minimum N required, using omni-directional antennas? Solution : In dB, we have S/I (dB) = 5(2 . 95)log 10 (3 N )- 10log 10 6. Testing with various N , I found: for N = 4, S/I (dB) = 8 . 14; for N = 7, S/I (dB) = 11 . 72; for N = 9, S/I (dB) = 13 . 33, and for N = 12, S/I (dB) = 15 . 17. Thus N = 12 is the minimum N required. (b) (7 pts) For your answer for (a), what are i and j ? Solution : For a choice of i = 2, j = 2 provides N = i 2 + ij + j 2 = 4 + 4 + 4 = 12. (c) (9 pts) What is the minimum N required, using 120 degree sectoring? Solution : Note that i = 2 for N = 4 , 7 , 9 , 12 so I used S/I (dB) = 5(2 . 95)log 10 (3 N )- 10log 10 2 and tested N 4....
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This note was uploaded on 02/25/2011 for the course ECE 5325 taught by Professor Dr.nealpatwari during the Spring '10 term at University of Utah.

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exam1-s10-solutions - ECE 5325/6325 Fall 2009: Exam 1...

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