exam2-s10-solutions

exam2-s10-solutions - ECE 5325/6325 Fall 2009: Exam 2...

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Unformatted text preview: ECE 5325/6325 Fall 2009: Exam 2 Solutions 1. (25 pts total) The total length of the direct path would be 200 2 + 29 2 = 202 m. Because of the symmetry, the two distances to the knife edge are d 1 = d 2 = 101 m, and the height h = 29 / 2 = 14 . 5 m. Since f c = 450 10 6 Hz, = 3 10 8 / 4 . 50 10 8 = 2 / 3 m. So = h radicalBigg 2( d 1 + d 2 ) d 1 d 2 = 14 . 5 radicalBigg 2(202) (2 / 3)(101) 2 = 3 . 53 . If you used the graph in Rappaport Figure 4.14, I read about G =- 23 . 5. Using Lees approximation, G = 20log 10 . 225 3 . 53 =- 23 . 9. The received power due the diffracted path is P r = P t + Gains- Losses where P t = 10 dBW. There are antenna gains of 10 dB and a diffraction gain of -23.9 dB, for a total of -13.9 dB. The free space loss on the line-of-sight is 20log 10 4 202 2 / 3 = 71 . 6 dB. So P r = 10 dBW- 13 . 9 dB- 71 . 6 dB =- 75 . 5 dBW =- 45 . 5 dBm 2. (20 pts total) 5325-only short answer question: For a given speed, what direction of travel results2....
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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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exam2-s10-solutions - ECE 5325/6325 Fall 2009: Exam 2...

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