homework3d_solution_0

# homework3d_solution_0 - of 4.0 since cellular systems are...

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ECE 5325/6325 Fall 2009: Homework 3(d) Solution (3d) Typical cell phones transmit 500mW maximum. BS receivers require -115 dBm of power. What is the maximum distance between a cell phone and base station that could work in an ideal case with matching base and mobile antennas: (a) isotropic antennas (b) dipole antennas (c) patch antennas with a gain of 7 dB? Use 1900 MHz for your initial center frequency. State the propagation model you use to solve this problem, and why you chose it. How do the values change when you go to higher center frequency? Solution : Here, you might use the free space propagation equation because the question asks for the “ideal case” and does not provide antenna heights or path loss exponent. However, for a cellular system, we would also want to use the log-distance model with path loss exponent
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Unformatted text preview: of 4.0, since cellular systems are designed to achieve higher path loss exponent. At 1900 MHz, λ = 3 × 10 8 / 19 × 10 8 = 3 / 19 m. Using a reference distance of 100 m,-115(dBm) = 27(dBm) + G t (dB) + G r (dB)-20 log 10 4 π 100 m 3 / 19 m-40 log 10 d 100 m (1) So 40 log 10 d 100 m = 142 + G t + G r-78 . (2) So d = 100 m 10 (64 . 0+ G t + G r ) / 40 (3) For G t = G r = 0 dB, d = 3 . 98 km. For G t = G r = 2 . 15 dB, d = 5 . 10 km. For G t = G r = 7 dB, d = 8 . 91 km. When we go up in center frequency from f old to f new , the path loss at the reference distance Π increases, by a factor of 20 log 10 λ old /λ new , or equivalently 20 log 10 f new /f old . So increasing the center frequency increases the path loss, and thus decreases the range....
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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 Utah.

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