Homework13BeamWidth - Actually, the dierence in the sky...

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ECE 3030: Electromagnetic Fields and Waves Fall 2009 Homework 13 SOLUTIONS for GRADER: Beam Widths and Radar Design Reminder: Date: Tuesday 12/1 1. Problem 17.6 This problem computes a more accurate answer for the simple approximation given by Equation (17.26) which for N = 10 yields ψ 1 3 π/N = 0 . 9425 radians = 54 . Iterating Equation (17.26) around 54 finds ψ 1 = 51 . 67 = 1 . 169 radians. 2. Problem 17.7 The approximation for the array factor at the first side- lobe is | F ( ψ 1 ) | 2 ± 2 N 3 π 2 = 4 . 503 or - 13 . 46 dB. Computing the value directly from Equation (17.26) yields | F ( φ 1 ) | 2 = sin 2 ( 1 / 2) sin 2 ( ψ 1 / 2) = 5 . 051 or - 12 . 97 dB, a difference of only half a decibel. 3. Problem 18.5 The sizing of an antenna for incoherent scatter radars posses some interesting questions. The difference between this problem and Problem 18.4, explicitly is in the frequency and the sky temperature.
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Unformatted text preview: Actually, the dierence in the sky temperature is really due to the dif-ference in the frequency, but giving the values just makes this problem easier. Accounting for just these two factors, S N A effective T sky So the size of the antenna required for 500 MHz would be A 500 = 300 x 300 50 5000 = 900 m 2 or a square antenna of 30 m by 30 m. Another point to be made here is that at the higher frequency, the signal bandwidth would be a factor of 10 narrower, which would reduce the antenna area by 10 = 3 . 16, or one with dimensions of 16.9 m by 16.9 m, an not insignicant reduction. CORNELL UNIVERSITY c WES SWARTZ (09/11/16) 131...
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