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HW#3_solutions

# HW#3_solutions - ECE594I Prof Brown Fall Quarter 2009 HW#3...

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ECE594I, Prof. Brown, Fall Quarter 2009 1 HW#3 Solutions 1) To evaluate the feedhorn, start with F( θ,φ ) = exp(-10 θ 2 ) , (a) Half-power beam width defined by ] ) 2 / ( 10 exp[ 5 . 0 2 β = where β is the full width at the half-power points. Solving f or β , we get 2 / 1 ] 10 / ) 5 . 0 ln( [ * 2 = 0.53 rad = 30.2 o . (b) The pattern solid angle is ( ) ,s i n p Fd d θ φθ φ Ω= ∫∫ But this is not integrable in θ (try mathematically to verify this using a symbolic integration code, such as Mathmatica or Maple). So we take advantage of fact that the pattern given has but a single main lobe, and approximate by spherical trigonometry Y X P . From part (a) β = β X = β Y = 0.527, so P = 0.277 steradians. But a better way to estimate P (as discussed in lecture) is possible since the beam is symmetric ∫∫ = π ρπ 2 0 2 / 0 2 00 sin sin ) , ( d d d d F where F( θ,φ ) is approximated as unity over the cone full angle β . This so called “ice-cream- cone” approximation leads to P 2 / 0 cos 2 = 2 π (1-cos β /2) = 0.217 steradian. (c) The antenna directivity is: D = 4 π / P 4 π / 0.277 = 45.4 (according to pencil beam approximation) D = 4 π / 0.217 = 57.9 (according to “ice-cream-cone” approximation) (d) S max at range of 10 cm is just P rad /(4 π r 2 ) D = 4.6x10 -5 W/cm 2 in “ice-cream-cone” approximation. 2) Dish with circular beamwidth of 1.5 o . (a) For pencil-beam case, D 4 π /( β x β y). For the dish, β = 1.5* π /180 o = 0.026 rad. So, D 4 π /(0.026) 2 = 1.84x10 4 or D = 10log10[1.84x10 4 ] = 42.6 dB (b) If dish area is doubled, we expect the directivity to also double consistent with D max = 4 π A/ λ 2 , and the beamwidth should decrease consistent with 4 π /( β X β Y ) = 4 π A/ λ 2 . So new directivity 4 3.66 10 45.6 Dd B . And the new beamwidth β = 1.5 o /(2) 1/2 = 1.06 o = 0.018 rad.

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HW#3_solutions - ECE594I Prof Brown Fall Quarter 2009 HW#3...

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