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450fall09hw6_Sol

# 450fall09hw6_Sol - ECE 450 Fall 2009 Homework 6 Solutions...

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ECE 450 Fall 2009 Homework 6 Solutions Due: Tuesday, Oct 6 th , 2009 1. Refer to figure below, which is also shown in the margins in lecture note #14 on pages 7-10, for graphical clarity (the Fresnal box). The goal is to determine the horizontal region about where the plane wave’s wave front has negligible error compared to the spherical wave front. The error is due the spherical spreading of the spherical wave as it propagates away from the source. On a plane perpendicular to the array normal at , the further you move away from in the plane the error between the two wave front increases. The error due to the delay between the spherical wave and the plane wave, respectively, is given as: 0 = r r r r 0 0 ( ) ( ) 2 2 0 0 2 0 2 0 2 1 2 1 2 2 4 16 k r x kr x kr r π λ π π ΔΦ = Φ − 2 0 2 0 0 1 4 p x r x r x r λ λ Φ = + = = = + If we take ( ) 2 1 which implies 4 0 0 4 x x r r λ λ < ± then we would have rad 16 π ΔΦ ± . And thus the maximum distance away from broadside that a 1

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plane wave approximation of the radiation field would be acceptable is 8 9 0 3 10 1 1 10 4.33 4 4 1 s m s km r x m λ × × = = = · · 2. a. The plane wave phasor form of the electrical field is given as 0 ˆ j j k r k r ² e eE e = = 0 E E , use the information given and determine each characteristic of the field. 0 2 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 r x y z kk xk yk zk k x y π π λ λ = + + = = = k ε , where 0 λ is the free space wavelength. The fact that V ˆ ˆ ˆ ˆ (0,0,0) 10 m z = E ² gives the polarization and amplitude of the field in general. With xx yy zz = + + r 0 2 2 2 2 4 2 2 2 2 3 2 2 ˆ ˆ ( ) 10 10 10 r j x y j x y z e z e z e π λ = = E r ² ε ² 4 2 2 3 2 2 V V ˆ m m j x y π π = To determine use the triad relationship: H 0 η η × × ² ² ˆ ˆ r k k = = E E H ² ε , where 0 η is the free space intrinsic impedance (120 π ) 4 2 2 3 2 2 2 2 2 2 ˆ ˆ ˆ 4 2 2 120 2 2 1 A ˆ ˆ 2 2 6 m j x y j x y x y z e x y e π π = × = − + H ² 4 2 3 2 10 π π 2
b. Utilize the general form: ( ) 0 0 ( , ) Re{ ˆ } A cos · A e e t t ω = + = A k A r j t ω r ² 4 2 2 V ˆ ( , ) 10 cos 3 2 2 m t z t x y π ω = E r 2 2 1 4 2 2 A ˆ ˆ ( , ) cos 2 2 6 3 2 2 m t x y t x y π ω π = − +

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450fall09hw6_Sol - ECE 450 Fall 2009 Homework 6 Solutions...

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