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 = rr r r 0 0 () 22 00 2 0 2 0 2 1 2 1 2 2 4 16 kr x k r x kr r π λ ΔΦ = Φ − ⎛⎞ ⎜⎟ 2 0 2 0 0 1 4 p x r x r x r Φ =+ = = = ≈+ If we take 2 1 which implies 4 0 0 4 x x r r < ± then we would have rad 16 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 31 0 1 11 0 4.33 44 1 s m s km r xm λ × × == = ·· 2. a. The plane wave phasor form of the electrical field is given as 0 ˆ j j −− kr ± ee E e 0 EE , use the information given and determine each characteristic of the field. 0 2 22 2 ˆˆ ˆ ˆ ˆ r xyz kk xk yk zk k x y π λλ ⎛⎞ =++= = = k ε , where 0 is the free space wavelength. The fact that V ˆ ˆ (0,0,0) 10 m z = E ± gives the polarization and amplitude of the field in general. With x xy yz z = + + r 0 2 2 4 322 ( ) 10 10 10 r jx y y ze ⎜⎟ ⎝⎠ Er ± ε ± 422 3 VV ˆ mm j x y ππ = To determine use the triad relationship: H 0 ηη ×× ±± r kk H ± ε , where 0 η is the free space intrinsic impedance (120 π ) 32 2 2 2 ˆ 4 2 2 120 1 A 6 m y y z e e ⎡⎤ =− × ⎢⎥ ⎣⎦ + H ± 42 32 1 0 2
b. Utilize the general form: ( ) 00 (,) R e{ ˆ }A c o s · A ee tt ω =+ =− Ak Ar jt r ± 42 2V ˆ (,) 10co s 32 2 m tz t x y π ⎛⎞ ⎜⎟ ⎝⎠ Er 22 1 4 22A ˆˆ co s 6 3 m txy t xy + Hr c. × = SEH 2 2 2 2 2 1 2 cos 2 2 2 6 2 5 4 W cos 3 3 22 m zt x y x y t x y t ππ ωω −− × + ˆ10 cos = Then, 2 522 W 3 1 6 ×− = == H SE GG 2 m One can also utilize another method, 1 2 * 0 } it's a plane wave 11 Re{ r kk ηη ×× = = EE EH S ±± ε 2 2 10 5 2 2 W 4 120 6 2 2 m 1 2 = S Note how in this case, S does not depend on position.

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## This note was uploaded on 02/21/2010 for the course ECE 450 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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

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