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450fall09hw5_Sol

# 450fall09hw5_Sol - ECE 450 Fall 2009 Homework 5 Solutions...

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ECE 450 Fall 2009 Homework 5 Solutions Due: Tuesday, Sep 29 th , 2009 1) Graphically, the currents related to each z -polarized short dipole can be represented in the x-y plane as shown in the figure below. Note the observation point is in the far-field at point ,s i n, 0 ) ( (, co 0,0), since 0 s rr r φφ φ == ° and symbols / × indicate the currents are 180 ° out of phase with each other. a) The key to determining the smallest non-zero d that gives a far-field null at the observation point is to recall that the sources are time-harmonic and thus have the sin(x) and cos(x) characteristics. We all can visualize how these functions add and subtract with each other. Furthermore, using superposition and symmetry arguments to determine the solution comes down to the path lengths from the each source to the observation point, each one is colored in the figure. In either domain, indifferent path lengths create a 1

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propagation delay: a time delay in the time-domain and a phase delay in the frequency domain. The result is the same, constructive and destructive interference at the observation point due the varying path lengths. Since elements 3 and 4 are symmetrically displaced about x , their path lengths are the same, i.e., the blue and red lines are of the same length. In the far-field you would expect these lines to be parallel to each other via the parallel approximation. Given the fact that they are 180 ° out of phase, the fields deconstruct at the observation point—think of a cosine graph adding to a its negative. Elements 1 and 2 on the other hand have different path lengths as indicated by the yellow and green curve hence the field at element 2 “picks up” additional phase by when it reaches element 1. Recall, they are 180 ° out of phase, imagine element 1 and 2 being at the exact same location, obviously they would cancel each other out, but they are not and since the field values repeat themselves every wavelength ( λ ) the next best thing is to have their separation 2 d = λ , or multiple integers thereof. In this way the field will deconstruct at the observation point. Thus, the smallest possible d = λ /2 b) Recall from last the last homework, we can write the total field pattern as: where the observation point unit vector is simply give a ˆ 3 12 ' ' ˆ ˆˆ ˆ · ·· · 01 02 03 '' 04 () [ ] [] , total ref jk jk jk jk Array Factor Array Factor Ie ⎡⎤ ∝+ + + ⎣⎦ r rr r r r Er E r ˆ sin cos sin sin cos xy z 4 ±± s, θ φθ ++ an ' ' 3 4 ˆ ˆ ,, , a n d dd d xd = r d the position vector of each source as v x yy == = = r r r ; 01 02 03 04 1 0 1 180 1 0 1 180 ,and II I I =∠ = = ° r each with input current phasors °° ° , respectively. Therefore, 2
[] 2 2 sin( cos ) sin( sin ) ,with 2 2 sin( cos ) sin( sin ) ArrayFactor e e e e jk d k d k d j πλ cos cos sin sin jkd jkd jkd jkd φφφφ φ φπ λ πφ ⎡⎤ =− + ⎣⎦ =+ = = cos ) sin( sin ) To determine the global maximum locations, one could look inside the array factor. Trying to solve analytically becomes difficult because the relationship is transcendental and the solution is best determined via graphical or numerical means.

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450fall09hw5_Sol - ECE 450 Fall 2009 Homework 5 Solutions...

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