450fall09hw4_Sol - ECE 450 Fall 2009 Homework 4 Solutions...

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ECE 450 Fall 2009 Homework 4 Solutions Due: Tuesday, Sep 22, 2009 1. The effective length of the -directed half-wave dipole is give by ˆ z 2 cos( cos ) 2 ) n ( si π θ λ πθ = A a. Directly, 0 ( 0) Indeterminate Form 0 == A ö , thus one must use l’Hospital’s rule as indicated. The rules stated that the 00 ' ( lim lim ) )) (( ff gg θθ ' ( ) →→ = . Thus we can determine the following: 0 sin cos s (l i in 0 22 0) still, do it once more. 2sin cos 0 m ⎛⎞ ⎜⎟ ⎝⎠ = A = ππ () 0 2 0 cos cos cos sin sin cos cos 2 2 0) 2 cos cos sin sin sin cos cos (for non-zero terms that exist) 2cos 2 2 (interesting, i the half-wavelength 4 m = ⎡⎤ −− + ⎢⎥ ⎣⎦ = = =♠ A = dipole looks only a quarter-wavelength long when viewing from 0 ) 1
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b. If you plotted from 01 8 0 θ °≤ ≤ ° your plot will resemble the right half of the plot. 0.00 90° 180° 0.05 0.10 0.15 0.20 0.25 0.30 0.35 -90° 2 cos ) 2 for 180 180 in cos( )1 s ( π λπ = −° ≤° A c. Likewise for the plot below 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 2 cos( cos ) 2 sin sin for 180 180 s ( in θθ = ≤≤° A 2
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***FYI: Now since you’ve learned about arrays, consider the following pattern multiplication analogy. The total pattern of an array of identical elements can be interpreted as the element factor multiplied by the array factor . Recall, the Hertzian dipole served as fundamental element from which all linear elements could be derived upon. The same concept works graphically just as well. Imagine the reference element being the Hertzian dipole, then imagine an infinite array of these elements stacked along the z -axis as a linear array, a half-wavelength in length. The current and phasing (see problem 3, 4 and 5) for each “array element” is such that the array factor equates to the effective length () θ A , the element factor being sin . Graphically, pattern multiplication shows: 0.0 90° -90° 180° 0.2 0.4 0.6 0.8 1.0 Element Factor: sin 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 ) Array Factor: ( λ A 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 ) Total Pattern: si ( n( ) A × = 2. The gain function of the antenna is given as: 2 ( sin for 0 2 ,) for 2 D D G π θφ θπ << =
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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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450fall09hw4_Sol - ECE 450 Fall 2009 Homework 4 Solutions...

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