LongPwrPtsCh6B

# LongPwrPtsCh6B - ECE 5318/6352 Antenna Engineering Spring...

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72 Chapter 6 Chapter 6 Part 7 Part 7 Schelkunoff Schelkunoff s s Polynomial Polynomial ECE 5318/6352 ECE 5318/6352 Antenna Engineering Antenna Engineering Spring 2006 Spring 2006 Dr. Stuart Long Dr. Stuart Long

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73 ± ± Schelkunoff Schelkunoff s s Polynomial Polynomial Representation Representation (for discrete arrays) (for discrete arrays) N = number of elements in array Note: each element can now have a different excitation coefficient A n β θ ψ + = cos kd = = 1 0 ) ( N n jn n e A AF
74 21 12 1 () jn N oN eA A A A ψ =⇒ = + + + + AF " zz z z z let So, array factor for a N So, array factor for a N -element array is a polynomial of degree element array is a polynomial of degree N-1 1 it has it has N-1 roots (zeros) roots (zeros) ± Schelkunoff Schelkunoff ’s (cont) (cont)

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75 Can write: ) ( ) ( ) ( ) ( 1 3 2 1 1 N- N A z z z z z z z z = " AF n z Where are the roots of the polynomial ± Schelkunoff Schelkunoff ’s (cont) (cont)
76 ± Advantage Advantage Shows explicitly the zeros in the radiation pattern (nulls) Can also think of it as the product of (N-1) two element arrays X Y ψ z-plane ± Schelkunoff Schelkunoff ’s (cont) (cont)

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77 ± How to make this representation useful??? How to make this representation useful??? Look at 2-element array z + = + = 1 1 ψ j e AF 1 , 1 1 = = A A o For 2 λ π d kd There are no sidelobes ! π - π AF ± Schelkunoff Schelkunoff ’s (cont) (cont)
78 Take the product of two of these arrays 2 AF z z z z + + = + + = 2 1 ) 1 ( ) 1 ( 1 , 2 , 1 2 1 = = = A A A o 3-element array with magnitudes of element array with magnitudes of 1 : 2 : 1 Sharper beam, but still no sidelobes ± Schelkunoff Schelkunoff ’s (cont) (cont)

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79 Continue n ) 1 ( z + = AF = k k A z AF Coefficients of this polynomial are the ! !( )! kk n n AC knk == Gives an N-element array w/ no sidelobes Similar to filters or transformers of type “Butterworth”, “Maximally flat”, or “Binomial” BINOMIAL COEFFICIENTS BINOMIAL COEFFICIENTS ± Schelkunoff Schelkunoff ’s (cont) (cont)
80 BINOMIAL COEFFICIENTS BINOMIAL COEFFICIENTS (Add the two numbers above each entry) Ν = 1 Ν = 2 Ν = 3 Ν = 4 Ν = 5 Ν = 6 Ν = 7 Ν = 8 1 1 1 1 1 1 1 1 1 1 1 1 1 2 33 4 46 55 10 10 66 15 15 20 7 72 1 21 35 35 1 1

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81 ± Placement of nulls Placement of nulls Use factored form of AF Just use amplitude shaping for null placement Place null at θ = 26º In phase, β = 0 3-element array λ /4 spacing (but not uniform) ± Example 1 Example 1 cos kd ψ θ = ± Schelkunoff Schelkunoff ’s (cont) (cont) (since β = 0)
82 π ψ 45 . 1 1 45 . j e = = = z () λ 45 . 9 . 2 26 cos 4 2 1 = = = D Let So one root of polynomial at One more zero arbitrary make it symmetric 45 . 2 2 45 . j e = = = z ± Example 1 Example 1 (cont) ± Schelkunoff Schelkunoff ’s (cont) (cont) cos kd θ =

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83 ( ) 2 cos0.45 sin0.45 1 jj ππ =− + + + zz 1 ; 313 . ; 1 2 1 = = = A A A o ( ) 1 45 . 45 . 2 + + = π j j e e z z () ( ) 2 1 z - z z - z = AF ( )( ) 45 . 45 . j j e e = z z 2 0.313 1 + ( ) 2 2cos0.45 1 + Amplitude coefficients Amplitude coefficients ± Schelkunoff Schelkunoff ’s (cont) (cont) ± Example 1 Example 1 (cont) 2 12 o AA A =+ + AF
84 θ = 26° θ = 154° There is really a “cheat” here. .

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## This note was uploaded on 09/09/2010 for the course ECE 4436 taught by Professor Staff during the Spring '08 term at University of Houston.

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LongPwrPtsCh6B - ECE 5318/6352 Antenna Engineering Spring...

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