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Ee522 Notes (5) Antenna Array

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Ee522 Notes (5) Antenna Array EE522 NOTES (5): ANTENNA ARRAY Array: a set of antennas working together to produce certain radiation pattern. Each antenna in an array is called an element antenna (or simply an element). The elements in an array can be the same or different. In most practical cases, they are identical in construction (with different feedings). Configuration of arrays: Linear (1D), planar (2D), and conformal (3D). Array analysis: to obtain array factor (AF) given array configuration and element feedings. Array synthesis: To determine the array configuration and/or element feedings to achieve desired array factor. For array made up of identical elements, (Array pattern) = (Array factor) X (element pattern) Array analysis: d Consider an array of two identical elements. When the element is placed in the origin of the coordinate, the radiation is e− jrβ EeCI= ˆ F(,)θφ r e If the two elements are placed in new positions, one at (0,0,d/2), and the other in (0,0,-d/2), and excited by I1 , and I2 , respectively, then the far-field from the elements are: − β jr1 = e θφ EeCI11ˆ Fe (,) r1 − β jr2 = e θφ EeCI22ˆ Fe (,) r2 == Using far-field approximation: rrr12 for the amplitude factors, and dd rr=+cosθθ , rr =− cos 1222 for the phase factors. The combined radiation is dd −+jrβθ(cos) −− jr βθ (cos) ee22 E=+ eCIˆˆ F(,)θφ eCI F (,) θφ 12rree − jrβ =+e jd(cos)/2βθ− jd (cos)/2 βθ θφ eCˆ I12 e I e Fe (,) r Hence the radiation pattern is θφ=+jd(cos)/2βθ− jd (cos)/2 βθ θφ =×() θφ FIeIeF(,)12 ee (,) AFF (,) where the array factor (AF) is given by =+jd(cos)/2βθ− jd (cos)/2 βθ AF Ie12 Ie It depends on: (1) the relative location of the elements (2) the relative excitation of the elements. Example: if the two elements are half-wave dipole, they are placed on z-axis and oriented parallel to z-axis. Find the array factor (AF) and the array pattern for the following cases: == =λ (1) II121, d / 2 == =λ (2) IIjd121, , 0.25 Solution: For both cases, the element pattern is θπθθ= Fe ( ) cos[( / 2)cos ]/ sin 2/2πλ π (1) βd /2==, λ 22 jj(/2)cosπθ− (/2)cos πθ AF=+ e e =2cos[(πθ /2)cos ] The array pattern is F(θφ , )=× cos[( π / 2)cos θ ] cos[( π / 2)cos θ ]/ sin θ = cos2 [(πθθ / 2)cos ]/ sin (2) βπd /2= /4, AF=+ ejjjj(/4)cosπθππθ e /2(/4)cos− = e (/4) π2cos[] (πθπ / 4)cos − / 4 Array pattern is F ()θφ,=−× cos[] ( π / 4)cos θ π / 4 cos[( π / 2)cos θ ]/ sin θ Example 2: Two identical isotropic source Equal excitation (for both amplitude and phase) Separation: λ /2 Location: z-axis. z z d (1) Sketch the polar radiation pattern by inspection (2) Derive the exact array factor. Solution: (1) Since the excitations are the same, θπ==−/2or θ π /2 have maximum radiation. At θ = 0 , the phase difference from the source to the observation point is 180 (deg) out of phase because of half-wave distance. Hence at this angle, the radiation is 0. The same is true for θπ= . The radiation at other angles are between 0 and maximum. Hence we have the sketch as shown. (2) AF=+ ejdβ (/2)cosθ e− jdβ (/2)cosθ Since dd==λβ/2, /2 (2 πλλπ / )( /2) = /2 AF = cos[] (πθ / 2)cos As expected, the radiation is 0 at θπ= 0, , and is maximum for θπ=± /2. Example 3: Two identical isotropic source Equal Amplitude 1 -j 90 (deg) phase difference, z Separation: λ /2 d Location: z-axis. (1) Sketch the polar radiation pattern by inspection (2) AF=? Solution: (1) As shown in the figure. 1,-1 (1,-j) (-j,-j) 1 -j (1,j) -1,-j 1,-1 (1,-j) (-j,-j) (2) =+jdβθπβ(/2)cos−− j /2 j ( d /2)cos θ AF I12 e I e =+−−−+jjdjπβ/4 ( /2)cos θπβ /4 jdj ( /2)cos θπ /4 Ie0 e e − π = e j /42Id cos[] (βθπ / 2) cos− / 4 0 =−− jπ /4 πθπ eI20 cos[] ( / 2)cos / 4 The exact plot is shown on next page. 90 0.99999 120 60 0.74999 150 0.49999 30 0.25 180 0 210 330 240 300 270 • Array factor of linear arrays. For a linear array, the array factor can be obtained similarly as the two-element array case. Assuming an N-element array, the element excitations are: II01,,! , IN − 1. If the location of the elements are zz01,,! , zN − 1, then the n-th β θ jzn cos element contribution to AF is given by Ien , and the AF is the summation of these terms: βθβθ β θ =+jz01cosjz1 cos + jzN − cos AF I01 e I e! IN − 1 e Equal space case: == = =− zzdzdzNd010, , 2 2 ,! ,N − 1 ( 1) NN−−11 βθ βθα+ ==jndcos jnd(cos)n AF∑∑ Inn e|| I e nn==00 Uniformly excited, equal spaced linear array with linear phase progression: ==jjαα =2 = jn α IIeIeIe011, , 2 ,n ,! The array factor in this case have a closed form NN−−11jNψ βθα+ ψ1− e AF===∑∑ ejn(cos) d e jn − jψ nn==001 e Sxxxx=+1 +23 + +! +N − 1 Reference: xS=++++ x x23 x! xNN− 1 + x (1−=−⇒=−−xS ) 1 xNN S (1 x ) /(1 x ) Consider 1−=eeejNψψψ jN/2− jN /2 − ee jN ψψ /2 jN /2 =−eejNψψψ/2()− jN /2 e jN /2 = ejNjNψ /22sin(ψ /2) Ignoring the phase, the uniform excited (UE), equal spaced linear array (ESLA) has the array factor sin(Nψ / 2) AF = N sin(ψ / 2) , UE, ESLA Observations: (1) Main lobe is in the direction so that ψ =+=βθαd cos 0 (2) The main lobe narrows as N increases. (3) Number of side lobes is N-1 in one period of AF()ψ . (4) SLL decreases as N increases. (5) AF()ψ is symmetric about ψ = π . • Electriconic beam scanning The maximum radiation direction for UE-ESLA array is in θ βθα−= direction 0 such that d cos0 0 . For fixed spacing (d), the main beam direction is controlled by the phase progression parameter α . Thus beam scan is realized by varying α . Pattern parameters for UE-ESLA (for isotropic elements) Half-Power beam width HP: When Nd >> λ , λ θ 0.886 csc0 , Near Broadside Nd HP = λ 2 0.886 , Endfire Nd Directivity: Need numerical integration. Distance difference Phase difference Beam forming in an arbitrary direction is realized by compensation of the distance differences with the phase differences in the excitations. Look at the sin(Nx)/(Nsinx) plot, there are two main beams in general (one at x=0, and the other at x = 2)π . In many practical applications, a single pencil beam is desired. How to select array parameters so that the array produces a single pencil beam? Ordinary endfire array: πλ1 22βπdord≤− , ≤ 1 − NN22 Non-uniformly excited, equally spaced linear array (NE,ESLA) Recall the AF for equally spaced linear array: =+jdβθcos + j β ( n− 1) d cos θ AF I01 I e! IN − 1 e For linear phase progression, we define ==+jnα ψ βθα IAenn,cos d =+jjNψψ ++ (1)− AF A01 A e! AN − 1 e Denote Ze= jψ , then =+ +21 ++ N − AF A01 A Z A 2 Z! AN − 1 Z Binomial distribution: AF=+(1 Z ) N −1 NAFZ==+2: 1 NAFZZ==++3: 1 2 2 N==+++4: AFZZZ 1 3 3 23 NAFZZZZ==++++5: 1 4 6234 4 The amplitude distribution is symmetric. The AF for binomial distribution has no sidelobes. 90 40.0087 90 40.0087 120 60 120 60 150 20.0043 30 150 20.0043 30 180 0 180 0 210 330 210 330 240 300 240 300 270 270 90 40.0087 90 40.0087 120 60 120 60 150 20.0043 30 150 20.0043 30 180 0 180 0 210 330 210 330 240 300 240 300 270 270 Two dimensional array Example: Four isotropic elements are placed on the four corners of a rectangular region as shown below. Find the AR for this array (Assuming that the excitations are identical for the four elements). y d d d 0 d x Solution: ββˆˆ⋅⋅β ˆ⋅ β ˆ ⋅ =+jrr12 jrr ++jrr3 jrr 4 AF I12 e I e I 34 e I e rxˆˆ=++sinθφ cos y ˆ sin θφ sin zˆ cos θ =+ ⋅=θφ + φ rxdydrrd11ˆˆ,sin(cossin) ˆ =− + ⋅=θφφ − + rxdydrrd22ˆˆ,sin(cossin) ˆ =− − ⋅=θφφ − − rxdydrrd33ˆˆ,sin(cossin) ˆ =− ⋅=θφ − φ rxdydrrd44ˆˆ,sin(cossin) ˆ AF=+ ejdβθφφsin (cos+− sin ) e jd βθφφ sin (cos sin ) ++eejdβθsin (−− cos φφ sin ) jd βθ sin ( −+ cos φφ sin ) =+ededjdβθφsin cos2cos()βθφ sin sin− jd βθφ sin cos 2cos () βθφ sin sin jdβθφsin cos− jd βθφ sin cos =+2cos()βθφdee sin sin = 4cos()βθφd sin sin cos()βθφd sin cos Generally, 2D rectangular arrays have the AF given by MN βαˆ⋅+' θφ = jrr()mn mn AF(,) ∑∑ Imn e mn==11 ' Where rmn is the location of the (m,n) element. ''=++ '' rxxyyzzmnˆˆ mn mnˆ mn rxˆˆ=++sinθφ cos y ˆ sin θφ sin zˆ cos θ ββθφθφθ⋅=++'' ' ' rrˆ mn() x mnsin cos y mn sin sin z mn cos θφ Main bean pointing direction (,)00 is given by αβ=−''' θφ + θφ + θ mn()xyz mnsin00 cos mn sin 00 sin mn cos 0.
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