Problem (a) θ 0 = 0

Transcription

1 Problem (a) θ 0 = 0 Beam Pattern: 3

2 For circular symmetric beam pattern, the weights a nm are given by ( ψ R J 1 ψ R n ) + m a(n, m) = π n + m In order to be consistent with Problem 4.1.5, we take ψ R to be BW NN = 0.497π, where BW NN is the null-to-null beam width in Problem Assume the array has a circular boundary and we want to apply a Kaiser window with β = 6 ) ] I 0 [β W kaiser (n, m) = 1 ( n +m I 0 (β) N, N 1 n, m N 1 where N=11. So the final weights for the array are { a(n, m) wkaiser (n, m), when n w(n, m) = + m N 0, otherwise (circular boundary) The beam pattern can be computed by B(ψ x, ψ y ) = N 1 n= N 1 The plot of B(ψ x, ψ y ) is shown in Figure 3. Directivity: For normalized beam pattern, N 1 m= N 1 D = [W H BW ] 1 w (n, m) e j(nψ x+mψ y ) where matrix B is exactly the same as we computed in Problem By plugging in B and W, we get D = HPBW: In this problem, we can NOT use equation (4..7) in the text to calculate θ H and ψ H, because we don t have any knowledge about the HPBW θx 0 and θy 0 corresponding to a(n, m) W kaiser (n, m). So we estimate HPBW ψ x and ψ y from simulated data. ψ x = ψ y =.03 u x = u y = (b) θ 0 = 45, ϕ 0 = 30 Beam Pattern: The new weight is W (n, m) = W (n, m) e j(nψx 0+mψy 0 ) 4

3 where ψ x0 = sin θ 0 cos ϕ 0, and ψ y0 = π sin θ 0 sin ϕ 0. The beam pattern is B (ψ x, ψ y ) = N 1 n= N 1 The plot of B (ψ x, ψ y ) is shown in Figure 4. Directivity: N 1 m= N 1 D = [W H BW ] 1 = W (n, m) e j(nψx+mψy) HPBW: ψ x = ψ y =.03 u x = u y = 0.646

4 B Matlab Code for Prob % Matlab code for problem in the text close all clear all % % % (a) theta0 = 0 % % N = 11; M = 11; theta0 = 0; psix0 = 0; psiy0 = 0; psir = 0.497*pi; psix = pi*(-1:1/100:1); psiy = pi*(-1:1/100:1); beta = 6; w = zeros(m,n); BXY = zeros(length(psiy), length(psix)); for n = -(N-1)/:(N-1)/ for m = -(M-1)/:(M-1)/ t = sqrt(n^ + m^); if (t<=(n-1)/) & (t>0) a_nm = psir*besselj(1,psir*t)//pi/t; wkaiser = besseli(0,beta*sqrt(1-(t/5.5)^))/besseli(0,beta); wnow = a_nm * wkaiser; w(m+(m-1)/+1,n+(n-1)/+1) = wnow; identicalvector = ones(1,length(psix)); V = exp(j*(n*psix. *identicalvector + m*identicalvector. *psiy)); BXY = BXY + conj(wnow) * V; elseif (t==0) wnow = pi*psir^/(4*pi^); w(m+(m-1)/+1,n+(n-1)/+1) = wnow; identicalvector = ones(1,length(psix)); v = exp(j*(n*psiy. *identicalvector + m*identicalvector. *psix)); BXY = BXY + conj(wnow) * v; end end end BXY = BXY. /max(max(abs(bxy))); figure(1) subplot(,,1); [x,y] = meshgrid(psix,psiy); meshc(x,y,0*log10(abs(bxy))); xlabel( \psi_x ); ylabel( \psi_y ); zlabel( B(\psi_x,\psi_y) (db) ); title( Circular symmetric beam pattern, when \theta_0=90^o ) % compute directivity B = zeros(n*m,n*m); for ii=1:n*m for jj=1:n*m deltapy = floor(ii/n) - floor(jj/n); deltapx = mod(ii,m)-mod(jj,m); rho = sqrt(deltapx^ + deltapy^); B(ii,jj) = sinc(rho); 7

5 end end % finish constructing matrix B n = -(N-1)/:(N-1)/; m = -(M-1)/:(M-1)/; identicalvector = ones(1,n); v0 = exp(j*(n. *psiy0*identicalvector + identicalvector. *m*psix0)); v0 = v0(:); w = w(:); disp( Directivity of broadside array: ) D = abs(w *v0)^/(w *B*w) % plot the contour of HPBW figure(1) subplot(,,) [x,y] = meshgrid(psix,psiy); contour(x,y,0*log10(abs(bxy)), [-3-3]); grid xlabel( \psi_x ); ylabel( \psi_y ); title( Contour plot of HPBW, when \theta_0=90^o ) figure(1) subplot(,,3) plot(psix, 0*log10( abs(bxy( (size(bxy,1)+1) /,:)) )) xlabel( \psi_r ); ylabel( B(\psi_x, \psi_y) (db) ) title( Cut of B(\psi_x,\psi_y), when \phi = 0^o ) figure(1) subplot(,,4) plot(psiy, 0*log10(abs(BXY(:, (size(bxy,1)+1) /)) )) xlabel( \psi_r ); ylabel( B(\psi_x, \psi_y) (db) ) title( Cut of B(\psi_x,\psi_y), when \phi = 90^o ) % % % (b) theta0 = 45 degrees, phi0 = 30 degrees % % N = 11; M = 11; theta0 = pi/4; phi0 = pi/6; psix0 = pi*sin(theta0)*cos(phi0); psiy0 = pi*sin(theta0)*sin(phi0); psir = 0.497*pi; psix = pi*(-1:1/100:1); psiy = pi*(-1:1/100:1); beta = 6; w = zeros(m,n); BXY = zeros(length(psiy), length(psix)); for n = -(N-1)/:(N-1)/ for m = -(M-1)/:(M-1)/ t = sqrt(n^ + m^); if (t<=(n-1)/) & (t>0) a_nm = psir*besselj(1,psir*t)//pi/t; wkaiser = besseli(0,beta*sqrt(1-(t/5.5)^))/besseli(0,beta); wnow = a_nm * wkaiser * exp(j*(n*psix0 + m*psiy0)); w(m+(m-1)/+1,n+(n-1)/+1) = wnow; identicalvector = ones(1,length(psix)); V = exp(j*(n*psix. *identicalvector + m*identicalvector. *psiy)); BXY = BXY + conj(wnow) * V; elseif (t==0) wnow = pi*psir^/(4*pi^) * exp(j*(n*psix0 + m*psiy0)); w(m+(m-1)/+1,n+(n-1)/+1) = wnow; identicalvector = ones(1,length(psix)); v = exp(j*(n*psiy. *identicalvector + m*identicalvector. *psix)); BXY = BXY + conj(wnow) * v; end end end BXY = BXY. /max(max(abs(bxy))); figure() subplot(,,1); [x,y] = meshgrid(psix,psiy); meshc(x,y,0*log10(abs(bxy))); xlabel( \psi_x ); ylabel( \psi_y ); zlabel( B(\psi_x,\psi_y) (db) ); title( Circular symmetric beam pattern, when \theta_0=45^o \phi_0=30^o ) % compute directivity B = zeros(n*m,n*m); for ii=1:n*m 8

6 for jj=1:n*m deltapy = floor(ii/n) - floor(jj/n); deltapx = mod(ii,m)-mod(jj,m); rho = sqrt(deltapx^ + deltapy^); B(ii,jj) = sinc(rho); end end % finish constructing matrix B n = -(N-1)/:(N-1)/; m = -(M-1)/:(M-1)/; identicalvector = ones(1,n); v0 = exp(j*(n. *psiy0*identicalvector + identicalvector. *m*psix0)); v0 = v0(:); w = w(:); disp( Directivity when theta0=45 phi0=30: ) D = abs(w *v0)^/(w *B*w) % plot the contour of HPBW figure() subplot(,,) [x,y] = meshgrid(psix,psiy); contour(x,y,0*log10(abs(bxy)), [-3-3]); grid xlabel( \psi_x ); ylabel( \psi_y ); title( Contour plot of HPBW, when \theta_0=45^o \phi_0=30^o ) figure() subplot(,,3) plot(psix, 0*log10( abs(bxy( (size(bxy,1)+1) /,:)) )) xlabel( \psi_r ); ylabel( B(\psi_x, \psi_y) (db) ) title( Cut of B(\psi_x,\psi_y), when \phi = 0^o ) lengthpsix = length(psix); n = -(lengthpsix-1)/:(lengthpsix-1)/; ypos = round(n*tan(phi0)) + (lengthpsix-1)/ + 1; BXY_Cut = []; for ii=1:length(ypos) BXY_Cut = [BXY_Cut, BXY(yPos(ii), n(ii) + (lengthpsix-1)/ + 1) ]; end figure() subplot(,,4) plot(psix/cos(phi0), 0*log10(abs(BXY_Cut) )) xlabel( \psi_r ); ylabel( B(\psi_x, \psi_y) (db) ) title( Cut of B(\psi_x,\psi_y), when \phi = 30^o ) 9

7 Figure 3: Beam pattern plots for Problem (a) 1

8 Figure 4: Beam pattern plots for Problem (b) 13

9 Problem 4..1 Consider an 8-element circular array with equal spacing between elements. It could also be viewed as two 4-element rectangular arrays with the first array oriented along the x-y axes and the second array rotated by 45. (a)using this model, find the beam pattern and compare your result to the result in the text. The distance between two elements on the same axis is d. Then, the elements for one rectangular array are at (d/, 0, 0), ( d/, 0, 0), (0, d/, 0) and (0, d/, 0). The weights are 1/8 each. So the beampattern is given by [e jk 0 d sin θ cos φ + e jk 0 d sin θ cos φ + e jk 0 d sin θ sin φ + e jk 0 d sin θ sin φ] where k 0 = π λ. B R1 (θ, φ) = 1 8 = 1 ( 4 cos d k 0 sin θ cos φ ) + 14 sin ( ) d k 0 sin θ sin φ Similarly, for the other array, rotated by 45 around the x-y axis, the beam pattern is given by B R (θ, φ) = B R1 (θ, φ π 4 ) = 1 ( 4 cos d k 0 (φ sin θ cos π ) ) + 1 ( 4 4 sin d k 0 (φ sin θ sin π ) ) 4 Thus, the beam patern is given by B(θ, ϕ) = (B R1 (θ, φ) + B R (θ, φ)) = 1 [ ( ) d cos k 0 sin θ cos φ 4 ( d + cos k 0 (φ sin θ cos π 4 ( d + sin k 0 ) sin θ sin φ ) ) + sin ( k 0 d sin θ sin (φ π 4 The beam pattern of circular array with N = 8 is given by (eq ) B(θ, φ) = N 1 m=0 1 B m (θ, φ) ) )] (1)

10 where B m (θ, φ) = w m e jkr sin θ cos(φ φ m)+jβ n β n = kr sin θ 0 cos(φ φ n ) Setting w m = 1/N = 1/8, β n = 0 since θ 0 = 0, φ m = πm N, and R = d/, the expression for beam pattern is and B(θ, φ) = Examining the first term: m=0 B m (θ, φ) = 1 8 ejk 0 d 7 B m (θ, φ) = n=0 sin θ cos(φ πm 8 ) 3 B m (θ, φ) + m=0 3 B m+1 (θ, φ) 3 B m (θ, φ) = 1 3 e jk 0 d sin θ cos(φ π m) 8 m=0 = 1 [e jk 0 d sin θ cos φ + e jk 0 d sin θ cos(φ π L ) + e jk 0 d sin θ cos(φ π) + e jk 0 d sin θ cos(φ 3π )] 8 = 1 [e jk 0 d sin θ cos φ + e jk 0 d sin θ sin φ + e jk 0 d sin θ cos φ + e jk 0 d sin θ sin φ] 8 = 1 [ ] d cos(k 0 4 sin θ cos φ) + cos(k d 0 sin θ sin φ) = B R1 (θ, φ) m=0 Similarly, 3 B m+1 (θ, φ) = B R (θ, φ) m=0 B(θ, φ) = B R1 (θ, φ) + B R (θ, φ) The same as Equation 1. (b) Assume d = λ in the rectangular arrays. Plot the resulting beam pattern. Substituting d = λ in Equation 1, we have B(θ, ϕ) = 1 4 [ ( ( cos(π sin θ cos φ) + sin(π sin θ sin φ) + cos π sin θ cos φ π )) ( ( + sin π sin θ sin φ π ))] 4 4 We plot the beam pattern cuts of this function in Figure 1. The matlab code is attached in Appendix A. We have normalized the beam pattern thus the peak value is 0 db, not near 10 db.

11 5 Circular Array Beam pattern Cut Beam pattern (db) φ = 0 45 φ = 45 φ= u r Figure 1: Beam pattern cuts for a uniformly weighted circular array (N=8) for different values of φ. A Matlab Code for Prob % File: t41b.m % Created: Wed 11/03/04 03:13: % Last Updated: Thu 11/04/04 03:13: % Description: ECE 51D - Fa04 - HW5-4..1b van Trees % : Beam Pattern Cuts - Circular Array % : N=8, d = lambda %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % We have developed upon the files provided by Prof. Bell (GMU) at % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % close all clear all N = 8; ur = -1:1/1000:1; theta = pi/ * ur; sintheta = sin(theta); PHI_SET = [0 pi/4 pi/]; B = zeros(length(phi_set), length(ur)); for k=1:length(phi_set) phi = PHI_SET(k); B(k,:) = B(k,:) + cos(pi * cos(phi) * sintheta); B(k,:) = B(k,:) + sin(pi * sin(phi) * sintheta); B(k,:) = B(k,:) + cos(pi * cos(phi-pi/4) * sintheta); B(k,:) = B(k,:) + sin(pi * sin(phi-pi/4) * sintheta); end B = B./max(max(abs(B))); figure; hold off plot(ur, 10 * log10(abs(b(1,:))), b- ); 3

12 hold on plot(ur, 10 * log10(abs(b(,:))), g-- ); plot(ur, 10 * log10(abs(b(3,:))), r-. ); axis([ ]); xlabel( \it u_r, Fontsize,14); ylabel( Beam pattern (db), Fontsize,14); title( Circular Array Beam pattern Cut ); h=legend( \phi = 0^\circ, \phi = 45^\circ, \phi=90^\circ, 3); set(h, Fontsize,1) 4

13 Problem - MPDR ˆ Recall that we had a 3-element ULA with six different plane waves impinging on the array. The snapshot vector: yrns 6 l1 s l rnsvpθ l q zrns, n 1,,..., M (1) where the angle of arrivals are θ 1 45 o, θ 60 o, θ 3 60 o, θ 4 90 o, θ o, and θ o. s 1 6rns are given as white noise sequences with variances as 0.5, 0.1, 0.1, 0.1, 0.15, and 0.1, respectively. zrns is a temporally and spatially zero mean white random vector with unit variance. ˆ We also have two different values of 5 o and 15 o. ˆ In this problem, we wish to plot the power spectrum of the MPDR beamformer which is given by the following equation: P pθq 1 v H pθqs x 1 vpθq where 0 θ π. Here S x is the (sample) covariance matrix of yrns and is given as: S x 1 N M n1 () yrnsy H rns (3) ˆ We plot the power spectrum P pθq for the two different values of. ˆ Figures 1 and show the required spatial spectra. 0.4 = 5 o P( ) o 45 o 60 o 60 o + 90 o 110 o 150 o 180 o Figure 1: Spatial spectrum with MPDR beamformer in a 3-element ULA with 5 o 1

14 0.35 = 15 o P( ) o 45 o 60 o 60 o + 90 o 110 o 150 o 180 o Figure : Spatial spectrum with MPDR beamformer in a 3-element ULA with 15 o The MATLAB code is given below: 1 % ECE 51D (Spring 017) HW #4 Problem 3 % Author: Govind Ravikumar Gopal (A ) 3 4 clear all; 5 close all; 6 clc; 7 8 % Initializations 9 N = 3; 10 M = 100; 11 L = 6; 1 13 l iter = (0:N 1).'; 14 y1 = zeros(n,m); 15 y = zeros(n,m); 16 delta = [5 15]; % in degrees v1 = exp(j*pi* (l iter (N 1)/) * cos(pi*45/180)); 19 v = exp(j*pi* (l iter (N 1)/) * cos(pi*60/180)); 0 v31 = exp(j*pi* (l iter (N 1)/) * cos(pi*(60+delta(1))/180)); 1 v3 = exp(j*pi* (l iter (N 1)/) * cos(pi*(60+delta())/180)); v4 = exp(j*pi* (l iter (N 1)/) * cos(pi*90/180)); 3 v5 = exp(j*pi* (l iter (N 1)/) * cos(pi*110/180)); 4 v6 = exp(j*pi* (l iter (N 1)/) * cos(pi*150/180)); 5 6 for n = 1:M 7 8 s1 = sqrt(0.5)*randn(1,m); 9 s = sqrt(0.1)*randn(1,m); 30 s3 = sqrt(0.1)*randn(1,m); 31 s4 = sqrt(0.)*randn(1,m); 3 s5 = sqrt(0.15)*randn(1,m); 33 s6 = sqrt(0.1)*randn(1,m); 34 z = randn(n,m); % random noise

15 35 y1(:,n) = v1.*s1(n) + v.*s(n) + v31.*s3(n) + v4.*s4(n) + v5.*s5(n) + v6.*s6(n) + z(:,n); 36 y(:,n) = v1.*s1(n) + v.*s(n) + v3.*s3(n) + v4.*s4(n) + v5.*s5(n) + v6.*s6(n) + z(:,n); 37 end % delta = 5 deg % calculate Sx 4 Sx = 0; 43 for ii = 1:M 44 Sx = Sx + y1(:,ii)*y1(:,ii)'; 45 end 46 Sx = Sx/M; theta = 0:0.001:pi; % array manifold 51 v theta = exp(j*pi * (l iter (N 1)/) * cos(theta)); 5 53 % spectrum 54 P = zeros(length(theta),1); 55 for ii = 1:length(theta) 56 P(ii) = 1./(v theta(:,ii)'*inv(sx)*v theta(:,ii)); 57 end % plot 60 figure; 61 plot(theta,abs(p),'linewidth',1.5); 6 xticks([0 pi*45/180 pi*60/180 pi*(60+delta(1))/180 pi*90/180 pi*110/180 pi*150/180 pi]); 63 xticklabels({'0ˆ{o}','45ˆ{o}','60ˆ{o}',' 60ˆ{o}+\Delta','90ˆ{o}','110ˆ{o}','150ˆ{o}','180ˆ 64 xlim([0 pi]); 65 xlabel('\theta'); 66 ylabel('p(\theta)'); 67 title('\delta = 5ˆ{o}'); 68 grid on; % delta = 15 deg 71 7 Sx = 0; 73 for ii = 1:M 74 Sx = Sx + y(:,ii)*y(:,ii)'; 75 end 76 Sx = Sx/M; P = zeros(length(theta),1); 79 for ii = 1:length(theta) 80 P(ii) = 1./(v theta(:,ii)'*inv(sx)*v theta(:,ii)); 81 end 8 83 figure; 84 plot(theta,abs(p),'linewidth',1.5); 85 xticks([0 pi*45/180 pi*60/180 pi*(60+delta())/180 pi*90/180 pi*110/180 pi*150/180 pi]); 86 xticklabels({'0ˆ{o}','45ˆ{o}','60ˆ{o}','60ˆ{o}+\delta','90ˆ{o}','110ˆ{o}','150ˆ{o}','180ˆ{o}'}); 87 xlim([0 pi]); 88 xlabel('\theta'); 89 ylabel('p(\theta)'); 90 title('\delta = 15ˆ{o}'); 91 grid on; 3