AASP HW 7

JASMEET SINGH

MATRIKEL - 53046

Question 1

Part (a)

%% Function to implement the standard ESPRIT algorithm %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function mu = ESPRIT(X,d,M)

%% Signal subspace estimation

[U,S,V]= svd(X);

Us = U(:,1:d);

%% Selection matrices for the subarrays with maximum overlap

J1 = [eye(M-1) zeros(M-1,1)];

%% Exchange matrix

exch_mat = @(i)fliplr(eye(i));

J2 = exch_mat(M-1)*J1*exch_mat(M);

%% Solving the invariance equation

epsi = pinv(J1*Us)*(J2*Us);

%% Estimation of the values of mu and sort

[U_epsi,D_epsi] = eig(epsi);

phi = diag(D_epsi);

phi = phi(1:d);

mu = sort(imag(log(phi)), 'ascend');

Part (b)

%% Checking the estimation accuracy of the ESPRIT function defined in Part (a) clear all;

clc all;

M = 12;

N = 100;

d = 4;

rho = 0;

mu = [-0.5; -0.3; 0.1; 0.2];

SNR1 = 10; % initial value of SNR in dB

SNR2 = 50; % increased value of SNR in dB

SNR3 = 100; % further increased value of SNR in dB

X1 = GetArrayOutput(M,mu,SNR1,N,rho); % measurement matrix for SNRl

X2 = GetArrayOutput(M,mu,SNR2,N,rho); % measurement matrix for SNR2

X3 = GetArrayOutput(M,mu,SNR3,N,rho); % measurement matrix for SNR3

mu1 = ESPRIT(X1,d,M); % estimate the spatial frequencies

mu2 = ESPRIT(X2,d,M); % using the function ESPRIT for SNR1, SNR2 and SNR3

mu3 = ESPRIT(X3,d,M);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Following are the outputs the estimates of the spatial frequency for different values of SNR.

mu1 = mu2 = mu3 =

-0.4772 -0.4997 0.5000

-0.2669 -0.3002 -0.3000

0.1687 0.0997 0.1000

0.5342 0.2008 0.2000

Comments:

From the above results, we can see that, as the value of SNR increases the estimate improves and

reaches the actual values when SNR tends to infinity (very high value). So it is confirmed that

the ESPRIT function is working correctly as expected.

Part (c) & Part (d)

%% Computation of squared estimation error for the ESPRIT function for different SNRs, with

%% and without FBA %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all;

close all;

clc;

rho = [0 0.9 0.95 0.99];

for k = 1:length(rho)

M = 12; N = 100; d = 4; % use the same parameters as in task 2

mu = [-0.5; -0.3; 0.1; 0.2];

N_runs = 100; % number of times the estimation process is run

SNR = linspace(0,50,20);

muest = zeros(length(mu),length(SNR));

muest_FBA = zeros(length(mu),length(SNR));

e_est = zeros(length(SNR),1); % estimation error

N_e = zeros(length(SNR),N_runs); % number of errors

N_e_FBA = zeros(length(SNR),N_runs); % Forward-backward averaging number of erros

for i = 1:length(SNR)

for j = 1:N_runs

X = GetArrayOutput(M,mu,SNR(i),N,rho(k)); % measurement matrix for different

%% Values of SNR

muest(:,i) = ESPRIT(X,d,M); % estimate the spatial frequencies

e_est(i) = norm(mu(:)-muest(:,i))^2;

N_e(i,j) = norm(mu(:)-muest(:,i))^2;

%% Forward-bakward averaging

exch_mat = @(i)fliplr(eye(i)); % exchange matrix

Z = [X,exch_mat(M)*conj(X)*exch_mat(N)];

muest_FBA(:,i) = ESPRIT(Z,d,M);

N_e_FBA(i,j) = norm(mu(:)-muest_FBA(:,i))^2;

end

end

%% plot Figures

figure(1000+k)

semilogy(SNR,e_est,'linewidth',1.65)

grid on

xlabel('SNR (dB)');

ylabel('Estimation error');

title(['Squared estimation error for the ESPRIT algorithm with' ' ',num2str(rho(k)),' ' 'correlation']);

figure(2000+k)

semilogy(SNR,mean(N_e,2),'linewidth',1.65)

hold on

semilogy(SNR,mean(N_e_FBA,2),'.-g','linewidth',1.65)

grid on

xlabel('SNR (dB)');

ylabel('Average squared estimation error');

title(['Average squared estimation error for for the ESPRIT algorithm with ' ' ',num2str(rho(k)),' ' 'correlation']);

legend('Without FBA','With FBA','Location','SouthEast');

end

To lower the estimation error by one decade (factor 10), how much does the SNR have to

be increased?

From the figure we can say that, SNR should be increased to about 8 dB in order to lower the

estimation error by one decade (factor of 10).

Question 2

Part (d) close all;

clear all;

theta1=30;

phy1=180;

theta2=45;

phy2=45;

u1=pi*sin(theta1);

u2=pi*sin(phy1);

v1=pi*sin(theta2);

v2=pi*sin(phy2);

A= [1 1;

exp(-1i*u1) exp(-1i*v1);

exp(-1i*2*u1) exp(-1i*2*v1);

exp(-1i*(u2+0)) exp(-1i*(v2+0));

exp(-1i*(u2+u1)) exp(-1i*(v2+v1)) ;

exp(-1i*(u2+2*u1)) exp(-1i*(v2+2*v1));

exp(-1i*(2*u2+0)) exp(-1i*(2*v2+ 0));

exp(-1i*(2*u2+u1)) exp(-1i*(2*v2+ v1));

exp(-1i*(2*u2+2*u1)) exp(-1i*(2*v2+ 2*v1))];

Q_9= (1/sqrt(2))* [1 0 0 0 0 1i 0 0 0; 0 1 0 0 0 0 1i 0 0; 0 0 1 0 0 0 0 1i 0; 0 0 0 1 0 0 0 0 1i; 0 0 0 0 sqrt(2) 0 0 0 0 ;

0 0 0 1 0 0 0 0 -1i;

0 0 1 0 0 0 0 -1i 0;

0 1 0 0 0 0 -1i 0 0;

1 0 0 0 0 -1i 0 0 0];

Rss=eye(2);

sigma_n=1;

Rxx = A*Rss*A' + (sigma_n)^2* eye(9);

Rxx_fb=(1/2)* ( Rxx + fliplr(eye(9))*conj(Rxx)*fliplr(eye(9)) );

psy_Rxx_fb= Q_9'*Rxx_fb*Q_9;

U=psy_Rxx_fb;

Us=U(:,1:2) ; % dominant eigen vectors

Output:

Part (e) close all;

clear all;

Q_6= (1/sqrt(2))* [1 0 0 1i 0 0; 0 1 0 0 1i 0; 0 0 1 0 0 1i; 0 0 1 0 0 -1i; 0 1 0 0 -1i 0; 1 0 0 -1i 0 0]; J_u2= [0 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0 0; 0 0 0 0 0 1 0 0 0; 0 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 1]; J_v2= [1 0 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0 0; 0 0 0 0 1 0 0 0 0; 0 0 0 0 0 0 1 0 0; 0 0 0 0 0 0 0 1 0]; k1 = Q_6'*J_u2*Q_9; k_u1= 2*real(k1); k_u2=2*imag(k1); k2= Q_6'*J_v2*Q_9; k_v1= 2*real(k2); k_v2=2*imag(k2);

Output:

Part (f)

Real Valued Invariance Equations are given as:

T_u_tls = (k_u1*Es)'*k_u2*Es; T_v_tls = (k_v1*Es)'*k_v2*Es;

Part (g)

Es= Q_9'*Us; T_u_tls = (k_u1*Es)'*k_u2*Es; T_v_tls = (k_v1*Es)'*k_v2*Es;

Output:

Part (h) close all;

clear all;

eig_val = eig(T_u_tls +1i*T_v_tls); for i=1:2 u(i)=2*atan(real(eig_val(i))) end; for i=1:2 v(i)=2*atan(imag(eig_val(i))) end;

Output:

u = 1.8028 2.9910

v = 1.8014 3.0992