Adaptive System homework 1: Homework

MATLAB Problem 1.2 – Gradient Descent

Matlab Code:

clear all;

close all;

clc;

for theta = [pi, 7/8*pi, 6/8*pi, 4/8*pi, 1/8*pi];

n = 0:1:100;

N = 2;

phi = 0;

%h = [1;5];

%h = [7;9];

h = [3;8];

varianz_nu = 0.5;

c0_r = -10:0.1:15;

c1_r = -10:0.1:15;

nu = sqrt(varianz_nu)*randn(size(n));

x = cos(theta*n + phi);

figure;

plot(x(1:min([2*ceil(2*pi/theta),length(x)])),'bx-');

X = [x;[0, x(1:end-1)]];

d = h'*X + nu;

rxx = xcorr(x, N-1,'unbiased');

rxx = rxx(N:end);

Rxx = toeplitz(rxx)

cond(Rxx);

p = xcorr(d,x,N-1,'unbiased');

p = p(N:end)';

J = zeros(length(c1_r),length(c0_r));

for i = 1:length(c0_r)

for j = 1:length(c1_r)

c_cur = [c0_r(i); c1_r(j)];

J(j,i) = mean((d-c_cur'*X).^2);

end

figure;

contour(c0_r,c1_r, J, 20);

grid on;

xlabel('c0');

ylabel('c1');

axis equal;

axis([c0_r(1) c0_r(end) c1_r(1) c1_r(end)]);

steps = 30;

c = zeros(2,steps+1);

c(:,1) = [0; 0];

mu = 0.8;

hold on;

plot(h(1),h(2),'xk','MarkerSize',20,'LineWidth',2);

for i = 1:steps

c(:,i+1) = c(:,i) + mu*(p-Rxx*c(:,i));

plot([c(1,i),c(1,i+1)],[c(2,i),c(2,i+1)],'ro-');

text(c(1,i+1), c(2,i+1),sprintf(' %d',i));

pause(0.1);

end

c_end = c(:,end);

end

Theta = л:

Theta = 7/8 * л:

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Theta = 6/8 * л:

Theta = 4/8 * л:

Theta = 1/8 * л:

Diskussion:

Bei der Frequenz л/2 wird Rxx zu einer Diagonalmatrix. Erhöht bzw. erniedrigt man ausgehend von л/2 die Frequenz wird Rxx immer „linear unabhängiger“.

Tap- input Vector „scant“ nicht mehr die ganze Ebene ( persistently exciting).

MATLAB Problem 1.3 – Least-Squares Tracking of a Time Varying System

Matlab Code:

clear all;

close all;

clc;

% a)

n = 0:999;

h = [ 1; -1; 1];

x = sin(2/16*pi*n) + cos(pi/4*n)+ sin(3/16*pi*n) - cos(10/16*pi*n);

d = conv(h,x);

d = d(1:end-2) + randn(size(n));

c = ls_filter(x, d, 3);

% b)

n = 0:999;

N = 3;

x = randn(size(n));

h = [ ones(size(n))

-1 + 0.002*n

+1 - 0.002*n ];

X = toeplitz([x(1); 0; 0], x);

d = zeros(size(n));

for i = 1:length(n)

d(i) = h(:,i)'*X(:,i);

end

figure;

waterfall(h);

for lambda = [0.1, 1.0]

for M = [10,50]

x_p = [zeros(1, M-1), x];

d_p = [zeros(1, M-1), d];

c = zeros(N, length(n));

for k = M:length(x_p)

x_cur = x_p((k-M+1):k);

d_cur = d_p((k-M+1):k);

c(:,k-M+1) = ls_filter(x_cur, d_cur, N, lambda);

end

figure;

plot(n, h(1,:), '-r'); hold on;

plot(n, c(1,:), '-b');

figure;

plot(n, h(2,:), '-r'); hold on;

plot(n, c(2,:), '-b');

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plot(n, h(3,:), '-r'); hold on;

plot(n, c(3,:), '-b');

figure;

waterfall(c);

end

function [ c ] = ls_filter( x, d, N, lambda )

% x ... input signal

% d ... desired output signal (of same length as x)

% N ... number of filter coeffcients

% lambda ... optional "forgetting factor" 0<lambda<=1 (default =1)

if(nargin < 4)

lambda = 1;

end

n0 = length(x)-1;

col_1 = x';

row_1 = [x(1), zeros(1,N-1)];

X = toeplitz(col_1, row_1);

w = lambda.^(n0:-1:0);

W = diag(w);

c = (W.^(1/2)*X)\(W.^(1/2)*d');

Lambda = 0.1

M = 10

Lambda = 0.1

M = 50

Lambda = 1.0

M = 10

Lambda = 1.0

M = 50

Diskussion:

Man sieht dass hier das „Vergessen“ notwendig ist da sich das System anscheinend relativ stark ändert.

MATLAB Problem 1.4 – Bonus Problem

Matlab Code:

clear all;

close all;

clc;

h = [1; 2];

N = length(h);

n = 0:100;

x1 = cos(0.5*pi*n);

x2 = cos(pi*n);

x3 = cos(pi*n)+2;

x4 = randn(size(n));

Signals = [x1;x2;x3;x4];

for i = 1:4

x = Signals(i,:);

X = toeplitz([x(1); 0], x)

d = h' * X;

steps = 10;

c = zeros(N, steps+1);

c(:,1) = [-1 -2];

for j = 1:steps

y = c(:,j)'*X(:,j);

c(:,j+1) = c(:,j) + 1/(X(:,j)'*X(:,j))*(d(j)-y)*X(:,j);

end

figure();

plot(h(1), h(2), 'xk', 'MarkerSize', 10, 'LineWidth', 2); hold on;

axis([-10 10 -10 10]);

plot(c(1,:), c(2,:), '-or');

end

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Cos(0.5*л*n)

Cos(л*n)

Cos(л*n)+2

Randn:

Diskussion:

Bei allem Signalen bis auf cos(л*n) kann das unbekannte System identifiziert werden. Bei diesem Signal liegt wieder eine lineare Abhängigkeit vor.

X = %für cos(л*n)

Columns 1 through 10

1 -1 1 -1 1 -1 1 -1 1 -1

0 1 -1 1 -1 1 -1 1 -1 1

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Adaptive System homework 1 .doc