For the below given input and output, the matrix A can be found out by pseudoinverse or mrdivision in MATLAB. Similarly, I would now like to know, how to determine A, if my output signal Y matrix contains additive zero mean, uncorrelated, Gaussian noise?
x1 = [1 1 1]';
x2 = [0 1 1]';
x3 = [0 0 1]';
x4 = [1 0 1]';
y1 = [1 2 0]';
y2 = [-1 0 3]';
y3 = [3 1 1]';
y4 = [5 3 -2]';
X = [x1 x2 x3 x4];
Y = [y1 y2 y3 y4];
A = Y/X
Also, I have modelled the unknown noisy output as below:
y1_n = y1 + sqrt(var(y1))*randn(size(y1));
y2_n = y2 + sqrt(var(y2))*randn(size(y2));
y3_n = y3 + sqrt(var(y3))*randn(size(y3));
y4_n = y4 + sqrt(var(y4))*randn(size(y4));
Y = [y1_n y2_n y3_n y4_n];
The statement A = Y/X solves the linear system of equations A*X = Y. If the system is overdetermined, as in your case, the solution given is the least squares solution. Thus, if you have additive, zero mean, uncorrelated, Gaussian noise on Y, then A = Y/X will give you the best possible, unbiased, estimate of A.
Note that the noise you add to your Y matrix is quite large, hence the estimate of A is far away from the ideal. If you add less noise, the estimate will be closer:
x1 = [1 1 1]';
x2 = [0 1 1]';
x3 = [0 0 1]';
x4 = [1 0 1]';
X = [x1 x2 x3 x4];
y1 = [1 2 0]';
y2 = [-1 0 3]';
y3 = [3 1 1]';
y4 = [5 3 -2]';
Y = [y1 y2 y3 y4];
for n = [1,0.1,0.01,0]
Y_n = Y + n*randn(size(Y));
A = Y_n/X;
fprintf('n = %f, A = \n',n)
disp(A)
end
Output:
n = 1.000000, A =
2.9728 -5.5407 2.8011
2.6563 -1.3166 0.6596
-3.3366 1.1349 1.5342
n = 0.100000, A =
2.0011 -4.0256 2.9402
1.9223 -1.0029 1.0921
-3.1383 1.9874 1.0913
n = 0.010000, A =
1.9903 -3.9912 2.9987
1.9941 -1.0001 1.0108
-3.0015 2.0001 1.0032
n = 0.000000, A =
2.0000 -4.0000 3.0000
2.0000 -1.0000 1.0000
-3.0000 2.0000 1.0000
Of course if you make X and Y larger by adding more vectors you'll get a better estimate too, and will be able to compensate more noisy data.