Matrix inversion and summation
Mostra commenti meno recenti
Hi all, can anyone help me to explain the mistake in my code below? The problem is as follow:
- The following parameter is assumed,
lma = 3;
ma = zeros(1,lma);
ma(2) = -0.2;
ma(lma) = -0.01;
- The parameter is then used to calculate the coordinate z using the following equation,

where z = x + i*y (complex number)
dumZ = 0;
sumdumZ = 0;
theta = linspace(0,2*pi,37);
xi = exp(1i.*theta);
for ii=1:length(theta)
for jj=1:lma
dumZ = ma(jj).*xi(ii).^jj;
sumdumZ = sumdumZ + dumZ;
end
zcon(ii) = 1./xi(ii)+sumdumZ;
dumZ = 0;
sumdumZ = 0;
end
zcon=zcon';
The following profile is obtained by plotting zcon,

- Now, using the hypothetical data above, i.e. the zcon, I want to back-calculate the variable "ma" or "alpha" as shown in the equation above. To do this, I first created a matrix A, which essentially a matrix containing all the constants. The following code is used to generate matrix A with 370 terms,
lx = length(zcon);
N = 10*37;
A = zeros(lx,N);
for ii=1:lx
for jj=1:N
A(ii,jj) = xi(ii).^jj;
end
end
A=[1./xi' A]; % insert a new colum to the left containing the 1/xi(ii) into the existing matrix A;
- Then the variable "ma" or "alpha" can be easily obtained using the following operation in Matlab,
maiter = A\zcon;
which give me a [370 x 1] matrix. To validate this, I again calculate the "predicted" z using the following code:
zdirect = A*maiter;
Which results an exactly the same curve (as expected) as shown below,

Now, coming to my question. Instead of using a direct calculation as shown above, i.e. the code "zdirect = A*maiter, instead I would like to manually use summation to calculate the "z" for each "theta" using the code below. Somehow, I could not get it right and cannot find the mistake. Below is the code that I used to calculate "z"
dumZ = 0;
sumdumZ = 0;
pow = [-1 1:1:N];
for ii=1:length(theta)
for jj=1:length(maiter)
dumZ = maiter(jj).*xi(ii).^pow(jj);
sumdumZ = sumdumZ + dumZ;
end
ziter(ii) = sumdumZ;
dumZ = 0;
sumdumZ = 0;
end
Instead, this is what I got,

Can someone please help me?
Risposta accettata
Più risposte (0)
Categorie
Scopri di più su Mathematics in Centro assistenza e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!