Matrix inversion and summation

Question

BeeTiaw 2018-6-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408078-matrix-inversion-and-summation

评论： BeeTiaw 2018-6-30

采纳的回答： David Goodmanson

在 MATLAB Online 中打开

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?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

David Goodmanson 2018-6-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408078-matrix-inversion-and-summation#answer_326895

Hi Bee,

Are you plotting the right variable? I ran your code, and zcon, zdirect and ziter all reproduce the same curve. Only the simple exponential variable xi gives the red circle in the second plot.

4 个评论
显示 2更早的评论隐藏 2更早的评论

BeeTiaw 2018-6-30

在 MATLAB Online 中打开

Hi @David Goodmanson,

Thanks for your reply. I am sure I plotted the correct one. Below is the numeric output of the three parameters that got using the same parameters.

One thing that I notice is that the real part of all the z is the same. However, the imaginary part of ziter is different with zcon and zdirect.

     zcon                  zdirect              ziter
0.7900 + 0.0000i   0.7900 - 0.0000i   0.7900 + 0.0000i
 0.7882 + 0.2471i   0.7882 + 0.2471i   0.7882 + 0.1002i
 0.7815 + 0.4792i   0.7815 + 0.4792i   0.7815 + 0.2048i
 0.7660 + 0.6832i   0.7660 + 0.6832i   0.7660 + 0.3168i
 0.7363 + 0.8484i   0.7363 + 0.8484i   0.7363 + 0.4372i
 0.6862 + 0.9680i   0.6862 + 0.9680i   0.6862 + 0.5641i
 0.6100 + 1.0392i   0.6100 + 1.0392i   0.6100 + 0.6928i
 0.5039 + 1.0633i   0.5039 + 1.0633i   0.5039 + 0.8161i
 0.3666 + 1.0446i   0.3666 + 1.0446i   0.3666 + 0.9251i
 0.2000 + 0.9900i   0.2000 + 0.9900i   0.2000 + 1.0100i
 0.0093 + 0.9077i   0.0093 + 0.9077i   0.0093 + 1.0619i
-0.1975 + 0.8061i  -0.1975 + 0.8061i  -0.1975 + 1.0733i
-0.4100 + 0.6928i  -0.4100 + 0.6928i  -0.4100 + 1.0392i
-0.6167 + 0.5741i  -0.6167 + 0.5741i  -0.6167 + 0.9580i
-0.8058 + 0.4545i  -0.8058 + 0.4545i  -0.8058 + 0.8311i
-0.9660 + 0.3368i  -0.9660 + 0.3368i  -0.9660 + 0.6632i
-1.0879 + 0.2221i  -1.0879 + 0.2221i  -1.0879 + 0.4619i
-1.1641 + 0.1102i  -1.1641 + 0.1102i  -1.1641 + 0.2371i
-1.1900 + 0.0000i  -1.1900 + 0.0000i  -1.1900 + 0.0000i
-1.1641 - 0.1102i  -1.1641 - 0.1102i  -1.1641 - 0.2371i
-1.0879 - 0.2221i  -1.0879 - 0.2221i  -1.0879 - 0.4619i
-0.9660 - 0.3368i  -0.9660 - 0.3368i  -0.9660 - 0.6632i
-0.8058 - 0.4545i  -0.8058 - 0.4545i  -0.8058 - 0.8311i
-0.6167 - 0.5741i  -0.6167 - 0.5741i  -0.6167 - 0.9580i
-0.4100 - 0.6928i  -0.4100 - 0.6928i  -0.4100 - 1.0392i
-0.1975 - 0.8061i  -0.1975 - 0.8061i  -0.1975 - 1.0733i
 0.0093 - 0.9077i   0.0093 - 0.9077i   0.0093 - 1.0619i
 0.2000 - 0.9900i   0.2000 - 0.9900i   0.2000 - 1.0100i
 0.3666 - 1.0446i   0.3666 - 1.0446i   0.3666 - 0.9251i
 0.5039 - 1.0633i   0.5039 - 1.0633i   0.5039 - 0.8161i
 0.6100 - 1.0392i   0.6100 - 1.0392i   0.6100 - 0.6928i
 0.6862 - 0.9680i   0.6862 - 0.9680i   0.6862 - 0.5641i
 0.7363 - 0.8484i   0.7363 - 0.8484i   0.7363 - 0.4372i
 0.7660 - 0.6832i   0.7660 - 0.6832i   0.7660 - 0.3168i
 0.7815 - 0.4792i   0.7815 - 0.4792i   0.7815 - 0.2048i
 0.7882 - 0.2471i   0.7882 - 0.2471i   0.7882 - 0.1002i
 0.7900 - 0.0000i   0.7900 + 0.0000i   0.7900 - 0.0000i

David Goodmanson 2018-6-30

在 MATLAB Online 中打开

Hi Bee,

Below are the first few rows of [zcon zdirect ziter]. The only change I made to your code was

zcon = zcon.' instead of zcon = zcon'

so that the complex conjugate of zcon is not taken. That means that the table I get is the complex conjugate of yours. I also added

ziter = ziter.'

so that ziter is a column vector. Of course neither of these changes/additions affects the shape of the plot, 'triangle' vs. circle.

7900 + 0.0000i   0.7900 + 0.0000i   0.7900 + 0.0000i
7882 - 0.2471i   0.7882 - 0.2471i   0.7882 - 0.2471i
7815 - 0.4792i   0.7815 - 0.4792i   0.7815 - 0.4792i
7660 - 0.6832i   0.7660 - 0.6832i   0.7660 - 0.6832i
7363 - 0.8484i   0.7363 - 0.8484i   0.7363 - 0.8484i
6862 - 0.9680i   0.6862 - 0.9680i   0.6862 - 0.9680i
6100 - 1.0392i   0.6100 - 1.0392i   0.6100 - 1.0392i
5039 - 1.0633i   0.5039 - 1.0633i   0.5039 - 1.0633i
3666 - 1.0446i   0.3666 - 1.0446i   0.3666 - 1.0446i
2000 - 0.9900i   0.2000 - 0.9900i   0.2000 - 0.9900i
0093 - 0.9077i   0.0093 - 0.9077i   0.0093 - 0.9077i
    -0.1975 - 0.8061i  -0.1975 - 0.8061i  -0.1975 - 0.8061i
    -0.4100 - 0.6928i  -0.4100 - 0.6928i  -0.4100 - 0.6928i
    -0.6167 - 0.5741i  -0.6167 - 0.5741i  -0.6167 - 0.5741i
    -0.8058 - 0.4545i  -0.8058 - 0.4545i  -0.8058 - 0.4545i

BeeTiaw 2018-6-30

Hi David Goodmanson,

Thanks for your answer!

I got it correct now after change it as per your suggestion. Thanks a lot!

It seems that it has something to do with consistency of the conjugate/non-conjugate of the complex number. Although I still do not clearly understand it.

But I would consider the problem is solved now.

请先登录，再进行评论。

Matrix inversion and summation

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

4 个评论
显示 2更早的评论隐藏 2更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

Matrix inversion and summation

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

4 个评论 显示 2更早的评论隐藏 2更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

4 个评论
显示 2更早的评论隐藏 2更早的评论