Create matrix from columns of variable (a)-for loop

Question

Wazy sky 2018-9-26

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420819-create-matrix-from-columns-of-variable-a-for-loop

回答： Wazy sky 2018-10-13

Hi I want to create a matrix TA from the columns EE generated in the last loop. I can display the wanted columns "disp (EE)' but when adding the matrix want to create from these columns, an error occurred, noted that the vectors/columns EE has a variable (a)

You may ignore the lines of the code before the line start with C = [I;A;B]; for getting the question

I=ones(20,2);
  for i=0:20
  theta=(pi/40)*i;
  r=2.5;x1=r*cos(theta);
  y1=r*sin(theta);s=i+1;
  I(s,:)=[x1,y1];
  end
    A=ones(20,2);
    for k=1:20
    r=(k-1)/10;
    A(k,:)=[1,r];
    end
    m=1:-(1/47):0;
    B=ones(length(m),2);
    B(:,1)=m';B(:,2)=2;
    C = [I;A;B];
    %%find the complex variable matrix, ZC connjgate Z
    Z=ones(length(C),1);
    for v=1:length(C)
    Z(v)=C(v,1)+1i*C(v,2);end 
    ZC=conj(Z);
    F=ones(89,2);syms Fb;
    for ii=1:89
      if ii<22 
        F(ii,1)=0;F(ii,2)=0;
      elseif ii<42
        F(ii,1)=1; F(ii,2)=0;
      else 
        F(ii,1)=(89-ii)/(89-42);F(ii,2)=0;
      end
    end 
    FF=F.*Fb;
    zzc=Z-ZC;
    QQ=ones(length(Z),21);SS=ones(length(Z),21);ss=0;
    T1=ones(length(Z),21);T2=ones(length(Z),21);syms a
for K=-6:14
  ss=ss+1;
  QQ(:,ss)=Z.^(2*K-1)+ZC.^(2*K-1);
  SS(:,ss)=(2*ss-1)*(Z.^(2*ss-2));
  T1(:,ss)=Z.^(2*K-2);T2(:,ss)=ZC.^(2*K-2);
 end
  %multiplying the T1,T2 BY THE FCTOR 
  Q=QQ.';S=SS.';
    TA=ones(89,21);TB=ones(89,21);
    Tz=sqrt(Z.^2-a^2); Tzc=sqrt(ZC.^2-a^2);
    for j=1:21
    CC=Tz.*T1(:,j);DD=Tzc.*T2(:,j);EE=CC+DD;
    TA(:,j)=EE;
    end

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KSSV 2018-9-26

You must specify the error and error line number.

Wazy sky 2018-9-26

Sorry missed that

Error The following error occurred converting from sym to double: DOUBLE cannot convert the input expression into a double array. If the input expression contains a symbolic variable, use VPA.

Error in filename (line 67)

TA(:,j)=EE;

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Answer 1

Walter Roberson 2018-9-26

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420819-create-matrix-from-columns-of-variable-a-for-loop#answer_338492

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syms a

so a is syms

Tz=sqrt(Z.^2-a^2); Tzc=sqrt(ZC.^2-a^2);

so Tz and Tzc are syms involving the unresolved variable a

CC=Tz.*T1(:,j);DD=Tzc.*T2(:,j);EE=CC+DD;

so CC and DD and EE all involve that unresolved variable a

TA=ones(89,21);

so TA can only store numeric values.

TA(:,j)=EE;

with TA only being able to store numeric values, the values in EE have to be converted from sym to numeric. But they cannot be converted, because they involve the unresolved symbolic variable a

Your EE entries are formulas, such as

      (5209627982755595*(7990063086789097/36028797018963968 - a^2)^(1/2))/137438953472 + (5252908098854583*(6546788930781455/18014398509481984 - a^2)^(1/2))/4398046511104

You cannot store a formula in a numeric array.

You can use vpa() to convert the formulas from using rational numbers to using floating point numbers; for example vpa() of the above gives the result

1194.3730212020871022104984149337*(0.36341979041573407860354905096756 - 1.0*a^2)^(1/2) + 37905.032388193612860050052404404*(0.22176880017902014796682408359629 - 1.0*a^2)^(1/2)

but this is still a formula and would need a symbolic array to be stored into.

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Wazy sky 2018-10-3

编辑：Wazy sky 2018-10-10

*Update!! you are absolutely correct one value should be used for (a) and that is the solution change for different values!

the same procedure followed now. For the new case (different input- [C] IN THE CODE ABOVE), I know from the source I am following that the solution of the variables is a complex number, not a real. Now I am not getting a solution with different errors/warning (no explicit solution, or the solution is an empty array). I used to solve, and vpasolve for solving the linear equations.* I am really in a very critical situation now. Need your help all. Thanks

Hi I assumed values for (a, Fb), and used the least squares method, got the solution for the variables (E, G).

When I tried another value of (a), the solutions of (E, G) changed! this is not correct as the solutions should be the same for whatever the condition (Fb, a).

E, & G satisfies the defined boundary conditions (matrices [FF], [C]). --- I know for sure that the solution for the variables E, G is not a function of any variables such as (a, Fb).

The solution of (E, & G) are vectors of real numbers.

I tried defining the a, Fb as variables,

Matlab warning was: Cannot find explicit solution.?

What am I missing here! is it the math or understanding the problem I am solving itself

Wazy sky 2018-10-11

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Hi Walter Roberson,

The question is: what is the expected mistake in my code to get this warning:.

[*Warning*: Cannot find explicit solution. > In solve (line 316)] when using solve function;

when I use vpasolve function to solve the linear equation, I get [The solution is sym (0x1)] which seems no solution exists.

knowing that there is a solution exist for this problem (the source is a published work showing: the procedure, the equation, the solution results. I am writing the code regenerate the data).

in these equations to be solved we know the solution of the variables is a complex number. the number of Linear equations = the number of variables. the functions used in Matlab are solve/vpasolve

my current code is

circle_pts=zeros(37,2);a=0.5; beta=45;
for j=0:36
theta=(pi/36)*j;
r=0.25;
x1=r*cos(theta);
y1=r*sin(theta);
circle_pts(j+1,:)=[x1,y1];
end
% Right-Side RS;x0=b/cos(pi/2-beta);x0=b/cos(pi/2-beta);b=1;
b=1;
x0=b/cos(deg2rad(90-beta));
h=2; RS=h-b*tan(deg2rad(90-beta));
de_RS=RS/20;RS_pts=zeros(20,2);
for jj=0:19
    x_RS=x0+jj*de_RS*cos(deg2rad(beta));
    y_RS=jj*de_RS*sin(deg2rad(beta));
    RS_pts(jj+1,:)=[x_RS,y_RS];
end
% Top-side TS_pts 58-78,23 De=2b/22;x1=RS*cos(90-beta)+xO, y1=RS*sin(beata)
TS=2; % TS the top side length
De_TS=2/20;TS_pts=zeros(21,2);% corner point frpm RS included.
for k=0:20
    X_TS=x0+RS*cos(deg2rad(beta))-k*De_TS*cos(deg2rad(90-beta));
    Y_TS=RS*sin(deg2rad(beta))+k*De_TS*sin(deg2rad(90-beta));
    TS_pts(k+1,:)=[X_TS,Y_TS];
end
% left side LS=H-RS, h=.5*H;h=2; Angle=beta,
LS=2*(h)-RS; 
LS_pts=zeros(20,2);
De_LS=LS/20;
for kk=1:20
    X_LS=TS_pts(21,1)-kk*De_LS*cos(deg2rad(beta));
    Y_LS=TS_pts(21,2)-kk*De_LS*sin(deg2rad(beta));
    LS_pts(kk,:)=[X_LS, Y_LS];
end
% geometry coordinates 
pt_1=[x0,0];
pt_2=[x0+RS*cos(deg2rad(beta)), RS*sin(deg2rad(beta))];
pt_3=[pt_2(1,1)-TS*cos(deg2rad(90-beta)), pt_2(1,2)+TS*sin(deg2rad(90-beta))];
pt_4=[pt_3(1,1)-LS*cos(deg2rad(beta)),pt_3(1,2)-LS*sin(deg2rad(beta))];
% the coordinate matrix [C]
C=[circle_pts;RS_pts;TS_pts;LS_pts]; 
Z=ones(length(C),1);
for v=1:length(C)
Z(v)=C(v,1)+1i*C(v,2);
end 
ZC=conj(Z);
Z_ZC=Z-ZC;
% Loading matrix. 
F=ones(98,2);Fb=1;
for n=1:98
  if n<38 
    F(n,1)=0;F(n,2)=0+1i*0;
  elseif n<57
    F(n,1)=-cos(deg2rad(beta)); F(n,2)=0+1i*sin(deg2rad(beta));
    elseif n==58
    F(n,1)=(-cos(deg2rad(beta))+1*sin(deg2rad(beta))); F(n,2)=0+1i*(sin(deg2rad(beta))+cos(deg2rad(beta)));
  elseif n<78 
    F(n,1)=((68-n)/(68-58))*sin(deg2rad(beta));F(n,2)=0+1i*((68-n)/(68-58))*cos(deg2rad(beta));
  elseif n==78
    F(n,1)=cos(deg2rad(beta))-1*sin(deg2rad(beta));F(n,2)=-(0+1i*(sin(deg2rad(beta))+cos(deg2rad(beta))));
  else 
    F(n,1)=cos(deg2rad(beta));F(n,2)=0-1i*sin(deg2rad(beta));
  end
end 
F(:,1)=F(:,1)*(-1);
FF=F.*Fb; 
QQ=ones(length(Z),29);SS=ones(length(Z),29);
T1=ones(length(Z),29);T2=ones(length(Z),29);
RR=ones(length(Z),29);
ss=0;
for K=-10:18
ss=ss+1;
QQ(:,ss)=Z.^(2*K-1)-ZC.^(2*K-1);
SS(:,ss)=(2*K-1)*(Z.^(2*K-2));
RR(:,ss)=((2*K-1)-((2*K-2).*(a./Z).^2)).*Z.^(2*K-1);
T1(:,ss)=Z.^(2*K-2);T2(:,ss)=ZC.^(2*K-2);
end
P=ones(98,29);
Tz=((Z.^2-a^2).^0.5); Tzc=((ZC.^2-a^2).^0.5);
for j=1:29
CC=(Tz.*T1(:,j));DD=(Tzc.*T2(:,j));EE=(CC+DD);
P(:,j)=(EE);
end
Tzz=1./Tz;
E=sym('E',[29 1]);G=sym('G',[29 1]);
H1=P*E; 
H2=QQ*G; 
H3=SS*G; 
H4=RR*E;
H5=Tzz.*H4;
H6=H5+H3; 
H7=conj(H6);
H8=Z_ZC.*H7;
H=H1+H2+H8;
H_Real= real(H);
%has Real&imag parts for both variables.
H_imag = imag(H);
t_H=toc; tic;
Eq1=(H_Real-FF(:,1)).^2+(H_imag-FF(:,2)).^2;
EmE=sym(ones(58,1));
for ij=1:29
    Em1=0;Em2=0;
    for jj=1:98
    Em1=Em1+diff(Eq1(jj),E(ij,1)); 
    Em2=Em2+diff(Eq1(jj),G(ij,1));nn=nn+1;
    end
    end
    EmE(ij,1)=Em1;EmE(ij+29,1)=Em2;
end
sol=vpasolve(EmE,[E;G]);
sol.E1; sol.E2; sol.E29; %call the solution for few variables for checking the solution.

Your suggestions are so much appreacited

Stephen23 2018-10-11

"Your suggestions are so much appreacited"

Your code is very badly aligned. Badly aligned code is one way that beginners hide bugs in their code. In this case, your badly aligned code has hidden several bugs and made them harder to identify. Correctly aligned code makes identifying some kinds of bugs easier. You should align your code correctly.

I recommend using the MATLAB Editor's default alignment settings, which occur automatically as you type. You can also align existing code: select the code text, press ctrl+i.

Wazy sky 2018-10-11

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Thanks, Stephen Cobeldick; I tried to simplify the question by presenting only the main part of the code, editing in text file led to losing its alignment. I am posting all the code here

use *a=0.5, beta=45* for example, please ignoreor texted part. Thanks!
   prompt={'a:','beta:'};
    % Create all your text fields with the questions specified by the variable prompt.
    title='Input';
    % The main title of your input dialog interface.
    answer=inputdlg(prompt,title);
    a = str2double(answer{1});
    beta = str2double(answer{2});
    circle_pts=zeros(37,2);
    tic
    for j=0:36
        theta=(pi/36)*j;
        r=0.25;
        x1=r*cos(theta);
        y1=r*sin(theta);
        circle_pts(j+1,:)=[x1,y1];
    end
    % Right-Side RS;x0=b/cos(pi/2-beta);x0=b/cos(pi/2-beta);b=1;
    b=1;
    x0=b/cos(deg2rad(90-beta));
    h=2; RS=h-b*tan(deg2rad(90-beta));
    de_RS=RS/20;RS_pts=zeros(20,2);
    for jj=0:19
        x_RS=x0+jj*de_RS*cos(deg2rad(beta));
        y_RS=jj*de_RS*sin(deg2rad(beta));
        RS_pts(jj+1,:)=[x_RS,y_RS];
    end
    % Top-side TS_pts 58-78,23 De=2b/22;x1=RS*cos(90-beta)+xO, y1=RS*sin(beata)
    TS=2; % TS the top side length
    De_TS=2/20;TS_pts=zeros(21,2);% corner point frpm RS included.
    for k=0:20
        X_TS=x0+RS*cos(deg2rad(beta))-k*De_TS*cos(deg2rad(90-beta));
        Y_TS=RS*sin(deg2rad(beta))+k*De_TS*sin(deg2rad(90-beta));
        TS_pts(k+1,:)=[X_TS,Y_TS];
    end
    % left side LS=H-RS, h=.5*H;h=2; Angle=beta,
    LS=2*(h)-RS;
    LS_pts=zeros(20,2);
    De_LS=LS/20;
    for kk=1:20
        X_LS=TS_pts(21,1)-kk*De_LS*cos(deg2rad(beta));
        Y_LS=TS_pts(21,2)-kk*De_LS*sin(deg2rad(beta));
        LS_pts(kk,:)=[X_LS, Y_LS];
    end
    % geometry coordinates
    pt_1=[x0,0];
    pt_2=[x0+RS*cos(deg2rad(beta)), RS*sin(deg2rad(beta))];
    pt_3=[pt_2(1,1)-TS*cos(deg2rad(90-beta)), pt_2(1,2)+TS*sin(deg2rad(90-beta))];
    pt_4=[pt_3(1,1)-LS*cos(deg2rad(beta)),pt_3(1,2)-LS*sin(deg2rad(beta))];
    % the coordinate matrix [C]
    C=[circle_pts;RS_pts;TS_pts;LS_pts]; t_coordinat=toc;
    Z=ones(length(C),1);
    for v=1:length(C)
        Z(v)=C(v,1)+1i*C(v,2);
    end
    ZC=conj(Z);
    Z_ZC=Z-ZC;
    % Loading matrix. assuming the load accumlating.
    % for m=1:98
    %   if m<38
    %     F(m,1)=0;F(m,2)=0+1i*0;
    %   elseif m<57
    %     F(m,1)=-((m-38)/20)*RS*cos(deg2rad(beta)); F(m,2)=0+1i*((m-38)/20)*RS*sin(deg2rad(beta));
    %     elseif m==58
    %     F(m,1)=-((m-38)/20)*RS*cos(deg2rad(beta))+1*sin(deg2rad(beta)); F(m,2)=0+1i*(((m-38)/20)*RS*sin(deg2rad(beta))+cos(deg2rad(beta)));
    %   elseif m<78
    %     F(m,1)=((68-m)/(68-58))*sin(deg2rad(beta));F(m,2)=0+1i*((68-m)/(68-58))*cos(deg2rad(beta));
    %   else
    %     F(m,1)=((m-78)/(89-78))*LS*cos(deg2rad(beta));F(m,2)=0-1i*((m-78)/(89-78))*LS*sin(deg2rad(beta));
    %   end
    % end
    % FF=F.*Fb;
    % the (-) in fy should be kept as fy calculated here for +Y
    %CHECK THAT FOR B=45 A ZERO FOR PT78; PT58.
    F=ones(98,2);Fb=1;
    for n=1:98
        if n<38
            F(n,1)=0;F(n,2)=0+1i*0;
        elseif n<57
            F(n,1)=-cos(deg2rad(beta)); F(n,2)=0+1i*sin(deg2rad(beta));
        elseif n==58
            F(n,1)=0.0*(-cos(deg2rad(beta))+1*sin(deg2rad(beta))); F(n,2)=0+1i*(sin(deg2rad(beta))+cos(deg2rad(beta)));
        elseif n<78
            F(n,1)=((68-n)/(68-58))*sin(deg2rad(beta));F(n,2)=0+1i*((68-n)/(68-58))*cos(deg2rad(beta));
        elseif n==78
            F(n,1)=0*cos(deg2rad(beta))-1*sin(deg2rad(beta));F(n,2)=-(0+1i*(sin(deg2rad(beta))+cos(deg2rad(beta))));
        else
            F(n,1)=cos(deg2rad(beta));F(n,2)=0-1i*sin(deg2rad(beta));
        end
    end
    % --->>> CORRECTION THE RS direction<<------
    % for n=1:98
    %   if n<38
    %     F(n,1)=0;F(n,2)=0+1i*0;
    %   elseif n<57
    %     F(n,1)=-cos(deg2rad(beta)); F(n,2)=0+1i*sin(deg2rad(beta));
    %     elseif n==58
    %     F(n,1)=0.0*(-cos(deg2rad(beta))+1*sin(deg2rad(beta))); F(n,2)=0+1i*(sin(deg2rad(beta))+cos(deg2rad(beta)));
    %   elseif n<78
    %     F(n,1)=((68-n)/(68-58))*sin(deg2rad(beta));F(n,2)=0+1i*((68-n)/(68-58))*cos(deg2rad(beta));
    %   elseif n==78
    %     F(n,1)=cos(deg2rad(beta))+1*sin(deg2rad(beta));F(n,2)=(0+1i*(sin(deg2rad(beta))-cos(deg2rad(beta))));
    %   else
    %     F(n,1)=-cos(deg2rad(beta));F(n,2)=0+1i*sin(deg2rad(beta));
    %   end
    % end
    F(:,1)=F(:,1)*(-1);
    FF=F.*Fb;
    QQ=ones(length(Z),29);SS=ones(length(Z),29);
    T1=ones(length(Z),29);T2=ones(length(Z),29);
    RR=ones(length(Z),29);
    ss=0; tic
    for K=-10:18
        ss=ss+1;
        QQ(:,ss)=Z.^(2*K-1)-ZC.^(2*K-1);
        SS(:,ss)=(2*K-1)*(Z.^(2*K-2));
        RR(:,ss)=((2*K-1)-((2*K-2).*(a./Z).^2)).*Z.^(2*K-1);
        T1(:,ss)=Z.^(2*K-2);T2(:,ss)=ZC.^(2*K-2);
    end
    P=ones(98,29);t_QSRT=toc;
    Tz=((Z.^2-a^2).^0.5); Tzc=((ZC.^2-a^2).^0.5);
    for j=1:29
        CC=(Tz.*T1(:,j));DD=(Tzc.*T2(:,j));EE=(CC+DD);
        P(:,j)=(EE);
    end
    Tzz=1./Tz;
    E=sym('E',[29 1]);G=sym('G',[29 1]);
    tic;
    H1=P*E;
    H2=QQ*G;
    H3=SS*G;
    H4=RR*E;
    H5=Tzz.*H4;
    H6=H5+H3;
    H7=conj(H6);
    H8=Z_ZC.*H7;
    H=H1+H2+H8;
    H_Real= real(H);
    %has Real&imag parts for both variables.
    H_imag = imag(H);
    t_H=toc; tic;
    Eq1=(H_Real-FF(:,1)).^2+(H_imag-FF(:,2)).^2;t_Eq1=toc;
    EmE=sym(ones(58,1));n=0;tic;Pluss=0;Minuss=0;
    % the inflection point diff(diff()) checked all points are Min
    for ij=1:29
        n=n+1;
        nn=0;
        Em1=0;Em2=0;
        for jj=1:98
            Em1=Em1+diff(Eq1(jj),E(ij,1));
            Em2=Em2+diff(Eq1(jj),G(ij,1));nn=nn+1;
        end
        if (diff((Em1),E(ij,1))>=0)
            Pluss=Pluss+1;
        else Minuss=Minuss+1;
        end
        EmE(ij,1)=Em1;EmE(ij+29,1)=Em2;
    end
    t_EmE=toc;
    tic;
    sol=vpasolve(EmE,[E;G]);
    t_sol=toc; ASol= real(sol.E1)
    sol.E1;sol.E2;sol.E3;sol.E4;sol.E5;sol.E6;sol.E7;sol.E8;sol.E9;sol.E10;sol.E11;sol.E12;sol.E13;sol.E14;sol.E15;sol.E16;sol.E17;sol.E18;sol.E19;sol.E20;sol.E21;sol.E22;sol.E23;sol.E24;sol.E25;sol.E26;sol.E27;sol.E28;sol.E29;
    FE=(2*sol.E1*(a)^(-22)+2*sol.E2*(a)^(-20)+2*sol.E3*(a)^(-18)+2*sol.E4*(a)^(-16)+2*sol.E5*(a)^(-14)+2*sol.E6*(a)^(-12)+2*sol.E7*(a)^(-10)+2*sol.E8*(a)^(-8)+2*sol.E9*(a)^(-6)+2*sol.E10*(a)^(-4)+2*sol.E11*(a)^(-2)+2*sol.E12*(a)^(0)+2*sol.E13*(a)^(2)+2*sol.E14*(a)^(4)+2*sol.E15*(a)^(6)+2*sol.E16*(a)^(8)+2*sol.E17*(a)^(10)+2*sol.E18*(a)^(12)+2*sol.E19*(a)^(14)+2*sol.E20*(a)^(16)+2*sol.E21*(a)^(18)+2*sol.E22*(a)^(20)+2*sol.E23*(a)^(22)+2*sol.E24*(a)^(24)+2*sol.E25*(a)^(26)+2*sol.E26*(a)^(28)+2*sol.E27*(a)^(30)+2*sol.E28*(a)^(32)+2*sol.E29*(a)^(34));

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Answer 2

Wazy sky 2018-10-11

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/420819-create-matrix-from-columns-of-variable-a-for-loop#answer_340831

编辑：Walter Roberson 2018-10-11

在 MATLAB Online 中打开

Hi Walter Roberson;

Thank you for your comment

Regards 1st comment:

I'm so sorry as I cleaned my code from extra lines, some left behind. should not be a problem there based on your comment & the original one below:

Regards 2nd comment:

Sorry again for that tic-toc, & n,nn.I should delete all them from my code when I was to post it. appears missed some of them. They are not part of the problem. I used them to check the timing tic-toc. For (nn) I used to check that the loop works well.

Pluss=0; Minuss=0; % used these to test/check that the points (Em1, Em2) are the minimum and they are all mimunm, as the 2nd diff >=0;  
    n=0;
  for ij=1:29
          n=n+1;
          nn=0;
          Em1=0;Em2=0;
          for jj=1:98
          Em1=Em1+diff(Eq1(jj),E(ij,1)); 
          Em2=Em2+diff(Eq1(jj),G(ij,1));nn=nn+1;
          end
           if (diff((Em1),E(ij,1))>=0)
              Pluss=Pluss+1;
          else Minuss=Minuss+1;
          end
          EmE(ij,1)=Em1;EmE(ij+29,1)=Em2;
      end

The issue with code is the warning for the explicit solution.

Thanks your comment is appreciated

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Walter Roberson 2018-10-11

If you dbstop at the vpasolve() and examine vpa(EmE) then you will see that it has a large number of real constants times real() or imag() of various variables, added together, with no obvious cross-products or exponentiation or exp() or the like. So each row is linear in real() and imag() of those variables. But each row also ends in a constant that has both a real and an imaginary coefficient. With all of those constants being real valued, and with real() and imag() also returning real-valued information, then the part involving the variables must be real-valued. That means there is no possibility of part of a term canceling out the imaginary part of the constant coefficient of the row. Therefore the rows cannot equal 0, because although you might balance the real portion, you are left with a non-zero imaginary portion.

Wazy sky 2018-10-12

Walter Roberson, Thank you for highlighting this point. I feel I have to check back the constructed matrices in case something misrepresented. Your point is very important to me and helpful to know at least that as I will not getting the expected results (complex number) there is a mistake here.

However, please bear with me to double check is that mean if the constant at the end of the equation was only real, then for this form of the equation, I will have a solution of a complex number

AS from my first practice with the code where I know the variables are real constant. the solution was real as expected. checking the equations there, they have the same form that you mentioned here [constant*real+/- another-constant*imag +/-another-constant with the only real part. does that mean there is something missing there, or it is still possible to get a real solution for all the variables? Thanks

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Answer 3

Wazy sky 2018-10-13

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https://ww2.mathworks.cn/matlabcentral/answers/420819-create-matrix-from-columns-of-variable-a-for-loop#answer_341198

在 MATLAB Online 中打开

Equation.pdf

Well I found where is the problem in the code?1

So when I extracted the real and imaginary part from my equations

H_imag = imag(H);
H_Real= real(H);

It seems that Matlab does not do it the right way, or as we may expect at this line when we run the code still see a mix of real & imaginary parts in both the H_imag, & H_Real, and here the problem

so the challenge is how to separate these parts. does Matlab, Mathematica or other software can solve the issue.

I found a source transform the same equation I have from the complex number in cartesian to polar coordinate. This step seems helped to separate the real and imaginary parts successfully

Please see the attached

I tried to convert the complex numbers to polar in MATLAB and separate the Real & imaginary parts but this did not work.

Thanks for your time and help in advance

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