Need Help Plotting Mode Shapes

Question

Amanda Lococo 2018-4-6

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/393190-need-help-plotting-mode-shapes

评论： Prajit T R 2018-4-10

I need this code to plot mode shapes, but my plots are coming up blank. Thanks in advance!

clear all;
format long;
im = sqrt(-1);
CellLength = 1;
ibeta = 1;
%Define materal properties 
CellLength = 1;
layers = 2;
d = [0.4;0.6];
dTotal = d(1,1)+d(2,1);
xc = [0;0.4];
Ef = 12;
pf = 3;
cf = sqrt(Ef/pf);
Em = 1;
pm = 1;
cm = sqrt(Em/pm);
w = 5;
T1 = [cos(.2*w) (1/(6*w))*sin(.2*w); -6*w*sin(.2*w) cos(.2*w)];
T2 = [cos(.6*w) (1/w)*sin(.6*w); -w*sin(.6*w) cos(.6*w)];
T = T2*T1;
Z1 = 6*w;
Z2 = w;
Z = [Z1;Z2];
%Solve eigenvalue problem for k
[V,D] = eig(T);                 %D = eigenvalues, %V = eigenvectors
k1 = log(D(1,1))/(im*dTotal);
k2 = log(D(2,2))/(im*dTotal);
k = [k1;k2];
for j = 1:layers
    B = @(j)([1 1; im*Z(j,1) -im*Z(j,1)]);
    B = B(j);
    C_a = @(j)([exp(im*k(j,1)*xc(j,1)) 0; 0 exp(-im*k(j,1)*xc(j,1))]);
    C_a = C_a(j);
      if j == 1
          a = (inv(B)*V(:,1));
          alpha = a;
          beta = B;
      else
          a = inv(C_a)*inv(B)*T*beta*alpha;
      end
      for x = 0:0.1:5
          C = @(x)([exp(im*k(j,1)*x) 0; 0 exp(-im*k(j,1)*x)]);
          C = C(x);
          y = @(x)(B*C*a);
          y = y(x);
      end
  end
plot(x,real(y(:,1)))

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Answer 1

Prajit T R 2018-4-9

1
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https://ww2.mathworks.cn/matlabcentral/answers/393190-need-help-plotting-mode-shapes#answer_314120

在 MATLAB Online 中打开

Hi Amanda

I am assuming that you wish to plot the variation of y against x for the values from 0 to 5 in steps of 0.5 as per the loop above. This can be done using the following code:

clear all;
format long;
im = sqrt(-1);
CellLength = 1;
ibeta = 1;
%Define materal properties 
CellLength = 1;
layers = 2;
d = [0.4;0.6];
dTotal = d(1,1)+d(2,1);
xc = [0;0.4];
Ef = 12;
pf = 3;
cf = sqrt(Ef/pf);
Em = 1;
pm = 1;
cm = sqrt(Em/pm);
w = 5;
T1 = [cos(.2*w) (1/(6*w))*sin(.2*w); -6*w*sin(.2*w) cos(.2*w)];
T2 = [cos(.6*w) (1/w)*sin(.6*w); -w*sin(.6*w) cos(.6*w)];
T = T2*T1
Z1 = 6*w;
Z2 = w;
Z = [Z1;Z2];
%Solve eigenvalue problem for k
[V,D] = eig(T);                 %D = eigenvalues, %V = eigenvectors
k1 = log(D(1,1))/(im*dTotal);
k2 = log(D(2,2))/(im*dTotal);
k = [k1;k2];
for j = 1:layers
  B = @(j)([1 1; im*Z(j,1) -im*Z(j,1)]);
  B = B(j);
  C_a = @(j)([exp(im*k(j,1)*xc(j,1)) 0; 0 exp(-im*k(j,1)*xc(j,1))]);
  C_a = C_a(j);
    if j == 1
        a = (inv(B)*V(:,1));
        alpha = a;
        beta = B;
    else
        a = inv(C_a)*inv(B)*T*beta*alpha;
    end
    L=[];
    M=[];
    for x = 0:0.1:5
        C = @(x)([exp(im*k(j,1)*x) 0; 0 exp(-im*k(j,1)*x)]);
        C = C(x);
        y = @(x)(B*C*a);
        y = y(x);
        L(end+1)=y(1);
        M(end+1)=y(2);
    end
end
plot(0:0.1:5,real(L))
hold on
plot(0:0.1:5,real(M))
hold off

In this code, L is the equivalent for y(:,1) that you were trying to achieve. M on the other hand is y(:,2). I hope this helps.

Cheers

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Amanda Lococo 2018-4-9

编辑：Amanda Lococo 2018-4-9

在 MATLAB Online 中打开

Thank you for this! Can you explain these two lines to me?

L(end+1)=y(1);
M(end+1)=y(2);

Thanks again!

Prajit T R 2018-4-10

y=y(x) inside the loop actually contains two elements. You can observe that by removing the semicolon next to the statement. So, what L(end+1) does is to append the first element of y to a list L, which is initially empty. Similarly, the second element of y is appended to M.

At the end of the iterations, L and M would contain all the complex values generated.

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