getting error as Matrix dimensions must agree." can someone help me resolve this

Question

asim asrar 2019-1-2

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/437904-getting-error-as-matrix-dimensions-must-agree-can-someone-help-me-resolve-this

评论： asim asrar 2019-1-2

getting error for my code as matrix dimension must agree "have tried to solve coupled wave equation using ode 45 command " error exaclty is as

Error using .*

Matrix dimensions must agree.

Error in odefun (line 58)

dE_omega_dz(1:length(E_omega)/2) =

fft(1i*conj(E1_t).*E2_t.*exp(1i*Dk.*z)

...

Error in odearguments (line 87)

f0 = feval(ode,t0,y0,args{:}); %

ODE15I sets args{1} to yp0.

Error in ode45 (line 113)

[neq, tspan, ntspan, next, t0,

tfinal, tdir, y0, f0, odeArgs,

odeFcn, ...

Error in odefun1 (line 54)

[z, E_omega] =

ode45(@odefun,zspan,E_omega);

code is as- "FUNCTION"

function dE_omega_dz = odefun(z, E_omega,~,~)
dE_omega_dz=zeros(length(E_omega),1);
z=z*10^4;
display(z);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
lambda1=800*10^-9;
c=(3*10^8)
lambda2=400*10^-9;
o=1;
y= 28.78076 %22.4431   %22.39002159      %20:0.25:50  22.39002159     22.443 
ne2=1.5687;
no2=1.6934;
no1=1.6614;
r22=2.1*10^-12                   %electro-optic coefficient in m/v
j=1;
L=1
t=1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for Ez=0:500*10^3:85*10^5
noE= no1*(1-0.5*no1^2*r22*Ez)
NEE= ((((sin(y)).^2)/((ne2)^2))+(((cos(y))^2)/((no2)^2)))^-0.5
deltak=-(((4*3.14*(NEE-noE))/ lambda1)) 
Dk(j)=(deltak);
% Dk=deltak;
V=(Ez*4)/10^6
V1(t)=V;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
l=800*10^-9;        % lambda
c=30*10^8;
pi=3.1415926535;
omega=2*pi*c/l;
n_2=6.6508*10^-20;
L_NL=6.8286e-18*1.5; 
LGVM=0.6*10^-3;
I0=0.4*10^11;
% you would have to split the fields back in two:
E1_omega = E_omega(1:end/2);
E2_omega = E_omega(end/2+1:end);
% go back to time space to calculate the nonlinear part:
E1_t = ifft(E1_omega);
E2_t = ifft(E2_omega);
figure(786)
x=-300*10^-15:1*10^-15:300*10^-15;
plot(x,fftshift(E1_t.^2));
% pause(0.2)
N=max(size(E1_t));
to=120e-15/1.655; % initial pulse widthin second
dt=1/120e-15;
dw=1/N/dt*2*pi;
%dw=2*pi*c/l;
 w=(-1*N/2:1:N/2-1)*dw;
% and calculate the derivatives:
dE_omega_dz(1:length(E_omega)/2) = fft(1i*conj(E1_t).*E2_t.*exp(1i*Dk.*z) ...
    + 1i*2*pi*n_2*I0*L_NL/l*(abs(E1_t.^2 + ...
    2*abs(E2_t.^2)).*E1_t));
dE_omega_dz(length(E_omega)/2+1:length(E_omega)) = 1i*w.*L_NL/LGVM * E2_t + fft(1i*E1_t.*E1_t.*exp(-1i*Dk.*z)....
    + 1i*4*pi*n_2*I0*L_NL/l*(2*abs(E1_t.^2 + ...
    abs(E2_t.^2)).*E1_t));
j=j+1;
t=t+1;
%k=k+1;
L=L+1;
o=o+1;
end
end

Calling function is as -

%first electric field envelop in fourier space
clc
clear all
close all
Po=0.4*10^11;
% Dk=-60;
Ao=sqrt(Po);
C=0.5;
pi=3.1415926535;
x=-300*10^-15:1*10^-15:300*10^-15;  
c=3e8;
n=1.655;
l=800*10^-9; 
w1=2*pi*c/l;
z=30*10^-3;
beta=-1.5*10^-29;       
i=sqrt(-1);
t=120*10^-15;
Wo=120*10^-15/1.655; %initial pulse width in second
u=Ao*exp(1i*(beta*z-w1*t)).*exp(-((1+i*(-C))/2)*(x./Wo).^2)
 figure(1)
 
plot(x,abs(u).^2,'b');
title('Input Pulse'); xlabel('Time'); ylabel('Amplitude');
hold on
grid on
E11_omega=fft(fftshift(u));
disp(E11_omega)
%second electric field envelop in fourier space
P1=0.4*10^11;
A1=sqrt(P1);
% C=5;
u1=A1*(exp(1i*(beta*z-2*w1*t)).*exp(-((1+i*(-C))/2)*(x./(Wo)).^2));
% figure(2)
% plot(x,abs(u1).^2,'b');
% title('Input Pulse'); xlabel('Time'); ylabel('Amplitude');
% hold on
% grid on
E22_omega=fft(fftshift(u1));
disp(E22_omega)
%  patching of the E1 and E2 electric fields with:
E_omega=[E11_omega,E22_omega]
disp(E_omega)
% figure(3)
% plot(abs(E_omega),'r');
%a=[1:0.1:3]
zspan=[0 30*10^-7];
% Z1=zspan*10;
E_omega_0=[0 0.0625]
[z, E_omega] = ode45(@odefun,zspan,E_omega);
E11_omega = E_omega(:, 1:end/2);
E22_omega = E_omega(:, end/2+1:end);
E1_t = ifft(E11_omega, [], 2);
E2_t = ifft(E22_omega, [], 2);
figure(101)
subplot(2, 2, 1)
[X, Y] = meshgrid(1:size(E11_omega, 2), z);
mesh(X, Y, abs(fftshift(E11_omega, 2)).^2);
subplot(2, 2, 2)
[X, Y] = meshgrid(1:size(E11_omega, 2), z);
mesh(X, Y,abs(fftshift(E22_omega, 2)).^2);
subplot(2, 2, 3)
[X, Y] = meshgrid(x, z);
mesh(X, Y,abs(fftshift(E1_t, 2)).^2);
subplot(2, 2, 4)
[X, Y] = meshgrid(x, z);
mesh(X, Y,abs(fftshift(E2_t, 2)).^2);
% qq(x,:)=E11_omega(end,:);
figure(1)
 plot(x,fftshift(E1_t(end,:)).^2);
% pause(0.1)
% figure(4)
%  plot(z,E_omega,'linewidth',1)
%     xlabel('z')
%    ylabel('phase (radian)')

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Answer 1

Walter Roberson 2019-1-2

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/437904-getting-error-as-matrix-dimensions-must-agree-can-someone-help-me-resolve-this#answer_354645

在 MATLAB Online 中打开

You have

dE_omega_dz(1:length(E_omega)/2) = fft(1i*conj(E1_t).*E2_t.*exp(1i*Dk.*z) ...
    + 1i*2*pi*n_2*I0*L_NL/l*(abs(E1_t.^2 + ...
    2*abs(E2_t.^2)).*E1_t));

This is in a loop in which you are doing

Dk(j)=(deltak);

where you are incrementing j for each iteration of the for Ez loop. Therefore after the first iteration of for Ez, Dk is not a scalar, and the 601 x 1 expression that you get from 1i*conj(E1_t).*E2_t is being .* by a 1 * j variable. In R2016a and earlier that is an error. In R2016b and later, it is permitted and would mean the same thing as if you had used

bsxfun(@times, 1i*conj(E1_t).*E2_t, exp(1i*Dk.*z)

producing a 601 x j result -- a size that is increasing in each iteration of the for Ez loop. But the output size dE_omega_dz(1:length(E_omega)/2) is fixed, leading to a problem of trying to output the wrong amount of information.

Perhaps you should be indexing Dk(j) in the statement.