I couldn't understand what's wrong in this.

Question

SAHIL SAHOO 2022-7-17

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1761960-i-couldn-t-understand-what-s-wrong-in-this

评论： Torsten 2022-7-18

%Unable to perform assignment because the left and right sides have a different number of elements.

clc
%defining constant
ti = 0; %inital time
tf = 100E-4;% final time
tspan=[ti tf];
o = 2E5;  % detuning frequency
tc = 30E-9;  %photon life time in cavity
tf = 230E-6;  %flouroscence lifetime
a1 = 0.1; %round trip loss
a2 = 0.1;
P1 = 0.2;  %pump strenght
P2 = 0.2;
kc = 3E-3; %critical coupling strength
%s =  0.17;
%k = s.*kc;
I = 1; %saturation intensity
V = 1;
% define function
%y(1) = A1
%y(2) = A2
%y(3) = G1
%y(4) = G2
%y(5) = phase difference
f = @(t,y,k) [  
    (((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;
    
    (((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;
    
    (P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;
    
    (P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;
    
    o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));
    
    ];
k =0:0.01E-3:3.2E-3;
Vf = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));
end
k =[0:0.01E-3 : 3.2E-3] ;
Vp = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;
end
%maybe here is the problem
k =[0:0.01E-3 : 3.2E-3] ;
T = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))
end
Unable to perform assignment because the left and right sides have a different number of elements.
subplot 221
plot((k./3E-3),Vf)
subplot 222
plot((k./3E-3),Vp)
subplot 223
plot((k./3E-3),T)

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Abderrahim. B 2022-7-17

编辑：Abderrahim. B 2022-7-17

在 MATLAB Online 中打开

Hi!

You can not assign a matrix of size 3501 x 321 to a matrix (row vector) of size 3501x1.

% The error is in this line:
T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));

I recommend to debug your code.

HTH

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Answer 1

Walter Roberson 2022-7-17

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https://ww2.mathworks.cn/matlabcentral/answers/1761960-i-couldn-t-understand-what-s-wrong-in-this#answer_1009140

在 MATLAB Online 中打开

for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))
end

Your tspan is a vector of length exactly 2. When tspan is a vector with exactly two elements, ode45() will return a vector of T values and a 2D array of Y values, with it internally deciding which times to emit results for. The number of entries returned is relatively unpredictable. The number of rows of Y will be the same as the number of entries in T..

You then index particular columns of the 2D array of Y values -- making calculations for all of the times. The number of results on the right side will be the same as the number of times.

You then try to overwrite one particular time entry with the vector calculated on the right hand side.

Note that before the for loop you pre-allocated T, but your call to ode45() then overwrites all of T. You probably want to use different variables there.

On the other ode45 loops, you were calculating mean() of values, and so only returned a single value rather than a vector. And in the other loops, you did not accidentally use the same variable for two different purposes.

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SAHIL SAHOO 2022-7-17

编辑：Walter Roberson 2022-7-17

在 MATLAB Online 中打开

ti = 0; %inital time
tf = 100E-4;% final time
tspan=[ti tf];
o = 2E5;  % detuning frequency
tc = 30E-9;  %photon life time in cavity
tf = 230E-6;  %flouroscence lifetime
a1 = 0.1; %round trip loss
a2 = 0.1;
P1 = 0.2;  %pump strenght
P2 = 0.2;
kc = 3E-3; %critical coupling strength
%s =  0.17;
%k = s.*kc;
I = 1; %saturation intensity
V = 1;
% define function
%y(1) = A1
%y(2) = A2
%y(3) = G1
%y(4) = G2
%y(5) = phase difference
f = @(t,y,k) [  
    (((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;
    
    (((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;
    
    (P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;
    
    (P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;
    
    o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));
    
    ];
k =0:0.01E-3:3.2E-3;
Vf = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));
end
k =[0:0.01E-3 : 3.2E-3] ;
Vp = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;
end
k =[0:0.01E-3 : 3.2E-3] ;
T = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
end
k =[0:0.01E-3 : 3.2E-3] ;
M = zeros(length(k),1) ;
for i = 1:length(k)
    M(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));
end
subplot 221
plot((k./3E-3),Vf)
subplot 222
plot((k./3E-3),Vp)
subplot 223
plot((k./3E-3),M)
%is this what you mean?

Walter Roberson 2022-7-17

编辑：Walter Roberson 2022-7-17

在 MATLAB Online 中打开

N = 50;

ti = 0; %inital time

tf = 100E-4;% final time

tspan = linspace(ti, tf, N);

o = 2E5; % detuning frequency

tc = 30E-9; %photon life time in cavity

tf = 230E-6; %flouroscence lifetime

a1 = 0.1; %round trip loss

a2 = 0.1;

P1 = 0.2; %pump strenght

P2 = 0.2;

kc = 3E-3; %critical coupling strength

%s = 0.17;

%k = s.*kc;

I = 1; %saturation intensity

V = 1;

% define function

%y(1) = A1

%y(2) = A2

%y(3) = G1

%y(4) = G2

%y(5) = phase difference

f = @(t,y,k) [

(((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;

(((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;

(P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;

(P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;

o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));

];

k = 0 : 0.01E-3 : 3.2E-3;

Vf = zeros(length(k),1) ;

Vp = zeros(length(k),1) ;

M = zeros(length(k),N) ;

for i = 1:length(k)

[T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);

Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));

Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;

M(i,:) = 1 - (k(i)./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));

end

subplot 221

plot((k./3E-3),Vf)

subplot 222

plot((k./3E-3),Vp)

subplot 223

plot((k./3E-3),M)

subplot 224

plot((k./3e-3), mean(M,2));

The third plot pretty much fills an area because for each k value, you have as many M values as you have time points (because you are not doing anything like mean() which you do for the other two calculations.) I added the 4th plot to show you what taking mean() of the M values over time might look like.

Notice that you do not need to keep redoing all those ode45, since you are running them under the same conditions (same k, same time span, same initial conditions.) So for each k value, you can run ode45 once and calculate Vf, Vp, and M values.

SAHIL SAHOO 2022-7-18

plot((k./3e-3), mean(M,2));

why you mean by mean(M,2) ?

Torsten 2022-7-18

From the documentation of "mean":

M = mean(A,dim) returns the mean along dimension dim. For example, if A is a matrix, then mean(A,2) is a column vector containing the mean of each row.

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Answer 2

KSSV 2022-7-17

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1761960-i-couldn-t-understand-what-s-wrong-in-this#answer_1009130

在 MATLAB Online 中打开

The error is simple. You are trying to save more number of elements in LHS than it is initialized for.

Demo

A = zeros(1,3) ;   % 1x3 array intialized 
A(1) = rand ;  % no error, you can save one element in one place 
A(2) = rand(1,2) ; % error, you cannot save two elements in a sinlge element place 
Unable to perform assignment because the left and right sides have a different number of elements.

Similarly, check the dimensions in your case, where error occurs.