"Index exceeds matrix dimensions."

clear all
clc
nodos=input('Ingresa la cantidad de nodos del sistema\n n = ');
noZ=input('Ingresa la cantidad de lineas del sistema \n NoZ = ');

for i=1:noZ  %Ciclo para ingresar los valores de la Z de rama.
    ne(i)=input('Número de referencia del nodo emisor = ');
    nr(i)=input('Número de referencia del nodo receptor = ');
    %clc
    fi=2%('Número de filas de la matriz = '); 
    co=2%2'Número de columnas de la matriz = ');
    clc
    disp(['Ingresa la matriz de impedancias de la linea #',num2str(i),':'])
    
    for j=1:fi%Filas
        for jj=1:co%Columnas
            disp(['El elemento (',num2str(j),',',num2str(jj),')'])
            Ztramo(j,jj,i)=input('');
        end
    end
    clc
    disp(['La matriz de impedancias del tramo #',num2str(i),' es:'])
    Ztramo% Muestra Z de tramo con ceros.
    disp(['La matriz de admitancias del tramo #',num2str(i),' es:'])
    Ytramo(:,:,i)=inv(Ztramo(:,:,i));
    Ytramo
    ybus=diag(0,(nodos*2)-1);
end

for i=1:noZ %Ciclo para la formación de YBUS
    k1=ne(:,:,i);
    k2=nr(:,:,i);
    ybus(k1,k1)=ybus(k1,k1)+Ytramo(:,:,i);
    ybus(k2,k2)=ybus(k2,k2)+Ytramo(:,:,i);
    ybus(k1,k2)=-Ytramo(:,:,i);
    ybus(k2,k1)=ybus(k1,k2);
end
clc
ybus

[per isakson]

Include at least the full error message in your question, see TUTORIAL: how to ask a question (on Answers) and get a fast answer

Maybe an array is empty? See my answer to index exceeds matrix dimensions

And see

• Debug a MATLAB Program and Examine Values While Debugging
• TUTORIAL: how to ask a question (on Answers) and get a fast answer

[Eduardo Rojas]

The full message is: Index exceeds matrix dimensions.

Assignment has more non-singleton rhs dimensions than
non-singleton subscripts

Code:

clear all
clc
nodos=input('Ingresa la cantidad de nodos del sistema\n n = ');
noZ=input('Ingresa la cantidad de lineas del sistema \n NoZ = ');

for i=1:noZ  %Ciclo para ingresar los valores de la Z de rama.
    
    ne(i)=input('Número de referencia del nodo emisor = ');
    nr(i)=input('Número de referencia del nodo receptor = ');
    %clc
    fi=2%('Número de filas de la matriz = '); 
    co=2%2'Número de columnas de la matriz = ');
    clc
    disp(['Ingresa la matriz de impedancias de la linea #',num2str(i),':'])
    
    for j=1:fi%Filas
        for jj=1:co%Columnas
            disp(['El elemento (',num2str(j),',',num2str(jj),')'])
            Ztramo(j,jj,i)=input('');
        end
    end
    
    clc
    disp(['La matriz de impedancias del tramo #',num2str(i),' es:'])
    Ztramo% Muestra Z de tramo con ceros.
    disp(['La matriz de admitancias del tramo #',num2str(i),' es:'])
    Ytramo(:,:,i)=inv(Ztramo(:,:,i));
    Ytramo
    ybus=diag(0,(nodos*2)-1);
 
end

for i=1:noZ %Ciclo para la formación de YBUS
    k1=ne(i);
    k2=nr(i);
    ybus(k1,k1)=ybus(k1,k1)+Ytramo(:,:,i);
    ybus(k2,k2)=ybus(k2,k2)+Ytramo(:,:,i);
    ybus(k1,k2)=-Ytramo(:,:,i);
    ybus(k2,k1)=ybus(k1,k2);
end
clc
ybus

[Walter Roberson]

What line is the error being shown for?

What are the inputs you are using?

You should post enough that we can reproduce the problem on our systems.

Answer 1

[Walter Roberson]

Ztramo is an fi by co by noZ matrix.

inv(Ztramo(:,:,i)) is a fi by co matrix if it works at all, which can only happen if fi and co are equal. Therefore

Ytramo(:,:,i)=inv(Ztramo(:,:,i));

makes Ytramo as a fi by co by noZ matrix.

Then in

k1=ne(i);
k2=nr(i);

k1 and k2 are scalars because i is a scalar and indexing by a scalar always gives a scalar.

ybus(k1,k1)=ybus(k1,k1)+Ytramo(:,:,i);

So the left hand size, ybus(k1,k1) is a scalar, and the right hand side of the equation needs to be a scalar. Looking at it, the ybus(k1,k1) part of the addition is a scalar, but Ytramo(:,:,i) we have determined must be fi by co . Adding a scalar to a matrix that is fi by co gives a matrix that is fi by co. Therefore the result of the addition is a matrix that is fi by co, and that is too big to store into the scalar area designated by the left side, ybus(k1,k1)

[Eduardo Rojas]

Sorry.

What's the meaning of "fi by co"?

[Walter Roberson]

    fi=2%('Número de filas de la matriz = '); 
    co=2%2'Número de columnas de la matriz = ');

You create a matrix which has fi rows and co columns.

[Eduardo Rojas]

Ok, thank you

I'm working on the solution