How to solve "Unable to perform assignment because the indices on the left side are not compatible with the size of the right side" error?

I am not understanding this indices error. Please, take a look at the code and help me with the solution. I have highlighted the line in the code in which the error is shown.

dis0=input('dis0=');
disd0=input('disd0=');
phi0=input('phi0=');
phid0=input('phid0=');
theta0=input('theta0=');
thetad0=input('thetad0=');
fmin=input('fmin=');
fmax=input('fmax=');
for f=fmin:fmax
    tp=1/f;
    dt=1/(25*f);
    for i=0:dt:2*tp
            t=i;
            j=i+2;
            f1=sin(2*pi*f*t);
            f2=a*sin(2*pi*f*t);
            f3=b*sin(2*pi*f*t);
            F=[f1; f2; f3];
            if i==0
%                 The error is shown in the below line.
                x(j)=[dis0; phi0; theta0];
                xd(j)=[disd0; phid0; thetad0];
                xdd(j)=M\(F-(K*x(j))-(C*xd(j)));
            else
                break
            end
            x(j-1)=x(j)-(dt*xd(j))+(0.5*(dt^2)*xdd(j));
            x(j+1)=((M/(dt^2))+(C/(2*dt)))\(F-(K-((2*M)/(dt^2)))*x(j)-((M/(dt^2))-(C/(2*dt)))*x(j-1));
    end
end

Answer 1

[Cris LaPierre]

x(j) is a single array element. You are trying to assign it 3 values, [dis0; phi0; theta0]. I'm unsure what the intent is, but one way to solve this is to specify row and column indices:

x(:,j)=[dis0; phi0; theta0];

You will need to double check everywhere you use x to take this change into account.

[Neeraj Kumar]

Cris, it is still showing the same error.

Since, you are unsure of the intent, let me give you some info.

x=[dis0; phi0; theta0] 

is a matrix and is changing in every iteration. There are several other matrices which are to calculated from this matrx.

So, i thought, i should add a counter "j",

x(j)=[dis0; phi0; theta0] 

But, the indices error is being shown.

[Cris LaPierre]

I get errors that variables a and b haven't been defined before I get any errors about assignment.

Making the change I suggested fixes the error for x. You will need to make a similar change to handle xd and xdd. You will then need to modify everywhere in your code you use x(j) and its variations: x(j-1) and x(j+1).

I also got an error that M, K and C are unrecognized.

[Neeraj Kumar]

Here is the complete program, please check it and help.

clc
clear all
n=input('3P or 4P system=');
m=input('m=');
l=input('l=');
w=input('w=');
h=input('h=');
Ix=((1/12)*m*((w^2)+(h^2)));
Iz=((1/12)*m*((l^2)+(h^2)));
M=[m 0 0; 0 Ix 0; 0 0 Iz];
a1=input('a1=');
a2=input('a2=');
b1=input('b1=');
b2=input('b2=');
k1=input('k1=');
c1=input('c1=');
k2=input('k2=');
c2=input('c2=');
k3=input('k3=');
c3=input('c3=');
for n=3
    k11=k1+k2+k3;
    k12=(-b1*k1)+(b2*k2);
    k13=(a1*k1)+(a1*k2)-(a2*k3);
    k21=k12;
    k22=(b1*b1*k1)+(b2*b2*k2);
    k23=(-a1*b1*k1)+(a1*b2*k2);
    k31=k13;
    k32=k23;
    k33=(a1*a1*k1)+(a1*a1*k2)+(a2*a2*k3);
    c11=c1+c2+c3;
    c12=(-b1*c1)+(b2*c2);
    c13=(a1*c1)+(a1*c2)-(a2*c3);
    c21=c12;
    c22=(b1*b1*c1)+(b2*b2*c2);
    c23=(-a1*b1*c1)+(a1*b2*c2);
    c31=c13;
    c32=c23;
    c33=(a1*a1*c1)+(a1*a1*c2)+(a2*a2*c3);
end
for n=4
    k4=input('k4=');
    c4=input('c4=');
    k11=k1+k2+k3+k4;
    k12=(-b1*k1)+(b2*k2)+(b2*k3)-(b1*k4);
    k13=(a1*k1)+(a1*k2)-(a2*k3)-(a2*k4);
    k21=k12;
    k22=(b1*b1*k1)+(b2*b2*k2)+(b2*b2*k3)+(b1*b1*k4);
    k23=(-a1*b1*k1)+(a1*b2*k2)-(a2*b2*k3)+(a2*b1*k4);
    k31=k13;
    k32=k23;
    k33=(a1*a1*k1)+(a1*a1*k2)+(a2*a2*k3)+(a2*a2*k4);
    c11=c1+c2+c3+c4;
    c12=(-b1*c1)+(b2*c2)+(b2*c3)-(b1*c4);
    c13=(a1*c1)+(a1*c2)-(a2*c3)-(a2*c4);
    c21=c12;
    c22=(b1*b1*c1)+(b2*b2*c2)+(b2*b2*c3)+(b1*b1*c4);
    c23=(-a1*b1*c1)+(a1*b2*c2)-(a2*b2*c3)+(a2*b1*c4);
    c31=c13;
    c32=c23;
    c33=(a1*a1*c1)+(a1*a1*c2)+(a2*a2*c3)+(a2*a2*c4);
end
K=[k11 k12 k13; k21 k22 k23; k31 k32 k33];
C=[c11 c12 c13; c21 c22 c23; c31 c32 c33];
a=input('a=');
b=input('b=');
dis0=input('dis0=');
disd0=input('disd0=');
phi0=input('phi0=');
phid0=input('phid0=');
theta0=input('theta0=');
thetad0=input('thetad0=');
fmin=input('fmin=');
fmax=input('fmax=');
A=-1;
B=(k11/m)+(k22/m)+(k33/m);
C=-((k11*k22)/(m*Ix))-((k11*k33)/(m*Iz))-((k22*k33)/(Ix*Iz))+(k12^2)+(k13^2)+(k23^2);
D=((k11*k22*k33)/(m*Ix*Iz))-((k11*(k23^2))/m)-((k22*(k13^2))/Ix)-((k33*(k12^2))/Iz)+(2*k12*k13*k23);
z=roots([A B C D]);
w1=sqrt(z(1))/(2*pi);
w2=sqrt(z(2))/(2*pi);
w3=sqrt(z(3))/(2*pi);
fprintf('Natural frequencies in Hz are %f, %f, %f \n',w1, w2, w3);
for f=fmin:fmax
    tp=1/f;
    dt=1/(25*f);
    fprintf('frequency= %f, tp= %f, dt= %f \n',f, tp, dt);
    for i=0:dt:2*tp
            t=i;
            j=i+2;
            fprintf('time= %f \n',t);
            fprintf('i= %f \n',i);
            fprintf('j= %f \n',j);
            f1=sin(2*pi*f*t);
            f2=a*sin(2*pi*f*t);
            f3=b*sin(2*pi*f*t);
            F=[f1; f2; f3];
            if i==0
%               Error is in the below line-
                x(j)=[dis0; phi0; theta0];
                xd(j)=[disd0; phid0; thetad0];
                xdd(j)=M\(F-(K*0)-(C*0));
            else
                break
            end
            x(j-1)=x(j)-(dt*xd(j))+(0.5*(dt^2)*xdd(j));
            x(j+1)=((M/(dt^2))+(C/(2*dt)))\(F-(K-((2*M)/(dt^2)))*x(j)-((M/(dt^2))-(C/(2*dt)))*x(j-1));
            fprintf('x= %f \n',x(j+1));
            xd(j)=(x(j+1)-x(j-1))/(2*dt);
            fprintf('xd= %f \n',xd(j));
            xdd(j)=(x(j+1)-(2*x(j))+x(j-1))/(dt^2);
            fprintf('xdd= %f \n',xdd(j));
            r1=[c1 -(c1*b1) c1*a1];
            r2=[k1 -(k1*b1) k1*a1];
            TF1(j)=(r1*xd(j))+(r2*x(j));
            fprintf('TF1= %f \n',TF1);
            s1=[c2 c2*b2 c2*a1];
            s2=[k2 k2*b2 k2*a1];
            TF2(j)=(s1*xd(j))+(s2*x(j));
            fprintf('TF2= %f \n',TF2);
            for n=3
                t1=[c3 0 -(c3*a2)];
                t2=[k3 0 -(k3*a2)];
                TF3(j)=(t1*xd(j))+(t2*x(j));
                fprintf('TF3= %f \n',TF3);
                TF(j)=TF1(j)+TF2(j)+TF3(j);
                fprintf('TF= %f \n',TF);
            end
            for n=4
                u1=[c3 c3*b2 -(c3*a2)];
                u2=[k3 k3*b2 -(k3*a2)];
                TF3(j)=(u1*xd(j))+(u2*x(j));
                fprintf('TF3= %f \n',TF3);
                v1=[c4 -(c4*b1) -(c4*a2)];
                v2=[k4 -(k4*b1) -(k4*a2)];
                TF4(j)=(v1*xd(j))+(v2*x(j));
                TF(j)=TF1(j)+TF2(j)+TF3(j)+TF4(j);
                fprintf('TF4= %f \n',TF4);
                fprintf('TF= %f \n',TF);
            end
    end
end

[Cris LaPierre]

You haven't made any of the changes i've already suggested yet.

[Neeraj Kumar]

That was the original code, without any changes, so that you can understand the intent better.

I tried doing this-

x(:,j)=[dis0; phi0; theta0];

and this-

x(y,z,j)=[dis0; phi0; theta0];

But the same error is coming.

[Steven Lord]

You have an egg carton and eggs that fit into the carton, one egg per cup in the carton. You're not allowed to break / scramble the eggs. How do you put three eggs into one cup of the egg carton?

That's effectively what you're trying to do with that line of code. In an expression A(ind) = B, A is the carton, ind selects some number of cups in the carton, and B is the collection of eggs you're trying to put in the cups.

Cris LaPierre suggested putting the eggs not in a single cup of the egg carton but in multiple cups, one per egg.

[Cris LaPierre]

This code does not create an error:

x(:,j)=[dis0; phi0; theta0];

It is the next line of code that does:

xd(j)=[disd0; phid0; thetad0];

You will need to make a similar change to handle xd and xdd. You will then need to modify everywhere in your code you use x(j) and its variations: x(j-1) and x(j+1), as well as xd(j) and xdd(j).

For xdd, you have a more complicated issue since M\(F-(K*0)-(C*0)) produces a 3x3 array. You might need to do something like

xdd(:,:,j)=M\(F-(K*0)-(C*0));

[Neeraj Kumar]

Thank you Cris for the solution and Thank you Steven for the analogy.