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?

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Neeraj Kumar 2020-9-15

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评论： Neeraj Kumar 2020-9-15

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

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Answer 1

Cris LaPierre 2020-9-15

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在 MATLAB Online 中打开

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.

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Neeraj Kumar 2020-9-15

在 MATLAB Online 中打开

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 2020-9-15

编辑：Cris LaPierre 2020-9-15

在 MATLAB Online 中打开

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 2020-9-15

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

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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?

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