Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 2-by-1. Error in linear2d (line 7) z(:,1)=[0;1];

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close all;
dt = 0.1;%time step
u1=1;
u2=1;
c = u2-u1;
u = sign(u2-u1);
z(:,1)=[0,1];
k=0.12;
m=2.21;%initial parameters
v=1;
a=1.4;
tend=1000;
t=0;
i=1
while t < tend-2*dt
vh=v(i)-dt*k*z(1,i)/ (2*m);
z(2,i) = z(1,i)+ dt*vh;
%a(i+1)= -k*z(i)/2;
v(i+1)= vh-dt*k*z(2,i)/ (2*m);% + dt*a(i+1)/2;
i = i+1;
t = t + dt
end
plot(z);

Risposte (2)

KALYAN ACHARJYA
KALYAN ACHARJYA il 25 Giu 2019
Modificato: KALYAN ACHARJYA il 25 Giu 2019
You defined v=1, as scalar, but you called the function as vector v(i)
vh=v(i)-dt*k*z(1,i)/ (2*m);
%....^.........
When i=1, then v(1)=??
When i=2, then v(2)=?? in iterations.
Also the while loops runs multiple times
As t < tend-2*dt, where t=0 and tend=1000, dt=0.1, On such case the size of z also not sufficient
vh=v(i)-dt*k*z(1,i)/ (2*m);
%...............^.........
Because z(:,1)=[0,1] allows only two iterations.
Requested to read about array indexing.
  2 Commenti
KALYAN ACHARJYA
KALYAN ACHARJYA il 25 Giu 2019
Yes, I have hinted where the error is, I have no idea what you doing inside the code. You have to properly define the v and z or you have to change the code. Please try to undestand why you getting the error? Probably you may solve it afterwards.

Accedi per commentare.


Advait Sinha
Advait Sinha il 25 Giu 2019
I am new to matlab so I’m finding it confusing and hard

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