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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];

2 ビュー (過去 30 日間)
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);

回答 (2 件)

KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 6 月 25 日
編集済み: KALYAN ACHARJYA 2019 年 6 月 25 日
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 件のコメント
Advait Sinha
Advait Sinha 2019 年 6 月 25 日
I am still getting the same error
KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 6 月 25 日
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.

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Advait Sinha
Advait Sinha 2019 年 6 月 25 日
I am new to matlab so I’m finding it confusing and hard

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