How to fix looping?

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Devia Rafika Putri
Devia Rafika Putri on 15 May 2013
this is my program:
v = 120;
r1 = 1.8;
x1 = 2.4;
r2 = 3.5;
x2 = 1.2;
xm = 60;
ns = 1800;
ws = 188.5;
s = (0.5:1:50)/50;
s (1) = 0.0001;
nm = (1-s)*ns;
for i = 1:51;
zf(i) =(((r2/s(i))+(j*x2))*(j*xm))/(((r2/s(i))+(j*x2))+(j*xm));
zb(i) = (((r2/(2-s(i)))+(j*x2))*(j*xm))/(((r2/(2-s(i)))+(j*x2))+(j*xm));
I(i) = v/(r1+(j*x1)+(0.5*zf(i))+(0.5*zb(i)));
PagF(i) = (abs(I(i)^2))*(0.5*real(zf(i)));
PagB(i) = (abs(I(i)^2))*(0.5*real(zb(i)));
Pag(i)=PagF(i)-PagB(i);
Tind(i) = Pag(i)/ws;
end
figure (1);
plot(nm,Tind,'Color','b','LineWidth',2.0);
grid on;
hold off;
*but after i run this program an error occurred with argument: Attempted to access s(51); index out of bounds because numel(s)=50.
Error in ==> fasbel3 at 15 zf(i) =(((r2/s(i))+(j*x2))*(j*xm))/(((r2/s(i))+(j*x2))+(j*xm));
how to fix it? please help thanks.. with sincerity :)
  4 Comments
Craig Cowled
Craig Cowled on 15 May 2013
Devia, Andrei's code is much neater. No need for a loop, just use element wise divide './'. And no, I'm not an electrical engineer. I'm a structural engineer working on experimental structural dynamics problems.
Devia Rafika Putri
Devia Rafika Putri on 15 May 2013
ok craig.. i see.. but some code from andrei, i still dont understand.. but i'll find it..
ooh, I thought you were an electrical engineer
because I'm having problems with programs relating to single phase motor..
thanks so much craig..you are very helpful..

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Accepted Answer

Andrei Bobrov
Andrei Bobrov on 15 May 2013
Edited: Andrei Bobrov on 15 May 2013
v = 120;
r1 = 1.8;
x1 = 2.4;
r2 = 3.5;
x2 = 1.2;
xm = 60;
ns = 1800;
ws = 188.5;
s = [.0001;(.5:50).'/50];
s1 = [2-s,s];
z =(r2./s1+1i*x2)*1i*xm./( r2./s1+1i*(x2+xm) );
I = v./(r1+1i*x1+mean(z,2));
Tind = diff(bsxfun(@times,abs(I.^2),real(z)*.5),1,2)/ws;
plot((1-s)*ns,Tind,'Color','b','LineWidth',2.0);
grid on;
  3 Comments
Andrei Bobrov
Andrei Bobrov on 15 May 2013
Hi Devia!
Please read about functions of the MATLAB: bsxfun, diff.
Rewrite code:
s = [.0001;(.5:50).'/50];
s1 = [2-s,s];
z =(r2./s1+1i*x2)*1i*xm./( r2./s1+1i*(x2+xm) );
I = v./(r1+1i*x1+mean(z,2));
Tind = diff(bsxfun(@times,abs(I.^2),real(z)*.5),1,2)/ws;
as:
s = [.0001;(.5:50).'/50];
s1 = [2-s,s];
z =(r2./s1+1i*x2)*1i*xm./( r2./s1+1i*(x2+xm) );
I = v./(r1+1i*x1+mean(z,2));
Ia2 = abs(I.^2);
r = real(z)*.5;
P = bsxfun(@times,Ia2,r); % as in your code: P = [PagB PagF];
Pag = diff(P,1,2); % as Pag = PagF - PagB;
Tind = Pag/ws;
Devia Rafika Putri
Devia Rafika Putri on 15 May 2013
Hi Andrei, thank you so much for your explain.. i'm so grateful.
but actually I have another problem. is about 1 phase motors, can you help me?
if you can, I will send my problem to your email..
thanks so much, Andrei..
with sincerity Devia

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More Answers (1)

Yao Li
Yao Li on 15 May 2013
It seems the length of array s is 50, but you wanna call s(51) in the for loop
  4 Comments
Yao Li
Yao Li on 15 May 2013
Sometimes, I act as an electrical engineer. lol.
Pls. feel free to contact me if u have other questions. Email preferred.
Devia Rafika Putri
Devia Rafika Putri on 15 May 2013
lol.. thanks so much Yao li, I've sent an email to you .. if you're not busy please check it..

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