how to repeat a loop?

3 visualizaciones (últimos 30 días)
arian hoseini
arian hoseini el 13 de Jun. de 2022
Comentada: Image Analyst el 16 de Jun. de 2022
LD = [1 2 0.004 1 0.05i 0 100
1 3 0.0057 2 0.0714i 0 70
3 4 0.005 3 0.0563i 0 80
4 5 0.005 4 0.045i 0 100
5 6 0.0045 5 0.0409i 0 110
2 6 0.0044 6 0.05i 0 90
1 6 0.005 7 0.05i 0 100];
PTR= 150e3/110;
CTR= [240 240 160 240 240 240 160 240 160 240 240 240 240 160];
for i=1:7
% for n=8:14
z(i,1)=(LD(i,3)+LD(i,5))*LD(i,7);
% z(n,1)=(LD(i,3)+LD(i,5))*LD(i,7);
% end
theta = angle(z(i,1));
z = abs(z)
end
z = 5.0160
z = 2×1
5.0160 5.0139
z = 3×1
5.0160 5.0139 4.5217
z = 4×1
5.0160 5.0139 4.5217 4.5277
z = 5×1
5.0160 5.0139 4.5217 4.5277 4.5261
z = 6×1
5.0160 5.0139 4.5217 4.5277 4.5261 4.5174
z = 7×1
5.0160 5.0139 4.5217 4.5277 4.5261 4.5174 5.0249
% for i= 1:14
% zsz1(1,i) = ((z(i,1))/(cos((theta(i,1)-45)*pi/180))*(CTR(1,i)/PTR));
% end
i need my loop to repeat again i want answers to be exactly like the first seven z(i,1)...thank u
  2 comentarios
Sam Chak
Sam Chak el 13 de Jun. de 2022
Do you mean to repeat the loop infinitely unless broken on Ctrl+C?
arian hoseini
arian hoseini el 13 de Jun. de 2022
no sir i need them like this(14 ans)
5.0160
5.0139
4.5217
4.5277
4.5261
4.5174
5.0249
5.0160
5.0139
4.5217
4.5277
4.5261
4.5174
5.0249

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Respuesta aceptada

Image Analyst
Image Analyst el 13 de Jun. de 2022
If you simply want to repeat the loop while printing out exactly the same values each time, then do this:
for k = 1 : 2
for i=1:7
z(i,1)=(LD(i,3)+LD(i,5))*LD(i,7);
theta = angle(z(i,1));
z = abs(z)
end
end

Más respuestas (1)

dpb
dpb el 13 de Jun. de 2022
Editada: dpb el 13 de Jun. de 2022
While you could, why not just duplicate the array as many times as needed once it's been generated --
z=repmat(z,2,1);
  4 comentarios
arian hoseini
arian hoseini el 16 de Jun. de 2022
well thank u all ...first why do another loop?the ans is i have to calculate line impedance so i have 1 to 2 line and 2 to 1 line ..thats why i need this to be repeated as u see here
%n fromline toline impedance
B=[1 2 1 1 z(1)
2 1 3 2 z(2)
3 3 4 3 z(3)
4 4 5 4 z(4)
5 5 6 5 z(5)
6 6 2 6 z(6)
7 6 1 7 z(7)
8 1 2 1 z(1)
9 3 1 2 z(2)
10 4 3 3 z(3)
11 5 4 4 z(4)
12 6 5 5 z(5)
13 2 6 6 z(6)
14 1 6 7 z(7)];
Image Analyst
Image Analyst el 16 de Jun. de 2022
If you have the left 4 columns of B already, you could just tack on two copies of z:
z = 1:7;
B=[1 2 1 1
2 1 3 2
3 3 4 3
4 4 5 4
5 5 6 5
6 6 2 6
7 6 1 7
8 1 2 1
9 3 1 2
10 4 3 3
11 5 4 4
12 6 5 5
13 2 6 6
14 1 6 7];
z2 = [z(:); z(:)];
B = [B, z2]

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