how to repeat a loop?

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

[Sam Chak]

Do you mean to repeat the loop infinitely unless broken on Ctrl+C?

[arian hoseini]

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

Answer 1

[Image Analyst]

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

Answer 2

[dpb]

While you could, why not just duplicate the array as many times as needed once it's been generated --

z=repmat(z,2,1);

[Image Analyst]

Or, once it's been generated once, just use z. Why do another loop just to do the very same thing? Or replicate another identical row? I don't see the point of either.

[dpb]

Well, agreed it's not clear "why" at all here, indeed, but... :)

I presumed there was something else going to happen later that needed the size to be 2X the initial size. It's not at all unusual to end up duplicating data to match some other array size for later array or vector or matrix operations.

Of course, we see lots of instances where beginners duplicate stuff needlessly, too, ...

[arian hoseini]

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]

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]