For Loop/Nested Loop - using solution to previous iteration in nested loop?

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Hello,
I'm trying to do a nested loop in which from the second iteration onwards the term zl of the function being iterated is substituted by the solution to the previous iteration. My code is underneath:
zl=(1/4.*((k.*a).^2))+(1j*0.6.*(k.*a));
%initiating loop
for i=2:5
for n=1:4
x(i-1)=(0.077/4)*(i-1);
S(i-1)=St*(exp(m*x(i-1)));
Z(i-1,n)=(zl(n)+B(i-1,n))./((1+zl(n))./S(i-1).*C(n));
end
end
So for the iteration 2 of Z, zl should corresponde to the first iteration of Z, for iteration 3 of Z, zl should correspond to the second iteration of Z and so forth. Hope I'm being clear. Any tips would be appreciated.

Answers (1)

Iman Ansari
Iman Ansari on 4 Apr 2013
Edited: Iman Ansari on 4 Apr 2013
Hello
You want for example in n=3 iteration zl be Z computed in n=2 iteration? But with this line your zl is the same for all iterations:
zl=(1/4*((k*a).^2))+(1j*0.6*(k*a));
You may need to write it before your loop:
%initiating loop
zl=(1/4*((k*a).^2))+(1j*0.6*(k*a));
for n=2:5
x(n)=(0.077/4)*(n-1);
S(n)=St*(exp(m*x(n-1)));
%terms for equation Z
A1=zl;
B1=(1j*(p0c./(S(n-1)))).*(tan(k*L));
C1=1;
D1=(1j*(zl./(p0c/(S(n-1))))).*(tan(k*L));
%equation Z composed of the previous terms
Z=(A1+B1)./(C1+D1);
zl=Z;
end
  3 Comments
Iman Ansari
Iman Ansari on 5 Apr 2013
With this code you use 1:4 elements of zl or Z but I think your zl was a vector with 10000 elements.
Before end of the loop you can use zl=Z to set the value of zl in next iteration, like my code.
catarina
catarina on 5 Apr 2013
Thanks for your help. Much simpler than I thought. The vector should be 10000 or any length really, I should be able to change that. Now I'm having trouble in plotting the results. I would like the final results of Z but the graph looks a bit weird. I end up with a Z matrix of Z(4,10000) so I thought I should plot(Z(1,:)) to plot the results of the first row since my code goes from the 4th iteration to the first. Any tips? cheers

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