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how can I create a matrix from calculated answers

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I have this answers and I want to put them in a diagonal matrix.Pls help me!
y =
0.0025 + 0.0155i
y =
0.0030 + 0.0171i
y =
0.0038 + 0.0191i
y =
0.0049 + 0.0216i
y =
0.0066 + 0.0248i
y =
0.0092 + 0.0289i
y =
0.0137 + 0.0344i
y =
0.0220 + 0.0414i
y =
0.0388 + 0.0487i
y =
0.0717 + 0.0450i
y =
1000000
y =
0.0717 - 0.0450i
y =
0.0388 - 0.0487i
y =
0.0220 - 0.0414i
y =
0.0137 - 0.0344i
y =
0.0092 - 0.0289i
y =
0.0066 - 0.0248i
y =
0.0049 - 0.0216i
y =
0.0038 - 0.0191i
y =
0.0030 - 0.0171i
y =
0.0025 - 0.0155i
  2 Comments
James Tursa
James Tursa on 26 Feb 2011
Can you show the code that generates these numbers? You can pre-allocate a result and then assign them to the appropriate spots in the result, or maybe there is a better vectorized solution, but we can't tell without seeing your code.
P
P on 26 Feb 2011
clear all;
f=1E3;
L1=1E-3;
L2=100;
Rs=1E-6;
R1=10;
w=2*pi*f;
k=10;
z1=eye(21);
for a = -11:k-1
omega = w*(a+1);
z=Rs+[j*omega*L2*((R1+j*omega*L1))]/[R1+j*omega*(L1+L2)];
y=1/z
end

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Accepted Answer

Matt Fig
Matt Fig on 26 Feb 2011
Don't you think it would help if you showed us how that "this answers" was generated? It really would.
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EDIT
O.k., here is one way, with your FOR loop:
clear all
f=1E3;
L1=1E-3;
L2=100;
Rs=1E-6;
R1=10;
w=2*pi*f;
k=10;
z1=eye(21); % I assume you want this to be the matrix.
cnt = 1;
for a = -11:k-1
omega = w*(a+1);
z1(cnt,cnt)=1/(Rs+(j*omega*L2*((R1+j*omega*L1)))/(R1+j*omega*(L1+L2)));
cnt = cnt + 1;
end
Or, simpler:
f=1E3;
L1=1E-3;
L2=100;
Rs=1E-6;
R1=10;
w=2*pi*f;
k=10;
omega = w*((-11:k-1)+1);
z1_2 = diag(1./(Rs+(j*omega*L2.*((R1+j*omega*L1)))./(R1+j*omega*(L1+L2))));
Now if you compare z1 and z1_2, you will see that they are equal.
  2 Comments
Matt Tearle
Matt Tearle on 26 Feb 2011
??? Undefined function or variable read_my_mind
Stupid MATLAB.

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More Answers (2)

P
P on 26 Feb 2011
here is the code, I've tried to multiply "y" with "z1" but it was a bad idea:)
clear all; f=1E3; L1=1E-3; L2=100; Rs=1E-6; R1=10; w=2*pi*f; k=10; z1=eye(21);
for a = -11:k-1 omega = w*(a+1); z=Rs+[j*omega*L2*((R1+j*omega*L1))]/[R1+j*omega*(L1+L2)]; y=1/z
end

Matt Tearle
Matt Tearle on 26 Feb 2011
M = zeros(n);
for k = 1:n
% calculate y
M(k,k) = y;
end
Or, preferably, if you can calculate all the y values in a vector, as explained here you can then do
M(1:n+1:end) = y;
Edit: fixed typo ( n+1 instead of n )
  4 Comments
P
P on 26 Feb 2011
Thanks a lot, finally I can go on with my project.
Matt Tearle
Matt Tearle on 27 Feb 2011
Oops, thanks James for catching the typo.

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