How to store value of for loop in a array

Hello everyone,

I am trying to store value of A in an array. But some how it stops after 3 itrations at z = 0.03.

Why So? If anybody can have a look where I going wrong.

A = zeros(1,31);
aes = 2;
counter= 1;
for z= 0:0.01:0.3
    A(1, counter) = 1/(1+((z/aes)^2));
    counter = counter+ 1;
end
A;

Thanks in Advance

Answer 1

[Adam Danz]

"some how it stops after 3 itrations at z = 0.03"

That's actually not true. In your script, z contains 4 values and the loop has 4 iterations

% First 7 values of A from your version
>> A(1:7)
ans =
            1      0.99998       0.9999      0.99978            0            0            0

You're initializeing A as a 1x31 vector of zeros which is why there are extra value in the output.

There's nothing wrong with your loop but I find it more intuitive to loop over intergers 1:n rather than looping over elements of a vector directly. Consider this version,

z= 0:0.01:0.03;
A = zeros(size(z));
aes = 2;
counter= 1;
for i= 1:numel(z)
    A(i) = 1/(1+((z(i)/aes)^2));
end

Result:

>> A
A =
            1      0.99998       0.9999      0.99978

If you intended to have 31 iterations, define z in the first line of my version as

z = linspace(0,.03,31)
% or 
z = 0 : 0.001: 0.03;

and then run the rest of my version.

Result:

A =
  Columns 1 through 11
            1            1            1            1            1      0.99999      0.99999      0.99999      0.99998      0.99998      0.99998
  Columns 12 through 22
      0.99997      0.99996      0.99996      0.99995      0.99994      0.99994      0.99993      0.99992      0.99991       0.9999      0.99989
  Columns 23 through 31
      0.99988      0.99987      0.99986      0.99984      0.99983      0.99982       0.9998      0.99979      0.99978

[Jay Talreja]

Thanks it worked.

Here in the code the range was from 0.0 to 0.3 with interval of 0.01.

But it worked for me. Thankyou!!!!

[Adam Danz]

Opps, maybe it's time for me to get new glasses. 🤓

Answer 2

[David Hill]

A = zeros(1,31);
aes = 2;
counter= 1;
for z= 0:0.001:0.03%I believe you want the interval to be .001 (otherwise loop only runs for 4 times)
    A(1, counter) = 1/(1+((z/aes)^2));
    counter = counter+ 1;
end

Answer 3

[madhan ravi]

 for z = linspace(0, 0.03, 31)

By the way you don’t need a Loop it’s simply:

 A = 1 ./ (1 + ((z / aes) .^ 2))