# forming a matrix with given entries

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Commentato: MADHVI il 20 Dic 2023
correct the use of if condition in the following code:
function [A,B] = fdm1(H,xi,a,b,N)
H = 1; xi = 0.1;
a= -1; b = 1; N=5;
h = (b-a)/N; h2 = h*h;
A = zeros(20*(N-1),20*(N-1));
B = zeros(20*(N-1),20*(N-1));
K = 10;
for k = 1:K;
A(2*(k-1)*(N-1)+1,2*(k-1)*(N-1)+1:2*(k-1)*(N-1)+2) = [-(2+(k*pi*h/H)^2) 1];
A(2*(k-1)*(N-1)+N,2*(k-1)*(N-1)+N:2*(k-1)*(N-1)+N+1) = [-(2+(k*pi*h/H)^2) 1];
A(2*(k-1)*(N-1)+N,2*(k-1)*(N-1)+2) = [h/2]
for n = 1:K;
if (n~=k) && any(n+k == 1:2:K) %ismember(n+k,1:2:K)
x = sum((2 * H * h * n * k * ((-1)^(n + k) - 1)) / ((n^2 - k^2)^2 * pi^2 * xi));
A(2*(k-1)*(N-1)+N,2*(n-1)*(N-1)+2) = [x]; end
B(2*(k-1)*(N-1)+1,2*(k-1)*(N-1)+2) = [h/2];
end
end
##### 2 CommentiMostra NessunoNascondi Nessuno
Dyuman Joshi il 4 Nov 2023
Modificato: Dyuman Joshi il 4 Nov 2023
"correct the use of if condition in the following code:"
How would we know what is the correct use?
You will need to specify more than just simply showing the code.
What is the objective here? What is supposed to be the "correct" use?
What is the expected output?
Thank you for the response.
The objective is attached in the file.

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### Risposte (1)

prabhat kumar sharma il 20 Dic 2023
I assume that you are trying to construct matrices `A` and `B` for a finite difference method (FDM) and are having trouble with the summation part of the second equation within the `if` condition. The code provided seems to have a few issues that need to be corrected, including the placement of the `if` block and the summation logic.
1. The `if` block should be properly nested within the `for` loops.
2. The `sum` function is not being used correctly. If you're trying to compute a sum based on the condition, you should accumulate the sum manually within the loop.
3.It's not clear what the indices for `A` and `B` should be in the summation part, but I'll assume you're trying to update the `(k,n)` element based on the condition.
function [A,B] = fdm1(H,xi,a,b,N)
% H = 1; xi = 0.1;
% a= -1; b = 1; N=5;
h = (b-a)/N; h2 = h*h;
A = zeros(20*(N-1),20*(N-1));
B = zeros(20*(N-1),20*(N-1));
K = 10;
for k = 1:K
A(2*(k-1)*(N-1)+1,2*(k-1)*(N-1)+1:2*(k-1)*(N-1)+2) = [-(2+(k*pi*h/H)^2) 1];
A(2*(k-1)*(N-1)+N,2*(k-1)*(N-1)+N:2*(k-1)*(N-1)+N+1) = [-(2+(k*pi*h/H)^2) 1];
A(2*(k-1)*(N-1)+N,2*(k-1)*(N-1)+2) = h/2; % Removed brackets for scalar assignment
B(2*(k-1)*(N-1)+1,2*(k-1)*(N-1)+2) = h/2; % Moved outside of the inner loop
for n = 1:K
if (n ~= k) && any(n+k == 1:2:K) % ismember(n+k,1:2:K)
x = (2 * H * h * n * k * ((-1)^(n + k) - 1)) / ((n^2 - k^2)^2 * pi^2 * xi);
A(2*(k-1)*(N-1)+N,2*(n-1)*(N-1)+2) = x; % Corrected the assignment
end
end
end
end
I hope it helps!
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Thanks for your help. This issue has been resolved.

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