Creating a Matrix Output from Element Integration Inside Matrix

조회 수: 15 (최근 30일)
Ammar
Ammar 2024년 7월 28일
편집: Ammar 2024년 7월 29일
EDIT:
I made an edit to the description to clarify what I am trying to do.
Hello everyone,
I am trying to output a square matrix that is N x N in size. In my case, N = 3.
I have a variable that's called "Kernel". This variable is calculated as an equation in terms of "z" and "z_prime".
My values in z_prime will vary from L / 2N to (2N-1)*(L/2N) , in steps of "L/N".
My "z" variable will be used as an "integration variable" as seen below.
I am trying to calculate an output called the "Z Matrix" which will integrate the "Kernal" variable across specific ranges.
I attached a .pdf file to clarify my end goal.
See the .pdf file attached:
Z_Matrix.pdf
I attached my MATLAB Code for reference.
See MATLAB .m file attached:
Z_Calculation.m
In my attempt, I have tried to use for loops and creating function handles to make it easier to generate this matrix.
I know that there is a way to use "nested" for loops to generate a square matrix of an "N X N" size. However, I wasn't quite sure on how to implement that in MATLAB for my case.
I have tried to look through different questions/answers on the Mathworks Forum regarding square matrices. However, I wasn't able to find anything that was relevant to my case.
%I'm thinking of attempting something like this
% NOTE: This is not all correct MATLAB syntax, this is mostly just logic written out/pseudocode
% I am trying to see how I can implement this by using nested for loops
% Values for N and L
N = 3;
% Since this will be a square matrix, it will be an "M X N" matrix.
M = N;
L = 1/2;
% Range for z_prime
z_prime_start = L / (2*N);
delta_z_prime = L / N;
z_prime_end = (2N-1)*z_prime_start;
z_prime = z_prime_start:delta_z_prime:z_prime_end;
%{
The Ranges for the "z" variable that will be used to
integrate the "Kernel" variable
%}
z_start = 0;
delta_z = L/N;
z_end = L/N;
% Each "Row" will have a specific "z_prime value"
% Row 1 = Kernal(z,z_prime_1)
for i = 1:N
for j = 1:M
loop_z_prime = z_prime(i,j)
Kernal(i,j) = Kernal(z,loop_z_prime)
Z(i,j) = integral(Kernal(i,j),z_start,z_end)
end
end
% The "z" boundaries for integration will keep increasing by "L/N" until
% the very last element in the "Z" matrix.
  댓글 수: 1
dpb
dpb 2024년 7월 28일
"showing the type of matrix I am trying to generate"
By that graphic, each row is a repitition of the same value for however many columns...each cell in each row is
K(z,z_prime_N)
where N in z_prime_N is 1, 2, 3, ... but is the same for all columns in each row 1:N and you've said above that z is fixed. Hence, so will be K for each n in 1:N
Something's missing.
Per usual show us and actual numeric case and cut the size down to, say, 5x5; enough to demonstrate the algorithm and input/outputs,but not too much to look at on a screen.

댓글을 달려면 로그인하십시오.

답변 (1개)

dpb
dpb 2024년 7월 28일
Rereading the Q? title, I guess the above supposition I thought an error is actually what you're looking to do.
If that is the case, then the simple thing is just
N=23; % set the desired size
L=???; % set L
z=???; % and z
z_p=[L/(2*N):L/N:(2N-1)*(L/2N); % compute z prime vector
M=K(z,z_p); % and evaluate K for inputs (assume vectorized, if not, use explicit loop)
M=[M repmat(M,1,N-1); % add replicated columns to the desired size...
The last uses a key feature of MATLAB -- automagic reallocation on demand. The [] puts the two pieces, the vector M and the (N-1) copies of it together into a new M.
  댓글 수: 2
Ammar
Ammar 2024년 7월 28일
Hi dpb,
Thank you for the quick response. I appreciate the insight on this.
I guess I should've just clarified what my "end goal" was from the beginning.
The reason why I'm trying to generate a "matrix" for my "K" variable is so that I can perform an integration on each individual element.
See the .pdf file attached:
Z_Matrix.pdf
See my MATLAB code attached for reference:
Z_Calculation.m
Essentially, objective is to calculate a "Z Matrix" that integrates the "K" variables across different ranges for different elements.
% Assuming:
N = 3;
L = 1/2;
z_prime_start = L / (2*N);
delta_z_prime = L / N;
z_prime_end = ( (2*N) - 1 ) / z_prime_start
z_prime = z_prime_start:delta_z:z_prime_end
z_start = 0;
delta_z = L/N;
z_end = L;
%{
What I am thinking of doing is using a for loop for the "z" variables to
keep "increasing" the boundaries by "L/N", since I eventually want to reach
"L"
%}
dpb
dpb 2024년 7월 29일
" that integrates the "K" variables across different ranges"
See cumsum and trapz

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by