Constructing a repeating array for a binary blazed diffraction grating

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Patrick Bevington
Patrick Bevington 2015-11-7
评论： Star Strider 2015-11-9
How can I build an array given the requirements below? The purpose of this is to construct a binary blazed diffraction grating
for an array NxM of A(i,j):
- for A(1,1), A(1,2), A(1,3) = 1 and A(1,4), A(1,5), A(1,6) = 0, repeat these 6 characters for A(1,M-5), A(1,M-4), A(1,M-3) = 1 and A(1,M-2), A(1,M-1), A(1,M) = 0.
- for A(2,1), A(2,2) = 1 and A(2,3), A(2,4), A(2,5), A(2,6) = 0, repeat these 6 characters for A(2,M-5), A(2,M-4) = 1 and A(2,M-3) A(2,M-2), A(2,M-1), A(2,M) = 0.
- for A(3,1) = 1 and A(3,2), A(3,3), A(3,4), A(3,5), A(3,6) = 0, repeat these 6 characters for A(3,M-5) = 1 and A(2,M-4), A(3,M-3), A(3,M-2), A(3,M-1), A(3,M) = 0
- Repeat the above 3 steps for N rows
i.e for a 12x12 array
A = [1 1 1 0 0 0 1 1 1 0 0 0;
1 1 0 0 0 0 1 1 0 0 0 0;
1 0 0 0 0 0 1 0 0 0 0 0;
1 1 1 0 0 0 1 1 1 0 0 0;
1 1 0 0 0 0 1 1 0 0 0 0;
1 0 0 0 0 0 1 0 0 0 0 0;
1 1 1 0 0 0 1 1 1 0 0 0;
1 1 0 0 0 0 1 1 0 0 0 0;
1 0 0 0 0 0 1 0 0 0 0 0;
1 1 1 0 0 0 1 1 1 0 0 0;
1 1 0 0 0 0 1 1 0 0 0 0;
1 0 0 0 0 0 1 0 0 0 0 0]

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Star Strider
Star Strider 2015-11-7
Use the repmat function:
Element = [1 1 1 0 0 0;
1 1 0 0 0 0;
1 0 0 0 0 0];
A = repmat(Element, 4, 2)
This takes the ‘Element’ array and duplicates it to create 4 copies vertically and 2 copies horizontally.
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Patrick Bevington
Patrick Bevington 2015-11-9
编辑：Star Strider 2015-11-9
Sorry I can see that my question was not clear, I meant can you think of a way to construct the repeating ' Element' matrix using loops?
The application of this is to make a saw toothed diffraction pattern, easy enough with a linear grating (example here, not yet sawtoothed: https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTsn5XzQAXdjDq6RffoF2yZOH_BEGAkZNqz2yEnwIzljOn1J8OV ), however I would like to do this for a forked grating (example here: https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcTPyhsftJxNmhnfT2XG248q_sK0Q_kve_KLl9haWQoyqTZBLFPg ). I can provide code for creating either pattern if you think this would be helpful.
Thank you for your continued assistance.
Star Strider
Star Strider 2015-11-9
This is much less efficient than using repmat, producing the same result:
RowRepeat = 4;
ColRepeat = 2;
Grating = [];
for k1 = 1:RowRepeat
Grating = [Grating; Element];
end
for k1 = 1:ColRepeat-1
Grating = [Grating Grating(:,1:size(Element,2))];
end
Grating % View Result
The code would be helpful because I have no idea what you’re doing.
If you already have the code for what you want to do, why not just use it?

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