フィルターのクリア

Constructing a repeating array for a binary blazed diffraction grating

4 ビュー (過去 30 日間)
Patrick Bevington
Patrick Bevington 2015 年 11 月 7 日
コメント済み: Star Strider 2015 年 11 月 9 日
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]

採用された回答

Star Strider
Star Strider 2015 年 11 月 7 日
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.
  4 件のコメント
Patrick Bevington
Patrick Bevington 2015 年 11 月 9 日
編集済み: Star Strider 2015 年 11 月 9 日
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 年 11 月 9 日
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?

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeMatrices and Arrays についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by