Constructing a repeating array for a binary blazed diffraction grating
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Patrick Bevington
il 7 Nov 2015
Commentato: Star Strider
il 9 Nov 2015
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]
0 Commenti
Risposta accettata
Star Strider
il 7 Nov 2015
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 Commenti
Star Strider
il 9 Nov 2015
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?
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Loops and Conditional Statements in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!