F 01 f 11 2 1 1 0 similarlywehaveh k l

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Unformatted text preview: 2 1 1 0 h[m, n] 1 1 • Let us compute F [k , l ] [0, 0] 2 2 F [k , l ] H [k , l ] 14 6 sample F [0, 0] f [0, 0] f [1, 0] f [0,1] f [1,1] 2 1 1 4 F [1, 0] f [0, 0] f [1, 0] f [0,1] f [1,1] 2 1 1 2 2 F [k , l ] F [0,1] f [0, 0] f [1, 0] f [0,1] f [1,1] 2 1 1 2 4 F [1,1] f [0, 0] f [1, 0] f [0,1] f [1,1] 2 1 1 0 • Similarly, we have: H [k , l ] 1 1 3 Ghassan AlRegib © Georgia Tech 0 2 • • • • 1 1 25 Ghassan AlRegib © Georgia Tech Example ECE 6258 Fall 2012 1 i 0 j 0 i 0 j 0 h[0, 0] f [0, 0] h[0,1] f [0, 1] h[1, 0] f [ 1, 0] h[1,1] f [1, 1] 20 2 h[0, 0] f [0,1] h[0,1] f [0, 0] h[1, 0] f [1,1] h[1,1] f [1, 0] 1 2 3 1 1 [1,1] h[i, j ] f [1 i,1 j ] 1 [1, 0] h[i, j ] f [1 i, j ] i 0 j 0 i 0 j 0 h[0, 0] f [1, 0] h[0,1] f [1, 1] h[1, 0] f [0, 0] h[1,1] f [0, 1] 1 0 2 3 Ghassan AlRegib © Georgia Tech 26 [0,1] h[i, j ] f [i,1 j ] 1 [0, 0] h[i, j ] f [i, j ] 1 1 i 0 j 0 1 1 4 2 Take the inverse DFT gives 6 2 [m, n] Compare this with ? NOT IDENTICAL Because of aliasing Next calculate [m, n] [m, n] h[i, j ] f [m i, n j ] (m, n) [0 :1;0 :1] Example ECE 6258 Fall 2012 Example ECE 6258 Fall 2012 27 h[0, 0] f [1,1] h[0,1] f [1, 0] h[1, 0] f [0,1] h[1,1] f [0, 0] 0 11 0 2 Ghassan AlRegib © Georgia Tech 28 7 9/13/2012 Example ECE 6258 Fall 2012 3 2 [m, n] 2 3 [m, n • Notice that ] g[m, n] M 1 N 1 mk F [k , l ] f [m, n]WM WNnl where j 2 WN exp m 0 n 0 N • Example: Repeat above example but by computing 4 4 g DFT to find [m, n] h h Ghassan AlRegib © Georgia Tech ECE 6258 Fall 2012 • Let us revisit the expression: M 1 N 1 mk F [ k , l ] f [m, n]WNnl WM m0 n0 N 1 • Define G[m, l ] f [m, n]WNnl M 1 M N • Repeating this for samples M N • One sample needs multiplications and M N addition M 2 N 2 is the computation cost of the above expression 29 Row‐Column Algorithm n 0 2‐D Fast Fourier Transform (FFT) ECE 6258 Fall 2012 the th column of is the 1‐D m G m DFT of the th column of f mk • Thus F [k , l ] G[m, l ]WM m 0 Ghassan AlRegib © Georgia Tech ECE 6258 Fall 2012 30 Row‐Column Algorithm • Now, we can compute the 2‐D DFT by either 1‐D DFT of Column, then 1‐D DFT of Row or 1‐D DFT of Row, the 1‐D DFT of Column • If M and N are powers of 2, then we can compute the 1‐D DFT using FFT MN # of multiplication: log 2 MN 2 # of addition: MN log 2 MN l l For a value of , the th row of is F the 1‐D DFT of the th row of G l Ghassan AlRegib © Georgia Tech 31 Ghassan AlRegib © Georgia Tech 32 8 9/13/2012 Example ECE 6258 Fall 2012...
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This note was uploaded on 02/14/2014 for the course ECE 6258 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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