# hw6s - N 1 k 1 n 1 e-j 2 π N 2 k 2 n 2 X k 1,k 2 = N 1-1 X...

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ECE 178: HW #6 Solutions Q1 . origlenna=imread( 'lena.gif' ); origlenna = im2double(origlenna); figure;imshow(origlenna); X = fft2(origlenna); figure; imshow(abs(ifft2(X./abs(X))),[]); title( 'Phase only reconstructed signal' ); figure; imshow(abs(ifft2(abs(X)))); title( 'Magnitude only reconstructed signal' ); Q2 . imshow(abs(ifft2(X. * lpfilter(filter type,size(X,1),size(X,2),D0)))); Here, you need to specify the filter_type by “ ideal ”, “ btw ” or “ gaussian ”. D0 should also be replaced with the cut-oﬀ frequencies provided in the question. Q3 . X 1 [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x 1 [ n 1 ,n 2 ] e - j 2 π N 1 k 1 n 1 e - j 2 π N 2 k 2 n 2 X 2 [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x 2 [ n 1 ,n 2 ] e - j 2 π N 1 k 1 n 1 e - j 2 π N 2 k 2 n 2 Y [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 ( αx 1 [ n 1 ,n 2 ] + βx 2 [ n 1 ,n 2 ]) e - j 2 π N 1 k 1 n 1 e - j 2 π N 2 k 2 n 2 Y [ k 1 ,k 2 ] = αX 1 [ k 1 ,k 2 ] + βX 2 [ k 1 ,k 2 ] Linear. 1

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Q4 . X [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x [ n 1 ,n 2 ] e - j 2 π N 1 k 1 n 1 e - j 2 π N 2 k 2 n 2 Y [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x [(( n 1 - m 1 )) N 1 , (( n 2 - m 2 )) N 2 ] e - j 2 π N 1 k 1 n 1 e - j 2 π N 2 k 2 n 2 Y [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x [ n 1 ,n 2 ] e - j 2 π N 1 k 1 ( n 1 - m 1 ) e - j 2 π N 2 k 2 ( n 2 - m 2 ) Y [ k 1 ,k 2 ] = X [ k 1 ,k 2 ] W k 1 m 1 N 1 W k 2 m 2 N 2 Q5 . X [ k 1 ,k 2 ] = N 1 - 1 X n 1 =0 N 2 - 1 X n 2 =0 x [ n 1 ,n 2 ] e - j 2 π
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Unformatted text preview: N 1 k 1 n 1 e-j 2 π N 2 k 2 n 2 ( X [ k 1 ,k 2 ]) * = N 1-1 X n 1 =0 N 2-1 X n 2 =0 ( x [ n 1 ,n 2 ]) * e j 2 π N 1 k 1 n 1 e j 2 π N 2 k 2 n 2 ( X [ k 1 ,k 2 ]) * = N 1-1 X n 1 =0 N 2-1 X n 2 =0 x [ n 1 ,n 2 ] e j 2 π N 1 k 1 n 1 e j 2 π N 2 k 2 n 2 X * [ k 1 ,k 2 ] = X [((-k 1 )) N 1 , ((-k 2 )) N 2 ] Q6 . By just looking at the magnitude of the Fourier transform we can determine the energy on speciﬁc frequency bands. If the energy contained in high frequency bands is high then we can say that there is an object with high frequency components. However, this information is not enough to localize it in the spatial domain. 2...
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## This note was uploaded on 06/12/2009 for the course ECE 178 taught by Professor Manjunath during the Winter '08 term at UCSB.

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hw6s - N 1 k 1 n 1 e-j 2 π N 2 k 2 n 2 X k 1,k 2 = N 1-1 X...

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