1.1) (15 pts.) Compute the DFT for the following two vectors with length N=4:

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1 Problem 1 (5 pts) 11) (15 pts) Compute the DFT for the following two vectors with length N=4: x = 1, x 2 = ) (15 pts) Compute the two-dimensional DFT with size M=N=4 for the following 4x4 stripe image x 3 5 = , as shown in (a) 5 13) (2 pts) Generate a 4x4 checker board pattern from the stripe image above by letting T x4 6x3^2 + 8x3 =, here ^ denotes element-wise square operation on matrix x 3, as shown in (b) Compute the Fourier transform of x 4 from that of x (a) N=4 (b) N=4 Solution: 11) 3 j2πuk / 4 1 ] = [1/ 4 1 e, k =,1,2,3] = u = DFT [ x [2 ] T

2 DFT [ x 2 ] = [1 1 ] note: all credits are given for correct calculation but different normalizing factors (eg 1/N instead of 1/sqrt(N)), same applies below 12) First perform DFT for each row: using DFT x ] we get [2 ] in the first and third row and zero elsewhere Then perform DFT for each column, using DFT x ] we get [ 1 [ ) note that x 3^2 = 5* x 3, then DFT 2[ x4] = DFT[6x3^2 + 8x3 T 11 ] = 3 8 The convolution/product property of FFT, but that s obviously more cumbersome

3 Problem 2 (5 pts) 21) (1 pts) An observed image is affected by convolution with a Point Spread Function h(, plus random additive noise n(, so that g( = f( * h( + n( Describe how the image is restored using the Inverse Filtering method Give a mathematical expression for the noise n'( in the restored image f'( ie f'( = f( + n'( The distorted image g( shown below is affected by constant speed motion plus noise Hence h( is a square impulse function in x Below right is shown the transfer function H( which is the Fourier transform of h(, also shown is a slice through this transfer function the first local minimum of this function is -217 and the first local maximum has value 128) To restore the image the Pseudo-Inverse method is used with two different values of the minimum amplitude of H, e= 25 and e= 15 Copy the H( ) slice and draw on the restoring filter slices ) for both values of e

4 22) (15 pts) After applying the Pseudo-Inverse filter for e=25 the restored image equals the true image convolved by a residual PSF plus noise Give a drawing of the residual PSF 23) (15 pts) After applying the Pseudo-Inverse filter for e=15 what is the shape of the residual transfer function? Give a drawing of the residual PSF with which the restored image is convolved 24) (1 pts) Compare the residual random noise in the three restorations in (a-c) above in terms of total noise power and justify your ranking Solution: 21) n '( = IFFT ( FFT ( n( ) / FFT ( h( ) ) 22) f '( = ( f ( H ( + n( ) pinv( H ( ) where pinv( H ( ) is the pseudo-inverse of H ( so f '( ( f ( H ( + n( ) / H (, when H ( > e, ( f ( + n( / H (, when H ( > e, We can rewrite it as: Here, R ( is: f '( = f ( + ( n( / H ( ) 1, if H ( > e, Note e = 25 And roughly estimated from the figure of H () So should be: When u < 2, H ( ) > e = 25, we find:

5 Hence 1, when u < 2, f '( = f ( * r( + n'(, where r ( = IFFT( ) ) r() 23) Similarly as in b), however, the only difference is e = 15 here So From the figure of H (, we know: 1, if H ( > 15, H ( > 15 when 55 < u < 45 or 25 < u < 25 or 45 < u < 55, So should be: The residual PSF: 1, when 55 < u < 45 or 25 < u < 25 or 45 < u < 55, r ( = IFFT( )

6 ) r() 24) From the discussion above, we know f '( = f ( + ( n( / H ( ) The total noise should be f '( f ( = f ( ( 1) + ( n( / H ( ) If only considering the power of the second part ( n ( / H ( ), it is: a) > c) > b) However, for the first part: f ( ( 1), the noise power is: a) < c) < b) (In fact, we can consider two extreme cases, e = and e = 1 When e = 1, is always equal to, so the second part ( n( / H ( ) is zero, but the first part f ( ( 1) would be large; When e =, is always equal to 1, hence the first part f ( ( 1) is zero, but the second part ( n ( / H ( ) would be large So in practice, the threshold e in pseudo inverse method should be neither too small nor too large, but carefully chosen with an appropriate value ) Problem 3 (5 pts, Optional program problem) In this homework, we want to analyze the energy distributions of different types of images A zip pack of the four images used for experiments can be downloaded here

7 31) [2%] First, convert the input M-by-N color image to the grayscale format Plot the 2-D log magnitude of the 2D DFT and DCT of the grayscale image, with center shifted Visually compare and comment on the similarity/differences among the images using the two transforms 32) [2%] Apply the truncation windows discussed in the class to keep 25% and 625% (1/4 and 1/16) of the DFT and DCT coefficients, ie two different ratios for each transform This truncation is done by keeping the coefficients of the lowest frequencies (those within a centered smaller rectangle of (M/2)x(N/2) and (M/4)x(N/4) on the shifted FFT, respectivel Apply the 2D inverse DFT to reconstruct the image for each of the truncated spectra Compute the Signal-to-Noise-Ratio (SNR) value for each of the reconstructed images Plot the reconstructed images visually examine and comment on the effects of truncation 33) [1% bonus] Discuss the differences between these two types of images, optionally extend or validate what you've observed here with new images of your own, ie, natural photos vs diagrams banboon monkeyking** sunflower* hexagon * from reused with the creative commons license ** from cartoon production by Shanghai Animation Studio 1961

8 31) FFT Baboon Monkey King Sunflower hex DCT We can see clear pattern in both FFT and DCT of the hex image The fft and DCT of Baboon is most close to a uniform distribution, because it contains much texture so that the high frequency part is also quite bright

9 32) FFT 25% Baboon Monkey King Sunflower Hex FFT 625% SNR = SNR = SNR = SNR = DCT 25% SNR = SNR = SNR = 2493 SNR = SNR = SNR = SNR = SNR = DCT 625% SNR = SNR = SNR = SNR = First of all, it is very obvious that the larger the truncation window is, the better the recovered image is The reason is simple: more information is kept when the window is larger We can also observe that the simpler (less texture) the image is, the higher quality (higher SNR) the recovered image is, because we only keep the low frequency part and texture usually causes high frequency Generally speaking, the recovery performance of FFT is a little better than that of DCT 33) Usually, natural photos contain much richer texture than diagrams, hence the high frequency part of natural objects would be brighter (larger) than that of diagrams Moreover, natural photos are

10 more comple ie, seldom contain objects with perfect symmetry, so usually we can not see obvious pattern in its fft or dct But we can find some perfect pattern in fft or dct of the diagrams sometimes because of their simple structure