EECS 556 Winter 2007 Exam2 Solutions

Transcription

1 EECS 556 Winter 2007 Exam2 Solutions 1. [10pts] From Wiener filter derivation you need P g and P fg. g = x **b 2 + v = (f** b 1 + u)** b 2 + v = f** b 1 ** b 2 + u** b 2 +v, [n,m] dropped for convenience. P g = P f B 1 2 B P u B P v, (ω x,ω y ) dropped for convenience. R fg [n,m] = E{ f[n,m]g * [0,0] } = E{f[n,m](f**b 1 **b 2 ) * [0,0] + f[n,m](u** b 2 ) * [0,0] + f[n,m]v * [0,0]} = b 1 * ** b 2 * ** R f [n,m], thus P fg = B 1 * B 2 * P f. H = P fg / P g = { B 1 * B 2 * P f } / { P f B 1 2 B P u B P v }, (ω x,ω y ) dropped for convenience. Most of you did cascaded approach, which yields the same answer. Some of you had P u B 1 2 denominator (-2pts). in the 2. (a) [5pts] MSE = (NN), (linear), (cubic). Cubic interpolation works best for the cameraman image (img1). (b) [5pts] MSE = (NN), (linear), (cubic). Linear interpolation works best for this image (img2). (c) [10pts] The answer is NO. For part (b) linear is better than cubic. If you inspect the shape of the second image, you can observe linearly increasing intensity from the corners towards the center, a cone shape if you will. Upsampling via interpolation is filling in the missing intermediate values. As the intensities are varying linearly, the best interpolator to use is the linear interpolator for the second image. Cubic interpolator is also capable of interpolating linearly varying intensities. Under noisy conditions, cubic interpolator may deviate from a perfect line while linear interpolator always produces a straight line. You cannot generalize this result to other images. In general, higher order interpolator (popular choice is cubic) is likely to be better than lower order interpolator.

2 % exam2 prob2 clear; load e2p2.mat img = img1; img = img2; sub_img = sub_img1; sub_img = sub_img2; figure; subplot(331);imagesc(img);colormap gray; axis image;colorbar;title('original Image'); subplot(332);imagesc(sub_img);colormap gray; axis image;colorbar;title('downsampled Image'); interp_img1 = interp2(x, y, sub_img, x_int, y_int, 'nearest'); mse1 = mean2( (interp_img1 - img).^2 ); subplot(334);imagesc(interp_img1);colormap gray; axis image;colorbar;title('nn interp'); subplot(337);imagesc(interp_img1 - img);colormap gray; axis image;colorbar;title('difference'); title(sprintf('nn interp error, MSE=%g', mse1)) interp_img2 = interp2(x, y, sub_img, x_int, y_int, 'linear'); mse2 = mean2( (interp_img2 - img).^2 ); subplot(335);imagesc(interp_img2);colormap gray; axis image;colorbar;title('linear interp'); subplot(338);imagesc(interp_img2 - img);colormap gray; axis image;colorbar;title('difference'); title(sprintf('linear interp error, MSE=%g', mse2))

3 interp_img3 = interp2(x, y, sub_img, x_int, y_int, 'cubic'); mse3 = mean2( (interp_img3 - img).^2 ); subplot(336);imagesc(interp_img3);colormap gray; axis image;colorbar;title('cubic interp'); subplot(339);imagesc(interp_img3 - img);colormap gray; axis image;colorbar;title('difference'); title(sprintf('cubic interp error, MSE=%g', mse3)) 3. (a) [5pts] As per template, take yy(1:10,1:10) and compute sample variance. MSE = See the figures for the filtered output; <estimated noise, no blur> (b) [5pts] MSE = See the figures for the filtered output, <true noise, no blur> (c) [5pts] MSE = See the figures for the filtered output, <true noise, estimated blur> (d) [5pts] MSE = See the figures for the filtered output, < true noise, true blur> (e) [5pts] In part (c), blur is estimated through periodogram (P y ), which is known to be very noisy. Other components of the formula, P s and P v, are derived analytically so there are no errors in those components. If you plot B(ω x,ω y ) and estimated blur, B_est(ω x,ω y ), then you will observe that estimated blur is quite noisy. Using this noisy blur function led to worsening of the MSE in part (c) compared to part (b) where there is no blur. Basically, you are better off using no information (i.e., no blur) than using unreliable information (i.e., noisy estimated blur). The other reason is that the true blur is a 3x3 filter with high center value which resembles a delta function a lot. Thus, assuming that the blur is a delta function is quite reasonable. In general, estimating a blur function is a difficult problem to tackle. Some of you kindly suggested that using windowed periodogram helps, which is a good suggestion.

4 % exam2 prob3 % wiener filter problem with blur % reused code from hw7 prob5 rand('state', 0); randn('state', 0); nx = 62; ny = 50; h1 = zeros(27,27); h1(10:18, 11:17) = 1; p = 0.004; ss_tmp = (rand(nx,ny) < p) - p; ss = conv2(ss_tmp, h1, 'same') / sqrt(p*(1-p)); % random rect signal blur = 1/9*[0 1 0;1 5 1;0 1 0]; yy = conv2(ss, blur, 'same') + 3 * randn(size(ss)); % add gaussian noise - what variance? px = 2*nx; py = 2*ny; % zero padding to reduce space aliasing n1 = size(h1,1)-1; ix = [-n1:n1]; n2 = size(h1,2)-1; iy = [-n2:n2]; Rs = zeros(px,py); % true autocorr func h2 = xcorr2(h1, h1); Rs(ix + end/2+1,iy + end/2+1) = h2; wx = [-px/2:px/2-1]'*2*pi/px; wy = [-py/2:py/2-1]'*2*pi/py; Ps = fftshift ( (Diric(wx,9) * Diric(wy,7)').^2 ); % Method 1: analytical power spectrum Ps = abs(fft2(h1, px, py)).^2; % Method 2: H_1 (the simplest here) % estimate noise Pv from background sub_yy = yy(1:10,1:10); Pv_est = var(sub_yy(:)); % estimate blur Py = abs( fft2( xcorr2(yy,yy)/(nx*ny), px, py) ); Py2 = Py -9; Py2(Py2 < 0) = 0; B_est = sqrt(py2./ps); % B_est is real valued between 0 and 1 B_est(B_est > 1) = 1; B_est(B_est < 0) = 0; B = fftshift( (1/9)*(5 + 2*cos(wx)*ones(1,py) + 2*ones(px,1)*cos(wy)') ); % analytical method B = abs(fft2(blur, px, py)); % magnitude of fft of blur H1 = Ps./ (Ps + Pv_est); % Wiener filter wo blur w estimated noise H2 = Ps./ (Ps + 9); % Wiener filter wo blur w true noise H3 = (Ps.*B_est)./ (Ps.*(B_est.^2) + 9); % Wiener filter with estimated blur H4 = (Ps.*B)./ (Ps.*(B.^2) + 9); % Wiener filter with true blur

5 Y = fft2(yy, px, py); s_hat1 = real(ifft2(y.* H1)); s_hat1 = s_hat1(1:nx,1:ny); % un-zero-pad s_hat2 = real(ifft2(y.* H2)); s_hat2 = s_hat2(1:nx,1:ny); % un-zero-pad s_hat3 = real(ifft2(y.* H3)); s_hat3 = s_hat3(1:nx,1:ny); % un-zero-pad s_hat4 = real(ifft2(y.* H4)); s_hat4 = s_hat4(1:nx,1:ny); % un-zero-pad clf, pl = 230; colormap(1-gray(256)), subplot(pl+1), imagesc(1:nx,1:ny,ss'), axis xy, axis image xlabel n, ylabel m, title 'True: s(n,m)', colorbar horiz subplot(pl+2), imagesc(1:nx,1:ny,yy'), axis xy, axis image xlabel n, ylabel m, title 'Noisy: y(n,m)', colorbar horiz subplot(pl+3), imagesc(1:nx, 1:ny, s_hat1'), axis xy, axis image xlabel n, ylabel m, colorbar horiz wiener_mse = mean2((s_hat1-ss).^2); title(sprintf('no blur, est. noise, s.hat1(n,m), MSE=%g', wiener_mse)) subplot(pl+4), imagesc(1:nx, 1:ny, s_hat2'), axis xy, axis image xlabel n, ylabel m, colorbar horiz wiener_mse = mean2((s_hat2-ss).^2); title(sprintf('no blur, true noise, s.hat2(n,m), MSE=%g', wiener_mse)) subplot(pl+5), imagesc(1:nx, 1:ny, s_hat3'), axis xy, axis image xlabel n, ylabel m, colorbar horiz wiener_mse = mean2((s_hat3-ss).^2); title(sprintf('est blur, true noise, s.hat3(n,m), MSE=%g', wiener_mse)) subplot(pl+6), imagesc(1:nx, 1:ny, s_hat3'), axis xy, axis image xlabel n, ylabel m, colorbar horiz wiener_mse = mean2((s_hat4-ss).^2); title(sprintf('true blur, true noise, s.hat4(n,m), MSE=%g', wiener_mse))

6 4. (a) [5pts] NPLS (1 st order penalty), MSE= 2.399, beta = 2 3 = 8. See plots for filtered output and difference image. (b) [5pts] NPLS2 (2 nd order penalty), MSE= 3.879, beta = 2 2 = 4. See plots for filtered output and difference image. (C) [10pts] Combined block by block approach, the left potion of the image has piecewise constant (step edges) signal, which incurs no penalty under NPLS. Thus, you choose NPLS for the left potion of the image. The right potion has linearly increasing shape, which incurs no penalty under NPLS2. Thus, you choose NPLS2 for the right potion of the image. MSE= 2.145, beta1 (NPLS, left potion) = 2 2 = 4, beta2 (NPLS2, right potion) = 2 5 = 32. See plots for filtered output and difference image. You need to vary beta1 with NPLS to minimize MSE of just the left portion of the image and repeat the same process with NPLS2 to find out beta2 for the right potion of the image. Above was the original intended answer applying different NPSL depending on the profile of the signal. However, if you apply NPLS to both potions beta1 = 4 (NPLS, left potion) and beta2 = 16 (NPLS, right potion) your will actually get lower MSE = This happened because you are restricted to beta values between If you are allowed larger beta values minimum MSE = 2.12 occurs at beta1 = 4 (NPLS, left potion), beta2 = 2 16 or above (NPLS2, right potion). Answers applying NPLS to both potions are also correct. In this case, the rationale for choosing NPLS for both potions is to minimize MSE. Grading: both answers allowed. You need to provide justification for choosing NPLS/NPLS2 (signal profile dependent) or NPLS/NPLS (minimize MSE) combination. Not having the justification -3pts.

7 % exam2 prob4 % adaptive NPLS % code reused from hw7 and hw9 rand('state', 0); randn('state', 0); % 1st part of the signal nx = 62; ny = 50; h1 = zeros(27,27); h1(10:18, 11:17) = 1; p = 0.003; ss_tmp = (rand(nx,ny) < p) - p; ss = conv2(ss_tmp, h1, 'same') / sqrt(p*(1-p)); % random rect signal % 2nd part of the signal xtrue = zeros(nx,ny); ix = -(nx-1)/2:(nx-1)/2; iy = -(ny-1)/2:(ny-1)/2;

8 [ix, iy] = ndgrid(ix, iy); xtrue = (1 - min(abs(ix/(nx/3)),1)) * 30; xtrue = (1 - min(abs(iy/(ny/3)),1)).* xtrue; xtrue = xtrue - mean2(xtrue) + mean2(ss); nx2 = 2*nx; ny2 = ny; ss2 = [ss; xtrue]; yy = ss2 + 5 * randn(size(ss2)); subplot(421), imagesc(ss2',[-15 35]), axis xy, axis image, colormap gray xlabel n, ylabel m, title 'True: ss2(n,m)', colorbar subplot(422), imagesc(yy',[-15 35]), axis xy, axis image, xlabel n, ylabel m, title 'Noisy: yy(n,m)', colorbar % apply NPLS niter=200; delta=1; % fixed beta1 = 2^3; % play with this number xhat1 = npls_sps(yy, niter, beta1, delta); % apply NPLS2 beta2 = 2^2; % play with this number xhat2 = npls2_sps(yy, niter, beta2, delta); subplot(423), imagesc(xhat1',[-15 35]), axis xy, axis image xlabel n, ylabel m, colorbar, title( sprintf( 'NPLS output, beta=%g', beta1) ) subplot(424), imagesc((xhat1-ss2)',[-10 10]), axis xy, axis image xlabel n, ylabel m, colorbar title(sprintf('npls error, MSE=%g', mean2((xhat1-ss2).^2))) subplot(425), imagesc(xhat2',[-15 35]), axis xy, axis image xlabel n, ylabel m, colorbar, title( sprintf( 'NPLS2 output, beta=%g', beta2) ) subplot(426), imagesc((xhat2-ss2)',[-10 10]), axis xy, axis image xlabel n, ylabel m, colorbar title(sprintf('npls2 error, MSE=%g', mean2((xhat2-ss2).^2))) % apply NPLS and NPLS2 block by block yy_part1 = yy(1:nx,1:ny2); yy_part2 = yy(nx+1:2*nx,1:ny2); beta1 = 2^2; beta2 = 2^5; % play with this number xhat3_part1 = npls_sps(yy_part1, niter, beta1, delta); xhat3_part2 = npls2_sps(yy_part2, niter, beta2, delta);

9 xhat3 = [xhat3_part1; xhat3_part2]; subplot(427), imagesc(xhat3',[-15 35]), axis xy, axis image xlabel n, ylabel m, colorbar, title( sprintf( 'combined output, beta1=%g beta2=%g', beta1, beta2) ) subplot(428), imagesc((xhat3-ss2)',[-10 10]), axis xy, axis image xlabel n, ylabel m, colorbar title(sprintf('combined NPLS error, MSE=%g', mean2((xhat3-ss2).^2)))