Homework Set 3 Solutions REVISED EECS 455 Oct. 25, Revisions to solutions to problems 2, 6 and marked with ***

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1 Homework Set 3 Solutions REVISED EECS 455 Oct. 25, 2006 Revisions to solutions to problems 2, 6 and marked with ***. Let U be a continuous random variable with pdf p U (u). Consider an N-point quantizer for U with levels ^x,..., ^x N and thresholds a,..., a N whose is denoted D U. Let V = bu + c where b>0 and c are constants. Thus, V is a scaled and shifted version of U. From random variable theory, we know that the pdf of V is related to that of U via p V (v) = b p U( v-c b ) That is the pdf of V is a scaled and shifted version of the pdf of U. Now consider a scaled and shifted quantizer for V with levels b ^x +c,..., b ^x N +c and thresholds b a +c,..., a N +c whose is denoted D V. (a) Sketch an example of p U (u) and the corresponding p V (v). U(u) = 2U + V(v) (b) Show that σ 2 V = b2 σ 2 U, where σ2 U and σ2 V denote the variances of U and V, respectively. σ 2 V = E (V-E[V])2 = E (bu+c - E[bU+c]) 2 = E (bu+c-be[u]-c) 2 since E[bU+c]=bE[U]+c = E (b(u-e[u])) 2 = b 2 E (U-E[U]) 2 = b 2 σ 2 U (c) Show that D V = b 2 D U. N b a i +c D V = (v - b ^x i -c) 2 N b a i +c p V (v) dv = (v - b ^x i -c) 2 b p U( v-c b ) dv i= b a i- +c i= b a i- +c N a i = (bu+c - b ^x i -c) 2 b p U(u) du with u = v-c b i= a i-, so v = bu+c = b 2 N a i (u - ^x i ) 2 i= a i- b p U(u) du = b 2 D V (d) Let D * U,N, respectively D* V,N, denote the least of any quantizer with N levels for U, respectively V. Show D * V,N b2 D * U,N Let ^x,..., ^x N and a,..., a N denote the levels and thresholds of an optimal N-point quantizer for U. It has distortion D U = D * U,N. Consider the scaled and shifted quantizer for V with levels b ^x +c,..., b ^x N +c and thresholds b a +c,..., a N +c whose is denoted D V. From Part (c) we know D V = b 2 D U. We also know that since D * V,N is the least distortion of any N point quantizer for V, D * V,N D V. Putting all these facts together we find D * V,N D V = b 2 D U = b 2 D * U,N

2 (e) Show D * V,N b2 D * U,N We note that U = b V - c b and we use the same approach as in Part (d) except b is replaced by /b and c is replaced by -c/b. Let ^x,..., ^x N and a,..., a N denote the levels and thresholds of an optimal N-point quantizer for V. It has distortion D V = D * V,N. Consider the scaled and shifted quantizer for U with levels ( ^x -c)/b,..., (^x N -c)/b and thresholds (a -c)/b,..., (a N -c)/b whose is denoted D U. From Part (c) we know D U = D V /b 2. We also know that since D * U,N is the least distortion of any N point quantizer for U, D * U,N D U. Putting all these facts together we find D * U,N D U = D U /b 2 = D * U,N /b2 Important Notes:. We conclude parts (c) and (d) that D * V,N = b2 D * U,N. 2. We also conclude that if we have a quantizer that is optimal for one pdf, then an optimal quantizer for a scaled and shifted pdf is simply a scaled and shifted version of the original quantizer. 2. Problem 7.9, p. 37. Note that P X is just the variance of the random variables being quantized. You will need to make use of the conclusions Problem. () We need to design an optimal 6-level uniform scalar quantizer for a Gaussian source with variance σ 2 = 0. The step size of this quantizer is σ = 0 times the step size The other issue is the offset, i.e. the location of the levels. Since the Gaussian density is symmetric about 0 we'll make the levels and thresholds symmetric about 0. The result is thresholds levels ***(2) The resulting distortion is σ 2 = 0 times the distortion of, which is ***(3) With 8 levels, = = The improvement in SNR is 0 log = 5.88 db Problem 7., p. 372 () We need to find an optimal 6-level nonuniform scalar quantizer for a Gaussian with variance σ 2 = 0, we take the levels and thresholds found in Table 7.2 and multiply them by σ = 0. thresholds levels (2) The resulting distortion is σ 2 =0 times the distortion of Table 7.2, namely, (3) With 8 levels, = = The improvement in SNR is 0 log = 5.62 db 2

3 4. Show that for a Gaussian density, β = 2# 3 3/2, where β is the term appearing in the high-resolution for the distortion of an optimal nonuniform quantizer with N levels, when N is large, namely D 2 N 2 β (this is called the Panter-Dite ) Let the Gaussian density be p(u) = 2" exp{-u2 2 } with variance σ2 =. Then β = σ 2 p /3 (u) du 3 = - - = (2") = 2 " 3 3/2 /6 2"3-2"3 (2")/6 exp{-u2 6 } du 3 u2 exp{- 2 3 } du 3 notice that the integral equals one because it is the integral of a Gaussian 5. Consider a Gaussian density with variance σ 2 = There were errors in the s given in for and D: Here are the correct s: For an optimum uniform scalar quantizer for a Gaussian source with variance, 4 σ ln N and D σ 4 ln N N 2 3 N 2 = 4 R σ2 3 (ln 2) 2 2R (a) For an optimal uniform scalar quantizer for this pdf, complete the following table: R 2 /2 % of col 3 % of col 5 % of col % 4.3% 36.4% % 8.2% 2.8% % 23.0% 5.7% % 24.2% 25.% % 23.7% 29.3% Comment on the of the s in columns 3, 5 and 6. The of Column 3 improves as R increases. For R = 5, it is accurate to about 5%. The Columns 5 & 6 do not show improving as R increases. The theory says the s become accurate for large R. Apparently, R must be larger than 5. (b) Complete the following table for the same situation: R db 2 /2 db db col3-col4 col5-col2 col6-col % % % % %. Comment again on the of the s in columns 3, 5 and 6. The of Column 3 improves as R increases. For R = 5, it is accurate to about.7db. The Columns 5 & 6 do not show improving as R increases. The theory says the s becomes accurate for large R. At R = 5, the of column 6 for SNR is about db. Apparently, R must be larger than 5 for greater. 3

4 (c) For an optimal nonuniform quantizer for the same pdf, complete the following table: R Table 7.2 col % % % % We used the Panter-Dite D 2 σ2 β 2-2R, with β = 2" 3 3/2 = Comment on the of the in column 3. The improves steadily as R increases. It's about 0% when R=4. (d) Complete the following table for the same situation: R db Table 7.2 col3-col2 db Comment again on the of the in column 3. Again, the improves steadily as R increases. It's about.5 db when R=4. 6. (a) CD recordings are made with a uniform scalar quantizer with rate 6 bits/sample. Assuming that music samples are Gaussian with variance σ 2 and assuming that the level spacing is chosen optimally for this source model, estimate the SNR of CD recordings. As in Problem 5, Part (a), we estimate SNR using SNR = 0 log σ 2 0 D, with D 4 R σ2 3 (ln 2) 2 2R. The result is SNR = 84.6 db. (b) Suppose instead that optimal nonuniform scalar quantization were used with the same rate (6 bits/sample). Estimate how much larger the SNR would be than in Part (a). As in Problem 5, Part (a), we estimate SNR using SNR = 0 log σ 2 0 D, with D 2 σ2 β 2-2R, with β = 2" 3 3/2 = The result is SNR = 92.0 db. (c) Suppose instead that optimal nonuniform scalar quantization were used with its rate chosen to give the same SNR as in part (a). Estimate how much more music could be stored on a CD. Express your answer in percent. Approach : Solving for the rate that gives SNR = 84.6 db with nonuniform scalar quantization gives R = 4.7, which is 8.% less than rate R = 6. Therefore, one can store 8.% more music. *** NOTE: Rounding up the rate to 5 is fine. This gives a 6.25% improvement instead of 8.%. Approach 2: At high rates SNR changes at 6 db per bit, if R = 6 gives 92.0 db and we wish to reduce SNR to 84.6 db, then the rate reduction is ( )/6 =.23. So we use R = , which is 7.5% less than rate R = 6. Therefore, one can store 7.5% more music. The difference between this and the previous answer is due to round off errors 4

5 7. In this problem you will use Matlab to perform uniform scalar quantization on an image. From the Homework Auxiliary Files webpage, download the template Matlab script uniformimagequant_template.m and the image peppers.tif. The Matlab script has a couple of lines that you must complete in order that it will do uniform scalar quantization of the image 'peppers' at a variety of rates and display the results. (Two other images are supplied in case you want to try the script on other images.) (a) Complete the script and include a copy of it and a printout of the displayed figure in your homework. (Among other things, the script requires one to compute. Since we don't have a probabilistic/statistical model (e.g. pdf) for the image, one needs to compute the empirical between the original and the quantized images. Specifically, one computes the sum of the squares of the pixel-bypixel differences between the original and quantized images, divided by the number of pixels.) The printout and script are given below. (b) Compare the 's computed by the script to the 2/2. (You can do this by hand, or modify the script so that it computes.) R / We see that 2 /2 is ver accurate until R = 2 or. Answers to the following questions are somewhat subjective and will depend on how large are the images on your screeen. It is recommended that you make them as large as possible by expanding the window as much as possible. (c) Estimate the size of the individual images on your screen. 2 /4 inch by 2 /4 inch (d) What is the smallest rate at which you cannot see evidence of quantization? What is the corresponding SNR? R = 5 (e) What is the smallest rate that you think would be acceptable, i.e. not too annoying, for posting on the internet? What is the corresponding SNR? R = 3 (f) What is the smallest rate at which someone can clearly recognize what is in the image? What is the corresponding SNR? R = 2 script: %%% uniformimagequant.m %%% a script to perform uniform scalar quantization of an image x = double(imread('peppers.tif','tiff')) ; R = [ ] ; % set of rates at which to quantize %% compute variance mean = sum(sum( x ) ) / prod(size(x)) variance = sum(sum( (x-mean).^2 )) / prod(size(x)) %% perform quantization and display the quantized image for k = :length(r) L = 2^R(k) ; delta = 255/L ; %compute step size of uniform quantizer y = ( floor(x/delta) +.5 ) * delta ; %compute quantized image mse = sum(sum( (x-y).^2 )); %compute mse 5

6 end mse = mse/prod(size(x)) ; SNR = 0 * log0( variance /mse ) ; %compute SNR subplot(3,2,k) ; imagesc(y); %display image axis image; colormap gray; xlabel(['r = ' num2str(r(k)) ', = ' num2str(mse,3) ', SNR = ' num2str(snr,3) 'db']) printout: 6

7 8. In this problem you will run a Matlab script that designs and runs nonuniform scalar quantization on an image. From the Homework Auxiliary Files webpage, download the Matlab script nonuniformimagequant.m and the image peppers.tif. Run the script. (a) Include a printout of the displayed figure in your homework. Printout is given below The answers to parts (b)-(e) are somewhat subjective. and will depend somewhat on how large the images are on your screeen. As before, it is recommended that you make them as large as possible by expanding the window as much as possible. (b) What is the smallest rate at which you cannot see evidence of quantization? What is the corresponding SNR? R = 4 (c) What is the smallest rate that you think would be acceptable, i.e. not too annoying, for posting on the internet? What is the corresponding SNR? R = 3 (d) What is the smallest rate at which someone can clearly recognize what is in the image? What is the corresponding SNR? R = 2 (e) Comment on the improvement of nonuniform scalar quantization over uniform. The SNR shows sizable improvements at rates less than 4. Gains:.9 db,.7 db,.3 db,.6 db at rates 4, 3, 2,, respectively. Qualitatively, we also see improvements at these rates (f) For the quantizer with 6 levels, find the widths of the largest and smallest quantization cells and their ratio. The thresholds are The largest cell is [83.2,255] which has width 7.8. The narrowest cells are [72.5,83.2] which has width 0.7. The ratio is 6.7 This indicates that the quantizer has a lot of nonuniformity. Notice that there is another nearly equally small cell, namley, [84.0,94.8] with width 0.8. It is no coincidence that this cell and the narrowest are located where the histogram is largest. Note: In practice, one would ordinarily design a scalar quantizer based on a training set of many images, and then apply it to images that were not in the training set. For example, you might modify the script so that it designs the quantizer based on the two images baboon and camerman and then run the quantizer on peppers. It will be interesting to see how much this reduces SNR. 7