Electrical Engineering Assignment Help

Electrical Engineering Assignment Help

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Problems

1. Is it Over Compressed or is it Modern A r t ?

Write your own video image compressor in MATLAB! This will be as simple as possible, so don’t sweat.

Write all your MATLAB commands in ps2p1.m. a. Start off by loading the flower image by typing the following in MATLAB. load imdemos fl ower; % Load fl ower matrix from image demos l i b ra r y fl ower = double(fl ower);% Convert the matrix to double precision colormap(’gray ’); % Grayscale for any pictures you want to display % (This creates a blank window that we ’ll use soon) % Display fl ower i n the window imshow(flower,[0 255]); b. Perform a 2D DCT operation on 8 × 8 blocks of the image matrix. You might find the com mands blkproc and dct2 useful. Use the first form of blkproc explained in its help page (help blkproc), where FUN is ‘dct 2 ’ (put single quotes around dct2). To keep help information from scrolling off the screen, type more on at the MATLAB prompt. c. Set to zero all coefficients with absolute value less than 10. An example to do this can be found in the help of dct2. We will later want to vary this cutoff value and count the number of nonzero values. After this step, many codecs apply run length and variable length encoding to transmit or store the image efficiently for later use. d. Perform a 2D Inverse DCT operation on each of the 8 × 8 blocks of the image matrix. The values of the resulting matrix will be in floating point, but you should round the elements to the nearest integer since image pixel values are integers. Use the blkproc and idct2 commands similarly to the way you performed the forwards DCT in Part b. Then use the round command to complete the reconstruction of the image. e. Determine the mean squared error between the original and reconstructed image. The definition of mean squared error to use is mean2((x y ) . ∧ 2 ). Values will appear larger than in the demo because our error definition does no normalization. The answer you should receive is 10.6891. Note that you can see your new image with the imshow command the same way you displayed the original image. If you want a new window for the image, so it doesn’t just paint over the previous one, type figure at the MATLAB prompt. matlabassignmentexperts.com

Now compose a graph, having along the xaxis the number of nonzero values of the matrix after the cutoff procedure (before you did the inverse DCT), and the mean squared error (after you did the inverse DCT and rounding) on the yaxis.

The number of nonzero values is the number of bytes required to store the image (ignoring overhead), so smaller means more compression. Range the cutoff value from 0 to 100 in increments of 4. The use of nnz and plot will do the trick. It will be easiest to use them if you create a vector with the numbers of nonzero values and another vector with the mean squared errors.

Write in ps2diary the largest byte size for which you can detect the difference between the original image and the reconstructed image by eye. Include any comments in the diary file. Be sure the script in ps2p1.m is executable in MATLAB.

2. Compression is Fun and Easy Now you can get your hands dirty with the LZW (LempelZivWelch) compression algorithm.1 Although this problem will not require the use of MATLAB, you may use it if you wish. A good reference to the LZW algo rithm is http://www.rasip.fer.hr/research/compress/algorithms/fund/lz/lzw.html. There are many deriva tives of LZW that are more powerful, but we’ll stay with what was explained in lecture and in the web link. a. Use the 7 bit ASCII table that can be found in Chapter 2 of the notes, handed out at the first lecture, for your basic alphabet. Notice that the ASCII table ends at HEX 7F (Hexadecimal is Base 16 using A F for values 11 15, so HEX 7F is 127). Your dictionary will therefore range from HEX 80 through HEX FF. Encode the following message in HEX using the LZW technique: peter piper picked a peck of pickled peppers The 2.110/6.050 staff would like to thank Joe Huang, the Spring 2000 class TA, for developing the image compression exercise used in this assignment. It is, in our opinion, really cool because it truly links what we discuss in class to the real world; once you’ve completed it, you will understand and have experience with the way J P E G images are created.

For this reason, the problem has not changed much since last two years, and we are trusting you to complete it without consulting last year’s solutions. We hope you enjoy it!

1

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Be sure that you can decode the message properly. Use the examples in Section 3.2 of the notes as a guide for formatting your solution. Put your results in ps2diary and include your dictionary entries. Note: HEX 20 is the space. For this exercise use HEX 7F (DEL) to signify the end of the message and HEX 00 (NUL) for the start of the message. b. It is proposed to improve this coding by using only the ASCII control characters HEX 10 1F to hold dictionary entries. In one paragraph, summarize the advantages and disadvantages of this proposal. Things to consider might include ability to handle various kinds of messages, efficiency, complexity of the coder and decoder, and adherence to widely accepted standards. Under what circumstances might this proposal be a good idea or a bad idea? c. Optional (just for fun): Implement the LZW coding and decoding operations of part above, using any computer language you know and have access to. Assume that any message you handle will be short so that the dictionary will not extend beyond HEX FF (naturally a robust implementation would need to handle dictionary overflow).

Tu r n i n g in Yo u r Soluti ons You should have two files (one Mfile and one diary). Name the Mfile ps2p1.m and name the diary ps2diary. Turn in this problem set by emailing your Mfiles and diary along with your answers to any problem(s) not done using MATLAB. You may do this either by attaching them to the email as text files, or by pasting their content directly into the body of the email (if you do this, please somehow indicate where each file begins and ends). The deadline for submission is the same no matter which option you choose.

Your solutions are due 5:00 PM on Friday, February 21, 2003. Later that day solutions will be posted on the web.

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Solutions 1.Is it Over Compressed or is it Modern Art? The following MATLAB code solves parts a to e: % Perform i n i ti a l i z ati o n as indicated i n the problem s et . load imdemos fl ower; fl ower=double(fl ower); colormap(’gray ’); % Since many fi gures w i l l be produced by t h i s s c r i p t , we use meaningful l a b e l s . set(gcf,’NumberTitle’,’off ’,’Name’,’Flower ’); imshow(flower,[0 255]); % Implement the comrpession scheme detailed i n the problem s e t . encoded=blkproc(fl ower,[8 8 ] ,’d c t 2 ’ ) ; encoded(abs(encoded)<10)=0; decoded=round(blkproc(encoded,[8 8 ] ,’ i d c t 2 ’ ) ) ; % Provide the error value to check against the expected value from the s e t . sprintf (’With cutoff =10, the mean squared error i s %. 4f ’, . . . mean2((flower decoded).∧2)) The mean squared error you should have obtained is 10.6891. The next piece of code produces the graph of file size versus mean. squared % I n i ti a l i z e the vectors that w i l l store the data for the graph. x = [ ] ; y=[]; % We need only encode the image once. Aft er t hat, since we w i l l be steadily % increasing the threshold, we need to reconvert again more because we w i l l be % simply zeroing out more elements with each iterati on through the for loop % (there i s no reason to recover a l l the o r igina l elements and star t from scratch % each time through the loop; we can progressively drop more and more data). encoded=blkproc(fl ower,[8 8 ] ,’d c t 2 ’ ) ; % Now we begin to co ll ect data for the graph. for cutoff =0:4:100, encoded(abs(encoded)<cutoff )=0; decoded=round(blkproc(encoded,[8 8 ] ,’ i d c t 2 ’ ) ) ; % We w i l l simply append to the vectors each time through t h i s loop.

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y=[y,mean2((fl ower

decoded).∧2)];

% The next three l i ne s can be commented out i f they are not desired. % w i l l produce a new window, label i t , and print a representati on of the % newly decoded image for each cutoff threshold.

This i s for comparison % with the o r ig in al image to answer the questi on i n the set that asks at % which point the diff erence between the o r igin al image and the compressed % image becomes percepti ble to the human eye. fi gu re ; s et(gcf,’Number Title’,’off ’,’Name’,spr intf (’cutoff =%d’,cutoff )); imshow(decoded,[0 255]); end % Now, plot the graph with a smooth curve and boxes around a l l the actual data points. fi g u re ; set(gcf,’NumberTitle’,’off ’,’Name’,’Graph for Problem 1 ’ ) ; p l o t ( x , y,’s ’ ) ti tle(’Comparison of F i l e Size and Image E r ro r ’ ) ; xlabel(’Non zero matrix values (number of bytes to s to r e ) ’ ) ; ylabel(’Mean squared e r ro r ’ ) ; You should have gotten something remotely resembling the graph in Figure 1. As you can see, there is a point where the MSE increases exponentially giving a quantitative value to the degradation of the reconstructed picture. Medical applications such as in Xrays tend to discourage the use of J PE G or similar lossy compression algorithms for saving images due to chances of distortion leading to an incorrect diagnosis. The largest byte size for which I could detect the difference between the original image and the recon structed image by eye was 1,785 bytes, which corresponds to a cutoff of 24 and a mean square error of 42.2132. This is rather remarkable because the image started out over 16kb long, which means that less that an eighth of the data recorded is actually necessary to create the visual effect we experience when looking at the picture. Beyond that, the quality is not horrible, either, but we can begin to detect bands forming due to the fact that the pixels were processed as distinct 8 × 8 blocks. By the end of the experiment, where the cutoff is 100, the picture is not good, yet there is still enough data that it is recognizable as a flower and shows a resemblance to the original image.

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2. Compression is F u n and Easy part a.

Table 1 represents the LZW analysis of the phrase “peter piper picked a peck of pickled peppers.”

The resulting data stream is: 80 70 65 74 65 72 20 70 69 82 86 88 63 6B 65 64 20 61 87 65 8D 20 6F 66 87 8C 6B 6C 8F 93 70 8A 73 81 This is 34 bytes long, whereas the original message was 46 bytes long (including the start and stop control characters), which amounts to 26.1% compression. If we count bits, the original message could have been sent in 7 bits per character (total 322 bits) whereas the LZW code requires 8 bits per character (total 272 bits) so the compression is 15.5%. Of course, this is a short example but contrived to make the dictionary grow quickly. For a sizeable selection of average English text, LZW typically yields 50% compression.

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Figure 1: Comparison of File Size and Image Error

part b.

This strategy has several advantages: we will stay within 7bit data, hence more compression and faster encoder/decoder; moreover, the dictionary can be searched quickly. There is a major drawback, though. The dictionary can only have 16 entries beyond the ASCII characters. Except for very short and repetitive sequences of characters, the dictionary will overflow. Another drawback is that the text cannot include any of the ASCII characters being used for the dictionary; among the characters excluded are three that are often used (DC1, DC3, and ESC). On the other hand, the characters HEX 80 FF would now be available since the dictionary no longer is there, and this includes many common accented letters of Western European languages. A practical implementation of the coder and decoder would not be either easier or harder by very much, but new programs would have to be written because this proposed arrangement is not standard, and new software always brings with it a cost and risk of bugs. matlabassignmentexperts.com

02 70

Input ST X p 65 74 65 72 20 70 69 70 65 72 20 70 69 63 6B 65 64 20 61 20 70 65 63 6B 20 6F 66 20 70 69 63 6B 6C 65 64 20 70 65 70 70 65 72 73 03 –

e t e r space p i p e r space p i c k e d space a space p e c k space o f space p i c k l e d space p e p p e r s ETX –

New dictionary entry –

– 82 83 84 85 86 87 88 89 – 8A – 8B – 8C 8D 8E 8F 90 91 92 – 93 94 – 95 96 97 98 – 99 – 9A 9B 9C – 9D – – 9E 9F – – A0 – –

pe et te er rspace spacep pi ip – per – rspacep – pic ck ke ed dspace spacea aspace – spacepe ec – ckspace spaceo of fspace – spacepi – ic ckl le – edspace – – spacepep pp – – pers – –

Transmission –80 – (start)

–

–

70 p 65 e 74 t 65 e 72 r 20 space 70 p 69 i – – 82 pe – – 86 rspace – – 88 pi 63 c 6B k 65 e 64 d 20 space 61 a – – 87 spacep 65 e – – 8D ck 20 space 6F o 66 f – – 87 spacep – – 8C pic 6B k 6C l – – 8F ed – – – – 93 spacepe 70 p – –to Problem 2, part a Table 1: Solution – – 8A per 73 s 81 (stop)

New dictionary entry – 80 81 82 83 84 85 86 87 – 88 – 89 – 8A 8B 8C 8D 8E 8F 90 – 91 92 – 93 94 95 96 – 97 – 98 99 9A – 9B – – 9C 9D – – 9E –

– pe et te er rspace spacep pi ip – per – rspacep – pic ck ke ed dspace spacea aspace – spacepe ec – ckspace spaceo of fspace – spacepi – ic ckl le – edspace – – spacepep pp – – pers –

Output ST X – p e t e r

space p i – pe – rspace – pi c k e d space a – spacep e – ck space o f – spacep – pic k l – ed – – spacepe pp – – per s ETX

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