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Digital Image Processing Exercise 1

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Exercises: - Questions : one week before classQuestions - Solutions : the day we have class Solutions - Slides along with Matlab code (if have) : after classSlidesMatlab code

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1 gigabyte = ( 8 × ) bits 1. One of the several HDTV formats is 1080p24, which means video stream of full frames of 1920×1080 pixels at frame rate 24 fps. If each pixel has 24 bits of intensity resolution (8 bits each for red, green and blue channels), how many gigabytes are needed for 2 hours of HDTV video without compression? HDTV (High-definition television) formats is 1080p24. Each pixel has 24 bits of intensity resolution. Frame rate 24 fps. Video size is n b × n f bits

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2. (Exam ) Perform connected component analysis of the follow- ing binary image. Use the two-scan labeling algorithm and represent results after each scan by using letters (a,b,c,... ) as labels. (a) Assume 4-connectivity. From left to right, top to bottom. Examine each pixel P and its neighbor pixels: left (x 1 ) and up (x 2 ). First scan:

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Second scan: Image is scanned and pixels are given final labels according to the equivalences found during the first scan.

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(b) Assume 8-connectivity. This picture is scanned in the similar way as with 4-connectivity, but now we examine four neighbors of P (also the diagonal neighbors). We notice that a is equal to b. They are given final label A in the second scan. (Matlab code)

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3. (Gonzalez & Woods 2007, Ex. 2.15) Consider the image segment shown. (a)Let V = {0, 1} and compute the lengths of the shortest 4-, 8-, and m-path between p and q. If a particular path does not exist between these points, explain why. i. There is no 4-path between p and q, as none of the 4-neighbors of pixel q have values from V.

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ii. The shortest 8-path from p to q, considering 8 neighborhood of one pixel (q) (p) 1012 V = {0, 1} p = (3; 0); (3; 1); (2; 2); (1; 2); (0; 3) = q The length is N - 1 where N is the number of pixels on the path. The length of the shortest 8-path is 4.

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iii. The shortest m-path from p to q. m- adjacency (Page 67) : Two pixels p and q with values from V are m-adjacent if: (i) q is in N 4 (p), or (ii) q is in N D (p) and the intersection set of N 4 (p) N 4 (q) has no pixels whose values are from V (q) (p) 1012 V = {0, 1} Intersection set of N 4 (p) and N 4 (q) is {1, 2} Thus, the length of this path is 5.

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(b) Let V = {1, 2} and compute the lengths of the shortest 4-, 8-, and m-path between p and q. If a particular path does not exist between these points, explain why. i. One possibility for 4-path: p = (3; 0); (2; 0); (2; 1); (2; 2); (2; 3); (1; 3); (0; 3) = q The length of this path is 6. ii. One possibility for the shortest 8-path: p = (3; 0); (2; 1); (1; 1); (0; 2); (0; 3) = q The length of the shortest path is 4. iii. One possibility for the shortest m-path: p = (3; 0); (2; 0); (2; 1); (1; 1); (0; 1); (0; 2); (0; 3) = q The length of this path is 6. Notice that these paths are not unique. It is easily verified that another path of the same length exists between p and q.

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Equalization of an image histogram is the cumulative density function. (a) Perform histogram equalization given the following histogram. (r=Gray level, n=number of occurrences) First, calculate the probability p k for each gray level: p k = n k /sum(n k ) 4.

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Second, compute the discrete cumulative density function s k. Finally, round to the nearest discrete value available: The equalized histogram is: x/7

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(b) Perform histogram specification of the previous histogram using the specified histogram shown in the following table. (r=Gray level, p=probability of occurrences) Transform the histogram into a given distribution. First, equalize the histogram. (in part (a) ) Second, change the equalized histogram into the given target distribution. (inverse transform z = G-1(s), where G(z) is a mapping that equalizes the target distribution) Compute this mapping: First, cumulate the probability.

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Next, apply the inverse transform z = G -1 (s), by finding the closest s k for each s k ’ computed in part (a). part (a) Thus, the histogram resulting from the transform is:

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5. An image is corrupted by additive uncorrelated, zero-average noise yielding How is the signal-to-noise ratio affected if you average these K images? Signal-to-noise power ratio: For single image:

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zero-average noise:

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The average image: Its signal-to-noise power ratio:

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Finally, the signal-to-noise ratio becomes: (Matlab code) P(x,y)

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6. (Exam ) Explain different methods for handling border pixels with mask operations. (a) operator modification operator is modified for exceptions where some of the necessary neighbors are missing often complex seldom used (b) adding zeroes easy to perform often used

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(c) reflecting usually better than adding zeroes often used

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(d) image is considered to be cyclic seldom used one should have some reason for assuming the image to be periodic

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(e) only the pixels that have all the necessary neighbors are processed the only ‘right’ way processed image is smaller than original

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