Question

Question 4: Image processing 12 marks 11) Briefly describe the quantization in a grey scale image. 2 marks 12) Consider the 1D image fragment below. Filter this image with: 4marks (1 mark each) () 1x3 median filter (ii State how the image boundaries are handled. 60 180 180 60 180 180 (iv) Give an example of when it would be beneficial disadvantage median filter, but also state a use 13) What is an image histogram and what information does it contain? Draw a labelled sketch. 2 marks 14) How might an 'edge' be defined n image. terms of image processing? Name a method for detecting edges 15) Describe the process 2 marks histogram equalization and the effect it has on an image. 37 PM Type here to search ENG 13/06/2019

Answer 1

Given 1-D Image                              = [160, 160, 60, 60, 160, 160]

1. 1x3 Mean Filter Output = Pi = (P(i-1) + P(i) + P(i+1) / 3

                  Therefore, Padding the 1-D Input Image with zeros =

                  [0        160,                     160,                    60,                   60,                     160,                      160 0      ]

                  [    (0+160+10)/3, (160+160+0)/3, (160+60+60)/3, (60+60+160)/3, (60+160+60)/3, (160+160+0)/3]

[         56.66                   106.6,               93.33,               93.33,                 93.33,                 106.66]

2. 1x3 median filter = Median of 3 Pixels

Therefore, Padding the 1-D Input Image with zeros =

                                            [0    160,                 160,                    60,                   60,                      160,                        160    0]

1x3 Median Filter Image = [(0, 160, 160), (160, 160, 60),     160, 60, 60),     (60, 60, 160),     (60, 160, 160),       (160,160,0) ]

                                        = [      160,                   160,                    60,                   60,                   160,                         160]

(iii)

Image boundaries are points where there is sharp change in pixel intensities. Therefore, the image boundaries exists at points (160,60) and (60, 160) pixel intensity pairs.

Image Histogram is defined as the graph between Gray color intensity and corresponding no. of pixels. The image histogram is basically probability distribution function (PDF). Image histogram provides color profile of the image.

Edges in images are defined as the pixel intensity derivatives. Where the intensity derivative with respect to pixel location is non-zero or vary high, those regions corresponds to edge points. There are different edge operators like, sobel, canny, Roberts, prewitt and logarithmic etc.

Histogram Equalization

Histogram is defined as the plot between No. of pixels (Pi) and grey level intensity L (0 – 255). Histogram equalization is used to enhance the given input image in spatial domain. And each pixel intensity in transformed domain is the related to input image as given by:

HistEq Value after Histogram Equalization = (CDFi- CDFmin)(N-1) x (L-1)

Where L is the gray level and CDFi is the cumulative distribution function at ith grey level intensity.

Therefore, it can be said that the pixel intensity in histogram equalized image for the pixel O(x,y) corresponds to input pixel I(x,y) .

Where O(x,y) and I(x,y) are the image pixels intensities in output and input images.

An example of the same is illustrated below:

Histogram Equalization Gray Value 1 2 3 4 5 7 r Pr 0 0 0 1 4 3 6 2 PDF P(r/16 Cumulative PDF C(r Normalized CDF =C(r) x L 0.0625 0.25 0.375 0 0 0.1875 0.125 0.0625 0.3125 0.5000 0.8750 1.0 0.4375 3.5 0 0 0 2.1875 6.125 7 Output Pixel 1 1 1 1 4 7 7 0.4375 0.4375 0.25 New PDF 0.0625 0.0625 0.0625 0.0625 0.1875 The above table shows the gray value and their respective probability density function along with cumulative PDF. The gray values are extracted from the histogram given in the problem. Also, the given histogram can be rearranged in pixel matrix form of 4x4 size as given below The cumulative PDF is obtained by summing the previous values using the followings: C(1) = P(0) And, C(r) C(r-1) + C) Input 3 Matrix 4 4 6 6 6 5 4 5 6 4 7 The normalized CDF is given by: C(r) x L Where Lithe Max. Gray Level =7, in the given problem) wwwwwwww Output Pixel 1 4 7 7 7 7 4 4 3 7 olo enenN