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 A02: Point Operations, Filteri ng, and Fre quency Domain Se pte mbe r 18, 2010  Assi gnmen t Brie f St udent name and no. Course EN4551: Fundamentalsof Machine Vision & Image Processing   A s se ss me n t A0 2: Poi nt Op e rat io ns, Filt e ring ,and Freq uen cy D om ai n  Weight This amounts to 10% ofthe module grade. Late penalty is 33% per week. Outc ome The st uden t must ex pe ri ment wi th si mple image pr oc essi ng al gori thms relat ed to poi nt ope rations, lt eri ng and fr equenc y doma in. Date handed out September 21, 2010 Date due October 19, 2010 Submi ssion date Student’ s Declaration I conrm that the wo rk submit te d fo r thisassi gnme nt is my own. Studentssignature Date Student’ s comments Feedback Grade Lecturer’s signature Date UniversityofMoratuwa,S riL anka Page 1of 2 Ranga Rodrigo T.N.Chandrapala 070059B 26-10-2010 Ref: http://www.mathworks.com/matlabcentral/ fileexchange/27023-log-polar-image-sampling 26-10-2010

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 A02: Point Operations, Filtering, and Frequency Domain

September 18, 2010

 Assignment Brief 

Student name and no.

Course EN4551: Fundamentals of Machine Vision & Image Processing  

  Assessment A02: Point Operations, Filtering, and Frequency Domain

  Weight This amounts to 10% of the module grade. Late penalty is 33% per week.Outcome The student must experiment with simple image processing algorithms

related to point operations, filtering and frequency domain.

Date handed out September 21, 2010 Date due October 19, 2010

Submission date

Student’s Declaration

I confirm that the work submitted for this assignment is my own.

Student’s signature Date

Student’s comments

Feedback 

Grade Lecturer’s signature Date

University of Moratuwa, Sri Lanka Page 1 of 2 Ranga Rodrigo

T.N.Chandrapala 070059B

26-10-2010

Ref: http://www.mathworks.com/matlabcentral/

fileexchange/27023-log-polar-image-sampling

26-10-2010

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A02.1

The object of interest was separated from the

image and its histogram was calculated. As a

similar sized object was to be located in a given

picture, the total image was scanned panningfrom one end to the other selecting similar

sized pixel blocks. Histograms of these image

blocks were calculated, and correlation

between the histogram of the original object

and these subsequent histograms were

calculated. The highest point of correlation was

taken as the point of detection.

 

Figure 1: Object identification using histogram

The program is able to draw a border aroundthe detected region if the correlation is high

enough to be an actual match.

A02.2

Spatial filtering was carried out as a convolution

of the image and the filter.

Gaussian Filter

The 2-D Gaussian function was used to create a

Gaussian filter. An image was added with noise

to create a noisy image, and it was convolved

with the filter.

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 Figure 2: Gaussian Filtering

The mean absolute difference (MAD) was

calculations are as follows.

•  Original and Image with noise: 14.41

•  Original and filtered image: 8.21

Average Filter

A simple 3x3 averaging kernel was convolvedwith the image with noise.

Figure 3: Mean Filtering

MAD calculations are as follows.

•  Original and Image with noise: 14.41

•  Original and filtered image: 8.27

Median Filter

The median filter is very effective against ‘salt

and pepper’ noise.

Figure 4: Median Filtering

The mean absolute difference (MAD) was

calculations are as follows.

•  Original and Image with noise: 11.3•  Original and filtered image: 8.4

Un-sharp filter

The theory behind this filter is the creation of a

sharpening mask by taking the difference

between an image that need to be sharpened

and its blurred version. Then the mask is added

to the image itself to create the sharp image.

‘img1’ has been created by convolving it with a

Gaussian kernel. It is referred as the blurred

image. Then it is recovered using the algorithm.

Figure 5: Sharpening image

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Sharpening using Laplace filter

A Blurred image is sharpened using a Laplace

mask. By convolving the Laplace filter with the

blurred image the edged become highlighted.

Then this mask is added to the blurred image to

get a sharper result.

Figure 6: Sharpening using Laplasian

This method seems to create sharper images

than unsharp masking.

A02.3

The image and its translated version weretransformed into the Fourier domain. The cross

power spectrum R was calculated. And the

inverse Fourier transformation was taken. The

peak value corresponds to the translation.

Figure 7: Original and translated images by (20,20)

Results:

The peak appeared in the point corresponding

to (20,20) as expected in the phase correlation

plot.

A02.4

The methodology in the paper “An FFT-Based

Technique for Translation, Rotation, and Scale-

Invariant Image Registration” was

implemented.

Figure 8: Origanal Vs Translated, rotated and scaled image

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An initial image was loaded and it was

subjected to translation, rotation and scaling.

Then the FFT of the two images was taken.

The highpass filter was designed according to

the specifications of the paper. After passing

through the highpass module, log-polar

transformation was done. This was done usinglogsample.m. The inputs to this function are the

image, the maximum and minimum of the

radius, the starting coordinates, the number of 

samples in the angle and radius. The natural

logarithm has been used.

After considering the two results as translated

images themselves, a phase correlation was

done to get the translations

Figure 9: Rotation and Scaling

The translation in the X axis gives the scalingand Y axis gives the rotation. To obtain the

actual values, the two translation values were

scaled as given in the code. (Theta and sc

values).

Accuracy:

Actual Result

Rotation 20 deg 19.6 deg

Scaling 2 1.99

Rotation 30 deg 29.6 deg

Scaling 1.25 1.23After obtaining the Rotations and Scaling

factors the effects were reversed to crate same

sized images with same orientation. But the

translation remained the same.

The two images after re-scaling and reverse

rotation was converted to the Fourier domain

and a phase correlation was carried out as

before to get the translation. The results are as

follows. (The translation in the given scenario is

zero when Tx=Ty=88)

Actual Result

X=18; Y=8 X=19; Y=9X=15; Y=15 X=16; Y=16

Figure 10: Translation