Activity 7: Enhancement in the Filter Domain

In this activity, the convolution theorem is used to enhance images. In the first part of the activity, we explored the convolution theorem. The succeeding parts deals with ridge enhancement, line removal, and canvas weave modeling and removal from images. All of which were done by exploiting the convolution theorem in the Fourier space.

A. Convolution Theorem
For this part, the Fourier transform (FT) of images were done. The images used are two dots of 1 pixel each, circles of different radii, squares of different area symmetric about the center. A Gaussian function of varying variance also was used instead of dots. The following are the images described above and their corresponding FTs.
It can be observed from the above images that with increasing area for circles and squares, the FT becomes smaller. It can also be seen that the FTs are sine and a sinc for circles and a squares, respectively. For FTs for the Gaussian functions decreases in area with increasing variance (or effectively increasing area).

B. Ridge Enhancement
Using the FT of an image, we can create a filter such that only the high frequencies can pass. This is what we did here. High frequencies in your image correspond to bright areas in your Fourier space. These high frequencies are repetitive in your image. Particularly for a fingerprint, these are the ridges. This is the reason why the particular filter below was used. The following shows the original fingerprint, the enhanced fingerprint, the filter used and the FT of the original image. Notice that in the reconstructed image, the blotches were minimized. they were acutally spread out, thus a more enhanced ridges were obtained. However, there is a downside with this enhancement. The reconstructed image was blurry on other areas. This maybe due to the filtered frequencies that should not be filtered out.

C. Line Removal
For this part, The inverse of part B was done. Here, we actually filtered out the bright areas in the image. This bright areas correspond to the lines you like to remove in the image. We designed a filter such that these high frequencies cannot pass. Indeed, we obtained an image without the lines. Again, the following are the image,its FT, the filter used, and the reconstructed image.
D. Canvas Weave Modeling and Removal
The same technique as that in C is applied here. The repetitive patterns of the canvas weave correspond to high frequencies in the Fourier space. To remove the canvas weave in the image, we designed a filter such that these high frequencies are blocked. In the canvas weave modeling, the filter (only the black dots) used is basically the canvas weave patterns. So taking its inverse FT is like modeling the canvas weave itself. However, results showed that the canvas weave model is angled either 90 or 180 degrees with respect to the original canvas. Results are shown below.

Canvas Weave Removal


Canvas weave Model

*** To enhance the image properly, make sure that the FT of the image coincides with the filter design. The problem I encountered was I cannot get an enhanced image. This is because my filter did not actually filtered the frequencies I wanted since it did not coincide with the FT of the image.

I give myself an 8/10 for this activity since I am not sure if I did the ridge enhancement properly. Remember that a blurry image was formed.

Thank you to Gary and Gilbert for suggesting to use "imwrite" instead of" imshow" in obtaining the FTs and designing the filters.

Activity 6: Properties of the 2D Fourier Transform

In this activity, the properties of the 2D Fourier Transform (FT) were investigated. For the first part, the FT of a square, donut, square annulus, two slits, and two dots were obtained. The following images show the images with its corresponding FT's.
The second part is to create a 2d sinusoids with different frequencies and obtain their FT's. Results showed that as you increase the frequency, the spacing between the dots in the FT's also increases. This is because at higher frequencies, the spacing between the sinusoids decreases. This is equivalent to an increase in the spacing of the delta function (or in this case, the dots) in the frequency domain. Note that the FT of a sinusoid is a delta function.

On the next part of the activity, we investigated the effect of adding a constant bias on the sinusoid. From the results, it can be seen that adding a negative or positive value of bias does not change the FT as long as the magnitudes are equal. Note that frequency = 4 for all was used.
When the sinusoids are rotated from 0-120, their FTs also rotates, which is in accrodance to the Introduction part of the manual. [1] (frequency = 4)
The product of two sinusoids is a checkerboard with circles or ellipses. Circles are obtained when both the coefficients for X and Y are the same, otherwise, an ellipse. The FTs produced are dots arranged in a square (for circles) and in rectangles (for ellipses) . Notice also that the distribution of the dots has the same shape as the component of the checkerboard. What follows are the images of the generated sinusoids and their corresponding FT's.
For the last part of the activity, the FTs of combinations of the sinusoids with varying rotating angle and the product of two sinusoids were first predicted. Superposition of the waves were basically done. From the linearity property of FT, the FT of the sum of two waves is just the FT of the first wave plus the FT of the second wave. Here, we expect to get the same result as the previous only this time they are rotated. Results verified this linearity property.
For this activity, I give myself an 8/10. Thank you to Raffy for introducing the 'mesh()' function. It was fun to look at the sinusoids in 3D! :D Thank you also to Ma'am Jing for the code :)

**Scilab code can be found in Reference 1.
Note that to see better results, open the image on a new tab :)

Reference:
Dr. Maricor Soriano. AP 186 Activity 6 Manual

Activity 5: Fourier Transform Model of Image Formation

For this activity, Fourier Transform was implemented on images. The first part of the activity is to get the FFT of a circle and the letter 'A'. Shown below are the original image, the shifted FFT of that image, and the FFT of its FFT. Notice that the third image is just a reconstruction of the original image, only this time it is inverted. This is a property of the inverse FFT - that the inverse FFT is just the same as the forward FFT with the image inverted. [1] This is not readily seen in the circle since it is symmetric in all directions.


The second part of the activity deals with convolution of two images. The system resembles an imaging system with the circle serving as an aperture and the text 'VIP' as the object. The radius of the circle is varied and the following results were obtained.

r=5

r=10
r=15
r = 48.5

As can be seen, the larger the radius of the aperture, the better the reconstruction. This is because the number of frequencies that can pass through is dependent on the size of the aperture. For larger apertures, more frequencies can pass through. Hence a better reconstruction is obtained as opposed to that of apertures with smaller radii.

The third part of the activity is to find the correlation of the letter 'A' in the sentence "THE RAIN IN SPAIN STAYS MAINLY ON THE PLANE." To do this, the FFT of 'A' is multiplied element per element with the FFT of the text. The result shows that the letter A's in the text are highlighted because of high correlation.
The figure above shows the text (1st), the letter 'A' with the same font style and font size, and the resulting image when their FFTs are multiplied element per element.

The last part of the activity shows edge detection using the convolution integral. Four patterns were used (horizontal, vertical, diagonal, and spot). The following results were obtained.
The pattern allows only certain components to be seen. For example, the reconstruction for the horizontal pattern allows only almost horizontal components. Most vertical components are missing in the horizontal pattern, most horizontal components are missing in the vertical pattern, other vertical and horizontal patterns are missing in the diagonal pattern, and most components are present in the spot pattern.



The code (taken from Reference 1):

//first part
I=imread("C:\Documents and Settings\2004-28298\Desktop\activity 5\circle.bmp");
Igray=im2gray(I);
FIgray= fft2(Igray);
fc=abs(fftshift(FIgray))/max(abs(fftshift(FIgray)));
ffc=abs(fft2(FIgray))/max(abs(fft2(FIgray)));
imwrite(fc,"C:\Documents and Settings\mimie\Desktop\activity 5\fc.bmp");
imwrite(ffc,"C:\Documents and Settings\mimie\Desktop\activity 5\ffc.bmp");


//2nd part
o=imread("C:\Documents and Settings\mimie\Desktop\activity 5\vip.bmp");
I=imread("C:\Documents and Settings\mimie\Desktop\activity 5\30r.bmp");
lgray=im2gray(I);
ogray=im2gray(o);
fl=fftshift(lgray);
fo=fft2(ogray);
col=fo.*(fl);
icol=fft2(col);
fimage=abs(icol);
fimagenorm=fimage/max(fimage);
imwrite(fimagenorm,""C:\Documents and Settings\mimie\Desktop\activity 5\vip30r.bmp");


//3rd part
t=imread("C:\Documents and Settings\mimie\Desktop\activity 5\text.bmp");
a2=imread("C:\Documents and Settings\mimie\Desktop\activity 5\A2.bmp");
tgray=im2gray(t);
a2gray=im2gray(a2);
ft=fft2(tgray);
fa2=fft2(a2gray);
cft=conj(ft);
cta2=fa2.*(cft);
fcta2=fft(cta2);
ta2=fftshift(abs(fcta2))/max(fftshift(abs(fcta2)));
imwrite(ta2,"C:\Documents and Settings\mimie\Desktop\activity 5\ta2.bmp");


//last part
i=imread("C:\Documents and Settings\mimie\Desktop\activity 5\vip.bmp");
igray=im2gray(i);
pattern1=[-1,-1,-1; 2,2,2; -1,-1,-1];
pattern2=[-1,2,-1; -1,2,-1;-1,2,-1];
pattern3=[-1,-1,-1; -1,8,-1;-1,2,-1];
pattern4=[2,-1,-1;-1,2,-1;-1,-1,2];

c1=imcorrcoef(igray,pattern1);
c2=imcorrcoef(igray,pattern2);
c3=imcorrcoef(igray,pattern3);
c4=imcorrcoef(igray,pattern4);

c1norm=c1/max(c1);
c2norm=c2/max(c2);
c3norm=c3/max(c3);
c4norm=c4/max(c4);

imwrite(c1norm,"C:\Documents and Settings\mimie\Desktop\activity 5\c1norm.bmp");
imwrite(c2norm,"C:\Documents and Settings\mimie\Desktop\activity 5\c2norm.bmp");
imwrite(c3norm,"C:\Documents and Settings\mimie\Desktop\activity 5\c3norm.bmp");
imwrite(c4norm,"C:\Documents and Settings\mimie\Desktop\activity 5\c4norm.bmp");

**Remember to comment the part of the code you don't need.

Many thanks to Ma'am Jing for providing the code which guided me through this activity. Thank you also to Kaye and Miguel for the discussions we had, and to Raffy for helping me. I give myself a 9/10 for this activity because I did what was expected from the activity.

[1] Dr. Maricor Soriano. Activity 5 Manual.

Activity 4: Enhancement by Histogram Manipulation

For this activity, a poor contrast image is enhanced by histogram manipulation by following the steps below. The code (if applicable) used and the results accompany the steps.

1. Get an image from the internet or from own collection. Crop if necessary.

I=imread('C:\Documents and Settings\mimie\Desktop\New Folder (2)\activity 4 images\crop.bmp');

2. Change image to grayscale and get its histogram and corresponding Cumulative Distribution Function (CDF).
I=im2gray(I);
gcrop=I/max(I);
imwrite(gcrop, "C:\Documents and Settings\mimie\Desktop\New Folder (2)\activity 4 images\gcrop.bmp");

xsize=size(I,1);
ysize=size(I,2);

x = matrix(0:255,[1,256]);
y = x.*0;
a = 1;

for i = 0:255
m = find(I== i);
y(a) = length(m);
a = a + 1;
end

cs=cumsum(y);
csnorm=cs/max(cs);


3. Backproject the CDF of original image to a linear CDF. Get resulting image, its histogram and its cdf.

//linear desired cdf

for i=1:xsize
for j=1:ysize
n=find(x==I(i,j));
newI(i,j)=cs(n);
end
end

4. Backproject to a nonlinear CDF and get resulting image, histogram, and CDF of reconstructed image.
//nonlinear

z = [0:255];
g = tanh(30.*(x-128)./255);
g = (g - min(g))./2;

plot(x,g);
for i = 1:xsize
for j = 1:ysize
n1 = find(x == I(i,j)); //get index of x from image
y1 = csnorm(n1); //get corresponding y value
n2 = find(g<=y1); //equate with y in desired cdf
newI2(i,j) = csnorm(n2(max(n2)));
end
end

The original image has poor contrast since its pixel values does not cover the entire 0-255 values. By backprojecting its original CDF to a linear CDF with pixel values from 0-255, a better contrast is obtained. When we say backproject, it means that it follows the new CDF, but not perfectly, of course. This can be seen from the CDF of the reconstructed image using a desired linear CDF. When the nonlinear CDF (for this case, hyperbolic tangent) is used, the quality of the image degraded. Notice that the histogram of the reconstructed image is narrower than that of original image. Hence a poorer image contrast was obtained. We can conclude in this activity that image contrast of an image can either be improved or degraded depending on the CDF to which these images are backprojected.

For this activity, I give myself an 7/10. Thank you to Ma'am Jing, Ma'am Gay, Raffy, Kaye, and Neil for the help.

***

• Note that to get the histogram and CDF of the reconstructed image, we have to multiply the image to 255 since we had it normalize to 255 before displaying it.
  
• to change the color of plot to green : plot(x,y,,['g']) may apply to other colors as well :)

References
1. Dr. Maricor Soriano. Activity 4 manual.
2. image taken from: http://yrekahigh.com/Portals/1/YrekahighPics/Misc_%20Pictures/Old%20Yreka%20High.bmp