Fourier Transform Model of Image Formation

     Fourier Transform (FT) is a powerful tool in image processing.  It gives the spatial frequency, that is, for a signal of dimension X, the FT gives 1/X.  This activity is divided into four parts, so let us go through each part.

A.  Familiarization of the discrete FFT.

     Let us first familiarize ourselves with the discrete FT.  In Paint, create a 128x128 image of a white circle with a black background (shown in Figure 1.a).  Then, load this image in Scilab and convert it into grayscale.  fft2() was applied to the grayscale image.  Note that the result is a complex number array, and to see the intensity you need to take the absolute value by using the abs() function (shown in Figure 1.b). 

Figure 1.a Image of a 128x128 white circle with a black background.

     Notice that the result of the FFT looks like a black picture, but don't be fooled!! Look closely at the edges!  There is a very small white spot on the corners.  This represents the spatial frequency of the image.  Application of fftshift() centers the FFT, as seen in Figure 1.c.

Figure 2.b FFT of the image in Figure 1.a

Figure 1.c Shifted FFT of the image in Figure 1.a (circle)

Figure 1.d Application of the FFT to the

 image in Figure 1.a twice.

Figure 1. Output images for the familiarization of discrete FFT.

     Notice that the application of the FFT twice gives back the original image (see Figure 1.d). Now, let us apply the same procedure to a 128x128 image of a letter A (shown in Figure 2.a).

Figure 2. a) 128x128 image of a letter A.

Figure 2. b) Fourier Transform of the image in Figure 2.a

Figure 2.c) Shifted FFT of the image in Figure 2.a

Figure 2.d)   Application of the FFT to the

 image in Figure 2.a twice

Figure 2.  Output images for the familiarization of FFT using the image of letter a.

     The resulting image is inverted! What happened?? To understand this, let us look at the formula of the Fourier Transform:

 

     Equations 1 and 2 gives the formula for the FT and the inverse FT.  Based from these two equations, the application of the FT twice would introduce a negative sign on the output.  This explains the resulting inverted image.  To get the original image, use the inverse FT.  The inversion of the resulting image for the circle is not apparent because it is symmetric everywhere.

B.  Simulation of an imaging device.

     Now, let us go to the concept of convolution.  Given two functions f and g, their convolution gives out a function that is similar to both f and g.  In this part of the activity, we will simulate an imaging system.  For the object, a 128x128 image of the letters "VIP" was used.  For the imaging aperture, three centered circles of different radii were used. 

Figure 3.  A 128x128 image of letters VIP.

Figure 4.  Circles of different radii to act as an aperture

of the imaging system.

     

Figure 5.  Convolution of the object to circle of increasing radii (left to right).

     As can be observed, as the radius of the aperture is decreased, the image becomes blurry.  In an imaging system, the aperture acts as the limiter of the amount of light that will enter it.  As the radius of the aperture is decreased, the amount of information from the object also decreases.  This explains the image becoming blurry when the aperture is decreased.

C.   Template matching using correlation.

     The FT also has applications on template matching. Template matching is a pattern recognition technique that finds exact identical patterns in a scene [1]. Figure 6 shows the template used and the pattern to be detected.  Their FFT was obtained. Then, the complex conjugate of the resulting FFT of the template was obtained using the function conj().  Then, it was multiplied element per element to the FFT of the letter A.  Then, the FFT was obtained, as shown in Figure 7.

Figure 6.  Image used for pattern recognition (left) and the pattern to be 

recognized (right).

Figure 7.  Output of the template matching.  

     The bright spots on the resulting image represents the positions of the letter A.  So, the pattern to be detected is indeed detected.  :)

 D.  Edge detection using the convolution integral.

     This is like template matching in the previous activity, but this time, the ones to be detected are the edges.  Using Scilab, horizontal, vertical, and diagonal patterns were generated.  3x3 matrices whose total sum is zero were used to generate this patterns, as shown in Figure 8.

Figure 8.  3x3 matrix for horizontal (left), vertical (middle), and

diagonal (right) patterns.  

     Using the function imcorrcoef(), the patterns were convolved with the VIP image.  Figures 9-11 shows the output images of the convolution of the edges to the VIP image.  

Figure 9.  Horizontal edge detection. 

     Notice that horizontal lines are more pronounced than vertical and diagonal lines.  More so, the vertical lines are completely not seen.  This is because the horizontal and vertical lines are not related in any way.  There is a hint of diagonal lines because it has a horizontal component.

Figure 10.  Vertical edge detection.

     Notice that the vertical lines are more pronounced than the other lines.  This time, the horizontal lines are missing, and can be explained using the same argument in the horizontal edge detection.   Again, there is a hint of diagonal lines because it has a vertical component.    

Figure 11.  Diagonal edge detection.

     For the diagonal edge detection, diagonal, horizontal, and vertical lines are all apparent.  As been explained earlier, the diagonal has horizontal and vertical components. 

And that is the end of this activity. Hurray!!! :)

     Though I posted this blog late, I would still give my self a grade of 10/10.  I produced all the required outputs and I'm able to explain the results.  I would like to thank Ma'am Jing and Joseph Raphael Bunao for the useful conversations regarding the FFT. :D

Image Enhancement by Histogram Manipulation

     Image can be enhanced by manipulating its histogram values.  This is what our activity is about.  So, let us go through the steps.

     First thing to do is to pick an image.  Figure 1 shows the image I used for this activity. Then, get the histogram of the image. To do this, I converted the image into grayscale and used the tabul() function in Scilab and get the number of the pixels ranging from 0 to 1.  Figure 2 shows the normalized histogram of the image.  From this normalized histogram, the cumulative distribution function was obtained using the function cumsum(), as shown in Figure 3.

Figure 1. A roundabout with a fountain sits between old buildings in Manila

(image taken here.)

Figure 2.  Probability Distribution Function (PDF) of the grayscaled image in 

Figure 1.

Figure 3.  Cumulative Distribution Function (CDF) of the

histogram in Figure 2.

     Now, using the technique called backprojecting, we would modify the image.  Based on the pixel values of the image, which is ranging from 0 to 1, we would create our desired CDF.  We would try first a linear CDF, as shown in Figure 4. Then, we would find the corresponding pixel values of the image to the desired CDF.  Using the function interp1(), the pixel values of the image were interpolated to its CDF.  Then, its output is interpolated again to the desired CDF.  This then gives the modified image, as shown in Figure 5.  It can be seen that the image has a darker tone.

Figure 4. Desired CDF (linear).

Figure 5. Histogram equalized image (linear CDF).

     To check if we have done the right thing, the CDF of the modified image should correspond to the desired CDF.  Figures 6 shows the CDF and the PDF of the modified image, respectively.

Figure 6 (a)

Figure 6 (b)

Figure 6. CDF (a) and PDF (b) of the modified image in Figure 5.

     Desired CDFs such as the quadratic and logarithmic functions were also tried.  Figure 7 shows the desired CDF for the two.

Figure 7 (a)

Figure 7 (b)

Figure 7.  Desired CDFs: quadratic (a), and logarithmic (b).

     Then, repeating the steps for backprojecting, the modified images were obtained, as shown in Figure 8.  It could be seen that the modified image from the quadratic CDF is lighter than the linear and logarithmic CDF.  The modified image from the logarithmic is the darkest of them all.

Figure 8 (a)

Figure 8 (b)

Figure 8.  Output images for quadratic CDF (a) and logarithmic CDF (b).

     To check, the CDF and PDF were again obtained for the modified images.

Figure 9 (a) CDF of the modified image (quadratic).

Figure 9 (b) PDF of the modified image (quadratic).

Figure 9 (c) CDF of the modified image (logarithmic).

Figure 9 (d) PDF of the modified image (logarithmic)

Figure 9.  The CDF and PDF of the modified images for the quadratic and logarithmic CDFs.

   Free software, such as GIMP, also allows you to modify the image by histogram manipulation.  To do this, open your image in GIMP then click IMAGE and choose the MODE as grayscale.  Then, go to COLORS and click CURVES.  Here, you can manipulate the histogram by dragging the line.  Figure 10 shows the manipulated histogram and the modified image using GIMP.

Figure 10 (a) Manipulated histogram.

Figure 10 (b) Modified image using GIMP's histogram manipulation.

Figure 10.  Manipulated histogram curve and the corresponding image output.

     I would like to thank Ma'am Jing and Joseph Raphael Bunao for clarifying important points to me.  I had trouble figuring out the correct algorithm for the backprojection of the image.

     I would give myself a score of 8/10.