..pics and physics

..on neural networks

2008-09-24T01:12:00.000-07:00

On the two previous activities, we applied pattern recognition to identify the class a given image belongs to. We have used the the minimum distance classification and the linear discriminant analysis. In this activity, we introduce another method: neural networks. Neural networks mimics the way how our brain neurons work. It classifies images by learning the rule of mapping using the training set. Below is a diagram of an artificial neural network. Xi would be the inputs and Z would be the output. The goal is to determine the values of the weights Wi such that the resulting output Z matches our desired output. To do this, we train the network using examples from the training set. After determining the values for the weights Wi, we can now classify the images from the test set.

We use the same set of images form the previous activities. We consider 10 (5 training set, 5 test set) images of leaves of each kind of two different plant: Mango and an ornamental plant (I dont know the name).

Mango Leaves

Ornamental Plant Leaves

We also enhance the images as to make the classification process easier. We used the Gray World Algorithm in white balancing our images. For each class, we divided it into two groups: the training set and the test set. We use the training set to get the characteristic features of the leaves that would easily differentiate it from the other class. In this case, we used the RGB values and the eccentricity of the shape of the leaves as our four distinguishing features. Below is a table showing the four features corresponding to each of the 10 training set. These would be our input X in the neural network.

From this training set, we determine the desired weights Wi of the network. We now classify the remaining images in the test set. The output would be 0 or 1 corresponding to the two classes (1 or 2). The table below is the summary of the results.

Like the two previous methods, neural networks have successfully classified all the 10 images. The success rate is 100%.

I was able to successfully accomplish the activity. I want to give myself a 10.

Acknowledgment to Jeric for the sample code and the discussion on neural networks.

..on probabilistic qualification

2008-09-18T01:11:00.000-07:00

Like the previous activity, our goal is to correctly classify an image to the class which it belongs. But this time, we use a different method: the linear discriminant analysis (LDA). We use the same set of images form the previous activity. We consider 10 (5 training set, 5 test set) images of leaves of each kind of three different plant: Mango, an ornamental plant (I dont know the name), and Indian tree.

Mango Leaves

Ornamental Plant Leaves

Indian Tree Leaves

We also enhance the images as to make the classification process easier. Discussed in the previous activity, we used the Gray World Algorithm in white balancing our images. For each class, we divided it into two groups: the training set and the test set. We use the training set to get the characteristic features of the leaves that would easily differentiate it from the other class. In this case, we used the RGB values and the eccentricity of the shape of the leaves as our four distinguishing features. Below is a table showing the four features corresponding to each of the 15 training set. This would be our matrix x.

We further group this matrix into their corresponding class: x1, x2, and x3.
x1:

x2:

x3:

We take the mean (u1, u2, u3) for each of these three matrix as well as the mean (u) of the entire matrix x. We then create the mean corrected (x1', x2',x3') matrix given by: xi'=xi-u for i=1,2,3.

We now compute for the covariance matrix given by: Ci=(xi'*xi)/ni for i=1,2,3 where ni is just the total number of trial images for class i. We further compute for the covariance matrix of the entire data given by:

where r and s corresponds for each row r and column s entry and n is just the total number of trial images. Finally, we compute for the linear discriminant function f for each class i. pi is ni/n.

The highest fi corresponds to the class where the image belongs. We do this for the 15 remaining images in the test set. The table below is the summary of the results.

Like the previous method, we have successfully classify all the 15 images. The success rate is 100%.

I was able to successfully accomplish the activity. I want to give myself a 10.

..on pattern recognition

2008-09-15T19:12:00.000-07:00

Given an image, our goal is to correctly determine which class it belongs. Specifically, we consider 10 images of leaves of each kind of three different plant: Mango, an ornamental plant (I dont know the name), and Indian tree.

Mango Leaves

Ornamental Plant Leaves

Indian Tree Leaves

We first enhance the images as to make the classification process easier. Discussed in the previous activity, we used the Gray World Algorithm in white balancing our images.

For each class, we divided it into two groups: the training set and the test set. We use the training set to get the characteristic features of the leaves that would easily differentiate it from the other class. In this case, we used the RGB values and the eccentricity of the shape of the leaves as our distinguishing features. This would be our feature vector x. We take the mean m of our feature vector for the entire training set. This is given by:

where j is the class index and N is the number of samples in our training set for each class . Below is the table for the mean values for each class.

To classify which class images belongs to, we use the minimum distance classification. Taking the feature vector x of an unknown image, we determine which class mean m it is nearest to. This is given by:

where

The smallest D corresponds to the class where the unknown image belongs. We do this for the 15 remaining images in the test set. The table below is the summary of the results.

We have successfully classify all the 15 test images. The success rate is 100%.

I was able to successfully accomplish the activity. I want to give myself a 10.

Acknowledgment to Rica for uploading the images.

..on basic video processing

2008-09-08T01:07:00.000-07:00

Our goal is to determine to moment of inertia (MOI) of a physical pendulum using only a video of its motion. We apply image processing techniques to extract necessary information needed to solve for the MOI. We used a flat metal ruler (mass=.0478kg, height=.635m, width=.03m) as the physical pendulum. Below is a sample video of its motion.

The MOI of a physical pendulum is given by:

I=mg(h/2)(T/2π)^2.
(source: http://cnx.org/content/m15585/latest/)

To solve for I, we have to determine the period of oscillation from the video. First, we convert the video into frames of images. In processing the images, we first apply the color image segmentation such that we pick out the moving physical pendulum away from the background. This technique has already been discussed from the previous activity. Below are the results.

Notice that the quality of the images are poor and distorted. We enhance the quality of the images using another technique previously discussed: morphological operations. We first binarized the images then apply dilation and erosion to enhance the shape of the physical pendulum.

Now that the images are enhanced, we determine the period T of the oscillation. Since the images are binarized, we could keep track of the motion of the pendulum by noting the x and y coordinates of the white pixels corresponding to the pendulum. We use the mean x and the mean y coordinate to approximate the motion of the center of mass. Below is the plot of the X axis coordinate as a function of time.

As the oscillation reaches a maximum, there would be a change in the directions of the coordinates of the center of mass. This periodic change in direction is the sinusoidal function we see in the plot above. Noting this and counting the number of frames between consecutive changes, we determine the number of frames corresponding to half the period. Since the frame rate of the video is 14.5 fps, the time interval per frame is 1/14.5=.069s. Thus the half period of the oscillation is just the number of frames times .069s.

From the processed images, the number of frames corresponding to half a period are: 10, 10, 9, 10, 9, 10, 9, 10, 10, 9. The mean value is: 9.6 frames corresponding to 0.6624s half period. Thus, the period of oscillation is 1.3248s. Using the equation above, we determine the MOI to be 6.612E-3.

To check, we determine to moment of inertia using another equation given as:

I=(m/2)(h)^2 + (m/12)(h)^2
(source: http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

Noting the value of the mass=.0478kg, height=.635m, and width=.03m, the theoretical MOI is 6.428E-3. Comparing with the experimental MOI above, this corresponds to 2.86% error. The method is successful in determining the MOI of a physical pendulum using only a video of its motion.

I've successfully accomplished the activity. I want to give myself a 10.

Acknowledgment to my collaborators on the physical pendulum video gathering: Cole and Jeric.

..on color image segmentation

2008-09-01T23:33:00.000-07:00

From a given image, our goal is to pick out the region of interest (ROI). To do this, we first have to get rid of the effect of brightness on the image (shadowing), and just consider the chromaticity. We represent the color space not by the RGB but by normalized chromaticity coordinates (NCC). Per pixel, we solve for the corresponding r,g,b values given by:

Since r+g+b=1, the chromaticity can be represented using only 2 values (r & g). Below is the normalized chromaticity space (x-axis is r and y-axis is g).

We now proceed with picking out the ROI. There are two methods: the Parametric and the Non Parametric.

Parametric

In this method, we first crop a subregion in the ROI. From this patch, we take the mean and standard deviation of r and g. We then compute for the distribution probability (we assume a Gaussian) using:

We do the same for p(g). The joint probability is just the product p(r)p(g). Below are sample the results:

Original Image

Parametric
ROI: Blue Tiles

Parametric
ROI: Yellow Tiles

Non Parametric (Histogram Backprojection)

In this method, we take the 2D r-g histogram of the crop subregion of the ROI. We then backproject the value to the pixel location of the original image. Below are sample the results:

Original Image

Parametric
ROI: Blue Tiles

Parametric
ROI: Yellow Tiles

Comparing between the two method, we could observe that the Parametric method has a better result.

Original Image

Parametric

Non Parametric

I've successfully accomplished the activity. I want to give myself a 10.

..on color image processing

2008-08-27T20:06:00.000-07:00

Below are images taken at different white balancing.

Daylight setting under flourescent light

Cloudy setting under flourescent light

Flourescent setting under daylight

We notice that the images are wrongly balanced. The background which is supposed to be white appears otherwise. The white background for the daylight setting under flourescent light appears greenish. For the second image, the background appears grayish while for the last image, the background appears bluish. In order to correct these, we apply two popular algorithms for achieving automatic white balance: the reference white algorithm and the gray world algorithm.

Reference White Algorithm (RWA)
This method utilizes a known white image and used its RGB values as the divider. For our images, we used the background as the reference white. Below are the results of the enhancement. Notice that we have improved the color of the background as well as the colorful patches.

RWA (Daylight setting under flourescent light)

RWA (Cloudy setting under flourescent light)

RWA
(Flourescent setting under daylight)

Gray World Algorithm (GWA)
This method assumes that the world is essentially gray. Thus, we just take the average of the red, green, and blue value of the captured image and use them as the divider. Below are the resulting enhanced images. Again, the colors of the resulting images are greatly enhanced.

GWA (Daylight setting under flourescent light)

GWA (Cloudy setting under flourescent light)

GWA
(Flourescent setting under daylight)

Finally, we take an image of an ensemble of leaves having the same hue (green). This image is taken in flourescent setting under daylight. Notice that the white background appears bluish. Also, the color of the dark leaves appears black.

Flourescent setting under daylight

We improve this image using the reference white algorithm and gray world algorithm. Below are the resulting images. Both methods were able to enhance the image. The background now appears white and the color of the leaves are enhanced. Between the two methods, the resulting image from the gray world algorithm is better. The background appears really white and the color of the leaves are really distinct and clear.

RWA

GWA

I was able to successfully perform the activity. I want to give myself a 10.

Acknowledgment to Jeric for the rubik's cube and Rica for uploading the images.

..on stereometry

2008-08-25T19:53:00.000-07:00

We try to reconstruct the image of a 3D object using images taken at different positions. Using a technique called stereometry, we will derive the depth z of the 3D object using 2D images (x,y). Consider the diagram below.

Given that the 2 images have the same y coordinates, we could solve for z using:

where b is just the traverse distance between the 2 images and f is just the focal length of the camera. We could solve for f using the calibration technique discussed on the previous activity.

Below are the two different images of a rubik's cube taken with b=5cm.

Using 25 different points (x,y), we calculated for the corresponding depth (z). Below are the 3D reconstruction using splin2d of scilab for not a knot, bilinear interpolation, natural, and monotone.

Enlarging further, we see that the reconstructed 3D object depicts a cube. Though the resulting rendition is not a perfect cube, we were able to show the general shape of the 3D object.

I want to give myself a 10.

..on photometric stereo

2008-08-07T05:47:00.000-07:00

We use photometric stereo to extract the 3D shape of an object using only the information from the shadow. We estimate the shape of the object using the shading obtained from different images of different light source location.

Consider a point source of light at infinity.

The intensity I of the image is related to the vector position of the camera given by:

where N is the number of images used. We now solve for g using:

to get the normal vector n, we simply normalize it:

to derive the shape from the normals, we note that the surface elevation f(x,y) is related to the normals by:

Finally, we solve for the surface elevation using:

We apply this technique using four images of a sphere. The resulting 3D rendition is:

Indeed, the resulting shape is a sphere.

I've successfully accomplished the activity. I want to give myself a 10.

..on correcting geometric distortions

2008-08-06T19:47:00.000-07:00

Below is an image of a "grid" capiz window. Notice that the image has a barrel distortion effect. The square grids located at the edges are much smaller than those found at the center. The center appears to be bloated while the sides are pinched. These are due to the "imperfect" lens of the camera that captured the image.

Our goal is to correct this distortion. We use the center square grid as our reference since it is less distorted. We then determine the transformation matrix that caused the barrel effect. Let f(x,y) be the coordinates of the ideal image while g(x',y') are the coordinates of the the distorted image. To determine the transformation matrix C, we map the coordinates of the ideal image in the distorted image.

We then compute for the transformation matrix C using:

Now that we have determined the transformation matrix, we just simply copy the graylevel v(x',y') of the pixel located in g(x',y') into f(x,y). But since the calculated g(x',y') coordinates are real (pixel coordinates should be integral), we use bilinear interpolation. The graylevel of an arbitrary point is determined by the graylevel of the 4 nearest pixels encompassing that point.

We can no solve for the graylevel using:

For the remaining blank pixels in the ideal image, again we use interpolation of the four nearest corners to determine its graylevel. Below is a comparison of the original distorted image and the enhanced(ideal) image. Notice that at the lower left level, the size of the square grid for the enhanced image increased. The grid lines also become more parallel. The resulting image has lessen the effect of the distortion. The image is no more bloated.

Original distorted image

Enhanced ideal image

I think I've performed the activity successfully. The distorted image was enhanced. I want to give myself a 10.

..on camera calibration

2008-07-29T23:12:00.000-07:00

A simple image is not an exact projection of the actual object. Some informations are lost. The task is to recover these lost details. And one of the important step is camera calibration. The goal is to know the transformation from the 3D world coordinates to the 2D image coordinates.

Below is an image of a 3D calibration checkboard. We designated an origin and using a right-hand coordinate system, we chose 25 square corners and measured their coordinates (Yi, Zi) with respect to the origin. We also determined their real world coordinates (Xo, Yo, Zo).

We now proceed with determining the transformation matrix a. The equation is given by

Finally, we solve for a using

To check for the accuracy, we randomly chose 4 points on the checkboard and using their real world coordinates, we calculated for their image coordinates. We measured the deviation from the theoretical value using the eucleadian distance. The results are: 1.12, 1.31, 1.44, and 0.98. These values shows that the calculated coordinates are less than 2 pixels away from the actual location. the accuracy is amazing.

I was able to finish the activity with high accuracy. I want to give myself a 10.

..on preprocessing handwritten text

2008-07-22T00:08:00.000-07:00

Below is a digitalized image of a document. Our goal is to extract the handwritten text and to remove the horizontal lines.

We remove the vertical lines using the same technique used on the previous activity(lunar image). The Fourier Transform of the image is shown below. To get rid of the horizontal lines, we have to block the frequencies corresponding to it. Thus, we introduced the filter below.

The resulting image is free of the horizontal lines. We then binarized this image and enhanced it using morphological operations.

Below is the final image. Though we are successful in getting rid of the horizontal lines and the extraction of the handwritten text, the quality is not enhanced. The letter count is 106 which is obviously higher than the actual number of letters in the text.

I was able to finish the activity. Though the lines were successfully removed, the extracted text is not well enhanced. I hope a 9 would be reasonable.

..on binary operations

2008-07-17T01:49:00.000-07:00

Below is an image of scattered punched paper. Our goal is approximate the size (pixel count) of a single punched paper.

Since the image is relatively large, we cut it down into twelve 256X256 subimages. We then convert them into binary images. Below are examples of the binary images.

These images are not ideal. We enhanced them further by applying opening and closing morphology operator. These clean the image of isolated pixel and small holes. Below are the results of the enhancement.

We now measure the pixel count for each blob. Since the enhancement was not able to perfectly separate individual punched paper, there are big blobs on the image. We need to filter out this outliers. We do this by getting the histogram of the pixel count.

For this particular plot, we consider only those values between 300-600 since it is in this range that we have an approximate Gaussian distribution. Finally, we take the mean of the pixel count within this range. Our result is 428.89 with a standard deviation of 71.53. To check, we processed images of individual punched paper and measured the pixel count. The average value is 456.1 with a standard deviation of 70.69. This is a 6% error.

Below is the Scilab code. I was able to implement the program such that it automatically outputs the mean pixel count and standard deviation for all images.

numt=[];
v=0;
for j=1:1:12
//upload image
dir('E:\acads\1st sem 08-09\186\activity9');
str=strcat(['C2_',string(j),'.jpg']);
image1=imread(str);

//convert to gray scale
image1=im2gray(image1);

//convert to binary
image1=im2bw(image1,229/255);
//f1=scf(1);imshow(image1);

//make structuring element
sq=ones(2,2);
se=mkfftfilter(sq,'binary',1);

//opening
C1=erode(image1,se,[1,1]);
//f2=scf(2);imshow(C1);
C2=dilate(C1,se,[1,1]);
//f3=scf(3);imshow(C2);

//closing
C3=dilate(C2,se,[1,1]);
//f4=scf(4);imshow(C3);
C4=erode(C3,se,[1,1]);
//f4=scf(2);imshow(C4);

//labeling
[mat,n]=bwlabel(C4);

//counting pixel size
val=[];
num=[];
counter=1;
for i=1:1:n
[x,y]=find(mat==i);
val(counter)=i;
num(counter)=length(x);
counter=counter+1;
end

for i=1:1:n
//mm=v+i
numt(v+i)=num(i);
end
v=v+n;

end

//histplot([10:30:800],numt);

//neglect outlier
numf=[];
c=1;
for i=1:1:length(numt)
if numt(i)>=300 & numt(i)<=600
numf(c)=numt(i);
c=c+1;
end
end

mean(numf)
stdev(numf)

I was able complete the activity. I was also able to implement the program automatically. I think I would give myself a 10 plus the bonus (5), thus a 15.=)

..on morphological operations

2008-07-15T01:00:00.000-07:00

Below are the results when different shapes are subjected to dilation and erosion using different structuring element (SE). Dilation results to an increased size of the original image while erosion reduces the size. The resulting shape is affected by the SE. The 1st column image are the original shapes, 2nd column are the results of dilation, and the third column are the results of erosion.

1. SE: 4x4 ones

2. SE: 2x4 ones

3. SE: 4x2 ones

4. SE: cross 5 pixel long 1 pixel thick

The results are consistent with the predicted sketches. As expected, the dilated image is bigger than the original while the eroded is smaller. The resulting image is affected by the SE used.

I was able to finish the activity and my results are consistent. I want to give myself a 10.

..on enhancement on the frequency domain

2008-07-09T19:42:00.000-07:00

Below are 2D sinusoidal patterns in increasing frequency (f=1,f=4,f=16).

Patterns

f=1

f=4

f=16

Taking the fourier transform (FT) of these patterns, we would notice that as the frequency increases, the separation between the two peaks increases. The two symmetric peaks represent the frequency of the pattern. As the frequency increases, the peaks deviates further from the center.

Fourier Transform of the Patterns

f=1

f=4

f=16

We now consider rotated patterns. These patterns were generated by:
z = sin(2*pi*f*(Y*sin(theta) + X*cos(theta))) where theta=30. The frequencies are f=4 and f=16.

Rotated Patterns

f=4

f=16

The resulting FT are also rotated. The rotation of the pattern results to the rotation of their FT's.

Fourier Transform of the Rotated Patterns

f=4

f=16

We now consider a combination of sinusoidal patterns on both X and Y axis, as well as rotated patterns.

Combined Patterns

fx=4
fy=4

fx=4
fy=16

f=8
theta=30

The resulting FT created multiple peaks. For fx=fy=4, the peaks are symmetric and are equidistant. When the frequency along one direction is increase, the peaks are stretch further along that direction. Again, we observe that the rotation of the pattern resulted to the rotation of the FT.

Fourier Transform of the Combined Patterns

fx=4
fy=4

fx=4
fy=16

f=8
theta=30

Fingerprint: Ridge Enhancement

Below is an image of a fingerprint in grayscale.

We take the FT of the image and filter out other frequencies to enhanced the ridges.

Lunar Landing Scanned Pictures: Line Removal

We also apply the same technique on the image below.

grayscale

Filter

Resulting Image

The lines are minimized.

I was able to complete the activity but the resulting images are not perfectly enhanced. I don't think I deserve a 10 on this. I hope a 9 would be justifiable.

Acknowledgment to Ed for the scanned fingerprint.

..on fourier transform model of image formation

2008-07-07T19:40:00.000-07:00

The Fourier Transform (FT) of an image is given by:

Below are 128x128 bitmap images of circles with different radius and letter A's in varying size.

We apply discrete 2D FFT on these images. The output of the FFT is a complex number. Getting the modulus gives us the intensity.

Big Circle

Big A

Medium Circle

Medium A

Small Circle

Small A

Applying fftshift() gives us:

Big Circle

Big A

Medium Circle

Medium A

Small Circle

Small A

These images are consistent with the analytical FT of the circle. As the radius of the circle decreases, we clearly observe the airy disc. The same for the letter A. As the size decreases, the airy patterns becomes more distinguishable. Applying FFT twice on the image results to the original image, but as we notice for letter A, the resulting images are inverted.

Convolution

The convolution of two 2D function f and g is given by:

This could be represented as h=f*g. If F and G are the FT of f and g, respectively, then the convolution is just simply H=FG. Below are the convolution images between the word VIP and the circles of varying radius. The circle serves as the aperture. As the radius of the circle decreases, the aperture becomes smaller and the resulting image is blurred. The bigger the aperture, the sharper the image.

Aperture 1

Image 1

Aperture 2

Image 2

Aperture 3

Image 3

Correlation

The correlation of two 2D function f and g is given by:

This could be represented as p=fog. If F and G are the FT of f and g, respectively, then the correlation is just simply P=F.G. The correlation measures the similarity between the two functions. Some of its application are as follows:

Template Matching

Below are the correlation images between the sentence "THE RAIN IN SPAIN STAYS MAINLY IN THE PLAIN", and the letter "A" and the word "IN". We notice that the maximum in the resulting images occurs on the part of the sentence (image) where there is a presence of A or a form like A. Same is true for the word IN.

A

IN

Edge Detection

Below are images resulting from the convolution of VIP and a 3x3 matrix pattern of an edge such that the total sum is zero. The first is a horizontal, the second is a vertical, and the third is a spot pattern. Notice that for the first pattern, the horizontal edges of the image are highlighted. For the second pattern, it is the vertical edges that are highlighted. And for the spot pattern, all the edges are highlighted.

Horizontal Pattern

Vertical Pattern

Spot Pattern

I was able to finish the activity with the results consistent with what were expected. I think I deserve a 10.

..on physical measurements from discrete fourier transforms

2008-07-02T18:08:00.000-07:00

The discrete fourier transform (DFT) is given by:

DFT could be used on getting the fourier transform (FT) of measured signal, numerically validating an analytic computation, and obtaining numerical results for signals whose FT have difficult or no analytic form.

To illustrate how DFT works, we first generate a sinusoidal signal (sin2pi()ft). Below is the plot of the generated signal with f=5 (frequency), T=2 (total signal time), N=256 (number of samples), and the sampling interval (dt) is given by dt=T/N.

We get the fourier transform of the signal. To get the frequency scale, we note that by Nyquist Theorem, the maximum frequency detectable is given by Fmax=1/(2dt). Since half of the output of the FT is just the mirror image of the other half, we just simply shift the half axis, and df=2Fmax/N. The resulting plot is:

The plot peaks at f=5/-5. This is consistent since the frequency of the generated signal is 5. The DFT could also be applied for images. Instead of a temporal signal (temporal domain), we have a spatial signal (spatial domain). The number of frequencies resulting from DFT represents the number of pixels on the original image. The method utilizes the 2 dimensional DFT given by:

The value of each point of X is obtained by multiplying the spatial image with the corresponding base function and summing the result. The base functions are sine and cosine with increasing frequencies. The resulting fourier transform is a complex valued number real and imaginary which represent the magnitude and the phase.
(source: http://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm)

For a fluorescent lamp that flickers around 120Hz, the threshold sampling rate is given by dt=1/(2F). This is around 0.004167 sec. Below is the resulting DFT generated with f=120. In order to have a dt that is below the sampling threshold, N is set to 500 with T=2. We see that peak occurred at around 120. For comparison, we set N=400 with T=2 and we could observe that DFT failed to detect it correctly (peak at 80).

F=120
N=500
T=2
Detected F=120

F=120
N=400
T=2
Detected F=80

Increasing the number of samples (N) increases the frequency range detectable. If the original N results to a sampling interval greater than the threshold indicated by the Nyquist Theorem, increasing N would increase the accuracy of the detected frequency. Below are the plots with increasing N (15,100,256). We see that the results are consistent with the analysis.

F=5
N=15
Detected F=2.3
Max F=4

F=5
N=100
Detected F=5
Max F=25

F=5
N=256
Detected F=5
Max F=80

As indicated by the Nyquist Theorem, decreasing the sampling interval (dt) increases the maximum frequency that could be detected by the FT. This widens the frequency range that could be detected which is appropriate for signals that contain higher frequencies. Below are plots of decreasing dt. We decreased dt by increasing N (since dt=T/N).

dt=.02
Max F=25

dt=.005
Max F=100

Since dt x N=T, increasing N with T fixed decreases dt. This results to a smaller sampling interval and thus, increasing its detection range to higher frequencies without aliasing. Below are plots of increasing N. We could see that the maximum F increases.

N=256
Max F=80

N=400
Max F=80

The analysis above are consistent with the results of the simulations. I want to give myself a 10.

..on enhancement by histogram manipulation

2008-06-30T19:01:00.000-07:00

Below is sample image w/ poor contrast. Our goal is to enhance the image by manipulating its histogram.

www.flickr.com

Original Image

The image was first converted into gray scale. Below is the corresponding probability distribution function(PDF) and the commulative distribution function (CDF) of image.

Original PDF

Original CDF

We now enhance the image by manipulating the original CDF. We do this by getting the T(r) of the CDF corresponding to the value of the pixel. The corresponding T(r) would be the new value of that pixel. We do this for every pixel of the original image. Below are the resulting enhanced images and their corresponding PDF and CDF. The resulting PDF is more "spread" than the original PDF. Also, The resulting CDF is now linear. We could observe that the contrast of the image was improved. The image of the rose petal becomes more distinct.

Resulting PDF

Resulting CDF

Resulting Image

Another Technique

We now apply another technique. In this method, we manipulate the CDF of the original image by "mapping" it on the desired CDF (G(z)). Specifically, we do this by getting the T(r) of the original CDF corresponding to the value of the pixel. We then map T(r) into G(z) and find the corresponding z. Below is an illustration of the "mapping".

We use G(z)=sqrt(z). Below is the original image and the resulting image with the corresponding CDF.

Original Image
theterroristsdictionary.blogspot.com

Original CDF

Desired CDF (G(z)=sqrt(z))

Resulting Image

Resulting CDF

The resulting CDF is verry much like the desired CDF. And we could see that the contrast of the resulting image is enhanced.

I want to give myself a 10. I have done the activity successfully.

Acknowledgment to Julie and Ed for the help.

..on image types and basic enhancement

2008-06-23T18:06:00.000-07:00

Below are the different types of images of Albert Einstein (binary, gray scale, true color, and indexed). Also shown are the properties of each image which was derived using imfinfo of scilab: "imfinfo("imagefilname",'verbose') ".

Binary
FileSize: 3971
Format: JPEG
Width: 118
Height: 145
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: inch
XResolution: 72.000000
YResolution: 72.000000
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Gray Scale
FileSize: 3621
Format: JPEG
Width: 109
Height: 150
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: inch
XResolution: 72.000000
YResolution: 72.000000

www.newz.dk

True Color
FileSize: 5236
Format: JPEG
Width: 145
Height: 122
Depth: 8
StorageType: truecolor
NumberOfColors: 0
ResolutionUnit: centimeter
XResolution: 72.000000
YResolution: 72.000000
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Indexed(Portable Network Graphics)
FileSize: 2938 Format:
JPEG Width: 130
Height: 113
Depth: 8
StorageType: truecolor
NumberOfColors: 0
ResolutionUnit: centimeter
XResolution: 72.000000
YResolution: 72.000000
us.geocities.com

Measuring the size of an image

Here is a scanned image of the ace of diamond card. The ruler serves as a reference measurement for the size of a pixel (1mm=13pixels or .005917sq. mm/pixel).

Below are the images of the center diamond (true color, gray scale, and binary) and the corresponding properties. To convert the image to binary, the histogram of the gray scale image was analyze (see plot below). We could see a clear separation at around 220. This was used as the threshold.

True Color
FileSize: 3549
Format: JPEG
Width: 160
Height: 211
Depth: 8
StorageType: truecolor
NumberOfColors: 0
ResolutionUnit: inch
XResolution: 96.000000
YResolution: 96.000000

Gray Scale
FileSize: 2466
Format: JPEG
Width: 160
Height: 211
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: inch
XResolution: 72.000000
YResolution: 72.000000

Binary

FileSize: 2584
Format: JPEG
Width: 160
Height: 211
Depth: 8
StorageType: indexed
NumberOfColors: 256
ResolutionUnit: inch
XResolution: 72.000000
YResolution: 72.000000

Histogram Plot

Finally, we measure the area of the diamond. We use the technique from the previous activity utilizing the follow command of scilab and the Green's Theorem. But before this, the image have to be inverted.

Inverted Image

The measured area is 10265 pixels. Using the conversion factor(.005917 sq. mm/pixel), the measured area is 60.74 sq. mm. The analytic area (using pixel counting) is 10464 pixels or 61.92 sq. mm. The percent error is just 1.9%.

I want to give myself a 10 since I was able to successfully finish the task .

Acknowledgment to Lei for the scanned images, Jeric and Ed for their help with Scilab code.

..on measuring area

2008-06-18T19:30:00.000-07:00

Below are some regular geometric shapes. Using Green's Theorem, the white areas were measured. This was done by utilizing Scilab's contour follower command (provided by the Sip toolbox library).

Rectangle
obtained area: 18387
analytic area: 18696

Circle
obtained area: 34636
analytic area: 33944

Square
obtained area: 41818
analytic area: 42228

Green's Theorem

For the rectangle, the percent deviation from the analytic area is 1.65%. For the circle it is 1.89%, and 0.97% for the square. These values are well below 5% acceptance threshold. The technique is accurate enough.

I want to give mysef a 10. I was successful in implementing the technique.

Thanks to Rica and Ed for the help with the Sip toolbox.

.. on digital scanning

2008-06-11T19:42:00.000-07:00

This hand-drawn plot was obtained from the paper Some Dynamic Applications of Liquid Manometrs (American Journal of Physics, 1935).

After scanning the image in grayscale, Paint was used to determine the locations of the points on the curve and replot them in MS Excel. To get the exact values for the curve, the ratio of the number of pixels per unit (x and y axis) was incorporated. Below is the superimposed plot of the original and the reconstructed.

Since I was able to accurately reconstruct the plot, i want to give myself a 10. Also, i get the bonus point for overlaying the two plots.