activity 17-

snapshots from the video of a white marble in a bowl.







after converting it to black and white and getting the position of the ball,information such as location of the ball(x and y), velocity at a certain time, frequency of oscillation can be extracted.



to the left is the graph of the balls position as a function of time. the ball ismoving along 1d (top view) that is why its why component as a function of time is almost contant(flat), while its speed varies because of the effect of gravity. this change in speed can be analyze by the sinusoidal graph.its shows the position of the ball along xaxis.







rating 10
collaborators: j dado

note: ma'am, inupload ko po ito nung saturday. hindi ko po alam kung bakit nabura pagcheck ko ngayon. i hope you would still consider. thank you.

activity 16-

the images on the right are the histogram of the corresponding images to their left. for this activity i used. a pure yellow ball and red ball with blue green background.



















the images below are the reconstructed images of the balls using parametric and non-parametic method.
(upper left - yellow ball, parametric
upper right - yellow ball, non-paramtric
lower left - red ball, parametric
lower right - red ball, non-paramtric)














rating: 10
collaborators: ed david, cole fabros, jeric tugaff

activity 15-

Activity 14-

rubix cube: moved to the left by 1cm:



























reconstructed images using the 4 methods in the scilab help example

rating 10
collaborators: j.dado, e.david, j.presto, b.narag

Activity 13-

images from the matlabfile:












reconstructed 3d image using photometric stereo technique:
















rating: 10
collaborators: j dado, m maallo.

Activity 12-

Activity10- processing Handwritten Texts

Original text:




















a portion of a text (under lower right description):








masked fft:













enhanced image (lines removed):













image after being closed:













lines are removed and words are readable:
rating:10/10
collaborators:marge maallo

Activity9b

2nd image(658x823)



















1 part of the image (1/12)(256 x 256 pixels)















processed image of the part above. the structuring element used is the same structuring element used in the first part (cross)

average area:

471.8495

pixel counting area:

435.8

error: 8.2 %









rating:9/10 -the error for the 2nd part is quite high, another structuring element might lower the uncertainty for this part.

collaborator: m. maallo

Activity9a

original image (748 x 618 pixels)















(1/9) one part of the original image (256x256, lower right)


















image processed result
the structuring element used was a cross with 5 pixels length and 1 pixel thickness

the process was done 9 times using 9 different parts of the image
average size:

519.8939

calculated average area of 10 full circles from different parts of the image using pixel counting:554.6


error: 6.25%

Activity8 – Morphological Operations (continuation)

hollow square dilated with the corresponding structuring elements.

the thickness of the width in the lower right and upper left images varied depending on the orientation of the corresponding structuring image used.















hollow square eroded with the corresponding structuring elements.














cross dilated with the corresponding structuring elements.




















cross eroded with the corresponding structuring elements.

Activity8 – Morphological Operations

Structuing elements:
upper left:4x4 pixels square
upper right: 2x4 pixels
lower left: 4x2 pixels
lower right: 5pixels length 1 pixel width cross















50 x 50 pixel square dilated with the structuring elements as above.

the 1st square has an area of 2809 (53x53), meaning the original image has expanded by 3 pixels.
the upper right and lower left both have areas of 2703 (53x51 or 51x53). its orientation is dependent on the dimensions of the structuring element.
















50 x 50 pixel square eroded with the structuring elements as above.















50 x 30 triangle dilated with the structuring elements as above.











50 x 30 triangle eroded with the structuring elements as above.
















25-pixel radius circle dilated with the structuring elements as above.
upper right circle elongated vertically while the lower left circle elongated horizontally.the lower right circle has both horizontal and vertical elongation.















25-pixel radius circle eroded with the structuring elements as above.
upper right circle looked like it has been elongated vertically because there more parts have been eroded along the horizontal. same principle applies to the lower left circle. the lower right circle has both horizontal and vertical erosion

Activity7 – Enhancement in the Frequency Domain

2d sinusoids (left) FT(right)

A:original 2d sinusoid with f=4
B:f=40, distance between 2 points in the FT increased
C:f=0.4, distance between 2 points in the FT decreased
D:rotated at 30 degrees: the orientation of the 2 points was tilted
E:rotated at 60 degrees: the orientation of the 2 points was tilted further
F:z = sin(2*%pi*4*X).*sin(2*%pi*4*Y): 4 points with equal distance (square)
G:z = sin(2*%pi*40*X).*sin(2*%pi*40*Y): distance between points increased (expected from (b))
H:z = sin(2*%pi*0.4*X).*sin(2*%pi*40*Y):vertical distance decreased but horizontal distance increased( expected from (b) and (c))








left:original indexed image
right:processed fingerprint









score: 8/10
collaborators:dado, maallo

activity6-Fourier Transform Model of Image

6A)


upper left: original image (circle)
upper right: fft2
lower left: fft shifted
lower right: fft shifted zoomed










upper left: original image ( bigger circle)
upper right: fft2
lower left: fft shifted
lower right: fft shifted zoomed











upper left: original image ("A")
upper right: fft2 (inverted image)
lower left: fft shifted
lower right: fft shifted zoomed









6B

upper left: image (VIP)
upper right: aperture (circle)
down: fft (inverted)















upper left: image (VIP)
upper right: aperture ( smaller circle)
down: fft (iimage not well reconstructed because of the smaller aperture size)






























SCore 8/10: failed to complete the last part of the exercise

Activity5 – Physical Measurements from Discrete Fourier Transforms

a) f=120 Hz signal and FT:
N=256
dt=2/256














b) N=512
dt=2/256
(dt fixed, N increased, T increased)














c) N=256
dt=1/256
(dt decreaased, N fixed, T decreased)















d) N=512
dt=T/256
(dt decreased, N increased, T fixed)














codes:
a)
T = 2;
N = 256;
dt = 2/256;
t = [0:dt:(N-1)*dt];
f = 120;
y = sin(2*%pi*f*t);
f1 = scf(1); plot(t,y);
FY = fft(y);
F = 1/(2*dt);
df = 2*F/256;
f = [-(df*(N/2)):df:df*(N/2 -1)];
f2 = scf(2); plot(f, fftshift(abs(FY)));

b)T = 2;
N = 512;
dt = 2/256;
t = [0:dt:(N-1)*dt];
f = 120;
y = sin(2*%pi*f*t);
f1 = scf(1); plot(t,y);
FY = fft(y);
F = 1/(2*dt);
df = 2*F/512;
f = [-(df*(N/2)):df:df*(N/2 -1)];
f2 = scf(2); plot(f, fftshift(abs(FY)));

c)T = 2;
N = 256;
dt = 1/256;
t = [0:dt:(N-1)*dt];
f = 120;
y = sin(2*%pi*f*t);
f1 = scf(1); plot(t,y);
FY = fft(y);
F = 1/(2*dt);
df = 2*F/256;
f = [-(df*(N/2)):df:df*(N/2 -1)];
f2 = scf(2); plot(f, fftshift(abs(FY)));

d)T = 2;
N = 512;
dt = T/512;
t = [0:dt:(N-1)*dt];
f = 120;
y = sin(2*%pi*f*t);
f1 = scf(1); plot(t,y);
FY = fft(y);
F = 1/(2*dt);
df = 2*F/512;
f = [-(df*(N/2)):df:df*(N/2 -1)];
f2 = scf(2); plot(f, fftshift(abs(FY)));

grade:10/10