Question

4.1 Tas 1 feature : Process a low resolution sensor image with an idealised You are given an idealised test image with a single circular feature. You are to develop an algorithm and implement that algorithmin the form of a MATLAB program that is capable of performing the following tasks Find the centre of the circular feature in pixels. 2. Calculate the radius of the feature in pixels. Calculate the area of the feature in pixels. Given that the distance from the camera to the surface is 1m and the camera sensor infor- mauion from Section 3 calculate the diagonal distance of the feature's center, in meters, relative to the top-left corner of the image (ie: pixel index 1,1), the radius of the feature in meters and the area of the feature in meters squared The distance should be a variable declared in your MATLAB script and the camera specification are given in Section 3. On demonstration day a previously-unknown distance will be used for marking purposes. Task 1: Low Resolution Sensor Image 10 20 30 50 60 10 20 50 60 80

can you just help with the first 3 parts?

Answer 1

code

solution

matlab code read the image readimg imread ('circle.png') readimgrgb2gray (readig [m1, n1] -size (reading ) ; &show image imshow (readimg %check the data image-b = zeros (ml, nl ); for i - 1:ml for j 1:nl if readimg (1,j) >100 image-b (1, j) =255; image-b(i,j) 0; end end end imshow (image_b) part c The area of the circle values 0; for i-1:m1 for j-1:nl if image-b (1.)-255 values values1; end end end

picture start-pointx 0; start pointy0 0; = stop-x % part a and b The radius of the circle and center of the circle for i=1:ml for j-2:n1-1 if image-b (1, j )一 start.Pointx = i; start _pointy ji end 255 image-b (i, j-1) -0 && if image b(i,j)255 &&image b(i,j+1) 0 stop_y end end x = stop-y - startpolnty; if x > picture picture =x; center-x = 3tartpointx ; center_y(stop_y + start pointy)/2; end end picture/2; radius % Display Data fprintf (' Center of Circle : (8.2f,%.2f)\n', center x, forintf("Radius of Circle : %.2fpixelsla', radius); fprintf( 'Area of Circle: %.2f pixels\n', values); center y);

//output

>>circle Center of circle: (174.00,189. 50) Radius of circle:85. 50 pixels Area of circle:24526.00 pixels

//copyable code

%matlab code

%read the image

readimg= imread('circle.png');

readimg = rgb2gray(readimg );

[m1,n1] = size(readimg );

%show image

imshow(readimg );

%check the data

image_b = zeros(m1,n1);

for i = 1:m1

for j = 1:n1

if readimg(i,j) > 100

image_b(i,j) = 255;

else

image_b(i,j) = 0;

end

end

end

imshow(image_b);

% part c The area of the circle

values = 0;

for i=1:m1

for j=1:n1

if image_b(i,j) == 255

values = values + 1;

end

end

end

picture = 0;

start_pointx = 0;

start_pointy = 0;

stop_x = 0;

stop_y = 0;

% part a and b The radius of the circle and center of the circle.

for i=1:m1

for j=2:n1-1

if image_b(i,j) == 255 && image_b(i,j-1) == 0

start_pointx = i;

start_pointy = j;

end

if image_b(i,j) == 255 && image_b(i,j+1) == 0

stop_x = i;

stop_y = j;

end

end

x = stop_y - start_pointy;

if x > picture

picture = x;

center_x = start_pointx;

center_y = (stop_y + start_pointy)/2;

end

end

radius = picture/2;

% Display Data

fprintf('Center of Circle: (%.2f,%.2f)\n', center_x, center_y);

fprintf('Radius of Circle: %.2f pixels\n', radius);

fprintf('Area of Circle: %.2f pixels\n', values);