![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/761311/image.png)
How can I design a correlator to detect the character ‘G’ from the google image?
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Rahul Shah
el 3 de Oct. de 2021
Comentada: Image Analyst
el 20 de Oct. de 2021
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/757079/image.jpeg)
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Respuesta aceptada
Image Analyst
el 8 de Oct. de 2021
If you know the G is there, you can get a mask for the G letter with the following code, which assumes the G will be the leftmost letter in the image:
rgbImage = imread('google.jpeg');
mask = ~imbinarize(rgb2gray(rgbImage)); % Get letters as foreground.
imshow(mask);
labeledImage = bwlabel(mask); % Give an ID# to each blob.
g = ismember(labeledImage, 1); % Extract blob #1, which will be the G
imshow(g);
If you're not sure if the G is there, then you can use the normalized cross correlation (as in my other answer) to see if it's there or not.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/761311/image.png)
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Más respuestas (3)
Mahesh Taparia
el 7 de Oct. de 2021
Hi
The letter 'G' in Google appear to be bigger than the rest of the letter. So 'G' can be selected based on its size. You can use regionprops function to find the bounding box coordinate with maximum area and then draw rectangle on it. For example consider the code below:
a=imread('google.jpeg');
b=rgb2gray(a)<200;
c=regionprops(b,'Area','BoundingBox');
area=zeros(length(c),1);
for i=1:length(c)
area(i)=c(i).Area;
end
[val idx]= max(area);
bbox = c(idx).BoundingBox;
imshow(a);
hold on;
rectangle('Position',bbox)
Hope it helps!
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Image Analyst
el 7 de Oct. de 2021
If you have the G as a template, you can use normalized cross correlation with the function normxcorr2(). See attached example.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/760886/image.png)
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yanqi liu
el 8 de Oct. de 2021
sir,please check the follow code to get some information
clc;
close all;
clear all;
img = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/757079/image.jpeg');
% get chars
hsv = rgb2hsv(img);
s = mat2gray(hsv(:,:,2));
s = imcomplement(s);
bw = im2bw(s, graythresh(s));
bw = ~bw;
figure; imshow(bw);
% gen G block
figure('Color', 'w'); hold on; axis off;
text(0.5,0.5,'G','FontSize',64);
f = getframe(gcf);
f = frame2im(f);
bw2 = ~im2bw(f);
[r,c] = find(bw2);
bw2 = bw2(min(r):max(r),min(c):max(c));
% make correlator
[L,num] = bwlabel(bw);
stats = regionprops(L);
rc = [];
for i = 1 : num
bwi = imcrop(bw, round(stats(i).BoundingBox));
bwi = imresize(bwi, size(bw2));
rc(i) = corr2(bwi,bw2);
end
% max corr
[~, ind] = max(rc);
figure; imshow(img);
hold on; rectangle('Position', round(stats(ind).BoundingBox), 'EdgeColor', 'r', 'LineWidth', 2, 'LineStyle', '-');
2 comentarios
Image Analyst
el 20 de Oct. de 2021
@Rahul Shah, youi can find the brightest point then get that line. If your correlation image is corrImage, do
maxValue = max(corrImage(:))
[rows, columns] = find(corrImage == maxValue);
verticalProfile = corrImage(:, columns(1));
plot(verticalProfile, 'r-', 'LineWidth', 2);
horizontalProfile = corrImage(rows(1), :);
plot(horizontalProfile, 'b-', 'LineWidth', 2);
ylabel('Correlation Value');
xlabel('Pixel Location')
title('Correlation Profile Through Peak')
grid on;
legend('Vertical', 'Horizontal')
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