Image Sharpening Using Laplacian Filter

Question

I was trying to sharpening on some standard image from Gonzalez books. Below are some code that I have tried but it doesn't get closer to the results of the sharpened image.

cvSmooth(grayImg, grayImg, CV_GAUSSIAN, 3, 0, 0, 0);

IplImage* laplaceImg = cvCreateImage(cvGetSize(oriImg), IPL_DEPTH_16S, 1);

IplImage* abs_laplaceImg = cvCreateImage(cvGetSize(oriImg), IPL_DEPTH_8U, 1);

cvLaplace(grayImg, laplaceImg, 3);

cvConvertScaleAbs(laplaceImg, abs_laplaceImg, 1, 0);

IplImage* dstImg = cvCreateImage(cvGetSize(oriImg), IPL_DEPTH_8U, 1);
cvAdd(abs_laplaceImg, grayImg, dstImg, NULL); 

Before Sharpening

My Sharpening Result

Desired Result

Absolute Laplace

Answer 1

I think the main problem lies in the fact that you do img + laplace, while img - laplace would give better results. I remember that img - 2*laplace was best, but I cannot find where I read that, probably in one of the books I read in university.

Answer 2

I think the problem is that you are blurring the image before take the 2nd derivate.

Here is the working code with the C++ API (I'm using Opencv 2.4.3). I tried also with MATLAB and the result is the same.

#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>

#include <iostream>

using namespace cv;
using namespace std;


int main(int /*argc*/, char** /*argv*/) {

    Mat img, imgLaplacian, imgResult;

    //------------------------------------------------------------------------------------------- test, first of all
    // now do it by hand
    img = (Mat_<uchar>(4,4) << 0,1,2,3,4,5,6,7,8,9,0,11,12,13,14,15); 

    // first, the good result
    Laplacian(img, imgLaplacian, CV_8UC1);
    cout << "let opencv do it" << endl;
    cout << imgLaplacian << endl;

    Mat kernel = (Mat_<float>(3,3) << 
        0,  1, 0,
        1, -4, 1,
        0,  1, 0); 
    int window_size = 3;

    // now, reaaallly by hand
    // note that, for avoiding padding, the result image will be smaller than the original one.
    Mat frame, frame32;
    Rect roi;
    imgLaplacian = Mat::zeros(img.size(), CV_32F);
    for(int y=0; y<img.rows-window_size/2-1; y++) {
        for(int x=0; x<img.cols-window_size/2-1; x++) {
            roi = Rect(x,y, window_size, window_size);
            frame = img(roi);
            frame.convertTo(frame, CV_32F);
            frame = frame.mul(kernel);
            float v = sum(frame)[0];
            imgLaplacian.at<float>(y,x) = v;
        }
    }
    imgLaplacian.convertTo(imgLaplacian, CV_8U);
    cout << "dudee" << imgLaplacian << endl;

    // a little bit less "by hand"..
    // using cv::filter2D
    filter2D(img, imgLaplacian, -1, kernel);
    cout << imgLaplacian << endl;


    //------------------------------------------------------------------------------------------- real stuffs now
    img = imread("moon.jpg", 0); // load grayscale image

    // ok, now try different kernel
    kernel = (Mat_<float>(3,3) << 
        1,  1, 1,
        1, -8, 1,
        1,  1, 1); // another approximation of second derivate, more stronger

    // do the laplacian filtering as it is
    // well, we need to convert everything in something more deeper then CV_8U
    // because the kernel has some negative values, 
    // and we can expect in general to have a Laplacian image with negative values
    // BUT a 8bits unsigned int (the one we are working with) can contain values from 0 to 255
    // so the possible negative number will be truncated
    filter2D(img, imgLaplacian, CV_32F, kernel);
    img.convertTo(img, CV_32F);
    imgResult = img - imgLaplacian;

    // convert back to 8bits gray scale
    imgResult.convertTo(imgResult, CV_8U);
    imgLaplacian.convertTo(imgLaplacian, CV_8U);

    namedWindow("laplacian", CV_WINDOW_AUTOSIZE);
    imshow( "laplacian", imgLaplacian );

    namedWindow("result", CV_WINDOW_AUTOSIZE);
    imshow( "result", imgResult );

    while( true ) {
        char c = (char)waitKey(10);
        if( c == 27 ) { break; }
    }

    return 0;
}

Have fun!

Answer 3

You need to do img - laplace instead of img + laplace.

laplace: f(x,y) = f(x-1,y+1) + f(x-1,y-1) + f(x,y+1) + f(x+1,y) - 4*f(x,y)

So, if you see subtract laplace from the original image you would see that the minus sign in front of 4*f(x,y) gets negated and this term becomes positive.

You could also have kernel with -5 in the center pixel instead of -4 to make the laplacian a one-step process instead of getting the getting the laplace and doing img - laplace Why? Try deriving that yourself.

This would be the final kernel.

Mat kernel = (Mat_(3,3) << -1, 0, -1, 0, -5, 0, -1, 0, -1);