CODEWITHSUNDEEP
c++image-processingfftfftw
Vektor88
Vektor88

Reputation: 4930

Implementing FFT low-pass filter in C with FFTW

I am trying to create a very simple C++ program that given an argument in range [0-100] applies a low-pass filter to a grayscale image that should "compress" it proprotionally to the value of the given argument. I am using the FFTW library.

I have some doubts about how I define the frequency threshold, cut. Is there any more effective way to define such value?

//fftw_complex *fft
//double[] magnitude
// . . . 

int percent = 100;
    if (percent < 0 || percent > 100) {
        cerr << "Compression rate must be a value between 0 and 100." << endl;
        return -1;
    }

double cut =(double)(w*h) * ((double)percent / (double)100);
    for (i = 0; i < (w * h); i++) {
        magnitude[i] = sqrt(pow(fft[i][0], 2.0) + pow(fft[i][1], 2.0));
        if (magnitude[i] < cut) {
            fft[i][0] = 0.0;
            fft[i][1] = 0.0;
        }
    }

Update1:

I've changed my code to this, but again I'm not sure this is a proper way to filter frequencies. The image is surely compressed, but non-square images are messed up and setting compression to 100% isn't the real maximum compression available (I can go up to ~140%). Here you can find an image of what I see now. 

int cX = w/2;
int cY = h/2;
cout<<"TEST "<<((double)percent/(double)100)*h<<endl;
for(i = 0; i<(w*h);i++){
    int row = i/s;
    int col = i%s;
    int distance = sqrt((col-cX)*(col-cX)+(row-cY)*(row-cY));
    if(distance<((double)percent/(double)100)*min(cX,cY)){
        fft[i][0] = 0.0;
        fft[i][1] = 0.0;
    }
}

Upvotes: 0

Views: 2603

Answers (1)

MSalters
MSalters

Reputation: 180205

This is not a low-pass filter at all. A low-pass filter passes low frequencies, i.e. it removes fine details (blurring). You obviously need a 2D FFT for that.

This code just removes random bits, essentially.

[edit] The new code looks a lot more like a low-pass filter. The 141% setting is expected: the diagonal of a square is sqrt(2)=1.41 times its side. Converting an index into a row/column pair should use the image width, not some random unexplained s.

I don't know where your zero frequency is located. That should be easy to spot (largest value) but it might be in (0,0) instead of (w/2,h/2)

Upvotes: 4

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