我正在使用快速傅里叶变换进行指纹图像增强。我从这个网站得到了这个想法。
我已经FFT使用 32*32 窗口实现了该功能，然后按照推荐站点的建议，我想power spectrum与FFT. 但我不明白，
如何计算图像的功率谱？或者 Power Spectrum 有什么理想的价值吗？
FFT 代码：
public FFT(int[] pixels, int w, int h) {
// progress = 0;
input = new TwoDArray(pixels, w, h);
intermediate = new TwoDArray(pixels, w, h);
output = new TwoDArray(pixels, w, h);
transform();
}
void transform() {
for (int i = 0; i < input.size; i+=32) {
for(int j = 0; j < input.size; j+=32){
ComplexNumber[] cn = recursiveFFT(input.getWindow(i,j));
output.putWindow(i,j, cn);
}
}
for (int j = 0; j < output.values.length; ++j) {
for (int i = 0; i < output.values[0].length; ++i) {
intermediate.values[i][j] = output.values[i][j];
input.values[i][j] = output.values[i][j];
}
}
}
static ComplexNumber[] recursiveFFT(ComplexNumber[] x) {
int N = x.length;
// base case
if (N == 1) return new ComplexNumber[] { x[0] };
// radix 2 Cooley-Tukey FFT
if (N % 2 != 0) { throw new RuntimeException("N is not a power of 2"); }
// fft of even terms
ComplexNumber[] even = new ComplexNumber[N/2];
for (int k = 0; k < N/2; k++) {
even[k] = x[2*k];
}
ComplexNumber[] q = recursiveFFT(even);
// fft of odd terms
ComplexNumber[] odd = even; // reuse the array
for (int k = 0; k < N/2; k++) {
odd[k] = x[2*k + 1];
}
ComplexNumber[] r = recursiveFFT(odd);
// combine
ComplexNumber[] y = new ComplexNumber[N];
for (int k = 0; k < N/2; k++) {
double kth = -2 * k * Math.PI / N;
ComplexNumber wk = new ComplexNumber(Math.cos(kth), Math.sin(kth));
ComplexNumber tmp = ComplexNumber.cMult(wk, r[k]);
y[k] = ComplexNumber.cSum(q[k], tmp);
ComplexNumber temp = ComplexNumber.cMult(wk, r[k]);
y[k + N/2] = ComplexNumber.cDif(q[k], temp);
}
return y;
}