/*
	This FFT has been proposed by Paul Bourke 
	http://paulbourke.net/miscellaneous/dft/
	This computes an in-place complex-to-complex FFT 
	x and y are the real and imaginary arrays of 2^m points.
	dir =  1 gives forward transform
	dir = -1 gives reverse transform 
	You MUST compute first the value m such that
	2^(m-1) < n (size of your signal) <= 2^m
	allocate a new signal of nm=2^m values
	then fill the n first values of this new signal 
 with your signal and fill the rest with 0
	WARNING : you must pass m, not nm !!!
	*/

int FFT(int dir,int m,double *x,double *y)
{
	int n,i,i1,j,k,i2,l,l1,l2;
	double c1,c2,tx,ty,t1,t2,u1,u2,z;
	
	/* Calculate the number of points */
	n = 1;
	for (i=0;i<m;i++) 
		n *= 2;
	
	/* Do the bit reversal */
	i2 = n >> 1;
	j = 0;
	for (i=0;i<n-1;i++) {
		if (i < j) {
			tx = x[i];
			ty = y[i];
			x[i] = x[j];
			y[i] = y[j];
			x[j] = tx;
			y[j] = ty;
		}
		k = i2;
		while (k <= j) {
			j -= k;
			k >>= 1;
		}
		j += k;
	}
	
	/* Compute the FFT */
	c1 = -1.0; 
	c2 = 0.0;
	l2 = 1;
	for (l=0;l<m;l++) {
		l1 = l2;
		l2 <<= 1;
		u1 = 1.0; 
		u2 = 0.0;
		for (j=0;j<l1;j++) {
			for (i=j;i<n;i+=l2) {
				i1 = i + l1;
				t1 = u1 * x[i1] - u2 * y[i1];
				t2 = u1 * y[i1] + u2 * x[i1];
				x[i1] = x[i] - t1; 
				y[i1] = y[i] - t2;
				x[i] += t1;
				y[i] += t2;
			}
			z =  u1 * c1 - u2 * c2;
			u2 = u1 * c2 + u2 * c1;
			u1 = z;
		}
		c2 = sqrt((1.0 - c1) / 2.0);
		if (dir == 1) 
			c2 = -c2;
		c1 = sqrt((1.0 + c1) / 2.0);
	}
	
	/* Scaling for forward transform */
	if (dir == 1) {
		for (i=0;i<n;i++) {
			x[i] /= n;
			y[i] /= n;
		}
	}
	
	return(1);
}