v_t norm()const{return*this*(1./sqrt(magsqr()));}
void b2a(double *b, int n, double *a)
{
double bmx, bm, p;
int i, j, nn, nnc;
char s[80];
/* next bigger power of 2 */
for (nn=1; nn<=n; nn*=2);
nn *=2;
/*nn = (int) (exp(log(2)*ceil(log((double)n)/log(2))))+1;*/
/* size of arrays used for computation */
nnc = nn*ZEROPAD;
for (i=0; i<n; i++) {bf[i*2] = b[i*2]; bf[i*2+1] = b[i*2+1];}
for (i=n; i<nnc; i++) {bf[i*2] = 0; bf[i*2+1] = 0.0;}
four1(bf-1, nnc, 1);
/* check to see (| Fourier (beta(x)) |) < 1 */
for (i=0, bmx=0; i<nnc; i++)
{
bm = magsqr(bf,i);
if (bm > bmx)
bmx = bm;
}
if (bmx >= 1.0)
for (i=0; i<nnc; i++)
{
bf[i*2] /= (sqrt(bmx)+MP);
bf[i*2+1] /= (sqrt(bmx)+MP);
}
/* compute |alpha(x)| */
for (i=0; i<nnc; i++) {
am[i*2] = sqrt(1.0 - magsqr(bf,i));
}
/* compute the phase of alpha, mag and phase are HT pair */
/* ABK - God, it took me forever to figure out what John did here...
he wants the hilbert transform of the log magnitude, so what
he does is to generate the analytic version of the signal knowing
that the imaginary part will be the negative of the hilbert transform
of the log magnitude.... this is why I took 261...
*/
for (i=0; i<nnc; i++) {
af[i*2] = log(am[i*2]);
af[i*2+1] = 0;
}
four1(af-1,nnc,1);
for (i=1; i<(nnc/2)-1; i++) { /* leave DC and halfway point untouched!! */
af[i*2] *= 2.0;
af[i*2+1] *= 2.0;
}
for (i=(nnc/2)+1; i<nnc; i++) {
af[i*2] = 0.0;
af[i*2+1] = 0.0;
}
four1(af-1,nnc,-1);
for (i=0; i<nnc; i++) {
af[i*2] /= nnc;
af[i*2+1] /= nnc;
}
/* compute the minimum phase alpha */
for (i=0; i<nnc; i++) {
p = af[i*2+1];
af[i*2] = am[i*2] * cos(-p);
af[i*2+1] = am[i*2] * sin(-p);
}
/* compute the minimum phase alpha coefficients */
four1(af-1, nnc, -1);
for (i=0; i<n; i++)
{
j = n-i-1;
a[j*2] = af[i*2]/nnc;
a[j*2+1] = af[i*2+1]/nnc;
}
}