void DrsCalibrateTime::CalcResult ( )

inline

Definition at line 1100 of file DrsCalib.h.

References end.

Referenced by GetResult().

    {
        for (int ch=0; ch<160; ch++)
        {
            const auto beg = fStat.begin() + ch*1024;
            const auto end = beg + 1024;

            // First calculate the average length s of a single
            // zero-crossing interval in the whole range [0;1023]
            // (which is identical to the/ wavelength of the
            // calibration signal)
            double s = 0;
            double w = 0;
            for (auto it=beg; it!=end; it++)
            {
                s += it->first;
                w += it->second;
            }
            s /= w;

            // Dividing the average length s of the zero-crossing
            // interval in the range [0;1023] by the average length
            // in the interval [0;n] yields the relative size of
            // the interval in the range [0;n].
            //
            // Example:
            // Average [0;1023]: 10.00  (global interval size in samples)
            // Average [0;512]:   8.00  (local interval size in samples)
            //
            // Globally, on average one interval is sampled by 10 samples.
            // In the sub-range [0;512] one interval is sampled on average
            // by 8 samples.
            // That means that the interval contains 64 periods, while
            // in the ideal case (each sample has the same length), it
            // should contain 51.2 periods.
            // So, the sampling position 512 corresponds to a time 640,
            // while in the ideal case with equally spaces samples,
            // it would correspond to a time 512.
            //
            // The offset (defined as 'ideal - real') is then calculated
            // as 512*(1-10/8) = -128, so that the time is calculated as
            // 'sampling position minus offset'
            //
            double sumw = 0;
            double sumv = 0;
            int n = 0;

            // Sums about many values are numerically less stable than
            // just sums over less. So we do the exercise from both sides.
            // First from the left
            for (auto it=beg; it!=end-512; it++, n++)
            {
                const double valv = it->first;
                const double valw = it->second;

                it->first  = sumv>0 ? n*(1-s*sumw/sumv) : 0;

                sumv += valv;
                sumw += valw;
            }

            sumw = 0;
            sumv = 0;
            n = 1;

            // Second from the right
            for (auto it=end-1; it!=beg-1+512; it--, n++)
            {
                const double valv = it->first;
                const double valw = it->second;

                sumv += valv;
                sumw += valw;

                it->first  = sumv>0 ? n*(s*sumw/sumv-1) : 0;
            }

            // A crosscheck has shown, that the values from the left
            // and right perfectly agree over the whole range. This means
            // the a calculation from just one side would be enough, but
            // doing it from both sides might still make the numerics
            // a bit more stable.
        }
    }

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