Pulse train timing

How long is each pulse, how long the gap, how many pulses fit in one gate, and what fraction of the time is the drive on?

p = \frac{1}{f} \qquad t = p \cdot \frac{D}{100} \qquad n = \left\lfloor \frac{G}{100} \cdot \frac{f}{g} \right\rfloor \qquad e = \frac{D \cdot G}{100}

Method

1. The pulse period is one over the pulse frequency. Kilohertz in, so the period comes out in fractions of a millisecond; it is reported in microseconds.
2. The on-time is the period times the pulse duty; the off-time is what is left of the period.
3. The gate is open for its own period times its duty. Divide that open window by the pulse period and round down: only whole pulses count, and a pulse the gate cuts in half is not a pulse the cell sees.
4. The effective duty is the two duties multiplied — the fraction of all time, gate and pulses together, that the drive is actually delivering. Average power is peak power times this.

Assumptions

• Both waveforms are ideal rectangles with instant edges and no jitter. A real gate has a rise time, and the first pulse after it opens is shorter than the rest.
• The pulse train and the gate are not synchronised. If they were locked — one gate edge on every Nth pulse edge — the count would be exact rather than rounded down. Meyer's circuits derived both from one clock, so on his bench it was exact.
• Nothing here is physics; it is bookkeeping for the calculations that follow. The interesting question — what the cell does with n pulses — is the step charging calculation's.