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Resonant build-up under a gated drive

Driven on resonance, how many cycles does the cell voltage take to climb toward Q times the drive — and how far does it get before the gate closes?

The formula
S=Q·VA=S(1−e−π·n/Q)m=ln10π·Q
A
Amplitude when the gate closes, V
S
Steady-state amplitude, V
m
Cycles to 90 %
Q
Q factor
V
Drive amplitude, V
n
Cycles in the gate
LaTeX 
S = Q \cdot V \qquad A = S \left( 1 - e^{-\pi \cdot n / Q} \right) \qquad m = \frac{\ln 10}{\pi} \cdot Q

Work it out

The loop's Q at resonance.

kHz

The frequency the loop is driven at, which is its resonance.

V

The voltage driving the loop.

How many cycles of drive arrive before the gate closes.

or see a worked example with typical numbers

Method

  1. The steady-state amplitude on resonance is Q times the drive.
  2. A driven resonator's envelope approaches that exponentially with time constant 2Q/ω₀ — the same constant its ringdown decays with. In cycles of the resonance that is Q/π, so after n cycles the amplitude is S(1 − e^{−πn/Q}).
  3. Setting that to 0.9 S gives n = Q·ln10/π cycles to nine tenths.

Assumptions

  • Linear loop, constant Q, drive exactly on resonance and switched on cleanly at a zero crossing. Off resonance the build-up beats as well as climbs.
  • The water's loss does not change as the voltage rises. It does — the leak grows with the field — so a real cell's Q falls as it climbs and the top of the curve is lower than S.
  • The build-up and the ringdown share one time constant. That is a property of a linear resonator, and the reason a gate cannot be faster than the loop it gates: pulses arriving after the ring has died start from nothing.