Damping and ringdown
After a pulse, does this loop ring or sag, at what frequency, and for how many cycles?
- ζ
- Damping ratio
- τ
- Decay time constant, µs
- d
- Ringing frequency, Hz
- m
- Cycles to 1 % energy
- L
- Inductance, mH
- C
- Capacitance, nF
- R
- Series resistance, Ω
LaTeX
ζ = \frac{R}{2} \sqrt{\frac{C}{L}} \qquad τ = \frac{2 \cdot L}{R} \qquad d = \frac{1}{2\pi} \sqrt{\frac{1}{L \cdot C} - \frac{1}{τ^{2}}} \qquad m = \frac{\ln 10}{\pi} \cdot \frac{1}{R} \sqrt{\frac{L}{C}}Result
ζ
Damping ratio
0.211
Below 1 the loop rings; at 1 it returns to rest as fast as it can without ringing; above 1 it sags back slowly.
τ
Decay time constant
665 µs
2L ÷ R. The ring's amplitude falls to 37 % in this time.
d
Ringing frequency
1.109 kHz
What the loop actually rings at — the undamped resonance pulled slightly down by the resistance. Absent when the loop is overdamped.
m
Cycles to 1 % energy
1.737
How many cycles of ring pass before 99 % of the stored energy has gone into the resistance. Roughly 0.73 × Q.
LaTeX
0.211 = \frac{0.5923\,\mathrm{Ω}}{2} \sqrt{\frac{100\,\mathrm{µF}}{196.9\,\mathrm{µH}}} \qquad 665\,\mathrm{µs} = \frac{2 \cdot 196.9\,\mathrm{µH}}{0.5923\,\mathrm{Ω}} \qquad 1.109\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{196.9\,\mathrm{µH} \cdot 100\,\mathrm{µF}} - \frac{1}{665\,\mathrm{µs}^{2}}} \qquad 1.737 = \frac{\ln 10}{\pi} \cdot \frac{1}{0.5923\,\mathrm{Ω}} \sqrt{\frac{196.9\,\mathrm{µH}}{100\,\mathrm{µF}}}This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.
What this looks like
The formula behind the curve
- ζ
- Damping ratio
- τ
- Decay time constant, µs
- d
- Ringing frequency, Hz
- m
- Cycles to 1 % energy
- L
- Inductance, mH
- C
- Capacitance, nF
- R
- Series resistance, Ω
LaTeX
ζ = \frac{R}{2} \sqrt{\frac{C}{L}} \qquad τ = \frac{2 \cdot L}{R} \qquad d = \frac{1}{2\pi} \sqrt{\frac{1}{L \cdot C} - \frac{1}{τ^{2}}} \qquad m = \frac{\ln 10}{\pi} \cdot \frac{1}{R} \sqrt{\frac{L}{C}}The formula behind the curve
- ζ
- Damping ratio
- τ
- Decay time constant, µs
- d
- Ringing frequency, Hz
- m
- Cycles to 1 % energy
- L
- Inductance, mH
- C
- Capacitance, nF
- R
- Series resistance, Ω
LaTeX
ζ = \frac{R}{2} \sqrt{\frac{C}{L}} \qquad τ = \frac{2 \cdot L}{R} \qquad d = \frac{1}{2\pi} \sqrt{\frac{1}{L \cdot C} - \frac{1}{τ^{2}}} \qquad m = \frac{\ln 10}{\pi} \cdot \frac{1}{R} \sqrt{\frac{L}{C}}Method
- Take √(L/C), the loop's characteristic impedance. The damping ratio is the resistance divided by twice that: ζ = (R/2)√(C/L). Everything in SI — henries, farads, ohms.
- The decay time constant is 2L/R. This is the time for the ring's envelope to fall to 1/e of where it started; it does not depend on the capacitance at all.
- The ringing frequency is the undamped resonance with the damping taken off: ω_d = √(ω₀² − α²) where ω₀ = 1/√(LC) and α = 1/τ. Divide by 2π for hertz. When α exceeds ω₀ the square root goes imaginary — the loop is overdamped and does not ring — and no frequency is reported.
- The energy in the loop falls as the square of the amplitude, so it reaches 1 % when the amplitude reaches 10 %. That takes ln(10) time constants, and dividing by the ring period gives ln(10)/π × Q cycles, with Q = (1/R)√(L/C).
Assumptions
- A linear series RLC with constant R, L and C. A cell full of water is not: its resistance depends on the voltage across it and its capacitance on the frequency, so the ring in a real VIC is less tidy than one exponential.
- The resistance is the only loss. Core loss in the chokes and dielectric loss in the water both take energy out too, and both make the real ringdown shorter than this.
- The choke's self-capacitance is ignored. Near its self-resonance a choke is not an inductor, and the numbers here stop meaning anything.