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Reading Linear Circuit Transfer Functions and one of the graphs got me curious.

I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7.

enter image description here

We have a peak of ~16.3 dB when Q is 7 @ 10Khz.

Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ?

Added in case its relevent enter image description here

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  • \$\begingroup\$ How did you measure the decay and value vs Q on this example?" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 22:59
  • \$\begingroup\$ @SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C \$\endgroup\$ – efox29 Apr 13 '19 at 23:08
  • \$\begingroup\$ because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 23:24
  • \$\begingroup\$ Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 23:38
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Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * \log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.

Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.

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  • \$\begingroup\$ It's always something simple. This has given me a items to explore deeper into. \$\endgroup\$ – efox29 Apr 13 '19 at 22:32
  • \$\begingroup\$ I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 22:42
  • \$\begingroup\$ @Dan Mills, I wonder where your informations are coming from (....among other definitions......voltage gain at resonance...) \$\endgroup\$ – LvW Apr 14 '19 at 12:20
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The above answer ("Q is.....the voltage gain at resonance") is definitely wrong.

There is only one single definition: The quality factor Q is the so called "pole-Q" - defined by the pole position in the complex frequency domain (s-plane). The relation between the quality factor Q and the magnitude peak in the frequency domain for a 2nd-order lowpass/highpass is as follows:

Amax=(Ao * Q)/sqrt[1-(1/4Q²)] with Ao=DC gain.

For a bandpass filter the Q value defines the 3-dB-bandwidth of the circuit.

TIME DOMAIN

In the time domain, the Q value determines the step response as follows:

(1) For Q>0.5 the step response shows an overshoot "gamma" above the final value (when the transient has settled). This "gamma" value is given in % about the final value.

"gamma"=100 * exp[-3.14/sqrt(4Q²-1)]

Examples (gamma values in brackets): Q=0.5(0%); Q=0.7071(4.3%); Q=1(16.3%); Q=10 (85.4%

(2) The oscillatory decay is determined by the real part ("sigma") of the pole position only: exp(-|sigma|t).

The relation between "sigma" and the Q value is |sigma|=wp/2Q with wp=pole frequency.

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I would give a slightly different definition for the quality factor \$Q\$:

\$Q=2\pi\frac{(stored\;energy)}{(energy\;dissipated\;per\;cycle)}\$

The series resistance represents the circuit losses. When the circuit is energized with the input stimulus, oscillations take place as an energy transfer swinging back and forth between \$L\$ and \$C\$. If the resistive term is 0, there are no losses and oscillations keep for ever: \$Q\$ is infinite and poles are imaginary with no real parts. As \$R\$ increases, you start dissipating energy in heat and you damp the circuit, bringing oscillations to 0 after a few cycles.

The measurement of the peaks and valleys via the logarithmic decrement \$\delta\$ lets you also compute the quality factor:

\$Q=\sqrt{(\frac{\pi}{\delta})^2+\frac{1}{4}}\$

These peaks and valleys are directly linked to the energy dissipated in \$R\$.

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