# High Q peak in frequency response means what in time domain?

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. 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 • How did you measure the decay and value vs Q on this example?" – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 22:59
• @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 – efox29 Apr 13 '19 at 23:08
• 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 – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 23:24
• 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 – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 23:38

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.

• It's always something simple. This has given me a items to explore deeper into. – efox29 Apr 13 '19 at 22:32
• 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) – Tony Stewart Sunnyskyguy EE75 Apr 13 '19 at 22:42
• @Dan Mills, I wonder where your informations are coming from (....among other definitions......voltage gain at resonance...) – LvW Apr 14 '19 at 12:20

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.

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\$$.