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The stored energy definition of the quality factor $Q$from wiki Q-factor from wiki is given by

$$ Q = 2\pi \frac{\text{energy stored}}{\text{energy dissipated per cycle}}= 2\pi f_0 \frac{\text{energy stored}}{\text{power loss}} $$

($f_0=$ resonance frequency) More generally and in the context of reactive component specification (especially inductors or caps), the frequency-dependent definition of $Q(f)$ is used:

$$Q(f)= 2\pi f \frac{\text{maximum energy stored}}{\text{power loss}}$$

Question: I not understand how to interpret what is meant here by "power loss" in the denominator in pure mathematical terms?

Note that in both cases we deal with formulas of the form $Q= \frac{A}{\text{ power loss }}$ resp the frequency dependent $Q(f)= \frac{A(f)}{\text{ power loss(f) }}$ where $Q, A $ and the "power loss" are regarded as "constants. (In frequency dependent case these are considered as constants too after having fixed an concrete $f$)

In LCR-oscillator case it's rather easy to write down the constant $A$ as $2 \pi f_0 \cdot E$ where $E:= \frac{1}{2}CV(t)^2+ \frac{1}{2}LI^2(t)$. Note, that although $\frac{1}{2}CV(t)^2$ and $\frac{1}{2}LI^2(t)$ aren't constant, their sum $E$ is; that's energy conservation.

In second case we have $A(f)= 2 \pi f \cdot \max\{E(t) \ \vert \ t \in [0, T]\}$.

Now the question is what is the explicit expression for "power loss"?

My conjecture is that the "power loss" here should be somehow related to the real power $P_R(t)$ discussed here, but how? Is it it's maximal value within $[0,T]$, or it's average value?

Note, here was asked another question about the definition of the Q-factor but it not treats the "power loss" interpretation; that's exactly the concern of this question.

user267839
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  • MY interpretation: $Q$ is a dimensionless quantity which is a ratio of energies in an oscillating system: total energy and energy lost per 1 complete oscillation. We can choose to multiply by the resonant frequency to give our solution some different context, and in the case of circuits simplify things because of the definition $\omega=2\pi f$. Ultimately though, I take "power loss" to mean the same thing as "energy dissipated per cycle", in terms of the wiki. – bleuofblue Mar 13 '22 at 22:28
  • at least there is an obvious connection "energy dissipated per cycle" = $T \cdot $ "power loss". But doesn't this imply that then "power loss" equals the avaraged real power $P_R(t)$ over a cycle? – user267839 Mar 14 '22 at 01:16
  • I think they mean that within circuits, taking the ratio of reactive power to real power will give an equivalent Q to if you measured the total energy of a system and its energy lost per cycle. I would be careful to equate "energy dissipated per cycle" as $T\cdot$"power loss". Take the system of a pendulum: stored energy is its initial potential energy, and across 1 cycle it loses a small fraction of energy to gravity. Multiplying by $2\pi f$ allows me to express $Q$ in terms of $\omega$. – bleuofblue Mar 14 '22 at 02:27
  • I think they mean that within circuits, taking the ratio of reactive power to real power will give an equivalent Q to if you measured the total energy of a system and its energy lost per cycle. Yes and no. In case of Q for "individual reactive components" (https://en.wikipedia.org/wiki/Q_factor#Individual_reactive_components) I agree with that $Q$ can be defined as the ratio of reactive power to real power $P_{reactive}/P_{real}$. But for resonator's definition of Q that's wrong. The reactive power at resonance frequency is zero, but Q can be non zero. – user267839 Mar 18 '22 at 23:34
  • I would be careful to equate "energy dissipated per cycle" as T⋅"power loss". But what is precisely "energy dissipated per cycle"? For an electric resonator that's the energy lost at resistence during a period $T$ and for a mechanic pendulum that the energy lost as heat does to irreversible friction again during a period. And how should we calculate it? Suggestion: The power at the resistence for a electric resonator should be equals to the real power $P_{real}(t)$. – user267839 Mar 18 '22 at 23:42
  • Because then
    • say our period beginns at $t_0$ - the

    "energy dissipated per cycle" for the cycle starting at $t_0$ should be the intergral $\int_{t=t_0}^{t_0+T}P_{real}(t)dt$. But this equals $T \cdot \frac1T \int_{t=t_0}^{t_0+T}P_{real}(t)dt =T \cdot (\text{ mean of } P_{real}(t))$. Since $f=\frac1T$ a factor comparison in the first equation of the question should give "$ \text{ mean of } P_{real}$"= "power loss". Or do I missing something?

     – user267839 Mar 18 '22 at 23:43

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