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

$$ Q = 2\pi \frac{\text{energy stored}}{\text{energy dissipated per cycle}}= 2\pi f_0 \frac{\text{energy stored}}{\text{power loss}} \tag{1}$$

(\$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:

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

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 \$ \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 mathematical expression for the "power loss"?

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

user267839
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  • You are confusing the notations, since both the quality factor and the reactive power use the same letter: Q. If you read carefully in your first link, right at the end it says: "*where it can be shown to be equal to the ratio of reactive power to real power*". So if Q in your first formula would be the same as the reactive power, it wouldn't make sense to equate Q=something*Q, unless that something is 1. Also, you may be overthinking it a bit with the time dependency. ;-) – a concerned citizen Mar 10 '22 at 19:45
  • @aconcernedcitizen: Thanks, I haven't noticed this notation conflict before. By \$Q \$ I indeed mean the *quality factor*, not the *reactive power*. To avoid notation confusions let call the from now the reactive power \$P_Q(t)\$ and the real power \$P_R(t)\$. These are both time dependent. And nevertheless, the question stays what is meant by *"power loss"**? – user267839 Mar 10 '22 at 19:56
  • I would have writen an answer but I lack the words now, it seems. I'll say that you don't need to involve the powers when talking about the quality factor. The power loss is the amount on imperfection you find everywhere in nature. An ideal LC will have purely imaginary roots, right on the jw axis. But in real life it will have some parasitic resistances, somewhere, and those will cause a tiny real part to appear. Try to make it oscillate by itself and it will decay, and that decay is directly proportional to the real part, the imperfection. – a concerned citizen Mar 10 '22 at 20:11
  • Ok, but my main concern here is if it's possble to write down as formula what is here meant by "power loss" precisely. For exmple in the numerator we can say precisely what the "stored energy" is. It's a constant value \$E= 1/2CV^2+ 1/2LI^2 \$ for an electrical oscillator. In the second case - for single reactive component -the stored energy is not constant over the time anymore; it's a function \$E(t)\$. But we set in the second case "maximal stored energy" \$= \max_{t \in [0,T]} E(t) \$. And I think there should be also an explicit mathematical expression for "power loss" here. – user267839 Mar 10 '22 at 20:31
  • You're already there: instead of L or C, use R, either as RI^2 or U^2/R. That's the power. If it's "per cycle" then you need to use the integral of it, which makes sense for oscillators, since otherwise you'd be having the instantaneous power. Same thing for the other energies. With these results, the reatios would be constant. Where frequency is part of the equation, it's different, but then the whole problem takes a new turn, see for example [this](https://electronics.stackexchange.com/q/603049/95619) for something close (out of many other examples on ee.se). – a concerned citizen Mar 10 '22 at 20:56
  • We have in both cases the expression for the quality factor as \$Q= \frac{A}{\text{ power loss }}\$ where \$Q \$ and \$A \$ are constants. So the "power loss" should be a constant too, right? But \$R I^2 \$ is a function, not a constant. \$I(t)\$ is sinoidal. That's the problem. How is "power loss" related to the real power \$RI^2\$? – user267839 Mar 10 '22 at 21:10
  • addenedum: in second case for \$Q(f)\$ with variable frequency we face the same problem after having fixed a frequency \$f\$. Indeed, we have \$Q(f)= \frac{A(f)}{\text{ power loss(f)}}\$ where for a fixed \$f\$ the terms \$A(f) \$ and "power loss(f)" are again considered as "constants". But what is precisely "power loss(f)" in mathematical terms? – user267839 Mar 10 '22 at 21:31
  • @katalaveino - Hi, I see you have now asked exactly the [same question at Physics.SE](https://physics.stackexchange.com/questions/698434). That is *[not cool](https://meta.stackexchange.com/questions/64068)*. As *Jeff Atwood* (SE founder) says at that link: "Just to be 100% clear, copy-pasting a question across sites with no changes is considered abusive behavior.". || I'm sorry that you didn't get the value you wanted here and felt that Physics.SE was better suited to the question. But to avoid wasting site member's time *here* duplicating the effort over there, this question is now closed. – SamGibson Mar 11 '22 at 00:44

1 Answers1

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It looks like you are determined to complicate the matter, but also that you didn't read everything in the comments. In the first equation you show, the energies are averaged over a cycle, therefore an integral is applied. This is because it's about a resonator and you can't use the instantaneous values in a meaningful value for Q. You can use \$\frac12 LI^2\$ and \$\frac12 CV^2\$ , but when AC is involved, integrals are used to average over a cycle.

I've also said what loss means: imperfections. Nothing in nature is 100% efficient (or more), which means there have to be losses somewhere. In the case of the oscillator, there can't be any passive system that, when left alone, will oscillate forever (perpetuum mobile). An ideal LC tank has the denominator of its transfer function \$s^2+\omega_p^2\$, meaning its roots are purely imaginary, on the \$j\omega\$ axis. In real life there are always parasitics which means the denominator turns into \$s^2+2\zeta\omega s+\omega^2\$, where \$\zeta\$ is the damping and is the reciprocal of Q. That's where the losses occur, and the expression is just like the definition of real power: \$RI^2\$, or \$V^2/R\$. Real, because there is a loss, it transforms into heat, whereas the rest of it will account for a reactive power (transferred back and forth between elements). And, just like above, if AC is involved then averaging is used. The result is a constant, Q.

For the frequency dependent part, the things complicate a bit. For example, an inductor or a capacitor will have a certain impedance which is frequency dependent (let's not get into high frequency behaviour). That means \$Z_L\$ will have a flat value until a certain point, then rise with frequency, and \$Z_C\$ will decrease after that point. Since Q doesn't make much sense to be a variable dependent on frequency, a convention is used: the Q is given at a certain frequency, usually 1 kHz or 1 MHz, and it takes the form \$|X_{[L,C]}|/R\$.

a concerned citizen
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  • `In the first equation you show, the energies are averaged over a cycle, therefore an integral is applied.` I'm not sure if I understand the point. Let me try to rephrase how I understood it: so firstly you are going to average over the sum \$ \frac{1}{2}CV(t)^2+ \frac{1}{2}LI^2(t) \$ over the cycle \$T\$? Then we would obtain the value \$K:=\frac{1}{T} \int_{}^T \frac{1}{2}CV(t)^2+ \frac{1}{2}LI^2(t)dt = \frac{1}{2}C\widehat{V}^2+ \frac{1}{2}L\widehat{I}^2dt \$ where \$\widehat{V}^2 \$ and \$\widehat{I}^2 \$ are averages over \$V(t)^2\$ and \$I^2(t)\$. – user267839 Mar 11 '22 at 00:47
  • I'm not pretty sure how this 'average value' $\K$\ is related to the the other terms in the equation (1). Do you mean it in the way that \$T \cdot K =\text{energy dissipated per cycle} \$ in the first equation? – user267839 Mar 11 '22 at 00:49
  • @katalaveino There's no `+`, the two energies for the L and C are equal, since it's an oscillating circuit. The integral over one period will give you a fixed value, and that value will represent the energy per cycle. Same with P. I also see on physics.se that you already linked to a much better answer about Q, in which it is talked about P, yet you say P is not discussed. It really looks like you're just skimming over. – a concerned citizen Mar 11 '22 at 12:33
  • just to avoid potential source of confusions: when you were writing about 'energies' or 'powers' above you meant there always the *averages* over cycle, not the time dependent functions, unless you mentioned explicitly that you were talking about functions with instantaneously changing values, right? (since in your answer you wrote that when AC is involved, integrals are used to average over a cycle; since we dealing here with AC only, so I should assume that therefore *all* here involved quanities like \$\frac12 LI^2, \frac12 CV^2, RI^2 \$ are implicitely averaged, right?) – user267839 Mar 11 '22 at 17:24
  • `There's no +, the two energies for the L and C are equal, since it's an oscillating circuit. The integral over one period will give you a fixed value, and that value will represent the energy per cycle.` Ok, so here you want to say that within the period \$[t_0, t_0+T]\$ the "energy per cycle" is structurally given by an integral of the form \$ \int_{t_0}^{t_0+T}A(t)dt \$ with appropriate integrand \$A(t) \$; and at this point you not say what the intergrand \$A(t) \$ really is; it's just about the nacked stetment that the "energy per cycle" is given by an integral over a period, right? – user267839 Mar 11 '22 at 18:03
  • So if the two comments I wrote above are correct -especially on this *'averaging quanities philosophy'* when working in AC setting - then the "power loss" durng \$[t_0, t_0+T]\$ is nothing but the averaged real power in this interval, ie \$=\frac1T \int_{t_0}^{t_0+T}P_R(t)dt \$. Is this the formally correct interpretation of "power loss" in the denominator of \$Q\$ (if we are only focussed on time interval \$[t_0, t_0+T]\$; surely if we dealing with damped HO the amplitude lowers due to losses, but assume we thinking only about what happens in period between \$t_0\$ and \$t_0+T\$)? – user267839 Mar 11 '22 at 18:03
  • @katalaveino Q does not vary with phase, it's a constant, therefore instantaneous values do not belong in the formula. What's with this fixation on time dependency? Everytime I said you don't need it, you brought it up again. Use your head: if the voltage is sin(x) and the current is cos(x), then V^2/R becomes 1/2+cos(2x)/2, and CV^2 ends up as 1/2-cos(2x)/2. What do you think happens when you divide those two? I won't discuss this anymore, it has become more than tedious. – a concerned citizen Mar 11 '22 at 18:25
  • where I should have said that \$ Q \$ should depend on phase? I nowhere claimed it. I was the whole time interested in determining mathematical expression for the "power loss" in the denominator. Ok, so your example shows that to calculate Q we can avaluate the "stored energy" and the "power loss" functions (which of course are phase dependent) at arbitrary phase; the phase terms are trivialy going to be cancled when we take the quotient. – user267839 Mar 11 '22 at 19:16
  • Fine, that's clear. Conclusion: as "power loss" in the denominator we can choose the real power \$P_R(t)\$ evaluated at any point as long as we evaluate the energy function representing the "stored energy" in the numerator at same point. – user267839 Mar 11 '22 at 19:16
  • But then, why were you going to talk about *averaging* energies over a cycle in first paragraph of your answer? I not understand how the 'taking average' procedure fits in the whole picture? I not understand where you have to make use of it, isn't it then redundant at all as your considerations in last comment show? – user267839 Mar 11 '22 at 19:25