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The quality factor is defined as the ratio of stored energy and dissipated energy (see below). The formula includes \$2\pi\$ as a constant coefficient. Why do we need \$2\pi\$ to calculate the stored and dissipated energy? I cannot understand the meaning of this constant.

$$ Q = 2\pi\frac{\textrm{Energy stored}}{\textrm{Energy lost}}$$

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  • \$\begingroup\$ I think it's because quality factor is just a unitless metric rather than a "real" physical quantity. So you can throw in whatever coefficient you want in front of it. You just have to handle it appropriately. They probably throw it into the definition because it streamlines the math somewhere or makes it more consistent with some other definition. \$\endgroup\$
    – DKNguyen
    Commented May 2, 2022 at 19:11
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    \$\begingroup\$ In your expression, energy loss should be energy loss per oscillation cycle which, in my opinion, justifies the 2\$\pi\$ presence. \$\endgroup\$
    – Verbal Kint
    Commented May 2, 2022 at 20:51
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    \$\begingroup\$ \$\text{d}\sin\omega = \cos\omega\:\text{d}\omega\$. But \$\omega=2\pi\,f\$. If you are thinking in terms of \$f\$ (cycles per second) and not radians per second, then \$\text{d}\sin\omega = \cos\omega\:\text{d}\omega=\cos\omega\:2\pi\:\text{d}f\$. And there's that co-efficient showing up. \$\endgroup\$
    – jonk
    Commented May 2, 2022 at 22:10

2 Answers 2

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There is no fundamental reason -- it's a convention, and happens to be convenient. 'Q' could have been defined to not use the 2π term, but then many calculations using it would require a 2π factor which becomes slightly cumbersome.

For example Q is also applicable to single inductors or capacitors, and the existing definition also allows Q = ωL/R for an inductor. Similarly in a LC resonator, the Q is given simply as F0/ΔF. Further, Q behaviour can be separated into over- or under-damped at Q == ½.

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  • \$\begingroup\$ Quote: "Similarly in a LC resonator, the bandwidth is given simply as F0/ΔF." I suppose you mean "......simply given as Fo/Q". However, this applies only when all the losses (from the inductor and the capacitor) are expressed as one single resistor (in series or in parallel). \$\endgroup\$
    – LvW
    Commented May 3, 2022 at 11:47
  • \$\begingroup\$ Thanks; fixed typo Yes, this only applies to a 'simple' LC resonator with one loss element. \$\endgroup\$
    – jp314
    Commented May 3, 2022 at 14:37
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Let me start with an example: Parallel connection of Rp and C (a lossy capacitance).

Energy: W=Integral (from 0 to t=To) over v(t)*i(t)dt.

Using i_c=C(dv(t)/dt) and i_r=v(t)/Rp with v(t)=Vmax*sin(wt) we arrive at

(1) Stored energy (capacitance): Wc=0.5(C*Vmax²)

(2) Losses (dissipated energy): Wr=(Vmax²/Rp)(Pi/w)

With

Q_C=2Pi(Wc/Wr)=wRpC

Q_C=Rp/(1/wC)=Real part/imag. part

Hence: The factor 2Pi allows us to find the Q value for a lossy capacitance (parallel resistor Rp) simply by finding the real-to-imag. ratio for the lossy complex impedance. A similar analysis is possible for a lossy inductance (series resistor Rs) resulting in Q_L=wL/Rs

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