Equivalency of $Q$ Factor Definitions

Question

The Q factor is defined (seemingly) as $$Q=2\pi\frac{\mathrm{energy \, \, stored}}{\mathrm{energy \, \,dissipated \, \, per \, \, cycle}}$$ however on Wikipedia is says that the Q factor can be described as $\frac{\omega_0}{\Delta\omega}$, or rather the resonant frequency over the bandwidth. Why are these two equivalent?

Answer 1

I'm answering the question in the familiar context of series resonance in an LCR circuit. The answer could be adapted to a more general system.

The definition you quote initially needs to be modified thus: $$Q'=2\pi \frac{\text{mean energy stored at resonance}}{\text{energy dissipated per cycle at resonance}}\ .$$

Writing $I$ for $I_{\text{rms}}$ at resonance, the rms pd across the capacitor at resonance is $I/\omega_{res} C=\omega_{res} LI.$

We therefore have $$Q'=2 \pi \frac{\tfrac12 C (\omega_{res} L I)^2+\frac12 L I^2}{\frac{2\pi}{\omega_{res}}I^2 R}=\frac{\omega_{res} L}{R}$$ (having remembered that $LC \omega_{res}^2 = 1$)

Now let's look at the definition in terms of bandwidth. We find the angular frequencies at which the current is $1/\sqrt 2$ of its value at resonance. It's easy to show from the impedance equation for series resonance that this is when $$R^2=\left(\omega L-\frac{1}{\omega C}\right)^2\ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ ±R=\omega L-\frac{1}{\omega C}$$ $\omega L-\frac{1}{\omega C}=R$ gives the value, $\omega_2$, for which $\omega_2>\omega_{res}.$

Re-arranging and solving as a quadratic in $\omega_2$ gives the positive root $$\omega_2=\frac{R+\sqrt{R^2 +4L/C}}{2L}.$$

$\omega L-\frac{1}{\omega C}=-R$ gives the value, $\omega_1$, for which $\omega_1<\omega_{res}.$ Re-arranging and solving as a quadratic in $\omega_1$ gives the positive root $$\omega_1=\frac{-R+\sqrt{R^2 +4L/C}}{2L}.$$

[Note the rather neat relationship, easily shown: $\omega_1 \omega_2 = \omega^2 _{res}.$]

By definition, $$\text{bandwidth}=\Delta \omega=\omega_2-\omega_1=\frac{2R}{2L}=\frac RL $$ So $$Q=\frac{\omega_{res}}{\Delta \omega}=\frac{\omega_{res} L}{R}$$ So $Q'$ and $Q$ are indeed the same! I know which I prefer as a definition ...

Answer 2

The Q factor is a measure of the sharpness of response of a cavity to an external stimulus. It is defined as: \begin{equation} Q=\omega_0 \frac{stored \, energy}{power \, loss} \end{equation} where $\omega_0$ is the resonant angular frequency, assuming no losses. When energy is conserved, the power dissipated in ohmic losses is the negative of the the change of the stored energy with respect to the time. This relation can be written as: \begin{equation} \frac{d}{dt} U(t)= -\frac{\omega_0}{Q} U(t) \end{equation} Thus $U(t)$ decays exponentially with a decay rate inversely proportional to Q. The time dependence of $U(t)$ implies that the electric field is the cavity is damped too. We can then write \begin{equation} E(t)= E_0 exp^{{-\frac{\omega_0}{2 Q_e} t}-j (\omega_0+\Delta \omega)} \end{equation} where $\Delta \omega$ is a shift of the resonant frequency as well as of the damping factor.$E(\omega)$ is related to $E(t)$ by the Fourier transform. This allows us to compute the magnitude squared of $E(\omega)$. When plotting the magnitude squared of $E(\omega)$ versus $\omega$ the full with at half maximum is $\frac{\omega_0}{Q}$. Thus the frequency separation between half power points cab be introduced into the definition to find the Wiki one.