Q as Energy Stored over Energy Dissipated

Yet another meaning for $Q$ is [20, p. 326]

$$Q \approxs 2\pi\frac{\hbox{Stored Energy}}{\hbox{Energy Dissipated in One Cycle}}$$

where the resonator is freely decaying (unexcited).

To analyze this, let's again consider a one-pole resonator as in §E.7.1. Then the impulse response is given by

$$h(t) \eqsp e^{\sigma_0 t}e^{j\omega_0 t} \isdefs e^{-\alpha t}e^{j\omega_0 t}$$

The total stored energy at time $t$ is given by the total energy of the remaining response at time $t$ :

$${\cal E}(t) \eqsp \int_t^\infty \vert h(t)\vert^2 dt \eqsp \int_0^\infty e^{-2\alpha t} dt \eqsp \frac{1}{2\alpha}.$$

The energy dissipated in the first period $P_0 = 2\pi/\omega_0$ is ${\cal E}(0)-{\cal E}(P_0)$ , where

$$\begin{eqnarray*} {\cal E}(P_0) &=& \int_{P_0}^\infty h^2(t)dt \eqsp \int_{P_0}^\infty e^{-2\alpha t}dt \\ &=& -\left.\frac{1}{2\alpha} e^{-2\alpha t} \right\vert _{P_0}^\infty \eqsp \frac{e^{-2\alpha P_0}}{2\alpha}\\ &=& \frac{e^{-2\alpha (2\pi/\omega_0)}}{2\alpha}. \eqsp {\cal E}(0)e^{-2\pi/(\omega_0/(2\alpha))} \end{eqnarray*}$$

Assuming $Q\gg 1/2$ as before, $\omega_0\approx\omega_p$ so that

$${\cal E}(P_0) \approxs {\cal E}(0)\,e^{-2\pi/Q}$$

Assuming further that $Q\gg 2\pi$ , we have $e^{-2\pi/Q}\approx 1 - 2\pi/Q$ so that

$$\frac{{\cal E}(0)}{{\cal E}(0)-{\cal E}(P_0)} \approxs \frac{Q}{2\pi}$$

or

$$Q \approxs 2\pi \frac{{\cal E}(0)}{{\cal E}(0)-{\cal E}(P_0)}$$

as claimed. Note that this rule of thumb requires $Q\gg 2\pi$ , while the one of the previous section only required $Q\gg 1/2$ .