Q as Energy Stored over Energy Dissipated

Yet another meaning for $Q$ is as follows [21, p. 326]

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

where the resonator is freely decaying (unexcited).

Proof. The total stored energy at time $t$ is equal to the total energy of the remaining response. After an impulse at time 0, the stored energy in a second-order resonator is

$${\cal E}(0) = \int_0^\infty h^2(t)dt \propto \int_0^\infty e^{-2\alpha t}dt = \frac{1}{2\alpha}.$$

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

$$\begin{eqnarray*} {\cal E}(P) &=& \int_P^\infty h^2(t)dt \propto \int_P^\infty e... ...P}}{2\alpha}\\ &=& \frac{e^{-2\alpha (2\pi/\omega_p)}}{2\alpha} \end{eqnarray*}$$

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

$${\cal E}(P) \approx \frac{e^{-2\pi/Q}}{2\alpha}.$$

Assuming further that $Q\gg 2\pi$, we obtain

$${\cal E}(0)-{\cal E}(P) \approx \frac{1}{2\alpha} \left(1-e^{-\frac{2\pi}{Q}}\right) \approx \frac{1}{2\alpha}\frac{2\pi}{Q}$$

This is the energy dissipated in one cycle. Dividing this into the total stored energy at time zero, ${\cal E}(0)=1/(2\alpha)$, gives

$$\frac{{\cal E}(0)}{{\cal E}(0)-{\cal E}(P)} \approx \frac{Q}{2\pi}$$

whence

$$Q = 2\pi \frac{{\cal E}(0)}{{\cal E}(0)-{\cal E}(P)}$$

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