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Q as Energy Stored over Energy Dissipated

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

$\displaystyle 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

$\displaystyle 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$ :

$\displaystyle {\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

{\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))}

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

$\displaystyle {\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

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


$\displaystyle 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$ .

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2018-02-16 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University