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Q as Energy Stored over Energy Dissipated

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

$\displaystyle 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

$\displaystyle {\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^{-2\alpha t}dt \\
&=& -\left.\frac{1}{2\alpha} e^{-2\alpha t} \right\vert _P^\infty\\
&=& \frac{e^{-2\alpha 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

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

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

$\displaystyle {\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

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

whence

$\displaystyle 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$ .


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA