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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}\\
&=& \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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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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