Decay Time is Q Periods

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Another well known rule of thumb is that the of a resonator is the number of ``periods'' under the exponential decay of its impulse response. More precisely, we will show that, for $Q\gg 1/2$ , the impulse response decays by the factor $e^{-\pi}$ in cycles, which is about 96 percent decay, or -27 dB.

The impulse response corresponding to Eq. $\,$ (E.8) is found by inverting the Laplace transform of the transfer function . Since it is only second order, the solution can be found in many tables of Laplace transforms. Alternatively, we can break it up into a sum of first-order terms which are invertible by inspection (possibly after rederiving the Laplace transform of an exponential decay, which is very simple). Thus we perform the partial fraction expansion of Eq. $\,$ (E.8) to obtain

$\displaystyle H(s) = \frac{g_1}{s-p_1} + \frac{g_2}{s-p_2}$

where

are given by Eq. $\,$ (E.9), and some algebra gives

$\displaystyle g_1$	$\displaystyle =$	$\displaystyle -g\frac{p_1}{p_2-p_1}$	(E.12)
$\displaystyle g_2$	$\displaystyle =$	$\displaystyle g\frac{p_2}{p_2-p_1}$	(E.13)

as the respective residues of the poles

The impulse response is thus

$\displaystyle h(t) = g_1 e^{p_1t} + g_2 e^{p_2t}.$

Assuming a resonator, , we have $p_2 = \overline{p}_1$ , where $p_1=\sigma_p +j\omega_p = -\alpha + j\omega_d$ (using notation of the preceding section), and the impulse response reduces to

$\displaystyle h(t) = g_1\,e^{p_1 t} + \overline{g}_1\,e^{\overline{p}_1 t} = A\,e^{-\alpha t} \cos(\omega_p t + \phi)$

where

and $\phi$ are overall amplitude and phase constants, respectively.^E.1

We have shown so far that the impulse response decays as $e^{-\alpha t}$ with a sinusoidal radian frequency $\omega_p=\omega_d$ under the exponential envelope. After Q periods at frequency $\omega_p$ , time has advanced to