Decay Time is Q Periods

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 with time-constant . More precisely, we will show that, for , the impulse response decays by the factor in cycles, which is about 96 percent decay, or -27 dB.

The impulse response corresponding to Eq.(E.7) 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.7) to obtain

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

(E.10) | |||

(E.11) |

as the respective residues of the poles .

The impulse response is thus

Assuming a resonator, , we have , where (using notation of the preceding section), and the impulse response reduces to

where and are overall amplitude and phase constants, respectively.

We have shown so far that the impulse response decays as with a sinusoidal radian frequency under the exponential envelope. After Q periods at frequency , time has advanced to

where we have used the definition Eq.(E.8) . Thus, after periods, the amplitude envelope has decayed to

which is about 96 percent decay. The only approximation in this derivation was

which holds whenever , or .

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