Predicting decay time of a resonant bandpass filter?
A band pass filter would typically have a transfer function like this: -

$$H(s) = \dfrac{s2\zeta\omega_n}{s^2 + s2\zeta\omega_n +\omega_n^2}$$

Where \$2\zeta\$ is 1/Q 

And \$\omega_n\$ is the natural resonant frequency of the filter. 

For an LC bandpass filter, \$\omega_n\$ = \$\dfrac{1}{\sqrt{LC}}\$

To convert this to the time domain (due to an impulse response) there will be three time-domain solutions: -


\$\zeta\$ &lt; 1 (under-damped)
\$\zeta\$ = 1 (critically-damped)
\$\zeta\$ > 1 (over-damped)


There is, unfortunately not a one-size-fits-all solution. Life gets easier if you set \$\omega_n\$ to 1 and then look at the inverse Laplace table for the three scenarios.



For a critically damped situation (\$\zeta = 1\$): -

$$H(s) = \dfrac{2s}{s^2 + 2s +1}$$

And, using inverse Laplace tables you find the solution as: -

$$h(t)=2e^{-t}-e^{-t}\cdot \:2t$$



For an underdamped situation with \$\zeta\$ = 0.5 you get: -

$$H(s) = \dfrac{s}{s^2 + s +1}$$

And this inverse transforms to: -

$$h(t)=e^{-\frac{t}{2}}\cos \left(\frac{\sqrt{3}t}{2}\right)-\frac{1}{\sqrt{3}}e^{-\frac{t}{2}}\sin \left(\frac{\sqrt{3}t}{2}\right)$$



For an overdamped situation where for example \$\zeta\$ = 2 you get: -

$$H(s)=\dfrac{4s}{s^2 + 4s+1}$$ 

And this inverse transforms to: - 

$$h(t)=4e^{-2t}\cosh \left(\sqrt{3}t\right)-\frac{8}{\sqrt{3}}e^{-2t}\sinh \left(\sqrt{3}t\right)$$



You can use this really good interactive calculator instead of tables should you wish: -




  What is the mathematical equation that would allow you to predict the
  decay time of the resonant filter based on its resonance setting?


The above formulas are the only ones I can recommend.