• QandDecayTime

Suppose: You have the transfer function of the damped oscillator with the quality factor of Q and the resonant frequency of f0.

Define s0 as

• s0 =2 pi f0

The filter transfer function of the system is (somewhat in general) described as

• H(s) = s0² / (s² + s s0/Q + s0²)

The impulse response of the system is obtained by the inverse Laplace transform of H(s)

• h(t) = Q s0 / Sqrt(1-4 Q²) * Exp[-s0 t/(2Q)] {Exp[-Sqrt((1-4Q²)s0²)t/(2Q)] - Exp[Sqrt((1-4Q²)s0²)t/(2Q)]}

If we assume Q is enough higher than the unity, 4Q²-1 can be approximated by 4 Q².

• h(t) = s0 / (2 i) * Exp[-s0 t/(2Q)] [Exp(-I s0 t) - Exp(I s0 t)]
• h(t) = - s0 Sin(s0 t) Exp[-s0 t/(2Q)]

Here the 2 Q/s0 = Q /(pi f0) is the decay time of the damped oscillation (1/e criteria).

If you rewrite the decay by the half life of the amplitude

• Exp[- Pi f0 t / Q] == 2^(-t/T_1/2)

• Q = Pi/log(2) f0 T_1/2 = 4.53 f0 T_1/2

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Relationship between the peak amplitudes of the damped oscillation:

h(t) = Cos(2 pi f0 t) Exp[- pi f0 t/Q]

t_j= j/(2 f0) (j=0, 1, 2, ...)

These t give us the peak amplitudes of ringdown oscillation if the Q is high. (Even if Q is as low as Q=0.5, the phase shift of the peak is 45deg, but no longer the oscillation can not be expressed by the above equations.)

==> h(t_j) = (-1)^j Exp[-j*pi/(2 Q)]

R = h(t_j+1)/h(t_j) = - Exp[-pi/(2 Q)]

R = 0.99 -> Q = 156
R = 0.98 -> Q = 77
R = 0.95 -> Q = 30
R = 0.9 -> Q = 15
R = 0.8 -> Q = 7
R = 0.7 -> Q = 4.4
R = 0.6 -> Q = 3
R = 0.5 -> Q = 2.2
R = 0.3 -> Q = 1.3