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I was wondering what would you classify the damping state (under damped, over-damped, critically damped) a second-order ode system with a negative damping ratio? To me it doesn't make much sense since a negative damping ratio results in an unstable system.

The system I have is:

$G(s) = \frac{5}{s^2-6s+16}$

So solving for the natural frequency and damping ratio:

$\omega_n = \sqrt{16} = 4$

$2 \zeta \omega_n = -6$

$\zeta = \frac{-6}{2 \omega_n} = -0.75$

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    Practically this is nonsense since such an oscillator will saturate the amplifier almost instantly and be of no practical use.2013-08-08

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I assume the G(s) which you give is the Laplace transform of the Green's function (please indicate next time what your symbols mean otherwise, a question can be hard to understand).

A negative damping rate is not damping but driving. Physically, it results from an external system which constantly pumps energy into your oscillator (I assume your 2nd order ode is an oscillator). In your case with $G(s) =\frac{5}{s^2-6s+16}$, the real-time Green's function is $G(t) = \frac{5}{\sqrt{7}} \sin(\sqrt{7} t) e^{3t}$, i.e., an oscillatory solution whose amplitude is exponentially growing.

By the way, why do you get that the frequency is 4 (I have $\sqrt{7}$). Do you use a different convention for the Laplace transform. My convention is $G(s) = \int_0^\infty \! dt \, G(t) e^{-s t}$.

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    Read (transfer function)=(Green's function): in your case the incoming signal will be amplified (not such an uncommon element in electrical engineering). I understand now our discrepancy regarding the frequency. You were calculating the **natural** frequency assuming the system is a damped harmonic oscillator whereas I did calculate the frequency with which the system responds. However, seeing that the signal gets amplified I'm not so sure how much sense it makes to try to fit it to that model.2011-02-24