Negative damping ratio for second order system? 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:
$\omega_n = \sqrt{16} = 4$
$2 \zeta \omega_n = -6$
$\zeta = \frac{-6}{2 \omega_n} = -0.75$
 A: 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 $G(s) = \int_0^\infty \! dt \, G(t) e^{-s t}$. 
A: If damping ratio is negative the poles of the system will clearly lie in the right half of the $S$ plane thus making the system unstable.
For a system to be stable it's poles must lie in the left half of the $S$ plane.
