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I have the following standard transfer function for a damped linear oscillator:

$$G(s) = \dfrac{\omega_0^2}{s^2 + 2\zeta\omega_0s + \omega_0^2}$$

Now I have two eigen values at locations $-100 \pm 100i$. I want to calculate the damping coefficient, $\zeta$, and natural frequency $\omega_0$.

I thought that the characteristic polynomial of matrix $A$ from the linear system in state space form is the denominator of $G(s)$. However, when solving $s^2 + 2\zeta\omega_0s + \omega_0^2=0$ for $s = -100 - 100i$, I get an equation with two unknowns right? What theoretical point am I missing here? how should I go about this?

Thanks in advance

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If you I understand you correctly your system has two complex conjugate poles at $p=-100+100i$ and $p^*=-100-100i$ (where $^*$ denotes complex conjugation). Consequently, your denominator polynomial can be written as

$$\begin{align}(s-p)(s-p^*)&=s^2-(p+p^*)s+|p|^2\\&=s^2-2\text{Re}\{p\}+|p|^2\\&=s^2-2|p|\cos\phi+|p|^2\end{align}\tag{1}$$

where $\phi$ is the pole angle: $p=|p|e^{i\phi}$. For the given poles you have $\phi=3\pi/4$. Comparing $(1)$ with the denominator in your question immediately gives $\omega_0=|p|$ and $\zeta=-\cos\phi$.

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