I was reading a script about state space representation and at one point it was mentioned that the eigenvalues of the state matrix, in that specific case the complex conjugate pair $\lambda_{1/2}= -0.68 \pm 1.63j $ has a damping factor of about $.4$ .

How can you get the damping factor from a (complex) eigenvalue?


I assume you are talking about continous systems.

To get the damping, draw a line from the eigenvalue to the origin. The slope of that line is the (absolute value of the) damping factor.

Or, as formula: given the eigenvalues $\lambda_i = a_i + j b_i$, the damping factors are

$$ D_i = \frac{-a_i}{\sqrt{a_i^2 + b_i^2}} \tag{1} $$

In your case: $D_1 = \frac{-(-0.68)}{\sqrt{(-0.68)^2 + 1.63^2}} = D_2 = \frac{-(-0.68)}{\sqrt{(-0.68)^2 + (-1.63)^2}} = \frac{0.68}{\sqrt{0.68^2 + 1.63^2}} = 0.385 \approx 0.4$.

If you have a discrete system with sample time $T$, you can first convert the discrete eigenvalues to a continous version from which you can get the damping factors.

The continous counterpart of the discrete eigenvalues $\tilde{\lambda}_i = \tilde{a}_i + j \tilde{b}_i$, are

$$ \lambda_i = a_i + j b_i = \frac{\log(\tilde{\lambda}_i)}{T} = \frac{\log(\tilde{a}_i + j \tilde{b}_i)}{T} \tag{2} $$

with $\log$ the natural logarithm. Then you can use the same formula $(1)$ as before, just with $(2)$.

  • $\begingroup$ Thank you for your answer! Where does this equation come from? $\endgroup$ – Pilotf4 Dec 1 at 17:38
  • $\begingroup$ @Pilotf4 If you mean who originally came up with that equation - I don't know, unfortunatly. I just know it from several lecture slides and books on control theory, as it is a very common formula. $\endgroup$ – SampleTime Dec 1 at 18:37
  • $\begingroup$ @Pilotf4 It essentially comes from the definition of the damping factor. This has the associated quadratic equation $s^2+2\,\zeta\,\omega_n\,s + \omega_n^2=0$ which has the roots $\lambda=\omega_n\,(-\zeta \pm i \sqrt{1 - \zeta^2})$. The other way around, so given the complex conjugate roots $\lambda=\rho\pm i\,\sigma$, this translates to $\omega_n = \sqrt{\rho^2 + \sigma^2}$ and $\zeta = -\rho / \sqrt{\rho^2 + \sigma^2}$. $\endgroup$ – Kwin van der Veen Dec 1 at 20:01

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