How is the damping equation obtained

control theoryordinary differential equations

Damping ratio

I modified some online notes in the internet and prepared this illustration for later use.

  1. Why does the formula fail?

If $B=0$ (it is a sinusoidal system with no exponential components, undamped oscillation), it gets the form $x'' = -Cx$. This is a sinusoidal movement with $\omega_0^2$ as C. Everything is okay here.

If $C=0$ (it is an exponential system with no sinusoidal components, damped without oscillation for positive B), it gets the form $x' = -Bx + k$ after an integral is applied. This has a decay rate $B$ (not $\frac{B}{2}$! Here the formula fails) with $\lambda = -\lambda_0 = B$.

$$x'' + 2\zeta\omega_0 x' + \omega_0^2 x = 0$$

Is not this a general formula applicable to any possible situation?

2.
$$\frac{\sqrt{B^2 – 4\omega_0^2}}{2} = \pm\omega_0i$$
$\omega_0^2 = C$ also produces mistaken results when $B \ne 0$ and I put it into the complex component of the quadratic formula. With some manipulation, it will take the form (multiplied by 2i):

$$\sqrt{-B^2 + 4\omega_0^2} = \pm2\omega_0$$
$$-B^2 + 4\omega_0^2 = 4\omega_0^2$$
$$B=0$$

But this contradicts our very assumption that $B \ne 0$. What am I confusing here?

  1. Why define $\zeta$ (the Damping Ratio) as $\frac{\lambda}{\omega_0}$?

I think this comes from Euler's equation for a complex number as an exponent of $e$:

$$e^{(\lambda_0 \pm \omega_0i)t} = e^{\lambda_0t}e^{(\pm\omega_0t)i}$$

Here the exponential with the imaginary power produces an oscillating, sinusoidal factor, while the other is about decay. So decay divided by oscillation should overall say us how neatly the system is stabilizing. A high $\zeta$ means more damping and less oscillation ($\zeta = 1$ being the best and more than it is overdamped). But now is not the formula Wikipedia gives ($\frac{\lambda}{\sqrt{\lambda^2 + \omega_0^2}}$) more formal and logical? For that, it compares the decay rate (the real component of $r$) to the magnitude of $r$.

I am quite a bit confused. Hopefully not too many questions. Thanks! I have this unfortunate habit to not progress in my course until every detail makes sense to me.

Best Answer

This answer is written assuming the control theory tag is appropriately assigned to this question and assumes this is the context. The control theory textbooks I've followed (e.g. [pg. 311, 1], [pg. 130, 2]) define the damping ratio first as an abstract parameter of a 2nd-order system that determines whether the (stable) system was:

  • Underdamped,
  • Critically damped,
  • Overdamped.

Texts often start with standard second-order system as a system that takes the form,

$$x'' + 2\,\zeta\,\omega_0\,x' + \omega_0^2 = 0$$

defining $\zeta$ and $\omega_0$ as parameters and giving them the names the damping ratio and the natural frequency. The leading coefficient can always be normalized to this form. After this, we can motivate these names with properties that can be analytically computed: e.g. what happens when $\zeta \in (0, 1)$ is that system response is underdamped. When the system is undamped, the system oscillates to an impulse at frequency $\omega_0.$ The decay rate is then defined afterwards when the step response to that system is derived for a specific case. The decay rate relationship $\lambda = \zeta \omega_0$ applies only to the underdamped case. This is the first treatment I have seen where the damping ratio is defined in terms of the decay rate. To be fair, the treatment in wikipedia refers to the underdamped case (decaying sinusoid). Given that, the short answer to your question is the damping ratio isn't defined in terms of the decay rate.

In control theory, the vast majority of systems we encounter (and even design) will be at least somewhat underdamped. Critically damped systems are practically irrelevant since you can't practically make $\zeta = 1$. Overdamped systems are rarely desireable since the slower dynamics dominate and, when you can, why not make that part faster and eliminate the overdamped behaviour? In fact, [pg. 139, 2] makes a case for why we'd want an underdamped response: the settling time (how fast the system converges upon steady-state) is optimized when $\zeta \approx 0.7$.

As a result, we just need a decay rate formula that works for the underdamped class of systems. Underdamped systems conveniently have the decay rate $\lambda = \zeta \omega_0$ and , better still, this directly corresponds to the real part of the eigenvalue (pole) of the system. This plays a critical role in the frequency domain control design methods, since settling time specifications (how quickly the system decays to converge upon a reference) directly translate to a specification on the real part of the system pole and also have a simple formula in terms of $\zeta$ and $\omega.$

Interestingly, $\zeta$ plays a useful role in overshoot specifications on the step response of the system. The overshoot is the amount that the system exceeds the target value of the input to the system before settling down and converging. The overshoot formula works out to just the phase of the eigenvalue/pole of the system and, most importantly, the overshoot is purely a function of $\zeta$ ($\omega_0$ plays no role). If you had defined $\zeta$ differently so that the coefficient of $x'$ wasn't $2\zeta\omega_0$, then $\omega_0$ will likely appear in the overshoot formula too. Given the current definition, overshoot specifications directly translate one-to-one into $\zeta$ specifications.

For instance, if I was told to design a controller to make a system have a decay rate of say $4$ and an overshoot of $10\%$ I know to first solve $\zeta$ from the overshoot specification (there is a formula). From this, I can then easily deduce what $\omega_0$ must be since,

$$\lambda = 4 = \zeta\omega_0$$

and I know $\zeta.$ This notion is even more important when you have specifications that compete and you instead place ranges on your specifications. For example, a more common specification would be:

  • A decay rate of at least $4$ and,
  • A maximum overshoot of $10\%$.

Anyway, there is a deeper reason to care about $\lambda = \zeta \omega_0.$ I would personally move away from viewing $\zeta$ as related to decay, and instead view it as an abstract parameter of a second order system telling you how damped it is or not. Formulas that compute the decay rate depending on the specific subclass of system (underdamped, overdamped) then derive from the damping ratio and the natural frequency.

[1]: Dorf, R. C. and Bishop, R. H. Modern Control Systems. 12 Ed.

[2]: Stefani, S. and Savant, H. Design of Feedback Control Systems. 4 Ed.

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