Let’s say I have a system 1/(T1s+1) or any other n-th order polynomial and a PI controller (KP and TI). I already know that the system is stable but for, let’s say, educational purposes (not homework), I want to test the stability using different stability criterions.

My question is:
Can I check if the system is stable or not by using Lyapunov stability criterion; similar to Nyquist or Routh-Hurwitz stability criterion? And how :) ? I was unable to find any examples on that or similar system.

I must admit that I am a beginner in the topic of stability, but as I see it Nyquist, Bode, Routh-Hurwitz they are more engineering orientated (i.e. in a few steps I can easily test the stability), while Lyapunov is not :(
Yet in almost every book Lyapunov’s direct method is considered one of the best for determining the stability of dynamical systems!

  • $\begingroup$ As far as I understood the description on wiki, Nyquist, Bode, and Hurwitz criteria work only for linear time invariant systems, while Lyapounov criterion works for non-linear systems as well. $\endgroup$ Dec 14, 2013 at 23:20

1 Answer 1


Short answer: Yes, you can.

Long answer: Here is how.

I remember we often used Lyapunov's direct method when studying nonlinear system analysis and control. It hinges on the following idea:

If the total energy of a system continuously decreases, then the system, whether linear or nonlinear, must eventually settle down to an equilibrium point.

Consider the linear time-invariant system from your example, i.e.

$$G(s) = \displaystyle\frac{1}{T_1s+1}, \quad T_1 \in \mathbb{R}_{>0}.$$

Then the time-domain description of the system is given by the following linear differential equation:

$$\dot{x}(t) + \frac{1}{T_1}x(t) = \frac{1}{T_1}u(t)$$

We would like to check the zero-input stability of the system with an arbitrary initial state, i.e. $u(t) = 0 \ \forall t \in \mathbb{R}_{\geq 0}$ and $x(0) = x_0$. Then we have

$$\dot{x}(t) = -\frac{1}{T_1}x(t)$$

with the equilibrium point at $x = 0$. Select a Lyapunov candidate function $V(x)$ as the square of the distance from the origin times a non-zero constant, e.g.

$$V(x) = \frac{1}{2}x^2.$$

Check that the above $V(x)$ satisfies the criterion for being a positive definite function:

$$V(x) \geq 0 \quad \forall x \in \mathbb{R}, \quad \text{and} \quad V(x) = 0 \iff x = 0.$$

Observe that $V(x)$ is a scalar function, which can be interpreted as the "energy function" of the system. Now, to check if the system is stable, we compute the "rate of change of energy", i.e. the derivative of $V(x)$:

$$\begin{aligned}\dot{V}(x) &= x\dot{x}\\ &= x\left(-\frac{1}{T_1}x\right)\\ &= -\frac{1}{T_1}x^2,\end{aligned}$$

which is negative definite. That is,

$$ \dot{V}(x) \leq 0 \quad \forall x \in \mathbb{R}, \quad \text{and} \quad \dot{V}(x) = 0 \iff x = 0.$$

So we may say that the system passively dissipates energy. Since

  • $V(x)$ is positive definite,
  • $\dot{V}(x)$ is negative definite, and
  • $V(x) \to \infty$ as $\|x\| \to \infty$,

then we conclude that the system is (globally) asymptotically stable at the origin in the Lyapunov sense.

  • $\begingroup$ What to do when I have second or third order polynomial? Then I have second (third) derivative on the “left” side! ___ I am actually trying to test the stability of an off-grid three phase inverter with cascade current and voltage controller. My initial idea was to determine the stability of d and q reference frame separately (as I did with Nyquist and Routh-Hurwitz). But now, I’m not sure! Is the Lyapunov stability of a three phase inverter usually determined in this way or by some other way? $\endgroup$
    – mikebuba
    Dec 17, 2013 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .