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Lyapunov functions are used to investigate stability of an eqilibrium (often the control reference $r(t) = 0$)... but what happens if tracking control is considered (i.e. $r(t)$ is non constant but a function of time)?

How would I treat the reference signal when checking whether $\dot{V} \leq 0$?

SampleTime
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    The prevalent idea is to use Lyapunov functions of the error $e(t)=y(t)-r(t)$ where $y$ is the output vector. – RTJ Nov 19 '16 at 21:47
  • Ok, but can't I always change the reference signal continously before $e(t)$ reaches zero such that the equilibrium is never reached? Then there should not exist a Lyapunov function even though the system tracks the reference signal? – SampleTime Nov 22 '16 at 21:43
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    I am not sure I understood your question but I will try to answer. You have to design the controller based on the error dynamics. For example, consider the simple integrator $\dot{y}=u$. Then, $\dot{e}=\dot{y}-\dot{r}=u-\dot{r}$. Choosing now $u=-ke+\dot{r}$ one has for the Lyapunov function $V=e^2$ that $\dot{V}=-ke^2\leq 0$. Thus, $0$ is an asymptotically stable equilibrium point for the error dynamics not the $y$ dynamics. – RTJ Nov 22 '16 at 22:09
  • Thanks, $y$ is asymptotic to the reference signal then, that answered my question. – SampleTime Nov 27 '16 at 00:22
  • Just as a comment, there can be constructed examples where the error dynamics are always zero but the states blow up - specifically in non-linear and adaptive control problems. So Lyapunov functions for the error dynamics don't always solve the problem. Just something to keep in mind, but I think the above comments answered your original question. – ITA Nov 28 '16 at 16:41
  • @IvanAbraham could you give reference to such a constructed example, sounds interesting. – SampleTime Nov 29 '16 at 18:37

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