This is a homework problem from my adaptive control course:
Given the plant $y_p = a_pu(t)$ ($a_p\neq 0$) and the reference model $y_m = a_mr(t)$, where $r(t)$ is bounded and continuous. Design a differentiator free control $u(t)$ such that all signals are bounded and $|y_p(t) - y_m(t)|\to 0$. Consider two different scenarios: (1) $|a_p|$ is unknown but its sign is known, and (2) $a_p$ is unknown.
What first occurred to me was to construct a PI-like controller, and below is my attempt for scenario 1:
Let $x_p, x_m$ be two signals such that $\dot{x}_p = y_p$ and $\dot{x}_m = y_m$. If $a_p$ is known, it is desired that $$ y_m(t) = a_mr(t) = y_p(t) = a_pu(t) \implies u(t) = \frac{a_m}{a_p}r(t) = a^*r(t). $$ Let $\Delta x = x_p - x_m$ and $u(t) = \theta(t)\Delta x(t) + a(t)r(t)$, where $a(t) = a^* + \Delta a(t)$ is an estimate of $a^*$. Then one has that \begin{align*} \Delta y(t) &=\Delta\dot{x}(t) = a_p[\theta(t)\Delta x(t) + a(t)r(t)] - a_mr(t)\\ &= a_p\theta(t)\Delta x(t) + a_p[a^* + \Delta a(t)]r(t) - a_mr(t) = a_p\theta(t)\Delta x(t) + a_p\Delta a(t)r(t). \end{align*} Let $V(\Delta x, \Delta a) = (|a_p|^{-1}\Delta x^2 +\Delta a^2)/2$. Then \begin{align*} \dot{V} &= \mathrm{sgn}(a_p)\theta(t)\Delta x^2 + \mathrm{sgn}(a_p)\Delta a r\Delta x + \Delta a\Delta\dot{a} = \mathrm{sgn}(a_p)\theta(t)\Delta x^2 + \Delta a[\mathrm{sgn}(a_p) r\Delta x + \Delta\dot{a}], \end{align*} which is negative definite w.r.t. $\Delta x$ when $$ \theta \equiv -\mathrm{sgn}(a_p),\quad \dot{a} = \Delta\dot{a} = -\mathrm{sgn}(a_p) r\Delta x\implies \dot{V} = -\Delta x^2. $$
From above I can show that all signals are bounded, and $\Delta x\in L^2\cap L^\infty$ (hence $\Delta x(t)\to 0$). But what I am stuck with is to show that $\Delta y = \Delta\dot x$ also tends to zero at large. I wanted to use Barbalat's lemma but looks like there isn't any guarantee that $\Delta\dot x$ is uniformly continuous. I wonder if there is any problem with the controller that I proposed? Any help will be greatly appreciated!
