Consider a system consisting of a cart to the top of which an inverted pendulum has been attached through a frictionless pivot. The cart is driven by a motor which at time \(t\) exerts a force \(u(t)\), taken as the control. (See Figure \(3.1(\mathrm{~b}) .)\) We assume that all motion occurs in a plane, that is, the cart moves along a straight line. We use \(\phi\) to denote the angle that the pendulum forms with the vertical, \(\delta\) for the displacement of the center of gravity of the cart with respect to some fixed point, \(F \geq 0\) for the coefficient of friction associated with the motion of the cart, \(g\) for the acceleration of gravity, \(l>0\) for the length of the pendulum, \(M>0\) for the mass of the cart, and \(m \geq 0\) for the mass of the pendulum, which we'll assume is concentrated at the tip. (If the mass is not so concentrated, elementary physics calculations show that one may replace the model by another one in which this does happen, using a possibly different length \(l\). We allow the case \(m=0\) to model the situation where this mass is negligible.) Newton's second law of motion applied to linear and angular displacements gives the two second order nonlinear equations and $$ l \ddot{\phi}-g \sin \phi+\ddot{\delta} \cos \phi=0 \text {. } $$ We shall only be concerned with a small angle \(\phi\), so we linearize the model about \(\phi=0\). This results, after taking $x_{1}=\delta, x_{2}=\dot{\delta}, x_{3}=\phi, x_{4}=\dot{\phi}\(, in a linear system \)\Sigma$ with \(n=4, m=1\) and matrices as follows: $$ A=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & -\frac{F}{M} & -\frac{m g}{M} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & \frac{F}{l M} & \frac{g(m+M)}{l M} & 0 \end{array}\right), \quad B=\left(\begin{array}{c} 0 \\ \frac{1}{M} \\ 0 \\ -\frac{1}{l M} \end{array}\right) . $$ Prove that \(\Sigma\) is controllable. (Controllability holds for all possible values of the constants; however, for simplicity you could take all these to be equal to one.) This example, commonly referred to as the "broom balancing" example, is a simplification of a model used for rocket stabilization (in that case the control \(u\) corresponds to the action of lateral jets).

The system \(\Sigma\) is (completely) controllable on the interval $[\sigma, \tau]\( if for each \)x, z \in X\( it holds that \)(x, \sigma) \sim(z, \tau) .$ It is (completely) controllable in time \(T\) if for each \(x, z \in X{X}\) it holds that \(x \underset{T}{\sim} z\). It is just (completely) controllable if $x \sim z\( for all \)x, z$.

Consider the system (with \(\mathcal{U}=\mathbb{R}, x=\mathbb{R}^{2}\) ) $$ \begin{aligned} &\dot{x}_{1}=x_{2} \\ &\dot{x}_{2}=-x_{1}-x_{2}+u \end{aligned} $$ which models a linearized pendulum with damping. Find explicitly the systems \(\Sigma_{[\delta]}\), for each \(\delta\). Characterize (without using the next Theorem) the \(\delta\) 's for which the system is \(\delta\)-sampled controllable. The example that we discussed above suggests that controllability will be preserved provided that we sample at a frequency \(1 / \delta\) that is larger than twice the natural frequency (there, \(1 / 2 \pi\) ) of the system. The next result, sometimes known as the "Kalman-Ho-Narendra" criterion, and the Lemma following it, make this precise.

Assume that \((A, B)\) is controllable and $\mathcal{U} \subseteq \mathbb{R}^{m}\( is a neighborhood of 0 . Then \)J_{k}^{\mathrm{R}} \subseteq \mathcal{R}_{u}(0)\( for all \)k$.

Let \(\mathcal{U} \subseteq \mathbb{R}^{m}\) and pick any two \(S, T \geq 0\). Then $$ \mathcal{R}_{u}^{T}(0)+e^{T A} \mathcal{R}_{u}^{S}(0)=\mathcal{R}_{u}^{S+T}(0) . $$

Consider, as an example, the system \(\Sigma\) corresponding to the linearized pendulum \((2.31)\), which was proved earlier to be controllable. In Appendix C.4 we compute $$ e^{t A}=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right) $$ Thus, for any \(\delta>0\), $$ \mathbf{R}\left(e^{\delta A}, B\right)=\left(\begin{array}{ll} 0 & \sin \delta \\ 1 & \cos \delta \end{array}\right) $$ which has determinant \((-\sin \delta)\). By Lemma \(3.4 .1, \Sigma\) is \(\delta\)-sampled controllable iff $$ \sin \delta \neq 0 \text { and } 2 k \pi i \neq \pm i \delta \text {, } $$ i.e., if and only if \(\delta\) is not a multiple of \(\pi\). Take, for instance, the sampling time \(\delta=2 \pi\). From the explicit form of \(e^{t A}\), we know that \(e^{\delta A}=I\). Thus, $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=0 $$ so \(G=0\). This means that the discrete-time system \(\Sigma_{[\delta]}\) has the evolution equation $$ x(t+1)=x(t) $$ No matter what (constant) control is applied during the sampling interval $[0, \delta]$, the state (position and velocity) is the same at the end of the interval as it was at the start of the period. (Intuitively, say for the linearized pendulum, we are acting against the natural motion for half the interval duration, and with the natural motion during the other half.) Consider now the case when \(\delta=\pi\), which according to the above Lemma should also result in noncontrollability of \(\Sigma_{[\delta]} .\) Here $$ F=e^{\delta A}=-I $$ and $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=-2 A^{-1}=2 A $$ so $$ G=2 A B=\left(\begin{array}{l} 2 \\ 0 \end{array}\right) . $$ Thus, the discrete-time system \(\Sigma_{[6]}\) has the evolution equations: $$ \begin{aligned} &x_{1}(t+1)=-x_{1}(t)+2 u(t) \\ &x_{2}(t+1)=-x_{2}(t) \end{aligned} $$ This means that we now can partially control the system, since the first coordinate (position) can be modified arbitrarily by applying suitable controls \(u\). On the other hand, the value of the second coordinate (velocity) cannot be modified in any way, and in fact at times \(\delta, 2 \delta, \ldots\) it will oscillate between the values \(\pm x_{2}(0)\), independently of the (constant) control applied during the interval.