Homework 2 Problem 8

#### Problem

Consider the control system

(1)
\begin{align} \frac{d}{dt} {\left[ \begin{matrix} a \\ b \\ c \end{matrix} \right]} = \left[ \begin{matrix} a \\ a b \\ 0 \end{matrix} \right] \tau + \left[ \begin{matrix}b \\ 0 \\ c \end{matrix} \right] \varsigma \end{align}

on $N$, where $\tau (t)$ and $\varsigma (t)$ are real-valued control inputs. Is this system locally accessible near $f(0, \frac{\pi}{6})$? Small-time locally controllable near $f(\frac{\pi}{2}, 0)$?

#### Solution

The first thing to note is that this system is drift-less, i.e. if you do not give it any control inputs it will not move. Because of this fact, the questions of whether or not the system is locally accessible or small-time locally controllable(STLC) are equivalent. To test for local accessibility (and STLC for the drift-less case), check:

(2)
\begin{align} span \left\{g_i,\left[g_i,g_j\right],\left[\left[g_i,g_j\right], g_k\right],...\right\} = \mathbb R^n \end{align}

For this problem, $n=3$, there is no term without a control input (no $g_0$), and there are two terms ($g_1$ and $g_0$) with control inputs control inputs ($\tau (t)$ and $\varsigma (t)$). For this problem:

(3)
\begin{align} g_0 = 0 , \ g_1 = \left[\begin{matrix}a \\ a b \\ 0\end{matrix}\right] , \ g_2 = \left[\begin{matrix}b \\ 0 \\ c\end{matrix}\right] \end{align}

The Jacobi-Lie bracket $g_1$ and $g_2$ is:

(4)
\begin{align} \left[g_1,g_2\right] = \left[\begin{matrix}0\ 1\ 0\\0\ 0\ 0\\ 0\ 0\ 1\end{matrix}\right] \left[\begin{matrix}a\\ab\\0\end{matrix}\right]-\left[\begin{matrix}1\ 0\ 0\\ b\ a\ 0\\ 0\ 0\ 0\end{matrix}\right] \left[\begin{matrix}b\\0\\c\end{matrix}\right]=\left[\begin{matrix}a b-b\\-b^2\\0\end{matrix}\right] \end{align}

So, in order to check for local accessibility (and STLC) we need:

(5)
\begin{align} span\left\{g_1,g_2,\left[g_1,g_2\right]\right\} = span\left\{\left[\begin{matrix}a\\a b\\0\end{matrix}\right] ,\left[\begin{matrix}b\\0\\c\end{matrix}\right],\left[\begin{matrix}a b-b\\-b^2\\0\end{matrix}\right]\right\} = \mathbb R^3 \end{align}

For this problem:

(6)
\begin{align} a = \cos\left(2\theta + \phi\right), \ b = \sin\left(\theta +3 \phi\right), \ c = 4 - 5\phi \end{align}

Plugging these functions into eq(5), we are left with:

(7)
\begin{align} span\left\{\left[\begin{matrix}\cos(2\theta +\phi)\\\cos(2\theta +\phi)\sin(\theta+3\phi)\\0\end{matrix}\right],\left[\begin{matrix}\sin(\theta+3\phi)\\0\\4-5\phi\end{matrix}\right],\left[\begin{matrix}\cos(2\theta +\phi)\sin(\theta+3\phi)-\cos(2\theta +\phi)\\-\sin^2(\theta+3\phi)\\0\end{matrix}\right]\right\} \end{align}

In order to check for local accessibility near $f\left(0,\frac{\pi}{6}\right)$, simply plug the values into equation 7 and check if the determinant of the matrix consisting of the three vectors is non-zero.

(8)
\begin{align} \left|\begin{array}{ccc}\frac{\sqrt3}{2} & 1 & \frac{\sqrt3}{2}-1\\ \frac{\sqrt3}{2} & 0 & -1\\0 & 4-\frac{5\pi}{6} & 0\end{array}\right| \approx 1.05 \end{align}

Since the determinant is non-zero, the three vectors are linearly independent and span $\mathbb R^3$. Therefore, the system is locally accessible (and small-time locally controllable since drift-less) near $f\left(0,\frac{\pi}{6}\right)$.

To check whether or not the system is small-time locally controllable near $f\left(\frac{\pi}{2},0\right)$, again plug in the values into equation 7 and check the determinant.

(9)
\begin{align} \left|\begin{array}{ccc} -1 &1 &-2 \\-1 & 0 & -1\\ 0 & 4 & 0\end{matrix}\right| = 4 \end{align}

Again, the determinant is non-zero and therefore the three vectors span $\mathbb R^3$ , so the system is small-time locally controllable (also locally accessible since drift-less) near $f\left(\frac{\pi}{2},0\right)$.

Scott agrees!

#### Discussion

page revision: 2, last edited: 03 Apr 2007 05:41