Thursday, February 21, 2013

Lecture 11

The idea of controllability is straightforward and we can understand it using ideas of linear algebra.
Consider again the broom balancing problem. It is not possible to stabilize two brooms at equilibrium simultaneously if they are the same lengths, but it is possible if the lengths are different. This is because of the forms:>
$A=\begin{pmatrix}0 & 1 & 0 & 0\\ \alpha & 0 & 0& 0\\ 0 & 0 & 0& 1\\ 0 & 0& \beta & 0\end{pmatrix}\quad B=\begin{pmatrix}0 \\-\alpha\\0\\-\beta\end{pmatrix}\quad M_4=\begin{pmatrix}0 & -\alpha & 0 & -\alpha^2\\ -\alpha & 0 & -\alpha^2& 0\\ 0 & -\beta & 0& -\beta^2\\ -\beta & 0& -\beta^2 & 0\end{pmatrix}$

where $\alpha = g/L_1$, $\beta=g/L_2$.

So $M_4$ is of rank 4 iff a and b are different. However, this assumes that as you move your hand you must keep it at constant height from the ground. What do you think might happen if you can move your hand up and down also?

In 1996, paper 2, question 10 there was this cute tripos question:
A fantasy kingdom has populations of x vampires and y humans. At the start of each of the four yearly seasons the king takes a census and then intervenes to admit or expel some humans. Suppose that the population dynamics of the kingdom are governed by the plant equations:
$x_{t+1} = x_t + (y_t – Y)$,
$y_{t+1} = y_t – (x_t – X) + u_t$,
where $x_t$ and $y_t$ are integers representing the respective populations of vampires and humans in the $t$-th season, $t=0,1,2,3,\dotsc$; $X$ and $Y$ represent equilibrium values in the population without regal intervention; ...
[ some more description]
Can the king regain equilibrium by only expelling humans?
Suppose, the census is taken during the day, so the king can only count humans, and thus $u_t$ is allowed only to depend on $y_t$. Show that ...

The last part of the question is about observability (see Lecture 12). The full question is in this file of past questions