Today will be on controllability of \({\displaystyle \sum } =\left ( A,B\right ) \). We talked before about controllability for LTI. We said that when rank of the controllability matrix is \(n\) then \(\left ( A,B\right ) \) is controllable. This is an algebraic view. When \(\left ( A,B\right ) \) is controllable, it means we can do some useful transformations. We talked about minimal realization. These are all algebraic properties. Today will talk about what physically it means for system to be controllable. This is the physical meaning to saying \(\rho \left ( \mathbb{C} \right ) =n.\) Also, if we want to manipulate the input we need physically controllability.
Physically controllability has to do with only \(A,B\), from \(x^{\prime }=Ax+Bu\). It is about the ability to steer the system with an input. What this means, for given state \(X\left ( 0\right ) \) we want to be able to transfer the system to new state \(X\left ( t\right ) \). This is called the target state.
So system is controllable at \(t_{0}\) if the following is true: (Note, we the ”at \(t_{0}"\,\ \)is important, since this is now LTV and system can change from time to time, so we always talk about controllability at some specific time with LTV).
Formal definition of physical controllability: Given any initial conditions \(x\left ( t_{0}\right ) =X_{0}\) any target state \(x^{\ast }\,\ \)then there exist a future time \(t_{1}>t_{0}\) and input \(u\left ( t\right ) \) over \(\left [ t_{0},t_{1}\right ] \) leading to \(x\left ( t_{1}\right ) =x^{\ast }.\)
Notice there is not constraint on \(u\left ( t\right ) \), it can be anything and as large as needed. (but the time interval to arrive at target state must be finite).
There can be many \(u\left ( t\right ) \) which will do the above, but we only need to find one. Why ”at \(t_{0}"\) is important? Looking at 2 extreme cases
There are cases in between the above extreme cases where it is not clear. For example, given\begin{align*} x_{1}^{\prime } & =x_{1}+u\\ x_{2}^{\prime } & =x_{2}+u \end{align*}
Not controllable. This is not coupled. We can control \(x_{1}\) state on its own, and \(x_{2}\) on its own, but not both at same time which would be necessary for complete controllability.
Reader: If system is controllable at \(t_{0}\), is it controllable at \(t_{0}<t?\) Yes. (but need to understand the argument given).
Suppose we have 2 vectors and they are time dependent. So we need to define what linear independent means in this case (when the vectors depend on time as well).
On linear independence of time vectors:
Let \begin{align*} f_{1}\left ( t\right ) & =\left [ f_{11}\left ( t\right ) ,f_{12}\left ( t\right ) ,\cdots ,f_{1p}\left ( t\right ) \right ] \\ f_{2}\left ( t\right ) & =\left [ f_{21}\left ( t\right ) ,f_{22}\left ( t\right ) ,\cdots ,f_{2p}\left ( t\right ) \right ] \\ & \vdots \\ f_{n}\left ( t\right ) & =\left [ f_{n1}\left ( t\right ) ,f_{n2}\left ( t\right ) ,\cdots ,f_{np}\left ( t\right ) \right ] \end{align*}
We say that \(f_{i}\) vectors are L.D. (Linear dependent) on time interval \(\left [ t_{0},t_{1}\right ] \) if the following occurs: There exist \(\alpha _{1},\alpha _{2},\cdots ,\alpha _{n}\) not all zero, such that \({\displaystyle \sum \limits _{i=1}^{n}} \alpha _{i}f_{i}\left ( t\right ) =0\) for all \(t=\left [ t_{0},t_{1}\right ] \). Otherwise they are L.I.
Examples: \(f_{1}\left ( t\right ) =-t,f_{2}\left ( t\right ) =t^{2}\) on \(\left [ 0,2\right ] \). Can we find \(\alpha _{1},\alpha _{2}\) such that \(\alpha _{1}f_{1}\left ( t\right ) +\alpha _{2}f_{2}\left ( t\right ) =0\). No. So L.I., notice that the same \(\alpha \) has to be used for all \(t\).
Reader: Show that \(f_{1}\left ( t\right ) =\left [ 1,t\right ] ,f_{2}\left ( t\right ) =\left [ t^{2},t^{3}\right ] \) is L.I. on \(\left [ -2,2\right ] \)
Reader: Show \(\left [ 1,t,t^{2},\cdots ,t^{n}\right ] \) are L.I. on any \(\left [ t_{0},t_{1}\right ] \) with \(t_{1}>t_{0}\)
Reader: Show \(\left [ e^{t},e^{-t},e^{3t},\cdots \right ] \) are L.I. on any \(\left [ t_{0},t_{1}\right ] \) with \(t_{1}>t_{0}\)
Reader: Does being L.I. on \(\left [ t_{0},t_{1}\right ] \,\ \)implies L.I. on \(\left [ t_{0}^{\prime },t_{1}^{\prime }\right ] \) where \(\left [ t_{0}^{\prime },t_{1}^{\prime }\right ] \supseteq \left [ t_{0},t_{1}\right ] \)? Yes. What about if \(\left [ t_{0}^{\prime },t_{1}^{\prime }\right ] \sqsubseteq \left [ t_{0},t_{1}\right ] \)? NO. Not necessarily.
Theorem: Given \(f_{i}\) and \(\left [ t_{0},t_{1}\right ] \), define matrix \(F\left ( t\right ) =\begin{pmatrix} f_{1}\left ( t\right ) \\ f_{2}\left ( t\right ) \\ \vdots \\ f_{n}\left ( t\right ) \end{pmatrix} \), so \(F\left ( t\right ) \) is \(n\times p\,\) size matrix. Now define the Gramian \[ W_{\left [ t_{0},t_{1}\right ] }={\displaystyle \int \limits _{t_{0}}^{t_{1}}} F\left ( t\right ) F^{T}\left ( t\right ) dt \] Then \(f_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \) iff \(W\left [ t_{0},t_{1}\right ] \) is not singular. Proof:
Necessity: Assume \(f_{i}\) are L.I. Show \(\Im \) not singular. Proof by contradiction. Assume \(W\) singular then \(\Im \vec{\alpha }=0\) for non-zero \(\vec{\alpha }\) vector. Also \(\vec{\alpha }^{T}\Im \vec{\alpha }=0\). Hence \({\displaystyle \int } \vec{\alpha }^{T}W\vec{\alpha }dt=0\) or \({\displaystyle \int } \vec{\alpha }^{T}F\left ( t\right ) F^{T}\vec{\alpha }dt\). Let \(F^{T}\vec{\alpha }=\xi \left ( t\right ) \) and \(\vec{\alpha }^{T}F\left ( t\right ) =\xi ^{T}\left ( t\right ) \). Then we have \({\displaystyle \int \limits _{t_{0}}^{t_{1}}}{\displaystyle \sum \limits _{i=1}^{n}} \xi _{i}^{2}\left ( t\right ) dt=0\) which implies \(\xi \left ( t\right ) =0\) identically. But this means \(F^{T}\vec{\alpha }=0\), which means \(f_{i}\) are L.I. But this is contradiction to assumption. Hence \(f_{i}\) are L.I. implies \(W\) not singular.
Now to proof the sufficiency: Assume \(\Im \) not singular, show \(f_{i}\) are L.I. Proof by contradiction. Assume \(f_{i}\) are L.D., then \(\vec{\alpha }\) \(\ \)exist such that \(\vec{\alpha }^{T}F\left ( t\right ) =0\) which implies \(\vec{\alpha }^{T}W=0\). But this means \(\Im \) is singular. Which is contradiction. This complete the proof that \(f_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \) iff \(W\left [ t_{0},t_{1}\right ] \) is not singular.