Went over second exam: Meaning of uniform convergence. Many had trouble with part (c) of first problem. For problem three use the commute property. Much easier that calculus.
Now back to lecture. We said before that
\begin{equation} \Im \left ( t\right ) =\begin{pmatrix} F\left ( t\right ) & F^{\prime }\left ( t\right ) & \cdots & F^{\left ( n-1\right ) }\left ( t\right ) & \cdots \end{pmatrix} \tag{1} \end{equation}
We were talking about Linear independence and we had these functions F\left ( t\right ) . We had the extra condition that they are analytic and wanted to check of there linearly independent on \left [ t_{0},t_{1}\right ] . Yes they are iff rank \Im \left ( t\right ) =n for all t\in \left [ t_{0},t_{1}\right ] . This was the stepping stone to physical controllability of LTI system. For LTI the Gramian matrix W\left ( t_{0},t\right ) simplifies to the controllability matrix \mathbb{C} =\begin{bmatrix} B & AB & \cdots & A^{n-1}B \end{bmatrix} .
Reader: If LTI system is controllable at t_{0} then it is controllable for every t. So in LTI, we do not need to keep saying at t_{0} and we drop it, and just saying that LTI is controllable, period. This implies for any t. Proof: Suppose \left ( A,B\right ) is controllable at t_{0}=0,. we need to show it is controllable at t_{0}^{\prime }>t_{0}. Argument: Since it is controllable at t_{0} we can find u\left ( t\right ) to steer the system to x\left ( t_{0}^{\prime }\right ) . Now shift u\left ( t\right ) by t_{0}^{\prime }, hence u\left ( t-t_{0}^{\prime }\right ) is applied again to show it will take the system from x\left ( t_{0}^{\prime }\right ) to x\left ( t_{1}\right ) . Note: I am not sure I follow this argument. Need to check with the prof. on this. I do not understand how applying u\left ( t-t_{0}^{\prime }\right ) makes it controllable at t_{0}^{\prime }.
Now we build \Im \left ( t\right ) for LTI. Instead of using \Phi \left ( t_{0},t\right ) B\left ( t\right ) for the F\left ( t\right ) functions, we now use e^{-At}B, since LTI. Hence (1) becomes\begin{equation} \Im \left ( t\right ) =\begin{pmatrix} e^{-At}B & -e^{-At}AB & e^{-At}A^{2}B & \cdots & \left ( -1\right ) ^{n}e^{-At}A^{n}B & \cdots \end{pmatrix} \tag{2} \end{equation}
The system is controllable iff rank \Im \left ( t\right ) =n Since e^{At} is non-singular, we factor it out
\Im \left ( t\right ) =e^{-At}\begin{pmatrix} B & -AB & A^{2}B & \cdots & \left ( -1\right ) ^{n}A^{n}B & \cdots \end{pmatrix}
The sign do not affect the rank, so we make all the signs the same \Im \left ( t\right ) =e^{-At}\begin{pmatrix} B & AB & A^{2}B & \cdots & A^{n}B & \cdots \end{pmatrix}
Example: Given A=\begin{bmatrix} 2 & 0 & 0\\ 1 & -4 & 0\\ 2 & -1 & 3 \end{bmatrix} , then \Delta \left ( \lambda \right ) =\det \left ( \lambda I-A\right ) =\left ( \lambda -2\right ) \left ( \lambda +4\right ) \left ( \lambda -4\right ) =\lambda ^{3}-\lambda ^{2}-14\lambda +24, hence by Cayley Hamilton we have A^{3}-A^{2}-14A+24I=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}
We need to show that columns from n to \infty do not contribute to rank of \Im \left ( t\right ) . This means A^{n} is linear combination of \left \{ I,A,A^{2},\cdots ,A^{n-1}\right \} . Using Cayley Hamilton applied for A^{k} we obtain that \begin{align*} 0 & =A^{n}+{\displaystyle \sum \limits _{i=0}^{n-1}} a_{i}A^{i}\\ A^{n} & ={\displaystyle \sum \limits _{i=0}^{n-1}} a_{i}A^{i} \end{align*}
Hence A^{n} is linear combination of \left \{ I,A,A^{2},\cdots ,A^{n-1}\right \} . We now do the same for A^{n+1} to show it is linear combination of A^{n} and all the other matrices, and so on. Hence all matrices A after n-1 do not contribute to rank of \Im \left ( t\right ) . This complete the proof that \Im \left ( t\right ) reduces to \mathbb{C} for LTI systems.
More on controllability: Differential controllability. We will start with LTV. We say {\displaystyle \sum } is differentially controllable at t_{0} if we can get to new state in as small time as we want. If given \varepsilon >0, arbitrary small, and any x\left ( t_{0}\right ) state, then there exist u\left ( t\right ) steering x\left ( t_{0}\right ) to x\left ( t_{0}+\varepsilon \right ) \,.
Reader: Give criteria and short cut for differential controllability at t_{0}. Use W\left ( t_{0},t_{0}+\varepsilon \right ) . For LTI, use short cut M.
For LTI: Reader: If {\displaystyle \sum } is controllable, then it is always differentially controllable. But this is not necessarily true for LTV. To show, let \begin{align*} x^{1} & =e^{At_{1}}x^{0}+{\displaystyle \int \limits _{t_{0}}^{t_{1}}} e^{\left ( t_{1}-\tau \right ) }Bu\left ( \tau \right ) d\tau \\ x^{\varepsilon } & =e^{A\varepsilon }x^{0}+{\displaystyle \int \limits _{t_{0}}^{\varepsilon }} e^{\left ( \varepsilon -\tau \right ) }B\tilde{u}\left ( \tau \right ) d\tau \end{align*}
Relate u\left ( t\right ) to \tilde{u}\left ( t\right ) so that we can get to x^{1} in as short time as we want. Note: The above is not clear to me, need to clean up.
Controllability with bounded control: i.e. we now have a bound on the magnitude of u\left ( t\right ) . There is whole theory on controllability with bounded input. Next we will do the same with observability, using duality to speed all the derivations by using results obtained from the controllability. We will later study decomposition, then stability.