Notes on first exam:
If one can put state space in controllable canonical form, then this implies it is controllable.
Lack of cancellation of poles/zero in a transfer function implies minimal system. Hence it is observable and controllable.
If \(H(s)\) is proper, then it is realizable (can obtain \(A,B,C,D\)). However, if it is not proper \(H(s)\) then we can’t decide. It might still be possible to obtain \(A,B,C,D\). Think of an example.
But if we are given \((A,B,C,D)\) then \(H(s) = C (sI-A)^{-1} B+D\) must come out to be proper by construction.
Next goal is to solve the state space equation. All our solutions live in the space of bounded functions \(B\left ( \left [t_{0},t_{1}\right ] , \Re ^{n}\right )\)
In these notes, I will use \(f^{\ast }\) for the uniform convergence limit and use \(f^{\ast }(t)\) for pointwise limit.
Uniform convergence: Lemma 1
Suppose \(f_{k}\rightarrow f^{\ast }\) convergence uniformly in \(B\), i.e. \(\left \Vert f_{k}-f^{\ast }\right \Vert \rightarrow 0\) (notice, when we write \(f^{\ast }\) and not \(f^{\ast }\left ( t\right ) \), then this means uniform convergence, then\[ \lim _{k\rightarrow \infty }{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) d\tau ={\displaystyle \int \limits _{t_{0}}^{t}} f^{\ast }\left ( \tau \right ) d\tau \] In words, if a sequence of functions converges to some limit, then the limit of the integral is the integral of the limit. This only applies for uniform converges. This does not hold (most of the time) for pointwise convergence.
Reader: proof this. Here is a case where the above fails for pointwise convergence. Note: A sequence of functions \(f_{k}\left ( t\right ) \), converges pointwise, if when we fix \(t\) to some specific \(t_{0}\), then the sequence \(f_{k}\left ( t_{0}\right ) \) converges to some limit. I.e. we have to fix \(t\) and only after that, generate the sequence and see if \(\left \vert f_{k}\left ( t_{0}\right ) \right \vert \) converges to some \(f^{\ast }\left ( t_{0}\right ) \) as \(k\rightarrow \infty \). This can be written as \(\left \vert f_{k}\left ( t_{0}\right ) -f^{\ast }\left ( t_{0}\right ) \right \vert <\epsilon \) wherever \(n>N\) where \(N\) is some integer. Given the function shown below
where in the above, \(f\left ( 0\right ) =0\) and \(f\left ( t\right ) =k\) for \(0<t\leq \frac{1}{k}.\) To find pointwise limit \(f^{\ast }\left ( t\right ) \): At \(t=0,f_{k}\left ( 0\right ) =0\), and at \(0<t\leq \frac{1}{k}\), \(f_{k}\left ( t\right ) =k\) and for \(t>\frac{1}{k}\) it is already zero. Hence as \(k\rightarrow \infty \) we see that \(f\left ( t\right ) \rightarrow 0\) everywhere. So \(f^{\ast }=0\) is the pointwise limit.
Back to the proof of the lemma above for uniform convergence.
proof of lemma:\begin{align*} error\left ( k\right ) & =\left \Vert{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) d\tau -{\displaystyle \int \limits _{t_{0}}^{t}} f^{\ast }\left ( \tau \right ) d\tau \right \Vert \\ & =\left \Vert{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) d\tau \right \Vert \\ & \leq{\displaystyle \int \limits _{t_{0}}^{t}} \left \Vert f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) \right \Vert d\tau \end{align*}
Since uniform convergence. But the above is\[{\displaystyle \int \limits _{t_{0}}^{t}} \left \Vert f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) \right \Vert d\tau \leq{\displaystyle \int \limits _{t_{0}}^{t}} \max \left \Vert f_{k}\left ( t\right ) -f^{\ast }\left ( t\right ) \right \Vert d\tau \] But now \(\max \left \Vert f_{k}\left ( t\right ) -f^{\ast }\left ( t\right ) \right \Vert \) is fixed, so we can take it out of the integral\begin{align*}{\displaystyle \int \limits _{t_{0}}^{t}} \left \Vert f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) \right \Vert d\tau & \leq \max \left \Vert f_{k}-f^{\ast }\right \Vert{\displaystyle \int \limits _{t_{0}}^{t}} d\tau \\{\displaystyle \int \limits _{t_{0}}^{t}} \left \Vert f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) \right \Vert d\tau & \leq \max \left \Vert f_{k}-f^{\ast }\right \Vert \left ( t-t_{0}\right ) \end{align*}
But since we assumed \(f_{k}\left ( t\right ) \) convergence uniformly then \(\lim _{k\rightarrow \infty }\left \Vert f_{k}-f^{\ast }\right \Vert =\max \left \Vert f_{k}-f^{\ast }\right \Vert =0,\) therefore RHS above is zero. Hence \({\displaystyle \int \limits _{t_{0}}^{t}} \left \Vert f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) \right \Vert d\tau =0\) or \(\left \Vert{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) d\tau \right \Vert =0\) or \({\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) -f^{\ast }\left ( \tau \right ) d\tau =0\) or \(\lim _{k\rightarrow \infty }{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) d\tau -{\displaystyle \int \limits _{t_{0}}^{t}} f^{\ast }\left ( \tau \right ) d\tau =0\) or\[ \lim _{k\rightarrow \infty }{\displaystyle \int \limits _{t_{0}}^{t}} f_{k}\left ( \tau \right ) d\tau ={\displaystyle \int \limits _{t_{0}}^{t}} f^{\ast }\left ( \tau \right ) d\tau \] Which is the lemma we wanted to proof.
Series in bounded spaces:
We will look at Series in \(B\left ( \left [ t_{0},t_{1}\right ] ,\Re ^{n}\right ) \)
We now look at series of functions \(f_{k}\left ( t\right ) \) in \(B\). And use Weierstrass M-test to see if the series converges or not. Looking at partial sum\[ S\left ( k\right ) ={\displaystyle \sum \limits _{i=1}^{k}} f_{i}\left ( t\right ) \] where \(f_{i}\left ( t\right ) \in B\). Does this converge? If we can find constants \(M_{i}\) s.t. \(\left \Vert f_{i}\right \Vert \leq M_{i}\), (\(M_{i}\) can for example be the \(\sup \) norm of \(f_{i}\left ( t\right ) \)), and if then we can determine that \({\displaystyle \sum \limits _{i=1}^{\infty }} M_{i}<\infty \) then we say that \(S\left ( k\right ) \underset{\text{uniform}}{\longrightarrow }S^{\ast }\in B\)
(need an example, see references)
Solving state space:
Now we start talking about solving the state space equation \(x^{\prime }=A\left ( t\right ) x\left ( t\right ) \). We start with the zero input case. Only initial conditions will drive this system. We look at using Picard iterations to solve it. By integration both sides of the above, we obtain\[ x\left ( t\right ) -x\left ( 0\right ) ={\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x\left ( \tau \right ) d\tau \] Define \(x^{0}=x\left ( 0\right ) \) and define this iteration scheme for \(k=0,1,2,\cdots \)\[ x^{k+1}=x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \] Reader:
For \(A\left ( t\right ) =\begin{pmatrix} 1 & t & -t^{2}\\ 0 & t+1 & t^{3}\\ 0 & 1 & -2 \end{pmatrix} \) and \(x^{0}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} \) find \(x^{3}\left ( t\right ) \)
Done in class. Direct integration.
Reader:
For scalar \(x^{\prime }=ax\) show that Picard iteration gives \(x=e^{at}x\left ( 0\right ) \).
Proof: \begin{align*} x^{1} & =x^{0}+{\displaystyle \int \limits _{0}^{t}} ax^{0}d\tau =x^{0}+ax^{0}t\\ x^{2} & =x^{0}+{\displaystyle \int \limits _{0}^{t}} ax^{1}d\tau \\ & =x^{0}+{\displaystyle \int \limits _{0}^{t}} a\left ( x^{0}+ax^{0}t\right ) d\tau \\ & =x^{0}+ax^{0}t+a^{2}x^{0}\frac{t^{2}}{2}\\ x^{3} & =x^{0}+{\displaystyle \int \limits _{0}^{t}} ax^{2}d\tau \\ & =x^{0}+{\displaystyle \int \limits _{0}^{t}} a\left ( x^{0}+ax^{0}t+a^{2}x^{0}\frac{t^{2}}{2}\right ) d\tau \\ & =x^{0}+ax^{0}t+a^{2}x^{0}\frac{t^{2}}{2}+a^{3}x^{0}\frac{t^{3}}{2\times 3} \end{align*}
and so on. Hence the result is \begin{align*} x^{\infty } & =x\left ( t\right ) =x^{0}\left ( 1+at+a^{2}\frac{t^{2}}{2}+a^{3}\frac{t^{3}}{3!}+\cdots \right ) \\ & =x\left ( 0\right ){\displaystyle \sum \limits _{k=0}^{\infty }} \frac{\left ( at\right ) ^{k}}{k!}\\ & =x\left ( 0\right ) e^{at} \end{align*}
Reader
Show the solution for scalar time varying \(x^{\prime }=a\left ( t\right ) x\) using Picard. This should become \(x\left ( t\right ) =x\left ( 0\right ){\displaystyle \sum \limits _{k=0}^{\infty }} \frac{1}{k!}\left ({\displaystyle \int \limits _{0}^{t}} a\left ( td\tau \right ) \right ) ^{k}\)
(need to work it out).
Convergence of Picard iterate \(x^{k}(t)\):
Helpful function is \(\Pi \left ( t\right ) ={\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert d\tau \). It has the following properties
The above is reader, need to show.
Lemma 1:
The Picard iterate \(x^{k}\left ( t\right ) \in B\left ( \left [ t_{0},t_{1}\right ] ,\mathbb{R} ^{n}\right ) \) for all \(k\). We want to show that for each \(k\), \(x^{\left ( k\right ) }\left ( t\right ) \) is bounded. This is done using induction.
proof:
For \(k=0\), it is clear that \(x^{0}=x\left ( 0\right ) \) is bounded, since initial conditions. Now assume that for some \(k\) it is true that \(x^{k}\left ( t\right ) \) is bounded, then we need to show that for \(k+1\) it is also bounded. Form \begin{align*} x^{k+1}\left ( t\right ) & =x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \\ \left \Vert x^{k+1}\left ( t\right ) \right \Vert & =\left \Vert x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \right \Vert \\ & \leq \left \Vert x^{0}\right \Vert +\left \Vert{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \right \Vert \\ & \leq \left \Vert x^{0}\right \Vert +{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) x^{k}\left ( \tau \right ) \right \Vert d\tau \\ & \leq \left \Vert x^{0}\right \Vert +{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \left \Vert x^{k}\left ( \tau \right ) \right \Vert d\tau \\ & \leq \left \Vert x^{0}\right \Vert +{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \left ( \sup \left \Vert x^{k}\left ( \tau \right ) \right \Vert \right ) d\tau \end{align*}
Since \(\sup \left \Vert x^{k}\left ( t\right ) \right \Vert \) is fixed, then we can remove it out of the integral\[ \left \Vert x^{k+1}\left ( t\right ) \right \Vert \leq \left \Vert x^{0}\right \Vert +\sup \left \Vert x^{k}\left ( t\right ) \right \Vert{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert d\tau \] But \({\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert d\tau =\pi \left ( t\right ) \) which is non-decreasing. So it was take its maximum value \(\pi \left ( t_{1}\right ) \) we can limit the above from below and write\[ \left \Vert x^{k+1}\left ( t\right ) \right \Vert \leq \left \Vert x^{0}\right \Vert +\sup \left \Vert x^{k}\left ( t\right ) \right \Vert \Pi \left ( t_{1}\right ) \] Therefore we just showed that \(x^{k+1}\left ( t\right ) \) is bounded. Since \(\sup \left \Vert x^{k}\left ( t\right ) \right \Vert \) is bounded by assumption.