We already talked about algebraic observability (observability matrix). We also talked about the observer design. Now we want to talk about physical observability. Starting with LTV system
\begin{align*} x^{\prime }\left ( t\right ) & =A\left ( t\right ) x\left ( t\right ) +B\left ( t\right ) u\left ( t\right ) \\ y\left ( t\right ) & =C\left ( t\right ) x\left ( t\right ) +D\left ( t\right ) u\left ( t\right ) \end{align*}
Formal definition of physical observability: System is observable at t_{0} if the following condition holds: With x\left ( t_{0}\right ) =x^{0} unknown, suppose u\left ( t\right ) and y\left ( t\right ) are known, then there exist time t_{1}\geq t_{0} such that x\left ( t_{0}\right ) can be determined from knowing u\left ( t\right ) and y\left ( t\right ) over \left [ t_{0},t_{1}\right ] . This is true for any x\left ( t_{0}\right ) .
Definition from Chen book: pair \left ( A\left ( t\right ) ,C\left ( t\right ) \right ) is observable at t_{0} iff there exists time t_{1}>t_{0} such that the n\times n matrix W_{o}\left ( t_{0},t_{1}\right ) ={\displaystyle \int \limits _{t_{0}}^{t}} \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) C\left ( \tau \right ) \Phi \left ( \tau ,t_{0}\right ) d\tau where \Phi \left ( t,\tau \right ) \,\ is the state transition matrix for x^{\prime }\left ( t\right ) =A\left ( t\right ) x\left ( t\right ) is nonsingular.
Controllability of the primal system equals the observability of the duel. If system is observable at t_{0} what about at t>t_{0}? The answer is not necessarily. What about for t<t_{0}? The answer is yes. We can always solve for x\left ( 0\right ) \begin{align*} y(t) & = C(t) \Phi (t,t_0) x^0+ \overbrace{C\left ( t\right ){\displaystyle \int \limits _{t_{0}}^{t}} \Phi \left ( t,\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau +D\left ( t\right ) u\left ( t\right ) }^{\text{not function of }x^{0}\text{ can be ignored}}\\ \tilde{y}(t) & = C(t) \Phi (t,t_0) x^0 \end{align*}
Pre-multiply by \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) and integrate{\displaystyle \int \limits _{t_{0}}^{t}} \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) \tilde{y}\left ( t\right ) d\tau =\overbrace{\left ({\displaystyle \int \limits _{t_{0}}^{t}} \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) C\left ( \tau \right ) \Phi \left ( \tau ,t_{0}\right ) d\tau \right ) }^{W\left ( t_{0},t\right ) }x^{0} Since observable we can invert W, hence x^{0}=W^{-1}\left ( t_{0},t\right ){\displaystyle \int \limits _{t_{0}}^{t}} \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) \tilde{y}\left ( t\right ) d\tau So system is observable iff W is non-singular for t>t_{0}. This means rows of \Phi ^{T}\left ( \tau ,t_{0}\right ) C^{T}\left ( \tau \right ) are linearly independent.
Reader express state transition matrix for the duel system \left ( \tilde{A},\tilde{B},\tilde{C},\tilde{D}\right ) where \begin{align*} \tilde{A} & =-A^{T}\\ \tilde{B} & =-C^{T}\\ \tilde{C} & =B^{T}\\ \tilde{D} & =D \end{align*}
Hence, the duel system\begin{align*} \tilde{x}^{\prime }\left ( t\right ) & =-A^{T}\left ( t\right ) \tilde{x}\left ( t\right ) -C^{T}\left ( t\right ) u\left ( t\right ) \\ \tilde{y}\left ( t\right ) & =B^{T}\left ( t\right ) \tilde{x}\left ( t\right ) +D^{T}\left ( t\right ) u\left ( t\right ) \end{align*}
Now we will establish the relation between the system transfer matrix \Phi for the primal and the duel. For the primal, we have \frac{d\Psi ^{-1}\left ( t\right ) }{dt}=-\Psi ^{-1}\left ( t\right ) A\left ( t\right ) Take the transpose of both sides gives\begin{equation} \frac{d\left [ \Psi ^{-1}\left ( t\right ) \right ] ^{T}}{dt}=-A^{T}\left ( t\right ) \left [ \Psi ^{-1}\left ( t\right ) \right ] ^{T} \tag{1} \end{equation} Now for the duel, we have \frac{d\tilde{\Psi }\left ( t\right ) }{dt}=\tilde{A}\left ( t\right ) \tilde{\Psi }\left ( t\right ) However \tilde{A}\left ( t\right ) =-A^{T}\left ( t\right ) therefore the above becomes\begin{equation} \frac{d\tilde{\Psi }\left ( t\right ) }{dt}=-A^{T}\left ( t\right ) \tilde{\Psi }\left ( t\right ) \tag{2} \end{equation} Comparing LHS and RHS of (1) and (2) we see that\begin{equation} \left [ \Psi ^{-1}\left ( t\right ) \right ] ^{T}=\tilde{\Psi }\left ( t\right ) \tag{3} \end{equation} The above also mean that \left [ \Psi ^{T}\left ( t\right ) \right ] ^{-1}=\tilde{\Psi }\left ( t\right ) Now taking the inverse of both sides\begin{equation} \Psi ^{T}\left ( t\right ) =\tilde{\Psi }^{-1}\left ( t\right ) \tag{4} \end{equation} (3) and (4) is all what we need to establish that \begin{align*} \tilde{\Phi }\left ( t,\tau \right ) & =\tilde{\Psi }\left ( t\right ) \tilde{\Psi }^{-1}\left ( \tau \right ) \\ & =\left [ \Psi ^{T}\left ( t\right ) \right ] ^{-1}\Psi ^{T}\left ( \tau \right ) \\ & =\left [ \Psi ^{-1}\left ( t\right ) \right ] ^{T}\Psi ^{T}\left ( \tau \right ) \\ & =\left [ \Psi \left ( \tau \right ) \Psi \left ( t\right ) ^{-1}\right ] ^{T}\\ & =\Phi ^{T}\left ( \tau ,t\right ) \end{align*}
Hence\begin{equation} \fbox{$\tilde{\Phi }\left ( t,\tau \right ) =\Phi ^T\left ( \tau ,t\right ) $} \tag{5} \end{equation} Now we describe the short cut method to determine observability. For n-1 differential system\begin{align*} L_{0}\left ( t\right ) & =C\left ( t\right ) \\ L_{k+1}\left ( t\right ) & =L_{k}\left ( t\right ) A\left ( t\right ) +\frac{d}{dt}L_{k}\left ( t\right ) \end{align*}
For k=0\cdots n-2. Then the system is observable at t_{0} iff there exist t_{1}>t_{0} such that \rho \left [ L\left ( t\right ) \right ] =\rho \begin{pmatrix} L_{0}\left ( t_{1}\right ) \\ L_{1}\left ( t_{1}\right ) \\ \vdots \\ L_{n-1}\left ( t_{1}\right ) \end{pmatrix} =n
Example: A\left ( t\right ) =\begin{pmatrix} 0 & 1 & t\\ t^{2} & -t & e^{t}\\ 1 & -2 & 1 \end{pmatrix} ,C\left ( t\right ) =\begin{pmatrix} 1 & t & e^{-t}\end{pmatrix} \begin{align*} L_{0}\left ( t\right ) & =\begin{pmatrix} 1 & t & e^{-t}\end{pmatrix} \\ L_{1}\left ( t\right ) & =\begin{pmatrix} 1 & t & e^{-t}\end{pmatrix}\begin{pmatrix} 0 & 1 & t\\ t^{2} & -t & e^{t}\\ 1 & -2 & 1 \end{pmatrix} +\begin{pmatrix} 0 & 1 & -e^{-t}\end{pmatrix} =\begin{pmatrix} e^{-t}+t^{3} & 2-t^{2}-2e^{-t} & t+te^{t}\end{pmatrix} \\ L_{2}\left ( t\right ) & =\begin{pmatrix} e^{-t}+t^{3} & 2-t^{2}-2e^{-t} & t+te^{t}\end{pmatrix}\begin{pmatrix} 0 & 1 & t\\ t^{2} & -t & e^{t}\\ 1 & -2 & 1 \end{pmatrix} +\begin{pmatrix} -e^{-t}+3t^{2} & -2t+2e^{-t} & 1+e^{t}+te^{t}\end{pmatrix} =\\ & \begin{pmatrix} t-e^{-t}-t^{2}\left ( 2e^{-t}+t^{2}-2\right ) +te^{t}+3t^{2} & 3e^{-t}-4t+t\left ( 2e^{-t}+t^{2}-2\right ) -2te^{t}+t^{3} & t+e^{t}+t\left ( e^{-t}+t^{3}\right ) -e^{t}\left ( 2e^{-t}+t^{2}-2\right ) +2te^{t}+1 \end{pmatrix} \end{align*}
Hence L=\begin{pmatrix} 1 & t & e^{-t}\\ e^{-t}+t^{3} & 2-t^{2}-2e^{-t} & t+te^{t}\\ t-e^{-t}-t^{2}\left ( 2e^{-t}+t^{2}-2\right ) +te^{t}+3t^{2} & 3e^{-t}-4t+t\left ( 2e^{-t}+t^{2}-2\right ) -2te^{t}+t^{3} & t+e^{t}+t\left ( e^{-t}+t^{3}\right ) -e^{t}\left ( 2e^{-t}+t^{2}-2\right ) +2te^{t}+1 \end{pmatrix} The rank of the above must be 3 for observable system.
Canonical decomposition theorem:
We want minimal realization. Canonical transformation: System {\displaystyle \sum } =\left ( A,B,C,D\right ) can be transformed to equivalent system via non-singular matrix T.\begin{align*} A^{\ast } & =TAT^{-1}\\ B^{\ast } & =TB\\ C^{\ast } & =CT^{-1}\\ D^{\ast } & =D \end{align*}
Which has the following structure. A^{\ast } is block structure of the form A^{\ast }=\begin{pmatrix} A_{c\bar{o}}^{\ast } & A_{12}^{\ast } & A_{13}^{\ast }\\ 0 & A_{co}^{\ast } & A_{23}^{\ast }\\ 0 & 0 & A_{\bar{c}}^{\ast }\end{pmatrix} ,B^{\ast }=\begin{pmatrix} B_{c\bar{o}}^{\ast }\\ B_{co}^{\ast }\\ 0 \end{pmatrix} ,C^{\ast }=\begin{pmatrix} 0 & C_{co}^{\ast } & C_{\bar{c}}^{\ast }\end{pmatrix} A_{c\bar{o}}^{\ast } means controllable but not observable, A_{co}^{\ast } means controllable and observable and A_{\bar{c}}^{\ast } means not controllable.
Hence, since x^{\ast }=Tx, then the {\displaystyle \sum ^{\ast }} is\begin{align*} x^{\ast \prime } & =A^{\ast }x^{\ast }+B^{\ast }u\left ( t\right ) \\ y & =C^{\ast }x^{\ast }+D^{\ast }u\left ( t\right ) \end{align*}
So the controllable and observable part is {\displaystyle \sum _{co}} =\left ( A_{co}^{\ast },B_{co}^{\ast },C_{co}^{\ast }\right ) . And The not controllable and not observable part is {\displaystyle \sum _{\bar{c}\bar{o}}} =\left ( A_{\bar{c}\bar{o}}^{\ast },0,C_{\bar{c}}^{\ast }\right ) . Notice that H_{co}^{\ast }\left ( s\right ) =H\left ( s\right ) Minimal realization is both controllable and observable.
Original A,B,C,D can be anything, and we can always find T to do the above transformation.