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\section{State space representation}
\begin{figure}[h]
\centering
\resizebox{\linewidth}{!}{
\input{tex/tikz/mimo}
}
\caption{Diagram of a LTI MIMO system with vector variables.}
\end{figure}
A system described by a system of linear differential equations of \(n\)-th order, can be equivalently be described by \(n\) first order differential equations. Which can be compactly written in matrix form as
\begin{align*}
\dot{\vec{x}} &= \mx{A}\vec{x} + \mx{B}\vec{u} \\
\vec{y} &= \mx{C}\vec{x} + \mx{D}\vec{u}.
\end{align*}
If the system is time \emph{variant} the matrices are functions of time.
\begin{table}
\centering
\begin{tabular}{ >{\(}c<{\)} >{\(}c<{\)} l }
\toprule
\text{\bfseries Symbol} & \text{\bfseries Size} & \bfseries Name \\
\midrule
\vec{x} & n & State vector \\
\vec{u} & m & Output vector \\
\vec{y} & k & Output vector \\
\midrule
\mx{A} & n\times n & System matrix \\
\mx{B} & m\times n & Input matrix \\
\mx{C} & n\times k & Output matrix \\
\mx{D} & k\times m & Feed forward matrix \\
\bottomrule
\end{tabular}
\caption{Matrices for a state space representation}
\end{table}
\subsection{Canonical representations}
\subsubsection{Controllable form}
\subsubsection{Observable form}
\subsubsection{Diagonalized or Jordan form}
The Jordan form diagonalizes the \(\mx{A}\) matrix. Thus we need to solve the eigenvalue problem \((\mx{A} - \lambda\mx{I})\vec{x} = \vec{0}\), which can be done by setting \(\det(\mx{A} -\lambda\mx{I}) = 0\), and solving the characteristic polynomial. The eigenvectors are obtained by plugging the \(\lambda\) values back into \((\mx{A} - \lambda\mx{I})\vec{x} = \vec{0}\), and solving an overdetermined system of equations. Tip: for a \(2\times2\) matrix \(\mx{A}\) the eigenvalues can be quickly calculated with
\[
m = \frac{1}{2} \tr \mx{A} = \frac{a + d}{2}, \quad
p = \det \mx{A} = ad - bc
\]
and then \(\lambda_{1,2} = m \pm \sqrt{m^2 - p}\).
The transformation to the eigenbasis \(\mx{T}\), obtained by using the eigenvector as columns of a matrix \(\mx{T} = \begin{bmatrix} \vec{v}_1 & \cdots & \vec{v}_n \end{bmatrix}\), is then used to compute
\begin{align*}
\mx{\hat{A}} & = \mx{T}\mx{A}\mx{T^{-1}} &
\mx{\hat{B}} & = \mx{T}\mx{B} \\
\mx{\hat{C}} & = \mx{C}\mx{T^{-1}} &
\mx{\hat{D}} & = \mx{D}.
\end{align*}
In this form the system is described with \(n\) decoupled states \(\xi_i\) with the equations \(\dot{\vec{\xi}} = \mx{\hat{A}}\vec{\xi} + \mx{\hat{B}}\vec{u}\) and \(\vec{y} = \mx{\hat{C}}\vec{\xi} + \mx{\hat{D}} \vec{u}\).
\subsection{Stability}
If \emph{all} eigenvalues \(\lambda\) are not zero and have a positive real part the system is asymptotically \emph{stable}. If \emph{all} eigenvalues are not zero but \emph{at least one} has a negative real part the system is \emph{unstable}.
\subsection{Controllability}
The state controllability condition implies that it is possible --- by admissible inputs --- to steer the states from any initial value to any final value within some finite time window. A LTI state space model is controllable iff the matrix
\[
\mx{Q} = \begin{bmatrix}
\mx{B} & \mx{A}\mx{B} & \mx{A}^2\mx{B} \cdots \mx{A}^{n-1}\mx{B}
\end{bmatrix}
\]
has \(\rank\mx{Q} = n\). For a SISO system, if all components of the vector \(\mx{\hat{B}}\) are not zero, then the system is controllable.
\subsection{Observability}
Observability is a measure for how well internal states of a system can be inferred by knowledge of its external outputs. A LTI state space mode is observable iff the matrix
\[
\mx{Q}^t = \begin{bmatrix}
\mx{C} & \mx{C}\mx{A} & \cdots & \mx{C}\mx{A}^{n-1}
\end{bmatrix}
\]
has \(\rank\mx{Q} = n\). For a SISO system it is also possible to infer observability from the diagonalized form: if all elements of the \(\mx{\hat{C}}\) are not zero, then the system is observable.
\subsection{Solutions in time domain}
%% TODO: solutions in the time domain
\subsection{Solutions in the \(s\)-domain}
By taking the Laplace transform of the system of differential equations we obtain
\begin{align*}
s\vec{X} - \vec{x}(0) &= \mx{A}\vec{X} + \mx{B}\vec{U} \\
\vec{Y} &= \mx{C}\vec{X} + \mx{D}\vec{U}.
\end{align*}
The first equation can be solved for \(\vec{X}\) giving
\[
\vec{X} = (s\mx{I} - A)^{-1}\left(\vec{x}(0) + \mx{B}\vec{U}\right).
\]
Substituting in the second equation results in
\[
\vec{Y} = \mx{C}(s\mx{I} - \mx{A})^{-1}\left(\vec{x}(0) + \mx{B}\vec{U}\right) + \mx{D}\vec{U}.
\]
Assuming that the initial conditions \(\vec{x}(0) = \vec{0}\), then
\[
\vec{Y} = \left(\mx{C}(s\mx{I} - \mx{A})^{-1}\mx{B} + \mx{D}\right)\vec{U}.
\]
from which can define the \emph{transfer matrix} \(\mx{H}\) to be the matrix that takes \(\vec{Y}\) to \(\vec{U}\), i.e.
\[
\vec{H} = \mx{C}(s\mx{I} - \mx{A})^{-1}\mx{B} + \mx{D},
\]
that we can use to compute \(y = \laplace^{-1}\left\{\mx{H}\laplace u\right\}\).
In the special case of a SISO system the transfer matrix \(\mx{H}\) is one dimensional and exactly equal to the transfer function \(H\).
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