From 08522b830cd1dd1d0d1d5a060574f50f388205a7 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Thu, 19 Aug 2021 18:46:28 +0200 Subject: Continue working --- tex/state-space.tex | 44 ++++++++++++++++++++++++++++++++++++++------ 1 file changed, 38 insertions(+), 6 deletions(-) (limited to 'tex/state-space.tex') diff --git a/tex/state-space.tex b/tex/state-space.tex index fadc6dd..3923b9c 100644 --- a/tex/state-space.tex +++ b/tex/state-space.tex @@ -1,11 +1,11 @@ \section{State space representation} -\begin{figure} +\begin{figure}[h] \centering \resizebox{\linewidth}{!}{ \input{tex/tikz/mimo} } - \caption{A LTI MIMO system.} + \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 @@ -41,7 +41,12 @@ If the system is time \emph{variant} the matrices are functions of time. \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. +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*} @@ -63,7 +68,7 @@ The state controllability condition implies that it is possible --- by admissibl \mx{B} & \mx{A}\mx{B} & \mx{A}^2\mx{B} \cdots \mx{A}^{n-1}\mx{B} \end{bmatrix} \] -has \(\rank\mx{Q} = n\). Or equivalently for a SISO system, if all components of the vector \(\mx{\hat{C}}_{i} \neq 0\). +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 @@ -72,8 +77,35 @@ Observability is a measure for how well internal states of a system can be infer \mx{C} & \mx{C}\mx{A} & \cdots & \mx{C}\mx{A}^{n-1} \end{bmatrix} \] -has \(\rank\mx{Q} = n\). +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} -\subsection{Solutions in frequency 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 \(H\). -- cgit v1.2.1