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Lecture 10: Special Transformation: Diagonalizing

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Schedule :: Perspective :: Transfer Functions :: Laplace Transform :: Poles and Eigenvalues :: Time Domain :: Linear Transformations :: Special Transformation :: Controllability :: Discrete-Time

For the state equations

\begin{displaymath}\xbfdot = A \xbf + B \fbf

\begin{displaymath}\ybf = C \xbf + D \fbf

let us create a new variable $\wbf = P \xbf$ for an invertible matrix $P$ . Then $\xbf = P^{-1} \wbf$ , and $\xbfdot = P^{-1}\wbfdot$ . Substituting we find

\begin{displaymath}P^{-1} \wbfdot = AP^{-1} \wbf + B \fbf


\begin{displaymath}\wbfdot = PAP^{-1} \wbf + PB \fbf = \Ahat \wbf + \Bhat \fbf


\begin{displaymath}\Ahat =PAP^{-1} \qquad \Bhat = PB.


\begin{displaymath}\ybf = \Chat \wbf + \Dhat \fbf


\begin{displaymath}\Chat = CP^{-1} \qquad \Dhat = D.

Instead of $(A,B,C,D)$ we have $(\Ahat,\Bhat,\Chat,\Dhat)$ .

Do these represent the same system?

\begin{displaymath}H(s) = C(sI-A)^{-1}B + D

\begin{displaymath}\Hhat(s) = \Chat(sI-\Ahat)^{-1}\Bhat + \Dhat.

(Work through details.)

Other observations: eigenvalues? Eigenvectors?

A special transformation: diagonalizing $A$

Given $\Ahat = PAP^{-1}$ , suppose that we want to find a transformation matrix $P$ such that $\Ahat$ is diagonal. (This is a convenient form, since it ``decouples'' all the modes.) How can we find such a $P$ ?

Let $\ebf_i$ be eigenvectors of $A$ , and $\lambda_i$ be the eigenvalues of $A$ , assumed (for our purposes) to be unique. Form

\begin{displaymath}Q = \begin{bmatrix}\ebf_1 & \ebf_2 & \cdots & \ebf_n \end{bmatrix}\end{displaymath}


\begin{displaymath}AQ = Q \Lambda

\begin{displaymath}\Lambda = \diag(\lambda_1, \lambda_2,\ldots, \lambda_n).

\begin{displaymath}\Lambda = Q^{-1}A Q

Identify: $\Lambda = \Ahat$ , $P = Q^{-1}$ .
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, June 08). Lecture 10: Special Transformation: Diagonalizing. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: This work is licensed under a Creative Commons License Creative Commons License