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Lecture 10: Laplace Transform of State Equations

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

When we talk about the Laplace transform of a vector, we will mean to apply the transform element by element. Thus, if

\begin{displaymath}\xbf(t) = \begin{bmatrix}x_1(t) \ x_2(t) \end{bmatrix}\end{displaymath}

\begin{displaymath}\Lc[\xbf(t)] = \begin{bmatrix}\Lc[x_1(t)] \ \Lc[x_2(t)]
\end{bmatrix} = \Xbf(s).

We find then that
\begin{displaymath}\Lc[\xbfdot(t)] = s \Xbf(s) - \xbf(0).

From the state equation $\xbfdot = A \xbf + B \fbf$ we obtain
\begin{displaymath}s \Xbf(s) - \xbf(0) = A \Xbf(s) + B \Fbf(s)

and from the output equation,

\begin{displaymath}\Ybf(s) = C\Xbf(s) + D \Fbf(s).

Let us solve for $\Xbf(s)$ from the first:

\begin{displaymath}(sI-A)\Xbf(s) = \xbf(0) + B\Fbf(s)

(Why the identity?) Watch the order!

\begin{displaymath}\Xbf(s) = (sI-A)^{-1}[\xbf(0) + B\Fbf(s)].

Let $\Phi(s) = (sI - A)^{-1}$ . We have

\begin{displaymath}\Xbf(s) = \Phi(s)[ \xbf(0) + \Phi(s) B \Fbf(s).]

Inverse transform:
\begin{displaymath}\xbf(t) = \Lc^{-1}[\Phi(s) \xbf(0)] + \Lc^{-1}[\Phi(s) B \Fbf(s)].

Identify zero-input components and zero-state components.


\begin{displaymath}Y(s) = C\Phi(s) \xbf(0) + [C \Phi(s) B + D] F(s)
Transfer function:
\begin{displaymath}H(s) = C \Phi(s) B + D

\begin{bmatrix}\xdot_1 \ \xdot_2\end{bmatri...
... & \frac{2(s^2+4s+2)}{(s+1)(s+2)}
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, June 08). Lecture 10: Laplace Transform of State Equations. 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