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Lecture 10: Time Domain Solution

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

We begin by defining a new function. For a square matrix $A$ (as in the state transition matrix) we define

\begin{displaymath}e^A = (I + A + \frac{a^2}{2} + \frac{A^3}{3!} + \cdots)
\end{displaymath}

(Taylor series). This is directly analogous to $e^a$ for scalars, except that all arithmetic is done using matrices. This is computed using the exmp function in M ATLAB , not exp . Note (show this)
\begin{displaymath}\frac{d}{dt} e^{At} = Ae^{At} = e^{At}A.
\end{displaymath}

The solution to the DE

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

is given by

\begin{displaymath}\xbf(t) = e^{At} \xbf(0) + \int_0^t e^{A(t-\tau)} B f(\tau) d\tau
\end{displaymath}

Show that it works by substitution.

Computing the matrix exponential: One way:

\begin{displaymath}e^{At} = \Lc^{-1} \Phi(s) = \Lc^{-1}[(sI-A)^{-1}]
\end{displaymath}
\begin{example}
$ A = \left[\begin{smallmatrix}-12 & 2/3 \ -36 & -1
\end{small...
...2 e^{-4t} & 1.6 e^{-9t} -0.6e^{-4t}
\end{bmatrix}\end{displaymath}\end{example}
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, June 08). Lecture 10: Time Domain Solution. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/node5_1.html. This work is licensed under a Creative Commons License Creative Commons License