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Lecture 1: Zero-Input Response

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Schedule :: Intro :: Signal Models for Discrete-Time :: Signal Operations :: Difference Equations :: Zero-Input :: Zero-State :: Natural & Forced :: System Stability

When there is no input, we can write
\begin{displaymath}Q[E] y_0[k] = 0 \end{displaymath}
or
\begin{displaymath}(E^n + a_{n-1}E^{n-1} + \cdots + a_{1} E + a_0)y_0[k] = 0 \end{displaymath}

(What happened for continuous time?) Similar. Let's try a simply case to get started:

\begin{displaymath}(E - \gamma)y_0[k] = 0 \end{displaymath}

Try a solution $y_0[k] = c \gamma^k$ . (Comment on the difference from earlier case.) Substitute in and show that it works.
\begin{displaymath}Ec\gamma^k = c \gamma^{k+1} \end{displaymath}
This works in the general case: subs. and show that it works. Substituting gives

\begin{displaymath}(\gamma^n + a_{n-1}\gamma^{n-1} + \cdots + a_1 \gamma + a_0)y_0[k] = 0 \end{displaymath}

or
\begin{displaymath}Q[\gamma]y_0[k] = 0 \end{displaymath}
For an interesting (nontrivial) solution, we will look for roots of
\begin{displaymath}Q[\gamma] = 0 \end{displaymath}

Write as
\begin{displaymath}(\gamma - \gamma_1)(\gamma-\gamma_2)\cdots(\gamma - \gamma_n) = 0
\end{displaymath}

$Q[\gamma]$ is thus the characteristic polynomial, and we look at its roots. The roots are $\gamma_1$ , $\gamma_2,\ldots,\gamma_n$ .

As before, we take all possible solutions in a linear combination:

\begin{displaymath}y_0[n] = c_1 \gamma_1^k + c_2 \gamma_2^k + \cdots + c_n \gamma_n^k
\end{displaymath}

Repeated roots: If

\begin{displaymath}Q[\gamma] = (\gamma - \gamma_1)^r(\gamma- \gamma_{r+1}) (\gamma -
\gamma_{r+2})\cdots(\gamma - \gamma_n) \end{displaymath}
then
\begin{displaymath}y_0[k] = (c_1 + c_2 k + \cdots c_r k^{r-1}) \gamma_1^k +
c_{r+1}\gamma_{r+1}^k + \cdots + c_n \gamma_n^k \end{displaymath}

So how do things behave as a function of root location. (0.8, -0.8, -.5, 1.2, $k$ (0.8), 1, etc.) Same old, same old.
\begin{example}Solve
\begin{displaymath}y[k+2] - 0.6 y[k+1] - 0.16 y[k] = 5f[k+2...
...get the zero-state solution
later.) Is this asymptotically stable?
\end{example}

Complex roots: With roots $\gamma = \vert\gamma\vert e^{j\pm \beta}$ , then

\begin{displaymath}y[k] = c\vert\gamma\vert^k \cos(\beta k + \theta) \end{displaymath}

with $c$ and $\theta$ arbitrary constants (same steps as before).

\begin{example}
\begin{displaymath}(E^2 - 1.56 E + .81)y[k] = (E+3)f[k] \end{dis...
...aymath}\theta = -.17 \text{ rad}\quad\quad c=2.34 \end{displaymath}\end{example}
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 16). Lecture 1: Zero-Input Response. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/1_4node4.html. This work is licensed under a Creative Commons License Creative Commons License