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Lecture 1: Zero-State Respone

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

The steps are similar to what we did before: introduce delta function, then the impulse response, then the convolution sum.

Discrete time impulse function:

\begin{displaymath}\delta[k] = \left\{\begin{array}{ll} 1 & k = 0 \ 0 & \mbox{otherwise}
\end{array}\right.
\end{displaymath}

(Plot). We have the discrete-time impulse response $h[k]$ :
\begin{displaymath}Q[E] h[k] = P[E]\delta[k] \end{displaymath}

Then $h[k]$ is the solution when initial conditions are all zero:
\begin{displaymath}h[-1] = h[-2] = \cdots = h[-n] = 0 \end{displaymath}

We can find a numerical solution by substitution.

\begin{example}$y[k] - 0.6 y[k-1] -.16 y[k-2] = 5 f[k]$:
\begin{displaymath}h[k]...
...] = A \delta[k] + y_0[k]u[k] \end{displaymath}where $A = b_0/a_0$.
\end{example}

\begin{example}$y[k] - 0.6 y[k-1] - 0.16y[k-2] = 5f[k]$
\begin{displaymath}(E^2 ...
...\end{displaymath}$h[0] = 5, h[1] = 3\rightarrow c_1 = 1, c_2 = 4$.
\end{example}

Zero-state response : Observe that we can write

\begin{displaymath}f[k] = f[0]\delta[k] + f[1]\delta[k-1] + \cdots = \sum_{m} f[m]
\delta[k-m] \end{displaymath}

Now we add up the response of the system to each of these outputs:
\begin{displaymath}f[0]\delta[k] \rightarrow f[0]h[k] \end{displaymath}
\begin{displaymath}f[1]\delta[k-1] \rightarrow f[1]h[k-1] \end{displaymath}

Adding these up,
\begin{displaymath}y[k] = \sum_m f[m]h[k-m] \end{displaymath}

Same sorts of properties as we had before:

Commutative
$f_1[k]*f_2[k] = f_2[k]*f_1[k]$
Distributive
$f_1[k]*(f_2[k] + f_3[k]) = f_1[k]*f_2[k] +
f_1[k]*f_3[k]$
Associative
$f_1[k]*(f_2[k]*f_3[k]) = (f_1[k]*f_2[k])*f_3[k]$
Shifting
$f_1[k-m]*f_2[k-n] = c[k-m-n]$
Convolution. with impulse
$f[k]*\delta[k] = f[k]$
Width:
If $f_1$ has length $m$ points and $f_2$ has length $n$ points, then $f_1*f_2$ has length $m+n-1$ points. (Note length given in points .)

For causal systems with causal inputs

\begin{displaymath}y[k] = \sum_{m=0}^k f[m] h[k-m] \end{displaymath}


\begin{example}Find $c[k] = f[k]*g[k]$ when $f[k] = (0.8)^k u[k]$ and $g[k] =
...
...(0.8/0.3)^k - 1} =
2[(.8)^{k+1} - (.3)^{k+1}]u[k]
\end{displaymath}\end{example}

\begin{example}The problem we have seen before: $y[k+2] -.6 y[k+1] -.16 y[k]
= 5...
...\
&=& [-1.26(.25)^k +.444(-.2)^k + 5.81(.8)^k]u[k]
\end{eqnarray*}\end{example}

Total response = zero-input component + zero-state component.

\begin{example}Take the same system as before, with $y[-1] = 0, y[-2] =25/4$.
Th...
...8)^k) + (-1.26(.25)^k + .444(-.2)^k +
5.81(.8)^k) \end{displaymath}\end{example}

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 16). Lecture 1: Zero-State Respone. 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_5node5.html. This work is licensed under a Creative Commons License Creative Commons License