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Lecture 2: Transfer Functions

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Schedule :: Intro :: Inverse :: Properties :: Solution :: Transfer Functions :: System Realization :: Bilateral :: Frequency Response :: Rubber Sheet Geometry :: FIR Filters

Under the assumption of zero initial conditions (the zero-state response) the general LTI difference equation

\begin{displaymath}Q[E]y[k] =P[e]f[k] \end{displaymath}
\begin{displaymath}(E^n + a_{n-1}E^{n-1} + \cdots + a_1 E + a_0) y[k] =
(b_nE^n + b_{n-1}E^{n-1} + \cdots + b_1 E + b_0) f[k]
\end{displaymath}

may be transformed to
\begin{displaymath}(z^n + a_{n-1}z^{n-1} + \cdots a_1z + a_0) Y[z] =
(b_nz^n + b_{n-1}z^{n-1} + \cdots b_1z + b_0) F[z]
\end{displaymath}
Solving for the output,
\begin{displaymath}Y[z] = \left(\frac{b_nz^n + b_{n-1}z^{n-1} + \cdots b_1z + b_0}
{z^n + a_{n-1}z^{n-1} + \cdots a_1z + a_0} \right) F[z] \end{displaymath}
We define
\begin{displaymath}H[z] = \frac{z^n + b_{n-1}z^{n-1} + \cdots b_1z + b_0}
{z^n + a_{n-1}z^{n-1} + \cdots a_1z + a_0} = \frac{P[z]}{Q[z]}
\end{displaymath}

as the transfer function . Note that

\begin{displaymath}H[z] = \frac{Y[z]}{F[z]} = \frac{\Zc[\mbox{zero-state response}]}
{\Zc[\mbox{input}]}
\end{displaymath}
and the output is obtained by
\begin{displaymath}Y[z] = F[z]H[z].\end{displaymath}
The poles of the transfer function are the roots of the characteristic equation, and we can determine the stability of the system by examination of the transfer function.

\begin{example}Unit delay: $y[k] = f[k-1]u[k-1]$. $Y[z] =
\frac{1}{z}F[z]$. $H[z] = z^{-1}$. (Remember this!)
\end{example}

\begin{example}
Find the transfer function for the system
\begin{displaymath}y[k...
...2}{3}(-.2)^k - \frac{8}{3}(-.8)^k + 2(-.5)^k]u[k] \end{displaymath}\end{example}

If the input is $f[k] = \delta[k]$ , then the output is

\begin{displaymath}Y[z] = H[z]F[z] = H[z]. \end{displaymath}
So
\begin{displaymath}h[k] \Leftrightarrow H[z]. \end{displaymath}

The transfer function is the Z-transform of the impulse response.

Nomenclature . A discrete-time filter which has only a numerator part (only zeros, except for possible poles at the origin which correspond to delays) is said to be a finite impulse response (FIR) filter.
\begin{example}What is the impulse response of a filter with
\begin{displaymath...
...^{-3} \end{displaymath}Note that {\bf all FIR filters are stable}.
\end{example}

A filter with poles is said to be an infinite impulse response (IIR) filter.


\begin{example}What is the impulse response of a filter with
\begin{displaymath}H[z] = \frac{z}{z-.5}. \end{displaymath}\end{example}
Note that there is no practical way of making an FIR filter for continuous time systems: this is available only for digital filters.

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 17). Lecture 2: Transfer Functions. 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.html. This work is licensed under a Creative Commons License Creative Commons License