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

$\begin{displaymath}Y[z] = H[z]F[z] = H[z]. \end{displaymath}$

$\begin{displaymath}h[k] \Leftrightarrow H[z]. \end{displaymath}$

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 November 24, 2009, 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.

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	Info Lecture 2: Transfer Functions Document Actions 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 November 24, 2009, 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.	Reuse Course Download this course

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