Convolution Representation of LTI Filters

GUIDE: Mathematics of the Discrete Fourier Transform (DFT) - Julius O. Smith III. Convolution Representation of LTI Filters

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NOTE: THIS DOCUMENT IS OBSOLETE, PLEASE CHECK THE NEW VERSION: "Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition", by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8. - Copyright © 2017-09-28 by Julius O. Smith III - Center for Computer Research in Music and Acoustics (CCRMA), Stanford University

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Convolution Representation of LTI Filters

If $y(n)$ is the output of an LTI filter with input $x(n)$ and impulse response $h(n)$, then $y$ is the convolution of $x$ with $h$,

\

Since convolution is commutative ($x\), we have also
\

Definition. The unilateral $z$ transform of the discrete-time signal $x(n)$ is defined to be

\ (B.9)

That $x(n)$ and $X(z)$ are transform pairs is expressed by writing $X(z)={\ or $X(z)\.

Theorem. The convolution theorem (Papoulis [21]) states that

\

In words, convolution in the time domain is multiplication in the frequency domain.

Taking the $z$ transform of both sides of Eq. (B.4.3) and applying the convolution theorem gives

\

where $H(z)$ is the $z$ transform of the filter impulse response. Thus the$z$ transform of the filter output is the $z$ transform of the input times the $z$ transform of the impulse response.

Definition. The transfer function $\ of a linear time-invariant discrete-time filter is defined to be the $z$ transform of the impulse response $\.

Theorem. The shift theorem [21] for $z$ transforms states that

\

The shift theorem can be derived immediately from the definition of the $z$ transformgiven in Eq. ([*]):
\


The general difference equation for an LTI filter appears as

$\ $\ $\ (B.10)
  $\ $\ (B.11)

Taking the $z$ transform of both sides, denoting the transform by ${\, gives
$\ $\ $\ (B.12)
    $\ (B.13)

using linearity and the shift theorem. Replacing ${\ by $Y(z)$, ${\ by $X(z)$, and moving the terms in $Y(z)$ to the left-hand side, we obtain
$\ $\ $\ (B.14)
  $\ $\ (B.15)

or
$\ $\ $\ (B.16)
  $\ $\ (B.17)

Defining the polynomials
$\ $\ $\ (B.18)
$\ $\ $\ (B.19)

the $z$ transform of the difference equation becomes
\

Finally, solving for $Y(z)/X(z)$ which equals the transfer function $H(z)$, yields
\

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