Convolution Representation of LTI Filters

GUIDE: Mathematics of the Discrete Fourier Transform (DFT). Convolution Representation of LTI Filters

It appears that you are using AdBlocking software. The cost of running this website is covered by advertisements. If you like it please feel free to a small amount of money to secure the future of this website.

<< Previous page  TOC  INDEX  Next page >>


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

\

<< Previous page  TOC  INDEX  Next page >>

 

© 1998-2017 – Nicola Asuni - Tecnick.com - All rights reserved.
about - disclaimer - privacy