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Let's look first at the case . We begin with the filter

$\displaystyle H(z) = \sum_{n=-\infty}^{\infty}h(n)z^{-n}$

Separate the even and odd terms to get

$\displaystyle H(z) = \sum_{n=-\infty}^{\infty}h(2n)z^{-2n} + z^{-1}\sum_{n=-\infty}^{\infty}h(2n+1)z^{-2n}$

Define the polyphase component filters as follows:

$\begin{eqnarray*} E_0(z)&=&\sum_{n=-\infty}^{\infty}h(2n)z^{-n}\\ [0.1in] E_1(z)&=&\sum_{n=-\infty}^{\infty}h(2n+1)z^{-n} \end{eqnarray*}$

and are the polyphase components of the polyphase decomposition of for .

Now write as the sum of the odd and even terms:

$\displaystyle \zbox {H(z) = E_0(z^2) + z^{-1}E_1(z^2)}$

As a simple example, consider

$\displaystyle H(z)=1 + 2z^{-1} + 3z^{-2} + 4z^{-3}.$

Then the even and odd terms are, respectively,

$\begin{eqnarray*} E_0(z) &=& 1 + 3z^{-1}\\ [0.1in] E_1(z) &=& 2 + 4z^{-1} \end{eqnarray*}$

Thus, can be written as the sum of the following two polyphase components:

$\begin{eqnarray*} E_0(z^2) &=& 1 + 3z^{-2}\\ [0.1in] z^{-1}E_1(z^2) &=& 2z^{-1} + 4z^{-3} \end{eqnarray*}$

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.

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