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The simplest nontrivial case is
channels. Starting with a
general linear time-invariant filter
![$\displaystyle H(z) \eqsp \sum_{n=-\infty}^{\infty}h(n)z^{-n},$](img1948.png) |
(12.6) |
we may separate the even- and odd-indexed terms to get
![$\displaystyle H(z) \eqsp \sum_{n=-\infty}^{\infty}h(2n)z^{-2n} + z^{-1}\sum_{n=-\infty}^{\infty}h(2n+1)z^{-2n}.$](img1949.png) |
(12.7) |
We define the polyphase component filters as follows:
and
are the polyphase components
of the polyphase decomposition of
for
.
Now write
in terms of its polyphase components:
![$\displaystyle \zbox {H(z) \eqsp E_0(z^2) + z^{-1}E_1(z^2)}$](img1953.png) |
(12.8) |
As a simple example, consider
![$\displaystyle H(z)\eqsp 1 + 2z^{-1} + 3z^{-2} + 4z^{-3}.$](img1954.png) |
(12.9) |
Then the polyphase component filters are
and
![$\displaystyle H(z) \eqsp E_0(z^2) + z^{-1}E_1(z^2) \eqsp (1 + 3z^{-2}) + (2z^{-1} + 4z^{-3}).$](img1956.png) |
(12.10) |
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