Two-Channel Case
The simplest nontrivial case is
channels. Starting with a
general linear time-invariant filter
![]() |
(12.6) |
we may separate the even- and odd-indexed terms to get
![]() |
(12.7) |
We define the polyphase component filters as follows:
![\begin{eqnarray*}
E_0(z)&=&\sum_{n=-\infty}^{\infty}h(2n)z^{-n}\\ [5pt]
E_1(z)&=&\sum_{n=-\infty}^{\infty}h(2n+1)z^{-n}
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/sasp2/img1958.png)
and
are the polyphase components
of the polyphase decomposition of
for
.
Now write
in terms of its polyphase components:
![]() |
(12.8) |
As a simple example, consider
![]() |
(12.9) |
Then the polyphase component filters are
![\begin{eqnarray*}
E_0(z) &=& 1 + 3z^{-1}\\ [5pt]
E_1(z) &=& 2 + 4z^{-1}
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/sasp2/img1963.png)
and
![]() |
(12.10) |
Next Section:
N-Channel Polyphase Decomposition
Previous Section:
Filtering and Downsampling