Digital Signal Processing Reference
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FIGURE 12.19
Commutative model for the polyphase interpolation filter.
For our example, L ¼ 2 and N ¼ 4, we have L 1 ¼ 1, and N=L 1 ¼ 1, respectively. Hence,
there are two filter banks, r 0 ðzÞ and r 1 ðzÞ , each having a length of 2, as illustrated in Figure 12.18 .
When k ¼ 0 and n ¼ 1, the upper limit of time index required for hðk þ nLÞ is
k þ nL ¼ 0 þ 1 2 ¼ 2. When k ¼ 1 and n ¼ 1, the upper limit of the time index for h(kþnL )is3.
Hence, the first filter r 0 ðzÞ has the coefficients 0 Þ and 2 Þ . Similarly, the second filter r 1 ðzÞ has
coefficients 1 Þ and 3 Þ . In fact, the filter coefficients of r 0 ðzÞ are a decimated version of h(n)
starting at k ¼ 0, while the filter coefficients of r 1 ðzÞ are a decimated version of h(n) starting at k ¼ 1,
and so on.
As shown in Figure 12.18 , we can reduce the computational complexity from eight multiplications
and six additions down to four multiplications and three additions for processing each input sample
xðnÞ . Generally, the computation can be reduced by a factor of L as compared with the direct process.
The commutative model for the polyphase interpolation filter is given in Figure 12.19 .
EXAMPLE 12.5
Verify y(1) in Table 12.1 using the polyphase filter implementation in Figures 12.18 and 12.19 , respectively.
Solution:
Applying the ployphase interpolation filter as shown in Figure 12.18 leads to
w 0 ðnÞ¼hð0ÞxðnÞþhð2Þxðn 1Þ
w 1 ðnÞ¼hð1ÞxðnÞþhð3Þxðn 1Þ
When n ¼ 0,
w 0 ð0Þ¼hð0Þxð0Þ
w 1 ð0Þ¼hð1Þxð0Þ
After interpolation, we have
y 0 ðmÞ : w 0 ð0Þ 0 /
and
y 1 ðmÞ : 0 w 1 ð0Þ 0 /
Note: there is a unit delay for the second filter bank. Hence
m ¼ 0; y 0 ð0Þ¼hð0Þxð0Þ; y 1 ð0Þ¼0
m ¼ 1; y 0 ð1Þ¼0; y 1 ð1Þ¼hð1Þxð0Þ
 
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