Digital Signal Processing Reference
In-Depth Information
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()
yn
0
()
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()
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f
s
Lf
s
yn
1
()
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
hð
0
Þ
and
hð
2
Þ
. Similarly, the second filter
r
1
ðzÞ
has
coefficients
hð
1
Þ
and
hð
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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