Let us also assume that we have chosen a block size of N for the
DFTs and the inverse DFTs.
Normally, we would choose
N
=
2
k
for some integer
k
to facilitate
the use of an FFT algorithm.
The blocks of input data are then chosen as follows.
1
The first
M

1 samples of the first data block will be zeros. The next
L
=
N

M
+
1 samples will be the first
L
samples of the input data.
2
The first
M

1 samples of the subsequent data blocks will come
from the last
M

1 data samples of the last data block. The next
L
samples will be the next
L
new samples of the data sequence.
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Overlap–Save Method
In this way, each block will involve
L
new samples of the input
sequence.
The output corresponding to each input block will be the last
L
samples of each output block.
The first
M

1 samples are discarded.
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Overlap–Save Example
Example (7.4)
Consider the use of the overlap–save procedure for filtering long
sequences when the input sequence is a speech sample with a
duration of 4 seconds and a sampling rate of 8000 samples per
second.
The FIR filter to be used has 100 filter coefficients.
In this example, we will consider the total number of complex
multiplication required to filter the sequence using the direct
method and compare it with the number required for the
overlap–save method.
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Overlap–Save Example
Example (7.4)
The number of complex multiplications required to compute a N
point FFT is given by
K
=
N
log
2
(
N
)
(64)
The total number of samples in the sequence is
L
1
=
4
(
8000
) =
32000
(65)
The minimum number of samples to use for the FFT in order to
obtain a linear convolution result is
N
1
=
32000
+
100

1
=
32099
(66)
The next number that is a power of 2 is
N
1
=
2
15
=
32768
(67)
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Overlap–Save Example
Example (7.4)
We need to add 32768

32000
=
768 zeros to the end of the
sample sequence and 32768

100
=
32668 zeros to the end of
the FIR filter.
We then compute the FFT of both sequences and multiply them
point by point to obtain the DFT of the output.
We then use the inverse FFT (IFFT) algorithm to obtain the output.
This means that we need to compute the FFT twice and the IFFT
once.
The point by point multiplication of the two DFTs requires 32768
complex multiplications.
The total number of complex multiplications is
K
1
=
3
(
32768
)(
15
) +
32768
=
1507328
(68)
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Overlap–Save Example
Example (7.4)
If we use a block size of
N
2
=
256
(
2
8
)
samples for the
overlap–save method, then each time we use
L
2
=
256

100
+
1
=
157
(69)
new samples in each input block of data.
We would have to use
B
2
=
32000
157
=
d
203
.
82
e
=
204
(70)
blocks.
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Overlap–Save Example
Example (7.4)
The total number of complex multiplications required is
K
2
=
204
[
2
(
256
)(
8
) +
256
] +
256
(
8
)
K
2
=
889856
(71)
It helps that we only need to compute the DFT of the filter once.
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 Digital Signal Processing, DFT, C. Williams