Digital Signal Processing Reference
In-Depth Information
Fig. 5
A combined
upsampling and
downsampling example
S
U
D
T
SU
UD
DT
2
−2
1
1
L =
M =
3
2
2
−2
It consists of a chain of four actors
S
,
U
,
D
,and
T
.
S
and
T
are source and sink
actors, respectively.
U
is an expander
2
−
2
32
=
L
and
D
is a compressor
11
2
M
=
−
2
The arcs are labeled
SU
,
UD
,and
DT
, respectively. Let
V
SU
be the identity matrix,
and
W
SU
=
the support matrix. The expander consumes the samples from
the source in row scan order, The output samples of the upsampler are ordered in
the way discussed above, one at a time (that is as (1,1)), releasing
diag
(
3
,
3
)
samples
at the output on each invocation. The decimator can consume the upsampler's
output samples in a rectangle of samples defined by a factorization
M
1
×
|
det
(
L
)
|
M
2
of
|
det
(
M
)
|
2
2 factorization, the (0,0) invocation of the downsampler consumes the (original)
samples (0,0), (
×
−
1,1), (0,1) and (
−
1,2) or equivalently, the natural ordered samples
(0,0), (0,1), (1,0), and (1,1).
Of course, other factorizations of
are possible, and the question
is whether a factorization can be found for which the total number of samples
output by the compressor equals the total number of samples consumed divided
by
|
det
(
M
)
|
in a complete cycle as determined by the repetitions matrix which, in
turn, depends on the factorizations of
|
det
(
M
)
|
|
det
(
L
)
|
and
|
det
(
M
)
|
. Thus, the question
N
W
UD
)
|
det
(
M
)
|
(
is whether
N
(
W
DT
)=
2
4 factorization,
though. Thus, denoting by
r
X
,
1
and
r
X
,
2
the repetitions of a node
X
in the
horizontal
direction and the
vertical
direction, respectively, the balance equations in this case
become
×
2 factorization of
|
det
(
M
)
|
. The condition is satisfied for a 1
×
⎛
⎞
3
×
r
S
,
1
=
1
×
r
U
,
1
⎝
⎠
3
×
r
S
,
2
=
1
×
r
U
,
2
5
×
r
U
,
1
=
1
×
r
D
,
1
2
×
r
U
,
2
=
4
×
r
D
,
2
r
D
,
1
=
r
T
,
1
r
D
,
2
=
r
T
,
2
