Digital Signal Processing Reference
In-Depth Information
Proof of property (vi)
By expanding the left-hand side of Eq. (1.23), we obtain
T
0
T
g
e
(
t
)d
t
=
g
e
(
t
)d
t
+
g
e
(
t
)d
t
.
−
T
−
T
0
integral I
integral II
Substituting
α =−
t
in integral I yields
0
T
T
integral I
=
g
e
(
−α
)(
−
d
α
)
=
g
e
(
α
)d
α =
g
e
(
t
)d
t
=
integral II
,
T
0
0
which proves Eq. (1.23).
1.1.6.2 Combinations of even and odd DT signals
Properties (i)-(vi) for CT signals can be extended to DT sequences. Consider
g
e
[
k
] and
h
e
[
k
] as even sequences and
g
o
[
k
] and
h
o
[
k
] are as odd sequences.
For the four DT signals, the following properties hold true.
(i) Multiplication of an even sequence with an odd sequence results in an odd
sequence. The DT sequence
x
[
k
]
=
g
e
[
k
]
g
o
[
k
]
,
for example, is an odd
sequence.
(ii) Multiplication of two odd sequences results in an even sequence. The DT
sequence
h
[
k
]
=
g
o
[
k
]
h
o
[
k
], for example, is an even sequence.
(iii) Multiplication of two even sequences results in an even sequence. The DT
sequence
z
[
k
]
=
g
e
[
k
]
h
e
[
k
]
,
for example, is an even sequence.
(iv) Due to its antisymmetry property, a DT odd sequence is always zero at
k
=
0. Therefore,
g
o
[0]
=
h
o
[0]
=
0.
(v) Adding the samples of a DT odd sequence
g
o
[
k
] within the range [
−
M
,
M
] is 0, i.e.
M
M
g
o
[
k
]
=
0
=
h
o
[
k
]
.
(1.24)
k
=−
M
k
=−
M
(vi) Adding the samples of a DT even sequence
g
e
[
k
] within the range [
−
M
,
M
] simplifies to
M
M
g
e
[
k
]
=
g
e
[0]
+
2
g
e
[
k
]
.
(1.25)
k
=−
M
k
=
1
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