Digital Signal Processing Reference
In-Depth Information
The integral on the right-hand side of the above equation includes several
impulse functions located at
Ω
=
0
,
2
π,
4
π,...
Only the impulse function
located at
Ω
=
0 falls in the frequency range
Ω
=
[
−π, π
]. Therefore,
x
1
[
k
]
can be simplified as follows:
x
1
[
k
]
=
π
δ
(
Ω
)d
Ω
=
1
.
−π
(ii) Using the synthesis equation, (11.28a), the inverse DTFT of
X
2
(
Ω
)is
given by
π
m
=−∞
δ
(
Ω
∞
1
2
π
1
2
π
X
2
(
Ω
)e
j
k
Ω
d
Ω
−
2
m
π
)e
j
k
Ω
d
Ω
.
x
1
[
k
]
=
=
2
π
−
Ω
0
2
π
−π
π
m
=−∞
δ
(
Ω
∞
−
2
m
π
)e
j
k
(
Ω
0
−
2
m
π
)
d
Ω
=
−
Ω
0
−π
π
m
=−∞
δ
(
Ω
∞
Ω
0
−
2
m
π
)e
j
k
Ω
0
e
j
k
2
m
π
=
−
d
Ω
=
1
−π
π
m
=−∞
δ
(
Ω
∞
=
e
j
k
Ω
0
−
Ω
0
−
2
m
π
)d
Ω
.
−π
The integral on the right-hand side of the above equation includes several
impulse functions located at
Ω
Ω
0
+
2
m
π
. Only one of these infinite num-
ber of impulse functions will be present in the frequency range
Ω
=
=
[
−π, π
].
Therefore, the integral will have a vaue of unity and the function
x
2
[
k
] can be
simplified as follows:
x
2
[
k
]
=
e
j
k
Ω
0
.
Table 11.2 lists the DTFT and DTFS representations for several DT sequences.
In situations where a DT sequence is aperiodic, the DTFS representation is
not possible and therefore not included in the table. The DTFT of the peri-
odic sequences is determined from its DTFS representation and is covered in
Section 11.4.
Table 11.3 plots the DTFT for several DT sequences. In situations where a
DT sequence or its DTFT is complex, we plot both the magnitude and phase
components. The magnitude component is shown using a bold line, and the
phase component is shown using a dashed line.
Example 11.7 illustrates the calculation of a DT function from its DTFT
using Eq. (11.28a). In many cases, it may be easier to calculate a DT
function from its DTFT using the partial fraction expansion and the DTFT
pairs listed in Table 11.2. This procedure is explained in more detail in
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