Digital Signal Processing Reference
In-Depth Information
D
n
0.13
0.11
0.13
0.11
0.09
0.07
0.06
0.050.050.050.050.050.05
0.06
0.07
0.09
0.13
0.11
0.11
0.11
0.11
0.05
0.06
0.07
0.09
0.09
0.07
0.06
0.050.050.050.050.050.05
0.06
0.07
0.09
0.09
0.07
0.06
0.05
n
−20
−18
−16
−14
−12
−10
−8
−6
−4
−2
20
0
2468 0 2 4
16
18
(a)
<
D
n
0.51
0.51
0.36
0.51
0.51
0.36
0.51
0.44
0.51
0.36
0.44
0.33
0.44
0.33
0.33
0.07
0.21
0.07
0.21
−14
−12
−10
−8
0
24 6
16
18
20
n
−20
−18
−16
−6
−4
−2
8 0 2 4
−0.2
−0.07
−0.2
−0.07
−0.33
−0.33
−0.33
−0.36
−0.44
−0.36
−0.44
−0.36
−0.44
−0.51
−0.51
−0.51
−0.51
−0.51
−0.51
(b)
Fig. 11.3. (a) Magnitude
spectrum and (b) phase
spectrum of the DTFS
coefficients in Example 11.3.
where the greatest common divisor between the fundamental period
N
and the
integer constant
m
is one.
Solution
We first show that the DT sequence
x
[
k
] is periodic and determine its fundamen-
tal period. It was mentioned in Proposition 1.1 that a DT complex exponential
sequence
x
[
k
]
=
exp(j(
Ω
0
k
+ θ
)) is periodic if 2
π/
Ω
0
is a rational number. In
this case, 2
π/
Ω
0
=
N
/
m
, which is a rational number as
m
,
K
and
N
are all
integers. In other words, the sequence
x
[
k
] is periodic. Using Eq. (1.8), the
fundamental period of
x
[
k
] is calculated to be
K
0
=
(2
π/
Ω
0
)
p
=
pN
/
m
,
where
p
is the smallest integer that results in an integer value for
K
0
. Note
that the fraction
N
/
m
represents a rational number, which cannot be reduced
further since the greatest common divisor between
m
and
N
is given to be one.
Selecting
p
=
m
, the fundamental period is obtained as
K
0
=
N
.
To compute the DTFS coefficients, we express
x
[
k
] as follows:
A
e
j
θ
e
j
0
mk
x
[
k
]
=
and compare this expression with Eq. (11.4). For 0
≤
n
≤
K
0
−
1, we observe
that
A
e
j
θ
=
m
if
n
D
n
=
(11.15)
=
m
.
i f
n
As a special case, we consider
A
=
2
,
K
0
=
6
,
m
=
5, and
θ = π/
4. The
magnitude and phase spectra for the selected values are shown in Figs.
11.4(a) and (b), where we have used the periodicity property of the DTFS
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