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( 0 k + θ )) is periodic if 2 π/ 0 is a rational number. In
this case, 2 π/ 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/ 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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