Digital Signal Processing Reference
In-Depth Information
Example 11.1 shows that both the real and imaginary components of the com-
plex exponential satisfy the frequency-periodicity property; therefore, the DT
complex exponential should also satisfy the frequency-periodicity property.
Example 11.1
Consider a CT sinusoidal function with a fundamental frequency of 1.4 Hz, i.e.
x
(
t
)
=
cos(2
.
8
π
t
+ φ
)
,
where
φ
is the constant phase. Sample the function with a sampling rate of
1 sample/s and determine the fundamental frequency of the resulting DT
sequence.
Solution
In the time domain, the DT sequence is obtained by sampling
x
(
t
)at
t
=
kT
.
Since the sampling interval
T
=
1s,
x
[
k
]
=
x
(
kT
)
=
cos(2
.
8
π
k
+ φ
)
,
which is periodic with a period
Ω
1
=
2
.
8
π
radians/s. Because the CT signal
x
(
t
)
is a sinusoid with a fundamental frequency of 1.4 Hz, the minimum sampling
rate, required to avoid aliasing, is given by 2.8 samples/s. Since the sampling
rate of 1 samples/s is less than the Nyquist sampling rate, aliasing is introduced
due to sampling. Based on Lemma 9.1, the reconstructed signal is given by
y
(
t
)
=
cos(2
π
(1
.
4
−
1)
t
)
=
cos(0
.
8
π
t
+ φ
)
.
Substituting
t
=
kT
, the DT representation of the reconstructed signal is given
by
y
[
k
]
=
cos(0
.
8
π
k
+ φ
)
,
which is periodic with a period
Ω
2
=
0
.
8
π
radians/s. From the above analysis, it
is clear that the DT sequences
x
[
k
]
=
cos(2.8
π
k
+ φ
) and
y
[
k
]
=
cos(0
.
8
π
k
+
φ
) are identical. This is because the difference in the fundamental frequencies
Ω
1
and
Ω
2
is 2
π
.
Proposition 11.2
A discrete-time periodic function x
[
k
]
with period K
0
can be
expressed as a superposition of DT complex exponentials as follows:
D
n
e
j
n
Ω
0
k
,
x
[
k
]
=
(11.4)
n
=<
K
0
>
where
Ω
0
is the fundamental frequency, given by
Ω
0
=
2
π/
K
0
, and the discrete-
time Fourier series (DTFS) coefficients D
n
for
1
≤
n
≤
K
0
are given by
k
=
K
0
x
[
k
]e
1
K
0
−
j
n
Ω
0
k
.
D
n
=
(11.5)
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