Digital Signal Processing Reference
In-Depth Information
C
a
se 2: f
c
= odd
n
umber
x
B=22
Hz, fs
=
2B=4
H
z
1
Phase
distortion
0.5
0
-0.5
-1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (s ec . )
Case 2: fc= odd number x B=22 Hz, fs=2B=4 Hz, with B=2 Hz oscillator
1
Message
signals
0.5
0
-0.5
-1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (s ec . )
FIGURE 12.43
Plots of the bandpass signals and sampled signals for Case 2.
Since
cos
ð
11
pnÞ
cos
ðpnÞ¼
1
(12.49)
it follows that
xðnTÞ
cos
ð
2
p
2
n=
4
Þ¼
cos
ðp
11
nÞmðnTÞ
cos
ðp nÞ¼mðnTÞ
(12.50)
which is the recovered message signal.
Figure 12.43
shows the sampled bandpass signals with the
reversed message spectrum and the corrected message spectrum, respectively, for a message signal
having a frequency of 0.5 Hz; that is,
mðtÞ¼
cos
ð
2
p
0
:
5
tÞ
(12.51)
Case 3
If
f
c
¼
noninterger
B
, we can extend the bandwidth
B
to
B
such that
f
c
¼
integer
B
and
f
s
¼
2
B
(12.52)
Then we can apply Case 1 or Case 2. An illustration of Case 3 is included in the following
example.
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