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
(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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