Multirate Digital Signal Processing, Oversampling of Analog-to-Digital Conversion, and Undersampling of Bandpass Signals - Digital Signal Processing: Fundamentals and Applications - page 605

C ase 1 f c = even n umber x 2 Hz = 20 Hz, f s =4 Hz

1

Sampled
signal

Bandpass
signal 0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Time (sec.)

FIGURE 12.42
Plots of the bandpass signal and sampled signal for Case 1.

Applying undersampling using a sampling rate of 4 Hz, it follows that

xðnTÞ ¼ cos ð 2 p 22 n= 4 ÞmðnTÞ ¼ cos ð 11 npÞmðnTÞ (12.45)

Since 11 np can be either an odd or an even integer multiple of p , we have

(

cos 11 pn ¼

1 n ¼ odd

(12.46)

1 n ¼ even

We see that Equation (12.46) causes the message samples to change sign alternatively with a carrier
frequency of 22 Hz, which is the odd integer multiple of the message bandwidth of 2 Hz. This in fact
will reverse the baseband message spectrum. To correct the spectrum reversal, we multiply an
oscillator with a frequency of B ¼ 2 Hz by the bandpass signal, that is
xðtÞ cos ð 2 p 2 tÞ¼ cos ð 2 p 22 tÞmðtÞ cos ð 2 p 2 tÞ

(12.47)

Then the undersampled signal is given by

xðnTÞ cos ð 2 p 2 n= 4 Þ¼ cos ð 2 p 22 n= 4 ÞmðnTÞ cos ð 2 p 2 n= 4 Þ
¼ cos ð 11 npÞmðnTÞ cos ðnpÞ

(12.48)