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Therefore, the energy in x ( n ) is
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A complex bandpass analog signal x a ( t )has a Fourier transform that is nonzero over the frequency range [ Q l , Q2] as shown in the figure below.
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The signal is sampled to produce the sequence x ( n ) = xa(nT,).
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( a ) What is the smallest sampling frequency that can be used so that x a ( t ) may be recovered from its samples x(n)
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SAMPLING
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(b) For this minimum sampling frequency, findthe interpolation formula for x,(t) in terms of x ( n ) .
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(a) Because the highest frequency in xa(t) is Q2, the Nyquist rate is 2Q2. However, note that if xa(t) is modulated
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with a complex exponential of frequency (Q2
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then y,(t) is a (complex) low-pass signal with a spectrum shown in the following figure:
Yu(jQ)
where Qo = (Q2 - Q1)/2. Thus, the Nyquist rate for yo(() is 2Q0 = Q2 - Q l , which suggests that x,(t) may be uniquely reconstructed from its samples x,(nT,) provided that
If xa(t) is sampled with a sampling frequency Q,, the spectrum of the sampled signal is
as illustrated below.
Q2-Qs
In order for there to be no interference between the shifted spectra, it is necessary that
If this condition is satisfied, xa(t) may be uniquely reconstructed from xs(t) using a bandpass filter with a frequency response as shown below.
(b) With a sampling frequency Q, = Q2 - Q l , the reconstruction filter is a complex bandpass filter with an impulse response
ha(!) = Ts s i n ( Q s t / 2 ) e - j ( ~ 2 + ~ ~ ~
CHAP. 31
SAMPLING Therefore, the output of the reconstruction filter, which produces the complex bandpass signal x,(t), is
Given a real-valued bandpass signal x , ( t ) with X a ( f ) = 0 for If l < f i and If 1 > f2, the Nyquist sampling theorem says that the minimum sampling frequency is fs = 2 f2. However, in some cases, the signal may be sampled at a lower rate.
(a) Suppose that fl = 8 kHz and f2 = 10 kHz. Make a sketch of the discrete-time Fourier transform of x ( n ) = x,(nTs) if fs = I /T, = 4 kHz. ( 6 ) Define the bandwidth of the bandpass signal to be
and the center frequency to be
Show that if f , > B / 2 and f2 is an integer multiple of the bandwidth B , no aliasing will occur if x a ( t ) is sampled at a sampling frequency fs = 2 8 .
(c) Repeat part (b) for the case in which f2 is not an integer multiple of the bandwidth B .
(a) Let x u ( [ )have a spectrum as shown in the figure below.
The spectrum of the sampled signal
which is formed by shifting X,(f ) by integer multiples of the sampling frequency and summing. With f, = 4 kHz, we have the spectrum sketched below.
Note that X,( f) is not aliased. Therefore, with the appropriate processing of x,(r), the signal x,(r) may be recovered from its samples. Finally, the DTFT of the discrete-time sequence x ( n ) = x,(nTs) is
SAMPLING
which is sketched below.
[CHAP. 3
x(c'")
(b) If fi is an integer multiple of B , we may express f i and f2 as follows:
With a sampling frequency of f, = 2 B , the sampled signal has a spectrum
Because X a ( f ) is nonzero only for (I - l ) B < If 1 < l B , there is only one term in the sum that contributes to X,( f ) in the frequency range O< f <B and only one term that contributes to the frequency range -B if < 0 (draw a picture as in part (a) to see this clearly). Therefore, there is no aliasing, and x,(t) may be sampled without aliasing if a sampling frequency f, = 2B.
(c) If f2 is not an integer multiple of B, we may always increase B until this is the case. Specifically, let
where
1.1 is defined to be the "integer part." Now, if we simply increase B to B' where
we have the case described in part ( b ) where fi is an integer multiple of the bandwidth. Thus, x,(t) may be sampled without aliasing a sampling frequency of
Determine the minimum sampling frequency for each of the following bandpass signals:
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