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Sampling and Reconstruction of Analog Signals 83
Referring to Figure 3.10, if π>Ω 0 T s —or equivalently, F s /2 >F 0 —
then
X(e jω )= 1 X
(j ω )
; − π < ω ≤ π (3.29)
T s T s T s T s T s
which leads to the sampling theorem for band-limited signals.
THEOREM 3
Sampling Principle
Aband-limited signal x a (t) with bandwidth F 0 can bereconstructed from
its sample values x(n) =x a (nT s ) if the sampling frequency F s =1/T s is
greater than twice the bandwidth F 0 of x a (t).
F s > 2F 0
Otherwise aliasing would result in x(n). The sampling rate of 2F 0 for an
analog band-limited signal is called the Nyquist rate.
Note: After x a (t) issampled, the highest analog frequency that x(n) represents
is F s /2Hz(orω = π). This agrees with the implication stated in
property 2 of the discrete-time Fourier transform in Section 3.1. Before
we delve into MATLAB implementation of sampling, we first consider
sampling of sinusoidal signals and the resulting Fourier transform in the
following example.
□ EXAMPLE 3.17 The analog signal x a(t) =4+2cos(150πt + π/3) + 4 sin(350πt) issampled at
F s = 200 sam/sec to obtain the discrete-time signal x(n). Determine x(n) and
its corresponding DTFT X(e jω ).
Solution
The highest frequency in the given x a(t) isF 0 = 175 Hz. Since F s = 200, which
is less than 2F 0, there will be aliasing in x(n) after sampling. The sampling
interval is T s =1/F s =0.005 sec. Hence we have
x(n) =x a(nT s)=x a(0.005n)
(
=4+2cos 0.75πn + π )
+4sin(1.75πn) (3.30)
3
Note that the digital frequency, 1.75π, ofthe third term in (3.30) is outside the
primary interval of −π ≤ ω ≤ π, signifying that aliasing has occurred. From
the periodicity property of digital sinusoidal sequences in Chapter 2, we know
that the period of the digital sinusoid is 2π. Hence we can determine the alias
of the frequency 1.75π. From (3.30) we have
x(n) =4+2cos(0.75πn + π 3
)+4sin(1.75πn − 2πn)
= 4+2cos(0.75πn + π 3
) − 4 sin(0.25πn) (3.31)
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