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FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
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Fig. 5-34 Ideal sampling.
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( h ) From Eq. (5.147) (Prob. 5.25) we have
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Let Then, according to the frequency convolution theorem (5.59), we have
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Using Eq. (1.261, we obtain
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which shows that X,(w) consists of periodically repeated replicas of X(w) centered about kw, for all k. The Fourier spectrum X,(w) is shown in Fig. 5-34 f ) for T, < r/w, (or w, > 2wM), and in Fig. 5-34( j ) for T, > r / w M (or w, < 2wM), where w, = 27~/T,. It is seen that no overlap of the replicas X(o - ko,) occurs in X,(o) for w, r 2wM and that overlap of the spectral replicas is produced for w,$ < 2wM. This effect is known as
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5.59. Let x ( t ) be a real-valued band-limited signal specified by Show that x ( t ) can be expressed as
sin wM(t - kT,)
x(kTs)
41) & -C =
where T,
= rr/w,.
- kT-)
From Eq. ( 5.183 we have
T,X,(w)
X ( o - ko,)
Then, under the following two conditions, (1) X(o)=O,IwI>o, and (2)
T,=-
we see from Eq. (5.1185 that
CHAP. 51
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
Next, taking the Fourier transform of Eq. (5.182), we have
Substituting Eq. (5.187) into Eq. (5.186), we obtain
Taking the inverse Fourier transform of Eq. (5.1881, we get
C k=
sin w M (t - kT,)
x(kT,)
W M ( -kTs) ~
From Probs. 5.58 and 5.59 we conclude that a band-limited signal which has no frequency components higher that f M hertz can be recovered completely from a set of samples taken at 2 the rate of f, ( 1 fM) samples per second. This is known as the uniform sampling theorem for low-pass signals. We refer to T, = X / W , = 1 / 2 fM ( o M 27r fM as the Nyquist sampling = interval and f, = 1/T, = 2 fM as the Nyquist sampling rate.
5.60. Consider the system shown in Fig. 5 - 3 5 ( a ) . The frequency response H ( w ) of the ideal low-pass filter is given by [Fig. 5 - 3 5 ( b ) ]
Show that if w ,
= 0J2,
then for any choice of T,,
Fig. 5-35
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
From Eq. (5.137) the impulse response h ( t ) of the ideal low-pass filter is given by sin w, t h ( t ) = T5----at From Eq. (5.182) we have
T5wc sin w,t w,t
By Eq. (2.6) and using Eqs. ( 2 . 7 ) and (1.261, the output y ( t ) is given by
Using Eq. (5.1891, we get
T p , sin w,(t - k c )
~ ( t=) If w,
=wJ2,
k = - cc
x(kT,)77
w,(t - kT,)
then T,w,/a
1 and we have
Setting t
= mT,
( m = integer) and using the fact that w,T,
2 ~we get ,
sin ~ X (kTS)
( -m ) k
Y ( ~ T=) Since
77(m - k )
we have
which shows that without any restriction on x ( t ) , y(mT5)= x(mT,) for any integer value of m . Note from the sampling theorem (Probs. 5.58 and 5.59) that if w, = 2 a / T 5 is greater than twice the highest frequency present in x ( t ) and w , = w J 2 , then y ( t ) = x ( t ) . If this condition on the bandwidth of x ( t ) is not satisfied, then y ( t ) z x ( t ) . However, if w, = 0 , / 2 , then y(mT,) = x(mT5) for any integer value of m .
CHAP. 51
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
Supplementary Problems
Consider a rectified sine wave signal
x(t)
defined by x ( l ) = IAsin.srt(
Sketch x ( t ) and find its fundamental period. (6) Find the complex exponential Fourier series of ( c ) Find the trigonometric Fourier series of x ( t 1.
x(!).
X(t)
is sketched in Fig. 5-36 and T,
(c) x ( t ) =
2A 4 A m 1 cos -- - C - k 2 r t .sr 4k2-1
Fig. 5-36
Find the trigonometric Fourier series of a periodic signal
x(t) =t2,
x(t)
defined by
-a < t < .rr
+ 2a) = x ( t )
Am. x ( t ) = 5.63.
(-ilk cos - kt
x(t)
Using the result from Prob. 5.10, find the trigonometric Fourier series of the signal in Fig. 5-37. A A " 1 2a A ~ S . ~ ( t= - - ) - sin ko,! wo= 2 .rr & = I k To
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