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5.16. ( a ) Verify the timeshifting property (5.50), that is, By definition (5.31)  to)H
ejw'~lX ) (o
By the change of variable
 t,,, we obtain
Hence, 5.17. Verify the frequencyshifting property (5.511, that is, x ( t ) eiwuJ X ( O H
 wO) By definition (5.31) CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Hence, ~ ( t e)i " ~ ' w X(w  w,) 518. Verify the duality property (5.54), that is, From the inverse Fourier transform definition (5.321, we have Changing t to  t, we obtain
Now interchanging r and o,we get
Since we conclude that
5.19. Find the Fourier transform of the rectangular pulse signal x ( t ) [Fig. 516(a)] defined by
By definition (5.31) (4 (b) Fig. 516 Rectangular pulse and its Fourier transform.
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
Hence, we obtain p u ( t ) 2sin o a
=2a sin ma wa
The Fourier transform X(o) of x(t) is sketched in Fig. 516(b). 5.20. Find the Fourier transform of the signal [Fig. 517(a)] x(t) = sin at 7Tt
From Eq. (5.136) we have sin o a pa(!) t ,  Now by the duality property (5.541, we have 2 sin at
257pa(  o ) Dividing both sides by 2~ (and by the linearity property), we obtain
pa(o) sin at
= pa(w) where pa(w) is defined by [see Eq. (5.135) and Fig. 51Xb)I
t Fig. 517 sin a t / ~ and its Fourier transform.
5.21. Find the Fourier transform of the signal [Fig. 518(a)] x ( t ) = ealfl
Signal x(t) can be rewritten as
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Fig. 518 eIal' and its Fourier transform.
Then
Hence, we get
The Fourier transform X ( w ) of x ( l ) is shown in Fig. 518(b). Find the Fourier transform of the signal [Fig. 519(a)] x(t) = a2 + t 2
From Eq. (5.138) we have
Now by the duality property ( 5 . 5 4 )we have
Fig. 519 l/(a2
+ 1 2 , and its Fourier transform.
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
Dividing both sides by 2 a , we obtain
The Fourier transform X ( w ) of
x(t) is shown in Fig. 519(b). 5.23. Find the Fourier transforms of the following signals: ( a ) x(t) = 1 (c) x(t)=eJWu' (e) x ( t ) = s i n o , t
( a ) By Eq. (5.43) we have
( b ) x(t) =eJWO' ( d ) x(t) = cos o,t
Thus, by the duality property (5.54) we get
Figures 520(a) and ( b ) illustrate the relationships in Eqs. (5.140) and ( 5 . 1 4 0 , respectively. ( b ) Applying the frequencyshifting property (5.51) to Eq. (5.1411, we get Fig. 520 ( a ) Unit impulse and its Fourier transform; ( 6 ) constant (dc) signal and its Fourier transform. FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
25 1 From Eq. (5.142) it follows that
ej"~'H2rTTS(o + w o ) From Euler's formula we have
mot = L ( e ~ " ' ~e'  ~ w ~ ' ) + 2
Thus, using Eqs. (5.142) and (5.143) and the linearity property (5.49), we get cos wot
~ [ 6 ( w w,,)  + 6(w + w o ) ] (5.144) Figure 521 illustrates the relationship in Eq. (5.144). Similarly, we have
and again using Eqs. (5.142) and (5.143), we get sin wot
 j.rr[S(o  w o )  6 ( w + w o ) ] (5.145) Fig. 521 Cosine signal and its Fourier transform.
5.24. Find the Fourier transform of a periodic signal
We express x(t ) as
x ( t ) with period T o .
Taking the Fourier transform of both sides and using Eq. (5.142) and the linearity property (5.49), we get which indicates that the Fourier transform of a periodic signal consists of a sequence of equidistant impulses located at the harmonic frequencies of the signal. FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
5.25. Find the Fourier transform of the periodic impulse train [Fig. 522(a)] From Eq. (5.115) in Prob. 5.8, the complex exponential Fourier series of 6,,,(t) is given by
Using Eq. (5.1461, we get
= W() S(w  kw,) = w,G,,(w) (5.147) Thus, the Fourier transform of a unit impulse train is also a similar impulse train [Fig. 522(b)]. Fig. 522 Unit impulse train and its Fourier transform.
5.26. Show that
x ( t ) cos o,,t
x ( t ) sin w,,t
From Euler's formula we have
i X ( o  o,,)+ i X ( o + o o ) [ : ~ ( w w,)  : ~ ( + w,)] o
Equation (5.148) is known as the modulation theorem.
cos wol = i ( e j W +
~ [ x ( t ) c owet] s Hence, X ( ~ ) C O S ) t $X(W w,) ~ ( c' Then by the frequencyshifting property (5.51) and the linearity property (5.491, we obtain
. ~ [ i x ( t ei"ut ) ;x(w
+ 'x(t) ejU~lt I  wo) + f x ( w + w,) + ; x ( w + w") CHAP. 5 1
FOURIER ANALYSIS OF TIME SIGNALS A N D SYSTEMS
In a similar manner we have
Y [ x ( t )sin w , t ] = x(t) eJ"ot  x ( t ) e '"o' =X(w
w,)  X(w
+ w,) +wo)] Hence, x ( t ) sin wot t* [ i ~ ( w wo)  ;X(O
5.27. The Fourier transform of a signal
x(t) is given by [Fig. 523(a)] X ( w ) = $pa(w  wo) + ; P ~ ( W + ~
Find and sketch
x(r). From Eq. (5.137)and the modulation theorem (5.148)it follows that sin at COS wot x(t) =Tf
which is sketched in Fig. 523(b). Fig. 523 FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
[CHAP. 5
5.28. Verify the differentiation property (5.55), that is, From Eq. (5.321 the inverse Fourier transform of X(w) is
Then
Comparing Eq. (5.151) with Eq. (5.150), we conclude that dr(t)/dt is the inverse Fourier transform of jwX(w). Thus, 5.29. Find the Fourier transform of the signum function, sgn(t) (Fig. 524), which is defined
The signum function, sgn(t), can be expressed as sgn(t) Using Eq. (1.301, we have
2u(t)  1 Fig. 524 Signum function.
CHAP. 51
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
Let sgn(t) X(w) Then applying the differentiation property (5.551, we have jwX(w) Hence, F[26(t)] +X(w) Note that sgn(t) is an odd function, and therefore its Fourier transform is a pure imaginary function of w (Prob. 5.41). 530. Verify Eq. (5.481, that is, ~ ( tH ) ~s(0.l)

