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Since we get
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which shows that the complex Fourier coefficients c, of f ( t ) equal Todkek.
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FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
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[CHAP. 5
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( b ) In Prob. 2.8(c), x , ( r ) = x 2 ( t )= x ( t ) , as shown in Fig. 2-12, which is the same as Fig. 5-8 (Prob. 5.5). From Eq. (5.105) we have
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Thus, by Eq. (5.133) the complex Fourier coefficients ck of
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Note that in Prob. 2.8(c), f ( t ) = x , ( r ) @ x , ( t ) , shown in Fig. 2-13(b), is proportional to x ( t ) , shown in Fig. 5-13(a). Thus, replacing A by A ~ T , /in the results from Prob. 5.9, ~ we get
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which are the same results obtained by using Eq. (5.133).
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FOURIER TRANSFORM
5.16. ( a ) Verify the time-shifting property (5.50), that is,
By definition (5.31)
- to)H
e-jw'~lX ) (o
By the change of variable
- t,,, we obtain
Hence,
5.17. Verify the frequency-shifting 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,) 5-18. 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. 5-16(a)] defined by
By definition (5.31)
(4 (b) Fig. 5-16 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. 5-16(b).
5.20. Find the Fourier transform of the signal [Fig. 5-17(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. 5-1Xb)I
t Fig. 5-17 sin a t / ~ and its Fourier transform.
5.21. Find the Fourier transform of the signal [Fig. 5-18(a)]
x ( t ) = e-alfl
Signal x(t) can be rewritten as
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Fig. 5-18
e-Ial'
and its Fourier transform.
Then
Hence, we get
The Fourier transform X ( w ) of x ( l ) is shown in Fig. 5-18(b).
Find the Fourier transform of the signal [Fig. 5-19(a)]
x(t) =
a2 + t 2
From Eq. (5.138) we have
Now by the duality property ( 5 . 5 4 )we have
Fig. 5-19
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. 5-19(b).
5.23. Find the Fourier transforms of the following signals:
( a ) x(t) = 1 (c) x(t)=e-JWu' (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 5-20(a) and ( b ) illustrate the relationships in Eqs. (5.140) and ( 5 . 1 4 0 , respectively. ( b ) Applying the frequency-shifting property (5.51) to Eq. (5.1411, we get
Fig. 5-20 ( 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
e-j"~'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 5-21 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. 5-21 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. 5-22(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. 5-22(b)].
Fig. 5-22
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 frequency-shifting property (5.51) and the linearity property (5.491, we obtain
. ~ [ i x ( t ei"ut ) ;x(w
+ 'x(t) e-jU~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. 5-23(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. 5-23(b).
Fig. 5-23
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. 5-24), which is defined
The signum function, sgn(t), can be expressed as sgn(t) Using Eq. (1.301, we have
2u(t) - 1
Fig. 5-24 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)
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