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As shown in Fig. 5-25, u(t) can be expressed as Note that $ is the even component of u(t) and sgn(t) is the odd component of d t ) . Thus, by Eqs. (5.141) and (5.153) and the linearity property (5.49) we obtain
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Fig. 5-25 Unit step function and its even and odd components.
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531. Prove t h e time convolution theorem (5.581, that is,
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-4)4 1 ) -X,(4X*(4 *By definitions (2.6) and (5.311, we have
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Changing the order of integration gives
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FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
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[CHAP. 5
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By the time-shifting property (5.50)
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Thus, we have
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5.32. Using the time convolution theorem (5.58), find the inverse Fourier transform of X ( w ) = l/(a j ~ ) ~ .
From Eq. ( 5 . 4 5 ) we have
Thus, by the time convolution theorem (5.58) we have
x ( r ) = e-"u(r)
* e-"u(t)
Hence,
5.33. Verify the integration property (5.57), that is,
From Eq. (2.60)we have
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Thus, by the time convolution theorem (5.58) and Eq. (5.154) we obtain
since X ( o ) S ( o )= X(O)S(w) by Eq. (1.25). Thus,
5.34. Using the integration property ( 5 . 5 7 ) and Eq. ( 1 . 3 1 ) , find the Fourier transform of
u(t).
From Eq. (1.31) we have
( t )=
S(r)dr
Now from Eq. (5.140)we have
S(t)
Setting x ( r ) = S(7) in Eq. (5.571, we have
x(t) =s ( t )- X ( o )
X(0) = 1
5.35. Prove the frequency convolution theorem (5.591, that is,
x d f)x2W
"% X I ( 4 * X 2 ( 4
By definitions (5.31) and (5.32) we have
. F [ x l ( t ) x 2 ( t )= ]
jm~ ~ ( t ) x , ( t ) e - ~ " ' d t m
X,(h)eJ" dA ~ , ( t ) e - ~ " ' d t
X,(A)
x,(t)e-j("-*)'dt
1 -/ ~-,( A ) X , ( o - A) d l X, 27
1 -X,(o) 27r
* X,(o)
Hence,
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
[CHAP. 5
5.36. Using the frequency convolution theorem (5.59), derive the modulation theorem (5.148).
From Eq. (5.144) we have cos w,t
a6(w - 0 , ) + a 6 ( 0 + w 0 )
By the frequency convolution theorem (5.59) we have
~ ( t cos 0 , t )
1 -X(0)*
[ a S ( w - 00)
+ d ( 0+w ~ ) ]
= + X ( O - 0,)
+ $X(w + 0 , )
The last equality follows from Eq. (2.59).
5.37. Verify Parseval's relation (5.63), that is,
From the frequency convolution theorem (5.59) we have
Setting w = 0, we get
By changing the dummy variable of integration, we obtain
5.38. Prove Parseval's identity [Eq. (5.6411 or Parseval's theorem for the Fourier transform,
that is,
By definition (5.31) we have
where
denotes the complex conjugate. Thus,
x*(t)-X*(-w)
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Setting x,(t) = x(r) and x2(t)=x*(t) in Parseval's relation (5.631, we get
5.39. Show that Eq. (5.61a), that is,
X*(O) = X ( - 0 )
is the necessary and sufficient condition for x ( t ) to be real.
By definition (5.31)
If x(t) is real, then x*(t) = x(t) and
X * ( o ) = X(- o). From the inverse Fourier transform definition (5.32)
Thus, X*(w) = X(-w) is the necessary condition for x(t) to be real. Next assume that
Then
which indicates that x(t) is real. Thus, we conclude that X * ( W )= X ( - W ) is the necessary and sufficient condition for x(t) to be real.
5.40. Find the Fourier transforms of the following signals:
(a) x ( t ) = u ( - t )
( b ) x(t)=ea'u(-t), a > O
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
From Eq. (5.53) we have
X(- t ) -X(-w)
Thus, if x ( t ) is real, then by Eq. ( 5 . 6 1 ~we have )
x( -1) - X ( - 0 ) ( a ) From Eq. (5.154) =X*(w)
Thus, by Eq. (5.158) we obtain
( b ) From Eq. (5.155) e-"'u(t)
a+jw
Thus, by Eq. (5.158) we get
5.41. Consider a real signal x ( t ) and let
X ( o ) = F [ x ( t ) ] =A ( w )
+j B ( o )
where x,(t) and x,(t) are the even and odd components of x ( t ) , respectively. Show that
x&) + N o ) x,( t ) - j B ( o
(5.161~)
(5.161b)
From Eqs. (1.5) and (1.6) we have
x , ( t ) = f [.(I)
+X( -t ) ] +
x,(t) = f [ x ( t ) - x ( - t ) ]
Now if x ( t ) is real, then by Eq. (5.158) we have
X(t) H X ( O ) = A ( @ ) jB(w) x ( - t ) H X ( - W ) = X * ( W )= A ( w ) - j B ( w )
Thus, we conclude that
x,(t) - ~ x ( w ) x,(t)
+ ;X*(O) =A(w)
=j B ( w )
~ x ( w ) $x*(o) -
Equations (5.161~) (5.161b) show that the Fourier transform of a real even signal is a real and function of o,and that of a real odd signal is an imaginary function of w, respectively.
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
5-42, Using Eqs. ( 5 . 1 6 1 ~ ) (5.1551, find the Fourier transform of e-"Itl ( a > 0). and
From Eq. (5.155)we have
By Eq. (1.5) the even component of e-"'u(t) is given by
Thus, by Eq. ( 5 . 1 6 1 ~ ) have we
which is the same result obtained in Prob. 5.21 [Eq. (5.138)l.
5.43. Find the Fourier transform of a gaussian pulse signal
By definition (5.31)
Taking the derivative of both sides of Eq. (5.162)with respect to o,we have
Now, using the integration by parts formula
and letting
-jwl
do =
we have
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
since a > 0. Thus, we get dX -- ( w ) - - -w ( w ) X dw
Solving the above separable differential equation for X ( w ) , we obtain
where A is an arbitrary constant. To evaluate A we proceed as follows. Setting w Eq. (5.162) and by a change of variable, we have
Substituting this value of A into Eq. (5.1631, we get
Hence, we have
Note that the Fourier transform of a gaussian pulse signal is also a gaussian pulse in the frequency domain. Figure 5-26 shows the relationship in Eq. (5.165).
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