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+ yI Scanning QR Code ISO/IEC18004 In VS .NET Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in .NET applications. Print QR Code In .NET Framework Using Barcode encoder for .NET Control to generate, create QR Code ISO/IEC18004 image in .NET applications. As shown in Fig. 525, 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 QRCode Scanner In .NET Using Barcode recognizer for .NET Control to read, scan read, scan image in Visual Studio .NET applications. Painting Bar Code In Visual Studio .NET Using Barcode creator for .NET framework Control to generate, create barcode image in Visual Studio .NET applications. Fig. 525 Unit step function and its even and odd components.
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Matrix 2D Barcode Generation In VS .NET Using Barcode printer for .NET framework Control to generate, create 2D Barcode image in .NET framework applications. Code 39 Full ASCII Creator In .NET Framework Using Barcode printer for VS .NET Control to generate, create ANSI/AIM Code 39 image in Visual Studio .NET applications. Changing the order of integration gives
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Generating Barcode In None Using Barcode printer for Font Control to generate, create barcode image in Font applications. Generating GTIN  128 In None Using Barcode creation for Font Control to generate, create EAN / UCC  14 image in Font applications. [CHAP. 5
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Draw Bar Code In Java Using Barcode drawer for Java Control to generate, create bar code image in Java applications. Reading Barcode In .NET Using Barcode scanner for .NET Control to read, scan read, scan image in VS .NET applications. 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)ej("*)'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
542, 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 526 shows the relationship in Eq. (5.165).

