2d barcode generator vb.net Verify the properties (6.62),(6.63~1, (6.63b);that is, if x [ n ] is real and and in VS .NET

Print QR in VS .NET Verify the properties (6.62),(6.63~1, (6.63b);that is, if x [ n ] is real and and

6.27. Verify the properties (6.62),(6.63~1, (6.63b);that is, if x [ n ] is real and and
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x [ n ] = x , [ n ] + x o [ n ] ++X(fl) =A(R)
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+jB(R)
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(6.140)
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where x , [ n ] and x o [ n ] are the even and odd components of x [ n ] , respectively, then X(-R) =X*(R)
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x,[n] x,[n]
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Re{X(R)} = A(i2)
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j Im{X(i2)} = j B ( f l )
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If x [ n ] is real, then x * [ n ] = x [ n ] , and by Eq. (6.45) we have
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x * [ n ] -X*(
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from which we get
x ( n )= x * ( - n )
x( - 0 ) = x * ( n )
(6.141)
Next, using Eq. (6.46) and Eqs. (1.2) and (1.3), we have
- n ] =x,[n] -x,[n]
c , -
X( - 0 ) = X * ( R ) = A ( R ) - jB(R)
= Re(X(R)}
Adding (subtracting) Eq. (6.141) to (from) Eq. (6.1401, we obtain
x,[n] x,[n]
-A(R)
jB( 0 )= j Im{ X ( R ) )
6.28. Show that
44 ++X(R>
Now, note that
s [ n ] = u [ n ] - u [ n - 11
Taking the Fourier transform of both sides of the above expression and by Eqs. (6.36) and (6.431, we have
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
Noting that (1 - e-jn) = 0 for R = 0, X ( R ) must be of the form
where A is a constant. To determine A we proceed as follows. From Eq. (1.5) the even component of u[nl is given by
u,[n]
$ + f6[n]
f - $[n]
Then the odd component of u [ n ] is given by
u o [ n ]= u [ n ] - u , [ n ] = u [ n ] -
y { u o [ n l ]= A
- e-jn
-aS(R) - 2
From Eq. (6.63b) the Fourier transform of an odd real sequence must be purely imaginary. Thus, we must have A = a , and
6.29. Verify the accumulation property (6.571, that is,
From Eq. (2.132)
Thus, by the convolution theorem (6.58) and Eq. (6.142) we get
6.30. Using the accumulation property (6.57) and Eq. (1.501, find the Fourier transform of
4nI.
From Eq. (1.50)
Now, from Eq. (6.36) we have
s[n] H 1
Setting x [ k l = 6 [ k ] in Eq. (6.571, we have
x [ n ] = 6 [ n ]H X ( R ) = 1
X(0)
FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
FREQUENCY RESPONSE
6.31. A causal discrete-time LTI system is described by
y [ n ] - + y [ n - 1 + $ y [ n - 21 = x [ n ] 1
( a ) Determine the frequency response H ( n ) of the system.
(6.143)
where x[n]and y[n] are the input and output of the system, respectively (Prob. 4.32).
( b ) Find the impulse response h[n]of the system.
( a ) Taking the Fourier transform of Eq. (6.1431, we obtain Y ( R )- i e - ' " ~ ( f l + ;e-j2'y ) or
(1 i e - i f l + Le - j 2 n ) Y ( R ) X ( R ) =
( a )= X ( R )
Thus,
( 6 ) Using partial-fraction expansions, we have H(R)= (1
- 1 - Ie-in - 1 - Le-in
Taking the inverse Fourier transform of H ( f l ) , we obtain
h [ n ] = [ 2 ( i l n- ( f ) " ] u [ n ]
which is the same result obtained in Prob. 4.32(6).
6.32. Consider a discrete-time LTI system described by
y [ n ] - ;y[n - 11 = x [ n ] + ix[n - 11
( a ) Determine the frequency response H ( n ) of the system.
( b ) Find the impulse response h[n] of the system. ( c ) Determine its response y[n] to the input
~ [ n=]cos-n 2
( a ) Taking the Fourier transform of Eq. (6.1441, we obtain Y ( R ) - i e - j n Y ( R )= X ( R ) + ; e - j n x ( R )
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
Thus,
Taking the inverse Fourier transform of H(R), we obtain
(c) From Eq.(6.137)
Then
Taking the inverse Fourier transform of Y(R) and using Eq. (6.1351, we get
6.33. Consider a discrete-time LTI system with impulse response
Find the output y[n] if the input x [ n ] is a periodic sequence with fundamental period No = 5 as shown in Fig. 6-17.
From Eq. (6.134) we have
and 7r/4, only Since R, = 27r/NO = 2 ~ / 5 the filter passes only frequencies in the range lRl I the dc term is passed through. From Fig. 6-17 and Eq. (6.11)
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSI'EMS [CHAP. 6
-2-10
1 2 3 4 5
Fig. 6-17
Thus, the output y[nl is given by ~ [ n= l
all n
634. Consider the discrete-time LTI system shown in Fig. 6-18.
(a) Find the frequency response H ( n ) of the system.
( b ) Find the impulse response h [ n ]of the system. ( c ) Sketch the magnitude response IH(n)I and the phase response N R ) . ( d ) Find the 3-dB bandwidth of the system.
From Fig. 6-18 we have y [ n ] = x [ n ] + x [ n - 11 Taking the Fourier transform of Eq. (6.145) and by Eq. (6.77), we have
(6.145)
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