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Printing Code 128 Code Set A in Software Dividing both sides by 2n gives the desired result.

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CHAP. 21
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Find the inverse DTFT of X(e1"') shown in the figure below:
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X(eJU)
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Because X ( e J U )is a piecewise constant function of w , finding the inverse DTFT may be easily accomplished by integration. Using the inverse DTFT, we have
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Rearranging the terms, we have
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which is the desired result. It is interesting to note that x ( n ) is expressed as the difference of two sequences, with the first being an ideal low-pass filter with a cutoff frequency of 3 x 1 4 , and the second an ideal low-pass filter with a cutoff frequency of n/4.This is a consequence of the fact that X(ei"') may be expressed as
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where
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X , ( e J " )=
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Iwl <
otherwise
X 2 ( c J W= )
Iwl <
otherwise
Another way to evaluate the inverse DTFT is to observe that x ( e i w )may be written as
X(eJ")= X ~ ( ~ J ' " ' + + X , e~'co- ) )
where X z ( e j w )is the ideal low-pass filter defined above. Thus, X(eJU') may be viewed as a modulated low-pass filter:
With
sin(na/4) x2(n) = nrc
x ( n ) may also be written as
This may be shown to be equivalent to the previous representation for x ( n ) by using the trigonometric identity 2sin A sin B = sin(A
+ B ) + sin(A
FOURIER ANALYSIS
[CHAP. 2
Find the inverse DTFT of x ( ~ J " )= cos2 w.
Recall that the DTFT of a delayed unit sample is a complex exponential:
S(n - no)
DTFT -inow e
Therefore, the inverse DTFT of X ( e j W )= cos2 w may be found easily if we expand it in terms of complex exponen-
Thus, it follows that x ( n ) is
x(n) = iS(n)
+ as(n + 2 ) + fS(n - 2 )
If h ( n ) is the unit sample response of a r e a l and c a u s a l linear shift-invariant system, show that the system is completely specified by the real part of its frequency response:
In other words, show that H ( e J W )may be uniquely recovered from its real part.
Recall from the symmetry properties of the DTFT that if h ( n ) is real, H (el") is conjugate symmetric. Therefore, if H ( e J w )is written in terms of its real and imaginary parts,
then the real part, HR(ei"), is the DTFT of the even part of h(n):
Therefore, given H R ( e J W )or h,(n), the question is how to recover h(n). Note that if h ( n ) is causal, h ( n ) = 0 for , n < 0 , and ih(n) n >0
n=O kh(-n)
As a result, h ( n ) may be recovered from h , ( n ) as follows:
If h ( n ) is real and causal, and if
HR(el") = ~ e { ~ ( e J "= j1 )
+ a cos 2 0
find h ( n ) .
Because the real part of H ( e J w )is
~ ~ ( e j1 + a)cos 2 w = I = ~
+ faeJ2" + ;aeFizw
the even part of h ( n ) , which is the inverse DTFT of H R ( e J w )is ,
h,(n) = 6(n)
+ iaS(n + 2 ) + $d(n
With h ( n ) a causal sequence, it follows from the results of Prob. 2.27 that
which gives
CHAP. 21
FOURIER ANALYSIS
DTFT Properties
Show that if x ( e j w )is real and even, x ( n ) is real and even. For x ( n ) we have
) If ~ ( e j " is real and even, then X(ej'") sin(nw) is real and odd. Therefore, when integrated from -n t o n , the integral is zero. Thus, x ( n ) may be written as
and it follows that x ( n ) is real. Finally, because X ( e j W ) cos(nw) is real and even, x ( n ) is real and even, that is,
Prove the convolution theorem. There are several ways to prove the convolution theorem. One way is by a straightforward manipulation of the DTFT sum. Specifically, if y ( n ) = h ( n ) * .u(n),
and the DTFT of y ( n ) is
Note that the expression in brackets is the DTFT of x(n - 1 ) . Using the delay property of the DTFT, this is equal to x(ejW)e-J'", and the right side of this equation becomes
) Factoring out X ( e J W from the sum, which does not depend on I, we have
which proves the theorem. Another way to prove the convolution theorem is to consider the following cascade of two LSI systems, one with a unit sample response of h ( n ) and the other with a unit sample response of x ( n ) :
If the input to this cascade is a complex exponential, ej"", the output of the first system is H ( e j w ) e j n wBecause this . complex exponential is the input to the second system, the output is H(ej")X(ej")eJn". Therefore, H ( e j " ) X ( e j w ) is the frequency response of the cascade, and because the unit sample response of the cascade is the convolution h ( n ) * x ( n ) , we have the DTFT pair h(n) * x(n) & T D H ( e j w ) x(dw) which establishes the convolution theorem.
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