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we see that a difference equation for this system is
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Performing Convolutions
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Because the DTFT maps convolution in the time domain into multiplication in the frequency domain, the DTFT provides an alternative to performing convolutions in the time domain. The following example illustrates the procedure.
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EXAMPLE 2.7.2 If the unit sample response of an LSI system is
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let us find the response of the system to the input x ( n ) = Bnu(n),where (a1< 1, the system is the convolution of x ( n ) with h(n),
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y(n) = h(n)* x(n)
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< 1, and a
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# B. Because the output of
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the DTFT of y ( n ) is
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Y ( e J W= H ( e J w W ( e J w= ) )
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Therefore, all that is required is to find the inverse DTFT of Y (el"). This may be done easily by expanding Y (ej") as follows:
FOURIER ANALYSIS
[CHAP. 2
where A and B are constants that are to be determined. Expressing the right-hand side of this expansion over a common denominator,
and equating coefficients, the constants A and B may be found by solving the pair of equations
The result is
Therefore, and it follows that the inverse DTFT is
2.73 Solving Difference Equations
In Chap. 1 we looked at methods for solving difference equations in the "time domain." The DTFT may be used to solve difference equations in the "frequency domain" provided that the initial conditions are zero. The procedure is simply to transform the difference equation into the frequency domain by taking the DTFT of each term in the equation, solving for the desired term, and finding the inverse DTFT.
EXAMPLE 2.7.3 Let us solve the following LCCDE for y ( n ) assuming zero initial conditions,
for x ( n ) = &).
We begin by taking the DTFT of each term in the difference equation:
Because the DTFT of x ( n ) is X ( e j w ) = 1,
Using the DTFT pair
the inverse DTFT of Y (ej") may be easily found using the linearity and shift properties,
2.7.4 Inverse Systems
The inverse of a system with unit sample response h(n) is a system that has a unit sample response g(n) such that
CHAP. 21
FOURIER ANALYSIS
In terms of the frequency response, it is easy to see that, if the inverse of H ( e J m exists, it is equal to )
Care must be exercised, however, because not all systems are invertible or, if the inverse exists, it may be noncausal. For example, the ideal low-pass filter in Example 2.2.3 does not have an inverse, and the inverse of the system
~ ( e j " '= 1 - 2e-jm )
which corresponds to a system that has a noncausal unit sample response
g ( n ) = -2Tnu(-n EXAMPLE 2.7.4
If the frequency response of an LSI system is
the inverse system is
which has a unit sample response g(n) = (0.25)"u(n)
+ OS(0.25)"-'u(n
- 1)
Solved Problems
Frequency Response
Let h ( n ) be the unit sample response of an LSI system. Find the frequency response when
(a) h ( n ) = 6 ( n )
+ 66(n - 1 ) + 3S(n - 2 )
( b ) h ( n ) = (T) u(n - 2 ) .
(a) This system has a unit sample response that is finite in length. Therefore, the frequency response is a polynomial in ej", with the coefficients of the polynomial equal to the values of h(n): ~ ( e j " )= 1 This may be shown more formally by writing
1 n+2
+ 6e-I" + 3e-'jW
Because
then
FOURIER ANALYSIS
[CHAP. 2
(6) For the second system, the frequency response is
Changing the limits on the sum so that it begins with n = 0,we have
Using the geometric series, we find
An Lth-order moving average filter is a linear shift-invariant system that, for an input x(n), produces the output
Find the frequency response of this system.
If the input to the moving average filter is x(n) = S(n), the response, by definition, will be the unit sample response, h(n). Therefore,
and Using the geometric series, we have
Factoring out a term e-j'L+')w/2from the numerator, and a term e - ~ " /from the denominator, we have ~
= H(eJW) ~ f l e
,,,,, sin(L + 1 1 4 2
sin 0 / 2
The input to a linear shift-invariant system is
Find the output if the unit sample response of the system is h(n) = 2 sin[(n - l)x/2] (n - 1)x
This problem may be solved using the eigenfunction property of LSI systems. Specifically, as we saw in Example 2.2.1, if the input to an LSI system is x(n) = cos(nq,), the response will be ~ ( n = lH(eiw0)lcos(nwo )
+ #dm))
CHAP. 21
FOURIER ANALYSIS
Therefore, we need to find the frequency response of the system. In Example 2.2.3, it was shown that the unit sample response of an ideal low-pass filter,
sin nw, h i @ )= 7 n Because h(n) = 2hl (n - 1) with w, = 1712,an expression may be derived for H ( e J W ) terms of H , (ej") as follows: in
Therefore,
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