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cos(mn -

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Therefore, the frequencies in the output are w = w f r j 4 , which are different from those in the input. This is o because the modulator is a shift-varying system. (c) The last system, called a down-sampler, produces the output

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thus creating frequencies in the output that are not present in the input. The down-sampler is a shift-varying system.

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For each of the following pairs of signals, x ( n ) and y(n), determine whether or not there is a linear shift-invariant system that has the given response, y(n), to the given input, x(n). If such a system exists, determine whether or not the system is unique, and find the frequency response of an LSI system with the desired behavior. If no such LSI system exits, explain why.

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x ( n ) = (i)"u(n), y(n) = (a)"u(n)

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(b) x ( n ) = e j n n f 4 , y(n) = 0 . 5 c j " " / 4 sin(nnf4) , y(n) - s i n ( n n f 2 ) (c) x ( n ) = 7

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x ( n ) = u(n), y(n) = 6(n)

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CHAP. 21

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FOURIER ANALYSIS

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(a) For the first input-output pair, we have

Because X(ejW)is nonzero for all w, the system that produces the response y(n) is unique and is given by

(b) For the second system, note that the input is a complex exponential with a frequency w = n / 4 . Therefore, if the system is LSI, the output must be a complex exponential of exactly the same frequency, that is, Y ( n ) = ~ ( ~ j )e h n / 4 W 4 Because the output is

y ( n ) = 0.5 ejnnI4

any LSI system with

H ( e J n l 4 = 0.5 )

will produce the given response. Thus, the system is not unique. One possible system is the low-pass filter

H (eJ") =

otherwise

(c) For the third system, recall that an ideal low-pass filter with a cutoff frequency w,. has a unit sample response given by (see Example 2.2.3) sin n o , h(n) = rn

Therefore, the DTFT of the input x ( n ) is

X(el") =

and the DTFT of the output y(n) is

otherwise

Y reJ") =

otherwise

Because X ( e j w ) = 0 for Iwl n/4, if the system is to be linear and shift-invariant, Y (el") must be equal to zero for lo1 > n / 4 (an LSI system cannot produce new frequencies). Because this is not the case, no LSI system will produce the given input-output pair. (d) For the last system, we are given x ( n ) = u(n) and y(n) = S(n). Therefore, and As in part (a),there is a unique LSI system that produces this input-output pair, and the frequency response of this system is

22 .3

Find the DTFT of the two-sided sequence

FOURIER ANALYSIS Note that we may write x ( n ) as the sum of a left-sided sequence and a right-sided sequence as follows:

[CHAP. 2

where the last term is included to remove the extra term that is introduced at n = 0 by the two exponential sequences. The DTFT of the first term is

and, using the time-reversal property, it follows that the DTFT of the second term is

Therefore,

I - 1--

+--I I

- Le,w

Use the orthogonality of the complex exponentials

to show that x ( n ) may be recovered from X(eju) as follows:

Given a sequence x ( n ) , the DTFT is defined by

To recover x ( n ) from X(eJm),it is necessary to "filter out" all of the terms in the sum except one (i.e., we must isolate a single term in the sum). This may be done by multiplying both sides of the equation by a complex exponential, elnw:

and integrating from -n to n,

Interchanging the order of the integral and the sum on the right gives

Using the orthogonality of the complex exponentials, it follows that the integral is zero when k # n , and it is equal to 2n when k = n. Therefore,