The response to a shifted input, x(n

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- no), is

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With the substitution 1 = k

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no this becomes

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From the expression for y(n) given in Eq. (1.24, we see that

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which is the same as yl(n). Therefore, this system is shift-invariant.

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CHAP. I]

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SIGNALS AND SYSTEMS

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( h ) For the second system, h I ( n ) is nor a function of n - k. Therefore, we should expect this system to be shift~ ( 1 1 ) 6(11), the =

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varying. Let us see if we can tind an example that demonstrates that it is a shift-varying system. For the input response is

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Because g l ( n ) # y(n

I ), the system is shift-varying.

( c ) Finally, for the last system, we see that although hk(n)is a function of n - k fork even and a function of (n - k) fork odd, 11k(n)# h k - ~ ( n- 1)

In other words, the response of the system to 6(n - k - 1) is not equal to the response of the system to 6(n - k) delayed by 1. Therefore. this system is shift-varying.

Let Tr.1 be a linear system, not necessarily shift-invariant, that has a response h k ( n )to the input 6 ( n - k). Derive a test in terms of k k ( n )that allows one to determine whether or not the system is stable and whether or not the system is causal.

(a) The response of a linear system to an input ~ ( nis)

Therefore. the output may be hounded as follows:

If x(n) is bounded, Ix(n)l 5 A < W, lywi i A As a result. if

I M ~ ) I

the output will be bounded, and the system is stable. Equation (1.23) is a necessary condition for stability. To establish the sufficiency of this condition, we will show that if this summation is not finite, we can find a bounded input that will produce an unbounded output. Let us assume that hk(n) is bounded for all k and n [otherwiue the system will be unstable. because the response to the bounded input S(n - k) will be unbounded]. With hi(tl) bounded for all k and n, suppose that the sum in Eq. (1.23) is unbounded for some n, say n = no. Let x(n) = s g n ( h , ( n ~ ) l that is,

For this Input, the response at time n = no is

which, by assumption, is unbounded. Therefore, the system is unstable and we have established the sufficiency of the condition given in Eq. (1.23).

SIGNALS AND SYSTEMS

[CHAP. 1

( b ) Let us now consider causality. For an input x ( n ) , the response is as given in Eq. (1.22). In order for a system to be causal, the output y ( n ) at time no cannot depend on the input x ( n ) for any n > no. Therefore, Eq. (1.22) must be of the form y(n) = k=-m

hk(n)x(k)

This, however, will be true for any x ( n ) if and only if

which is the desired test for causality. Determine whether o r not the systems defined in Prob. 1 .I5 are (a)stable and (b) causal.

(a) For the first system, h k ( n ) = ( n

- k)u(n - k ) , note that h k ( n )grows linearly with n . Therefore, this system cannot be stable. For example, note that if x ( n ) = S(n), the output will be

which is unbounded. Alternatively, we may use the test derived in Prob. 1 .I6 to check for stability. Because

this system is unstable. On the other hand, because h,(n) = 0 for n < k , this system is causal. ( b ) For the second system, h k ( n ) = S(2n - k), note that h l ( n ) has, at most, one nonzero value, and this nonzero value is equal to I . Therefore,

for all n , and the system is stable. However, the system is not causal. To show this, note that if x ( n ) = &(n- 2 ) , the response is y ( n ) = h 2 ( n )= 6(2n - 2 ) = &(n- I) Because the system produces a response before the input occurs, it is noncausal.