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(c) For the last system, note that
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A=-A add
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Su(n - k ) =
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which is unbounded. Therefore, this system is unstable. Finally, because h k ( n ) = 0 for n < k , the system is causal. Consider a linear system that has a response to a delayed unit step given by
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That is, s k ( n ) is the response of the linear system to the input x ( n ) = u ( n - k ) . Find the response of this system to the input x ( n ) = 6 ( n - k ) , where k is an arbitrary integer, and determine whether o r not this system is shift-invariant, stable, o r causal. Because this system is linear, we may find the response, h k ( n ) ,to the input &(n- k ) as follows. With &(n- k ) = u(n - k ) - u(n - k - I), using linearity it follows that
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which is shown below:
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CHAP. 11
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SIGNALS AND SYSTEMS
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From this plot, we see that the system is not shift-invariant, because the response of the system to a unit sample changes in amplitude as the unit sample is advanced or delayed. However, because h k ( n )= 0 for n < k, the system is causal. Finally, because h k ( n )is unbounded as a function of k, it follows that the system is unstable. In particular, note that the test for stability of a linear system derived in Prob. 1.16 requires that
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For this system,
Note that in evaluating this sum, we are summing over k. This is most easily performed by plotting h k ( n )versus n as illustrated in the figure below.
Because this sum cannot be bounded by a finite number B, this system is unstable. Because this system is unstable, we should be able to find a bounded input that produces an unbounded output. One such sequence is the following:
The response is
y ( n ) = n ( - l)"u(n)
which is clearly unbounded.
Consider a system whose output y ( n ) is related to the input x ( n ) by
Determine whether o r not the system is (a) linear, (b) shift-invariant, ( c ) stable, (d) causal.
SIGNALS AND SYSTEMS
[CHAP. I
(a) The first thing that we should observe about y(n) is that it is formed by summing products of .r(n) with shifted versions of itself. For example,
y(O) =
I=-w
.r2(k)
We expect, therefore, this system to be nonlinec~r. Let us confirm this by example. Note that if .r(n) = 6(n), y(n) = S(n). However, if x(n) = 2S(n), y(n) = 46(n). Therefore. the system is not homogeneous and, consequently, is nonlinear. (b) For shift-invariance, we want to compare y(n - no) = to the response of the system to x l ( n ) = x(n
C x(k)x(n
I=-n;
+ k)
rill). which is
where the last equality follows with the substitution k' = k - ncl. Because y , ( n ) # y(n not shift-invariant.
nu), this system is
(c) For stability, note that if x(n) is a unit step, y(0) is unbounded. Therefore, this system is unstable.
(d) Finally, for causality, note thal the output depends on the values of .t (11) for all n. For example, y(O) is the sum of the squares of x(k) for all k. Therefore, this system is not causal.
Given that x ( n ) is the system input and y ( n ) is the system output, which of the following systems are causal
(d) y ( n ) = r ( n ) - x ( n 2
y(n) = n x ( n - k )
(a) The system y(n) = r2(n)u(n) is rnernoryless (i.e.. the response of the system at time n depends only on the
input at time n and on no other values of the input). Therefore, this system is causal. when (b) The system y(n) = x(ln1) is an example of a noncausal system. This may be seen by looking at the outpu~ n < 0. In particular, note that y(- I) = s ( l ) . Therefore. the output of the system at time 11 = -1 depends on the value of the input at a future time.
(c) For this system, in order to compute the output y(n) at time n all we need to know is the value of the input x ( n ) at times n, n - 3, and n - 10. Therefore. this system must be causal.
(d) This system is noncausal, which may be seen by evaluating v(n) for 11 < 0. For example,
Because y(- I) depends on the value of .r(2), which occurs after time n = - I , this system is noncausal
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