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SIGNALS AND SYSTEMS The output of this system at time n is the product of the values of the input x ( n ) at times n - 1, . . . , n Therefore, because the output depends only on past values of the input signal, the system is causal.
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( f ) This system is not causal, which may be seen easily if we rewrite the system definition as follows:
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Therefore, the input must be known for all n 5 0 to determine the output at time n . For example, to find y ( - 5 ) we must know x(O), x ( - I), x(-2), . . .. Thus, the system is noncausal.
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Determine which of the following systems are stable:
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(b) y ( n ) = ex(")/ x ( n - 1)
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(a) Let x(n) be any bounded input with Ix(n)l c M .Then it follows that the output, y ( n ) = x2(n),may be bounded by I ~ ( n ) = lx(n)12 < M 2 l Therefore, this system is stable,
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(b) This system is clearly not stable. For example, note that the response of the system to a unit sample x ( n ) = S(n) is infinite for all values of n except n = 1.
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(c) Because Icos(x)l 5 1 for all x, this system is stable.
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(d) This system corresponds to a digital integrator and is unstable. Consider, for example, the step response of the system. With x ( n ) = u ( n ) we have, for n 2 0 ,
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Although the input is bounded, (x(n)l 5 1, the response of the system is unbounded.
This system may be shown to be stable by using the following inequality:
Specifically, if x ( n ) is bounded, Ix(n)l < M ,
Therefore, the output is bounded, and the system is stable.
( f ) This system is not stable. This may be seen by considering the bounded input x(n) = cos(nrr/l). Specifically, note that the output of the system at time n = 0 is
which is unbounded. Alternatively, because the input-output relation is one of convolution, this is a linear shift-invariant system with a unit sample response
h ( n ) = cos
SIGNALS AND SYSTEMS Because a linear shift-invariant system will be stable only if
[CHAP. 1
we see that this system is not stable.
Determine which of the following systems are invertible:
To test for invertibility, we may show that a system is invertible by designing an inverse system that uniquely recovers the input from the output, or we may show that a system is not invertible by finding two different inputs that produce the same output. Each system defined above will be tested for invertibility using one of these two methods.
( a ) This system is clearly invertible because, given the output y ( n ) , we may recover the input using x ( n ) = 0 . 5 y ( n ) . ( h ) This system is not invertible, because the value of x ( n ) at 11 = 0 cannot be recovered from y ( n ) . For example, the response of the system to X ( R )and to x l ( n ) = x ( n ) a & n ) will be the same for any a .
(c) Due to the differencing between two successive input values, this system will not be invertible. For example, note that the inputs x ( n ) and x ( n ) c will produce the same output for any value of c.
(6) This system corresponds to an integrator and is an invertible system. To show that i t is invertible, we may construct the inverse system, which is .u(n) = y ( n ) - y ( n - I) To show that this is the inverse system, note that
(e) Invertibility must hold for complex as well as real-valued signals. Therefore, this system is noninvertible because
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