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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,

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

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Path: euler, 295 hamiltonian, 296 Peek at a stack top, 103 Permutation recursive method, 177 Pivot element, 263 Polymorphism, 14 inclusion, 14 overloading, 15 universal, 15 Polynomial, 141 Pop from a stack, 103 Postcondition, 256 Postfix representation, 109, 113, 210 Postorder traversal algorithm, 208 Precondition, 256 Prefix representation, 113, 210 Preorder traversal algorithm, 208 Primes class, 35 Primitive type, 5 Print queue, 87 Priority queue, 90, 247 PriorityQueue class, 69 private access, 19 private modifier, 8, 9 Program, 19 Promotion, 98 protected access, 19 protected modifier, 8, 9 public access, 19 public modifier, 8, 9 Push onto a stack, 103 Queue, 87, 99, 247 priority, 247 Queue class, 69 Queue interface, 69, 99 Quick Sort, 263 Quick sort algorithm, 263 Radix sort algorithm, 270 Ragged array, 27 Recursion: direct, 179 indirect, 179 Recursive algorithm, 165 Recursive binary search algorithm, 30, 168 Recursive call, 177 Recursive function, 165 Recursive part of recursive algorithm, 166 Reference type, 5 Reverse Polish notation, 109, 210 RPN, 109 RPN calculator, 109

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( 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)

Natural mapping, 206, 219, 245 Nested class, 5, 59 62 No-arg constructor, 7, 143 Nonstatic method, 10 null value, 27 Object, 2

java.util.Arrays, 15 LinkedHashMap, 70 LinkedHashSet, 69 LinkedList, 69 Map, 69

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.

- 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

Object.clone() method, 28

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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,

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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.

Rewriting this system function as a ratio of polynomials, it follows that the locations of the zeros of H ( z ) will depend on the locations of poles as well as the zeros of H,,(s), and there is no way 10 guarantee that the zeros lie inside the unit circle. A simple example showing that a minimum phase continuous-time filter will not necessarily be mapped to a minimum phase discrete-time tilter is the following:

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