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, sn 1} is a sorted sequence of n values of the same type as x) (Postcondition: either the index i is returned where si = x, or 1 is returned) 1 Let ss be a subsequence of the sequence s, initially set equal to s 2 If the subsequence ss is empty, return 1..

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Then by definition, log b (uv ) = v y = v log b u f Let y = log b x Then by definition, x = b y, so log c x = log c (b y ) = y log c b, by Law e above Thus log b x = y = (log c x)/(log c b) a b c d e f g h i j k l True True True False True True False False False True False True + ar , so S rS.

H (z) =

0.3249 0.3249(1 + Z-I) 0.2452(1 + z-I) 1 - z-' 1 - 0.5095~-I (1 - 2-I) O.3249(1 z-I) 0.3249 1 z-'

Program to test the De Moivre s formula for Fibonacci numbers:

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Proof of Theorem A.5 on page 323: 2 3 n 1 2 3 4 + ar Let S = a + ar + ar + ar + . Then rS = a r + ar + ar + ar + n, (1 r)S = a(1 r n ), and thus S = a(1 r n )/(1 r). = a ar

( a ) Determine the locations of the poles and zeros of H ( z ) if the filter is designed using the bilinear transformation with T, = 2.

(b) Repeat part (a)for the impulse invariance technique, again with T, = 2.

public class Fibonacci { public static void main(String[] args) { final double SQRT5 = Math.sqrt(5.0); final double PHI = (1 + SQRT5)/2; final double PSI = (1 - SQRT5)/2; long f0, f1 = 0, f2 = 1; for (int n = 2; n < 32; n++) { f0 = f1; f1 = f2; f2 = f1 + f0; double fn = (Math.pow(PHI, n) - Math.pow(PSI, n))/SQRT5; System.out.print("%4d%12.2", f2, fn); } } }

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Proof of Theorem A.6 on page 323: If 1 < r < 1, then as n increases without bound, rn shrinks down to zero. Let rn =0 in the formula in Theorem A.5. Proof of Theorem A.7 on page 323: Let S = 1 + 2 + 3 + + 3 + 2 + 1 also. Add these two equations, + n . Then S = n + summing the 2n terms on the right pairwise: (1 + n), (2 + (n 1)), etc. There are n pairs, and each pair has the same sum of n + 1. So the total sum on the right is n(n + 1). Then, since the sum on the left is 2S, the correct value of S must be n(n + 1)/2.

z2 = I-h To find the zeros, it is necessary to combine the two terms in the system function over a common denominator. Doing this. we have

Finding the roots of the numerator may be facilitated by noting that H,(s) has a zero at s = co,which gets mapped to z = - 1 with the bilinear transformation. Therefore. one of the factors of the numerator is ( 1 z-I). Dividing the numerator by this factor, we obtain the second factor, which is [(2 - a - b) - 2z-'1. Thus, H ( ; ) has zeros at

Abstract data type, 3 abstract modifier, 8 AbstractList class, 69 Access package, 9 private, 19 protected, 19 public, 19 Accessor method, 7, 106 Ackermann function, 178 Acyclic graph, 288 Adel son-Velskii, G.M., 237 Adjacency list, 291, 293 Adjacency matrix, 290, 293 Aggregation, 10 Algorithm: binary search, 31, 46, 168, 241 bubble sort, 256, 320 bucket sort, 271 Dijkstra, 297 Euclidean, 171 heap sort, 265 insertion sort, 258 merge sort, 260 quick sort, 263 radix sort, 270 selection sort, 257 shell sort, 259 Almost complete binary tree, 207 Amortized constant time, 86 API, 16 Array, 26 associative, 148 generic, 106 Array component, 26, 33 Array duplication, 28 Array element, 26, 33 ArrayDeque class, 69

The output is:

z,, = i(e2"

(a) If this filter was designed using impulse invariance with T, = 2, find the system function, H&), of an analog filter that could have been the analog filter prototype. Is your answer unique

ArrayIndexOutOfBoundsException, 26, 33

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