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of the array, putting them in ascending order, as we can see from the output from line 8. At line 9, the Arrays.binarySearch() method is invoked. The second argument, 44, is the search target. The method returns the index 2, which is assigned to k at line 9. Line 11 verifies that 44 is indeed the value of a[2]. The method is invoked again at line 13, this time searching for the target 45. The value is not found in the array, so the method returns a negative number, k = 4. When this happens, the index i = k 1 will be the position in the array where the target element should be inserted to maintain the ascending order of the array. Note that, in this case, i = k 1 = 3, and 45 should be inserted at a[3] since there are three elements in the array that are less than 45. The output from line 17 shows how the Arrays.fill() method works: It filled the eight-element array b[] with the argument 55.

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With the window design method, the amplitudes of the ripples in each band of a multiband filter will be approximately equal, and the transition bands will have approximately the same width. Because the requirements on the peak ripple in the three bands of this bandpass filter are not the same, it is necessary to design the filter so that it has the smallest ripple in all three bands, which, in this case, requires that we set 6, = 0.0025. In addition, because the transition bands do not have the same width, it is necessary to set the desired transition width, Aw, equal to the smaller of the two (Aw = 0 . 1 5 ~ ) . With a, = -20 log 6, = 52 dB. it follows that we may use a Hamming window, and with NAf = 3.3

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For a type I1 filter, however, N must be odd, so we set N = 45. Now we must find the unit sample response of the ideal bandpass filter that is to be windowed. Because the , width of both the upper and lower transition bands will be approximately Aw = 0 . 1 5 ~for the ideal filter we set the lower cutoff frequency equal to

Finally, line 18 shows how the Arrays.equals() method works. It will return true only if the two arrays have the same element type (as a[] and b[] do: int[]), the same length (as a[] and b[] do: 8), and the same values at each element (a[] and b[] do not).

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public class TestBinarySearch { public static void main(String[] args) { int[] a = {22, 33, 44, 55, 66, 77, 88, 99}; ch02.ex02.DuplicatingArrays.print(a); System.out.println("search(a, 44): " + search(a, 44)); System.out.println("search(a, 50): " + search(a, 50)); System.out.println("search(a, 77): " + search(a, 77)); System.out.println("search(a, 100): " + search(a, 100)); } public static int search(int[] a, int x) { // POSTCONDITIONS: returns an integer i; // if i >= 0, then a[i] == x; for (int i=0; i<a.length; i++) { // INVARIANT: x is not among a[0]...a[i-1] if (a[i] == x) { return i; } } return -1; } }

Repeating the steps in the derivaticn of the unit sample response of an ideal bandpass fi lter given in Prob. 9.8, using the given cutoff frequencies and a delay of N j 3 = 22.5. we have

The java.util.Arrays class is outlined in more detail in page 95. THE SEQUENTIAL SEARCH ALGORITHM The sequential search (also called the linear search) is the simplest search algorithm. It is also the least efficient. It simply examines each element sequentially, starting with the first element, until it finds the key element or it reaches the end of the array. If you were looking for someone on a moving passenger train, you would use a sequential search. Here is the sequential search algorithm: (Postcondition: either the index i is returned where si = x, or 1 is returned.) 1. Repeat steps 2 3, for i = 0 to n 1. 2. (Invariant: none of the elements in the subsequence {s 0...s i 1} is equal to x.) 3. If s i = x, return i . 4. Return 1. It is implemented in Example 2.4. EXAMPLE 2.4 The Sequential Search

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