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2.12 Write and test this method:

m. If Ial < 1,

lirn h N ( n )= -- a"

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2.1 2.2 2.3 2.4 Run a test program to see how the Arrays.fill() method handles an array of objects. If the sequential search took 50 ms to run on an array of 10,000 elements, how long would you expect it to take to run on an array of 20,000 elements on the same computer If the binary search took 5 ms to run on an array of 1,000 elements, how long would you expect it to take to run on an array of 1,000,000 elements on the same computer The interpolation search is the same as the binary search except that in step 4 the element si is chosen so that the proportion of elements less than si in the subsequence ss equals the proportion that would be expected in a uniform distribution. For example, looking up the name Byrd in a phone book of 2,600 pages, one would open first near page 200 because one would expect about 2/26 of all the names to precede it. The interpolation search can be shown to run in O(lglgn) time. If it took 5 ms to run on an array of 1,000 elements, how long would you expect it to take to run on an array of 1,000,000 elements on the same computer

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void shuffle(Object[] a) // randomly permutes the elements of a[]

-I\'

which is the inverse filter, that is,

2.5 2.6

2.13 Write and test this method:

lim h N ( n )= a n u ( n )= g - l ( n )

lirn H N ( z )= I - az-'

Run a test driver for the binary search method in Example 2.5 on page 32 on an array of 10,000 elements and count the number of iterations. Write and test this method:

int[] tally(String string) // returns an array a[] of 26 integers that count the frequencies // of the (case insensitive) letters in the given string

> 1,

lirn h N ( n )= -= a-n-2

boolean isSorted(int[] a) // returns true iff a[0] <= a[1] <= ... <= a[a.length-1]

2.14 Write and test this method:

@N+ a -2 1

n 1 0

Write and test this method:

double innerProduct(double[] x, double[] y) // returns the algebraic inner product (the sum of the component// wise products) of the two given arrays as (algebraic) vectors

lirn HN( z ) =

int minimum(int[] a) // returns the minimum element of a[]

-a-Iz- 1

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2.15 Write and test this method:

which is not the inverse filter. Note that although &n) = h N ( n )* g ( n ) does not converge to S(n) as N + m, taking ) the limit o ~ B N ( zas N + co,we have

Write and test this method:

which is an all-pass filter, that is,

double[][] outerProduct(double[] x, double[] y) // returns the algebraic outer product of the two given arrays // as (algebraic) vectors: p[i][j] = a[i]*b[j]

The first five samples of the unit sample response of a causal filter are h(0) = 3 h(1) = - 1 h(2) = 1

double mean(double[] a) // returns the average value of all the elements in a[]

h(3) = 2

2.16 Write and test this method:

h(4) = 0

Write and test this method:

If it is known that the system function has two zeros and two poles, determine whether or not the filter is stable. The system function of this filter has the form

double[][] product(double[][] a, double[][] b) // returns the matrix product of the two given arrays a matrix: // p[i][j] = Sum(a[i][k]*b[k][j]:k)

FILTER DESIGN To determine whether or not this system is stable, it is necessary to find the denominator polynomial,

2.10 Write and test this method:

[CHAP, 9

2.17 Write and test this method:

and check to see whether or not the roots of A(z) lie inside the unit circle. Given that H(z) has two poles and two zeros, we may use the Pad6 approximation method to find the denominator coefficients:

void reverse(int[] a) // reverses the elements of a[]

double[][] transpose(double[][] a) // returns the transpose ta of the specified array as a matrix: // ta[i][j] = a[j][i]

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