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x = ABCDE y = ABCDE (x == y) = false x.equals(y) = true Here, the two objects x and y (or, more precisely, the two objects that are referenced by the reference variables x and y) are different objects, occupying different memory locations, so they are not identically equal: (x == y) evaluates to false at line 7. But they do both have the same contents, so they are mathematically equal, and x.equals(y) evaluates to true at line 8.

Therefore, the homogeneous solution has the form yh(n) = A1(0.5)" Thus, the total solution is y(n) = $

+ A2(-0.5)"

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The distinction between identical equality and mathematical equality exists in Java only for reference variables (i.e., only for objects). For all variables of primitive types, the equality operator tests for mathematical equality.

+ Al(0.5)" + A2(-0.5)"

The constants A, and A2 must now be found so that the total solution satisfies the given initial conditions, y(-1) = 1 and y(-2) = 0. Because the solution given in Eq. (1.17) only applies for n 0,we must derive an equivalent set of initial conditions for y(0) and y(1). Evaluating Eq. (1.16) at n = 0 and n = 1. we have

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public class TestStringEquality { static public void main(String[] args) { String x = new String("ABCDE"); String y = new String("ABCDE"); System.out.println("x = " + x); System.out.println("y = " + y); System.out.println("(x == y) = " + (x == y)); System.out.println("x.equals(y) = " + x.equals(y)); } }

Data structures have both content and structure. So it is possible for two data structures to have equal contents (i.e., have the same contents) but be organized differently. For example, two arrays could both contain the three numbers 22, 44, and 88, but in different orders. EXAMPLE 11.7 Testing Equality of Arrays

Solving for A and A2 we find A, = -I Thus, the solution is y(n) =

- (0.5)""

The output is:

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+ d(-0.5)"

Although we have focused thus far on linear difference equations with constant coefficients, not all systems and not all difference equations of interest are linear, and not all have constant coefficients. A system that computes a running average of a signal x ( n ) over the interval [0, n ] , for example, is defined by

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public class TestArraysEquality { public static void main(String[] args) { int[] x = { 22, 44, 88 }; int[] y = { 88, 44, 22 }; ch02.ex02.DuplicatingArrays.print(x); ch02.ex02.DuplicatingArrays.print(y); System.out.println("Arrays.equals(x, y) = " + Arrays.equals(x, y)); Arrays.sort(x); Arrays.sort(y); ch02.ex02.DuplicatingArrays.print(x); ch02.ex02.DuplicatingArrays.print(y); System.out.println("Arrays.equals(x, y) = " + Arrays.equals(x, y)); } }

This system may be represented by a difference equation that has time-varying coefficients:

Although more complicated and difficult to solve, nonlinear difference equations or difference equations with time-varying coefficients are important and arise frequently in many applications.

The output it:

Determine whether or not the signals below are periodic and, for each signal that is periodic, determine fundamental period.

(a) To what analog frequency does the index k = 150 correspond What about k = 800 (b) What is the spacing between the spectral samples

{22, 44, 88} {88, 44, 22} Arrays.equals(x, y) = false {22, 44, 88} {22, 44, 88} Arrays.equals(x, y) = true This shows that the java.util.Arrays.equal() method requires not only the same contents for

(a) With a sampling frequency R, = 2n/T, = 2rr(20 . I@), the discrete frequency w is related to the analog fre-

quency R by w = QT,

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