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page 50) does not initialize it. In Java, every class field that is an object reference (i.e., its type is either a class or an interface) is automatically initialized to null, unless it is initialized by its constructor to some existing object. In the figures that follow, each Node object is shown as a box with two parts: the left side contains the integer data, and the right side contains the next reference. This simply abbreviates the versions shown in Figure 3.9. Continuing the code in Example 3.4, at line 2, the start node s next field is assigned to a new Node object containing the data 33. Now the list has two nodes, as shown in Figure 3.12. Figure 3.12 Adding a node

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Using the symmetry of r , ( n ) , we have, for n < 0,

with initial conditions y ( - I ) = 0.75 and y(-2) = 0.25.

Now, instead of an array a[], we need only keep track of the single start reference. The Java runtime system does all the rest of the bookkeeping. The code is given in Example 3.3. EXAMPLE 3.3 A Node Class

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The next node is added to the end of the list at line 3. To do that, we have to assign it to the next field of the node that contains 33. But the only node to which we have external access (i.e., the only node that has a variable name) is the first node. Its name is start. So we have to use the expression start.next.next to refer to the next field of the node that contains 33. Similarly, the fourth node is added at line 4 using the expression start.next.next.next, and the fifth node is added at line 5 using the expression start.next.next.next.next. That finally gives us the five-node list shown in Figure 3.13.

The first step in solving this difference equation is to find the particular solution. With x ( n ) = (OS)"u(n), assume we a solution of the form y,(n) = C l ( 0 . 5 ) " n 20 Substituting this solution into the difference equation, we have

c, = ;

1 2 3

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The code in Example 3.4 is clumsy and unsuited for generalization. Obviously, if we wanted to build a linked list of 50 nodes, this approach would be unworkable. The solution is to use a local reference variable that can walk through the list, locating one node after the Figure 3.13 The five-node list other and thereby giving local access to the nodes. Traditionally, the variable p (for pointer ) is used for this purpose. Since it will refer to individual nodes, it should be declared to be a Node reference, like this:

Therefore, the form of the homogeneous solution is yh(n) = and the total solution becomes y(n) = (0.5)"+' ein"l> ~~~-1nnl3

The constants A1 and A2 must now be found so that the total solution satisfies the given initial conditions, y(-1) = 0.75 and y(-2) = 0.25. Because the solution given in Eq. (1.25) is only applicable for n > 0, we must derive an equivalent set of initial conditions for y(0) and y(l). Evaluating the difference equation for n = 0 and n = I, we have y(0) = y(- 1) - y(-2) and y(l) = y(0) - y(-1)

class Node { int data; Node next;

Node p;

+ 0 . 5 ~ ( 1+ OSx(0) = 1 - 0.75 + 0.25 + 0.5 = 1 )

Now, substituting these derived initial conditions into Eq. (1.25), we have y(O) = 0.5 Al + A2 = I y( I) = 0.25 A ,ei"I3 ~

CHAP. 3]

And since our only access to the nodes is from the start node, we should initialize p like this:

in matrix form,

4 5 6 7 8

$~jn/3- 2

Node p=start;

= I T [-teinp+;]

Node(int data) { this.data = data; } }

An important observation to make about this solution is that. because the difference equation has real coefficients, the roots of the characteristic polynomial are in complex-conjugate pairs. This ensures that the unit sample response is real. With a real-valued input x(n), the response must be real and, therefore, it follows that A2 will be the complex conjugate of A 1 :

p = p.next;

(h) Find the system's response to the input x ( n ) = u(n) - u(n - 10) with zero initial conditions.

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