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Given that there is a single a particle in the reactor at time n = 0, find an expression for the total number of particles within the reactor at time n. Let a ( n ) and B ( n ) be the number of a particles and B particles within the reactor at time n. The behavior within the reactor may be described by the following pair of coupled difference equations:

p.next = new Node(44); p = p.next; p.next = new Node(55); p = p.next; p.next = new Node(66);

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Before we can solve these difference equations, we must uncouple them. Therefore, let us derive a single difference equation for B(n). From the first equation we see that a ( n ) = # (n - I). Substituting this relation into the second difference equation, we have

The third part gives the update expression, advancing the control variable to the next element:

B(n+ 1)=8B(n1)+2B(n)

This code may not seem much better than the other version in Example 3.4. But one big advange is that it is easily managed within a loop. For example, the same list can be built with the three lines of code in Example 3.6.

or, equivalently,

Example 3.9 shows a test driver for a simple external Node class. EXAMPLE 3.9 Testing the Node Class

- 2)

1 2 3 4

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With the initial conditions a ( 0 ) = I and B(0) = 0, we may solve for A1 and A2 as follows:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

a ( n ) = i(4)" B ( n ) = :(4)"

Node start = new Node(22), p = start; for (int i=0; i<4; i++) { p = p.next = new Node(33+11*i); }

+ !(-2)"

public class TestNode { public static void main(String[] args) { Node start = new Node(22); Node p = start; for (int i = 1; i < 5; i++) { p = p.next = new Node(22 + 11*i); } for (p = start; p != null; p = p.next) { System.out.println(p.data); } for (p = start; p != null; p = p.next) { System.out.println(p); } } } class Node { int data; Node next; Node(int data) { this.data = data; } }

- (-2)" 3

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Obviously, this form could just as easily build a linked list of 50 nodes. Each step in the execution of this code is shown in Figure 3.16.The reference variable p is analogous to an array index i: It

The input to a linear shift-invariant system is periodic with period N . (a) Show that the output of the system is also periodic with period N. (b) If the system is linear but shift-varying, is the output guaranteed to be periodic (c) If the system is nonlinear but shift-invariant, is the output guaranteed to be periodic

The output is:

1.43 1.44

advances through the nodes of a linked list just as i advances through the elements of an array. Consequently, it is natural to use p in a for loop, just as we would use the array index i. For example, compare Example 3.7 with Example 3.8.

If x(n) = 0 for n < 0, and the odd part is x,,(n) = n(0.5)1n1, x(n) given that x(0) = 1. find Find the conjugate symmetric part of the sequence

22 33 44 55 66

1.45 1.46

1 2 3

CHAP. 3]

If x(n) is odd, what is y(n) = x2(n)

If x(n) = 0 for n < 0, Pe is the power in the even part of x(n), and Po is the power in the odd part, which of the following statements are true (a) Pc Po (b) Po 2 Pe (c) Pe = Po

for (Node p = start; p != null; p = p.next) { System.out.println(p.data); }

The first node is constructed at line 3. Then the for loop at lines 5 7 constructs the other four nodes. The second for loop at lines 8 10 prints the node data in the first five lines of output. The third for loop at lines 11 13 gives the actual memory addresses of the five Node objects. When you use an object reference like p in a string expression such as

Express the sequence x(n) = as a sum of scaled and shifted unit steps. (-1 (0

as a sum of scaled and shifted pulses,

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