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Table 15.1 shows a trace of Algorithm 15.2 on the graph shown in Figure 15.29. The start vertex is v0 = A.

(d) This system is shift-varying, which may be shown with a simple counterexample. Note that if x(n) = S(n), the response will be y(n) = 6(n). However,ifxl(n) = 6(n-2). the response will be yl(n) = xl(n2)= 6(n2-2) = 0, which is not equal to y(n - 2). Therefore, the system is shift-varying.

which is the same as the response tox(n). Because yl (n) # y(n- N), ingeneral, this system isnot shift-invariant.

CHAP. 15]

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Q A B B, E E, C E, C, F C, F F, D F, D, G D, G G F D G A, B, E, C, F A, B, E, C, F, D A, B, E, C, F, D, G

(f) This system may easily be shown to be shift-varying with a counterexample. However, suppose we use the direct approach and let x(n) be an input and y(n) = x(-n) be the response. If we consider the shifted input, x l (n) = x(n - no), we find that the response is

However, note that if we shift y(n) by no,

Table 15.3 shows a trace of Algorithm 15.4 on the same graph as in Example 15.26 (Figure 15.29 on page 301). The start vertex v0 = A. The result, of course, is the same as that in Example 15.27. The only real difference is that the explicit

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The resulting BFS order of visitation is returned in the list L = (A, B, E, C, F, D, G), and the resulting BFS spanning tree (Figure 15.30) is returned in the set T = {AB, AE, BC, BF, CD, CD}.

A linear discrete-time system is characterized by its response h k ( n ) to a delayed unit sample S(n - k). For each linear system defined below, determine whether or not the system is shift-invariant.

Here is a Java implementation of the two traversal algorithms for the Network class introduced in Example 15.25 on page 299:

(a) hk(n) = ( n - k)u(n - k )

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Algorithm 15.3 The Depth-First Search (DFS) Algorithm (Preconditions: G = (V,E) is a graph or digraph with initial vertex v0; each vertex has a boolean visited field initialized to false; T is an empty set of edges; L is an empty list of vertices.) (Postcondition: L lists the vertices in DFS order, and T is a DFS spanning tree for G.) 1. Initialize an empty stack S for temporary storage of vertices. 2. Add v0 to L. 3. Push v0 onto S. 4. Mark v0 visited. 5. Repeat steps 6 8 while S is not empty. 6. Let x be the top element on S. 7. If x has any adjacent unvisited vertices, do steps 9 13. 8. Otherwise, pop the stack S and go back to step 5.

(6) hk(n) = S(2n - k )

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

k - 1)

9. 10. 11. 12. 13.

(a) Note that hk(n)is a function of n - k . This suggests that the system is shift-invariant. To verify this, let y(n) be the response of the system to x(n):

public class Network { Vertex start; private class Vertex { Object object; Edge edges; Vertex nextVertex; boolean visited; } private class Edge { Vertex to; int weight; Edge nextEdge; } public static void visit(Vertex x) { System.out.println(x.object); }

The stopband cutoff frequency is w, = 0.4n, and the passband cutoff frequency is o,,= 0 . 5 ~ ~ . In addition, the stopband ripple is 8, = 0.0574, and the passband ripple is 6, = 0.1722. used Determine the weighting function, W ( e J W ) , to design this filter, and find the length of the unit sample response. Describe approximately where the zeros of the system function of this filter lie in the z-plane.

Let y be an unvisited vertex that is adjacent to x. Add the edge xy to T. Add y to L. Push y onto S. Mark y visited.

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