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Find the relationship between the parameters 6, and 6, for the discrete-time filter and between the parameters c and A for the continuous-time filter.

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15.19 For the weighted digraph G1 shown in Figure 15.44 on page 311. a. Draw the adjacency matrix. b. Draw the adjacency list. 15.20 Perform the indicated traversal algorithm on the graph shown in Figure 15.45 on page 311. Give the order of the vertices visited and show the resulting spanning tree: a. Trace the breadth-first search starting at node A. b. Trace the depth-first search of the graph, starting at node A and printing the label of each node when it is visited.

15.4 15.5 15.6

and With a stopband ripple of &, the normalization of the peak passband gain to I produces a peak stopband ripple of

+ 8,)

CHAP. 15]

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The graph G1 cannot be eulerian because it has odd degree vertices. But the hamiltonian cycle shown in Figure 15.51 on page 314 verifies that it is hamiltonian. The graph G2 is neither eulerian nor hamiltonian. a. b. c. d. e. f. g. h. i. j. k. l. Disconnected, cyclic, and spanning. Disconnected, acyclic, and spanning. Disconnected, cyclic, and spanning. Disconnected, cyclic, and not spanning. Connected, acyclic, and spanning. Connected, acyclic, and spanning. Connected, cyclic, and spanning. Connected, acyclic, and not spanning. Disconnected, cyclic, and spanning. Connected, cyclic, and not spanning. Disconnected, acyclic, and not spanning. Connected, acyclic, and not spanning.

As the order of an analog Butterworth filter is increased, the slope of J H , ( J Q ) ~ the 3-dB cutoff at ~ increases. Derive an expression for the slope of I H , ( J Q ) at~ as a function of the filter I R,. frequency, Q,., order, N.

To evaluate the slope of IH a ( jSt)I2 at St = St,., we may set St, = 1 and evaluate the derivative at St = I. Therefore, with 1 l H a ( j ~ )= l~ I

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15.1 15.2 15.3 15.4 A graph is simple if it has no loops or repeated edges. No: In an undirected graph, an edge cannot be a path because an edge is a set of two elements (i.e., an unordered pair) while a path is a sequence (i.e., an ordered list of vertices). Two vertices are connected if there is a path from one to the other. Two vertices are adjacent if they form an edge. Using only the definition of graph isomorphism, it is easier to prove that two graphs are isomorphic because it only requires finding an isomorphism and verifying that it is one. Proving from the definition that two graphs are not isomorphic would require verifying that every one of the n! one-to-one functions is not an isomorphism. No: The five conditions of are not sufficient for two graphs to be isomorphic. It is possible for all five conditions to be true for two nonisomorphic graphs. (See Problem 15.7.) The reason that the natural definition of a graph prohibits loops is that an edge in a graph is a two-element set, and that requires the two elements to be different. In the natural definition of a digraph, an edge is an ordered pair, and that allows both components to be the same. a. b. c. d. e. f. g. True True True True False True True

The two graphs shown in Figure 15.52 on page 314 are not isomorphic because the one on the left has a 4-cycle containing two vertices of degree 2 and the one on the right does not. Yet, all five conditions of Theorem 15.4 on page 289 are satisfied.

Show that the frequency response of an Nth-order low-pass Butterworth filter is maximallyflat at R = 0 in the sense that the first 2N - I derivatives of J H , ( J R )are equal to zero at R = 0. I~

Without any loss in generality, we may assume that Q, = 1 and evaluate the derivative of the function G(Q) = at Q = 0. Multiplying both sides of this equation by (1

15.5 15.6

15.8 15.9

Again setting R = 0, we have G"(Q)],=, = 0 If we continue to differentiate k times, where k 5 2N - I, we have an equation of the form

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