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2 n e r, where r number of regions, e number of edges, and n number of nodes. A region is an area enclosed (or de ned) by arcs. Using algebra, this can be converted to r e n 2. Therefore, the number of regions on a planar graph equals the cyclomatic number.

Graphically z f(x, y) represents a surface, called the probability surface, as indicated in Fig. 2-4. The total volume bounded by this surface and the xy plane is equal to 1 in accordance with Property 2 above. The probability that X lies between a and b while Y lies between c and d is given graphically by the shaded volume of Fig. 2-4 and mathematically by

(20)

and L 2 H, has an applied voltage v 10 e 2t cos 10t 308 . Obtain the current i by an s-domain analysis. v 10 Since i Iest , 10 308 est 10Iest 2sIest Substituting s 2 j10, I 10 308 10 308 0:48 43:38 10 2 2 j10 6 j20 or I 10 308 10 2s 308 est Ri L di di 10i 2 dt dt

Fig. 2-4

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and C 0:2 F, has the same applied voltage as in Example 8.6. Obtain the current by an s-domain analysis. As in Example 8.6, 1 i dt 10i 5 i dt v 10 308 est Ri C Since i Iest , 10 308 est 10Iest 5 st Ie s from which I 10 308 1:01 10 5=s 32:88

More generally, if A represents any event, there will be a region 5A of the xy plane that corresponds to it. In such case we can find the probability of A by performing the integration over 5A, i.e., P(A) 33 f (x, y) dx dy

Then, i 1:01e 2t cos 10t 32:88 (A).

(21)

Label the regions in the control flow graph from Example 5.6 with Roman numerals. As shown in Fig. 5-2, there are five regions. Region I is the outside of the graph.

F(x, y)

Note that the s-domain impedance for the capacitance is 1= sC . a series RLC circuit will be Z s R sL 1= sC

(22)

A driving voltage of the form v Vest applied to a passive network will result in currents and voltages throughout the network, each having the same time function est ; for example, Ie j est . Therefore, only the magnitude I and phase angle need be determined. We are thus led to consider an s-domain where voltages and currents are expressed in polar form, for instance, V , I , and so on. Figure 8-12 suggests the correspondence between the time-domain network, where s j!, and the

It follows in analogy with (11), page 38, that '2F 'x 'y f (x, y)

(23)

Fig. 8-12

i.e., the density function is obtained by differentiating the distribution function with respect to x and y. From (22) we obtain

Fig. 5-2.

CHAP. 8]

F1(x)

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s-domain where only magnitudes and phase angles are shown. In the s-domain, inductances are expressed by sL and capacitances by 1= sC . The impedance in the s-domain is Z s V s =I s . A network function H s is de ned as the ratio of the complex amplitude of an exponential output Y s to the complex amplitude of an exponential input X s If, for example, X s is a driving voltage and Y s is the output voltage across a pair of terminals, then the ratio Y s =X s is nondimensional. The network function H s can be derived from the input-output di erential equation an dny d n 1 y dy d mx d m 1 x dx a0 y bm m bm 1 m 1 b1 b0 x a1 n an 1 n 1 dt dt dt dt dt dt

F2( y)

When x t Xest and y t Yest , an sn an 1 sn 1 a1 s a0 est bm sm bm 1 sm 1 b1 s b0 est Then, H s Y s a sn an 1 sn 1 a1 s a0 nm X s bm s bm 1 sm 1 b1 s b0

(25)

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