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The problem might also have been solved by assigning mesh currents and solving simultaneous di erential equations.

p 6

23 R 3

For the time functions listed in the rst column of Table 8-2, write the corresponding amplitude and phase angle (cosine-based) and the complex frequency s.

3 X 4.

Fig. 5-4.

See columns 2 and 3 of the table. Table 8-2 Time Function i t 86:6 A 3 i t 15:0e 2 10 t A v t 25:0 cos 250t 458 V v t 0:50 sin 250t 308 V i t 5:0e 100t sin 50t 908 A i t 3 cos 50t 4 sin 50t A A 8 86:6 08 A 15:0 08 A 25:0 458 V 0:50 608 V 5:0 08 A 5 53:138 A s 0 2 103 Np/s j250 rad/s j250 rad/s 100 j50 s 1 j50 rad/s

f (x)

0 0 e

For each amplitude and phase angle in the rst column and complex frequency s in the second column in Table 8-3, write the corresponding time function.

F(2)

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See column 3 of the table. Table 8-3 A 8 10 08 2 458 5 908 15 08 100 308 s j120 j120 2 j50 5000 j1000 0 Time Function 10 cos 120 t 2 cos 120 t 458 5e 2t cos 50t 908 15e 5000t cos 1000t 86.6

e 4. Hence, 2)

(1 3 1

CHAP. 8]

The cfg is the same as earlier examples, except that the c-h arc has been replaced by an h-c arc. This will not change the counts of nodes, edges, or regions. Thus, the first two methods of counting the cyclomatic number will not change. However, decision d has been eliminated, so the third method will not give the same answer. However, this is not a valid cfg, since there is now no stopping node.

e 4)

p An amplitude and phase angle of 10 2 458 V has an associated complex frequency s 50 j 100 s 1 . Find the voltage at t 10 ms.

P( 3

2u 4 0

A passive network contains resistors, a 70-mH inductor, and a 25-mF capacitor. Obtain the respective s-domain impedances for a driving voltage (a) v 100 sin 300t 458 V , (b) v 100e 100t cos 300t V .

30 2e

(a) Resistance is independent of frequency. At s j300 rad/s, the impedance of the inductor is sL j300 70 10 3 j21 and that of the capacitor is 1 j133:3 sC (b) At s 100 j300 s 1 , sL 100 j300 70 10 3 7 j21 1 1 40 j120 sC 100 j300 25 10 6

P( 3

An important aspect of a metric is guidance about when the values are reasonable and when the values are not reasonable. McCabe analyzed a large project and discovered that for modules with cyclomatic number over 10, the modules had histories of many more errors and many more di culties in maintenance. Thus, 10 has been accepted as the threshold value for the cyclomatic number in a module. If the cyclomatic number is greater than 10, e orts should be made to reduce the value or to split the module.

P(X F(4) (1

For the circuit shown in Fig. 8-20, obtain v at t 0:1 s for source current a i 10 cos 2t (A), (b) i 10e t cos 2t (A).

Joint distributions and independent variables 2.8. The joint probability function of two discrete random variables X and Y is given by f(x, y) c(2x x and y can assume all integers such that 0 x 2, 0 y 3, and f(x, y) 0 otherwise.

Zin s 2 (a) At s j2 rad/s, Zin j2 3:22 7:138

(a) Find the value of the constant c. (b) Find P(X 2, Y 1). (c) Find P(X 1, Y 2).

V IZin 10 08 3:22 7:138 32:2 7:138 V and v 0:1 32:2 cos 18:598 30:5 V. (b) At s 1 j2 s 1 , Zin 1 j2 3:14 11:318

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