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For the network of Fig. 9-46 assign two di erent sets of mesh currents and show that for each, z 55:9 26:578 2 . For each choice, calculate the phasor voltage V. Obtain the phasor voltage across the 3 j4 impedance and compare with V. Ans: V V3 j4 22:36 10:308 V For the network of Fig. 9-47, use the mesh current method to nd the current in the 2 j3 impedance due Ans: 2:41 6:458 A; 1:36 141:458 A to each of the sources V1 and V2 .

(36)

Fig. 9-47

fX(v)

(continuous variable)

In the network shown in Fig. 9-48, the two equal capacitances C and the shunting resistance R are adjusted until the detector current ID is zero. Assuming a source angular frequency !, determine the values of Rx and Lx . Ans: Rx 1= !2 C 2 R ; Lx 1= 2!C

1 heads _

(37)

Fig. 9-48

Since u eivx u 1, the series and the integral always converge absolutely. The corresponding results (31) and (32) become fX(v) 1 imv mr 2 v2 2! c irmrr vr r! c (38)

CHAP. 9]

(39)

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When no confusion can result, we often write f(v) instead of fX(v). Theorems for characteristic functions corresponding to Theorems 3-8, 3-9, and 3-10 are as follows. Theorem 3-11 If fX(v) is the characteristic function of the random variable X and a and b (b 2 0) are constants, then the characteristic function of (X a) > b is f(X Theorem 3-12

Ans:

a)>b(v)

1 heads _

908

If X and Y are independent random variables having characteristic functions fX (v) and fY (v), respectively, then fX Y (v) fX (v) fY (v) (41)

(40)

Fig. 9-49

More generally, the characteristic function of a sum of independent random variables is equal to the product of their characteristic functions. Theorem 3-13 (Uniqueness Theorem) Suppose that X and Y are random variables having characteristic functions fX (v) and fY (v), respectively. Then X and Y have the same probability distribution if and only if fX (v) fY (v) identically.

Show that Ztransfer;31 Ztransfer;13 .

An important reason for introducing the characteristic function is that (37) represents the Fourier transform of the density function f (x). From the theory of Fourier transforms, we can easily determine the density function from the characteristic function. In fact, f (x) 1 ` e 2p 3 `

(42)

Fig. 6-1

In the network of Fig. 9-50, obtain the voltage ratio V1 =V2 by application of the node voltage method. Ans: 11 1:61 12 29:88 Ans: 5:59 17:358

which is often called an inversion formula, or inverse Fourier transform. In a similar manner we can show in the discrete case that the probability function f(x) can be obtained from (36) by use of Fourier series, which is the analog of the Fourier integral for the discrete case. See Problem 3.39. Another reason for using the characteristic function is that it always exists whereas the moment generating function may not exist.

For the network of Fig. 9-50, obtain the driving-point impedance Zinput;1 .

The results given above for one variable can be extended to two or more variables. For example, if X and Y are two continuous random variables having joint density function f(x, y), the means, or expectations, of X and Y are

E(X)

Fig. 9-50

3 ` 3 `xf (x, y) dx dy,

Fig. 9-51

E(Y)

Risk mitigation is the proactive strategy of trying to nd ways to either decrease the probability of the risk event happening or the impact of it happening. While there are no magic ways to reduce risk, a common approach is to try to resolve risks related to uncertainty early. For example, if there is concern about particular systems that need to be used, investigating those systems early is better. Often, a prototype can be built that would identify problems early.

(43)

Obtain the Thevenin and Norton equivalent circuits at terminals ab for the network of Fig. 9-51. Choose the polarity such that V 0 Vab . Ans: V 0 20:0 08 V; I 0 5:56 23:068 A; Z 0 3:60 23:068

mX )2] mY)2]

Obtain the Thevenin and Norton equivalent circuits at terminals ab for the network of Fig. 9-52. Ans: V 0 11:5 95:88 V; I 0 1:39 80:68 A; Z 0 8:26 15:28

Note that the marginal density functions of X and Y are not directly involved in (43) and (44). Another quantity that arises in the case of two variables X and Y is the covariance defined by sXY Cov (X, Y) E[(X mX)(Y mY)] (45)

(46)

Fig. 9-52

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