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The second moment about the origin is E(X2) (a b) 1 x2[xa 1(1 (a) (b) 30 (a b) 1 xa 1(1 (a) (b) 30 (a b) (a (a) (b) (a (a b) (a) (b) (a (a Then using Problem 4.32, the variance is s2 E(X2) [E(X )]2 (a a(a b)(a 1) b 1) a a a b b

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2) (b) 2 b) (a b 1)a (a) (b) 1)(a b) (a 1) b)

which are identical with the Y-parameters obtained in Problem 3.5 for Fig. 13-19. Thus, the two networks are equivalent.

The chi-square distribution 4.34. Show that the moment generating function of a random variable X, which is chi-square distributed with v degrees of freedom, is M(t) (1 2t) v/2.

Apply the short-circuit equations (10) to nd the Y-parameters of the two-port network in Fig. 13-21.

1 x(v (v>2) 30

u in the last integral, we find M(t)

Fig. 13-21 1 1 1 V or Y11 12 12 1 6 V2 V2 1 1 1 V or Y12 4 12 2 6 4 12 V1 1 or Y21 12 12 V2 V2 1 1 5 or Y22 V 3 12 2 12 3 12

2v>2 (1

CHAP. 13]

Fig. 8-15.

2t) v>2 u(v>2) 1e (v>2) 3 0

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4.35. Let X1 and X2 be independent random variables that are chi-square distributed with v1 and v2 degrees of freedom, respectively, (a) Show that the moment generating function of Z X1 X2 is (1 2t) (v1 v2)>2, thereby (b) show that Z is chi-square distributed with v1 v2 degrees of freedom.

Apply KCL at the nodes of the network in Fig. 13-21 to obtain its terminal characteristics and Yparameters. Show that two-port networks of Figs. 13-18 to 13-21 are all equivalent.

X2 )]

Input node: Output node: V1 V1 V2 V2 12 12 4 V2 V2 V1 I2 3 12 1 1 1 5 I2 V1 V2 I1 V 1 V 2 6 6 12 12 I1

X2 is (1 2t)

v1>2 (1

The Y-parameters observed from the above characteristic equations are identical with the Y-parameters of the circuits in Figs. 13-18, 13-19, and 13-20. Therefore, the four circuits are equivalent.

11. Draw a state diagram for the dental o ce problem (see Problem 4.2). See Fig. 8-16.

E(etX1) E(etX2)

Z-parameters of the two-port network N in Fig. 13-22(a) are Z11 4s, Z12 Z21 3s, and Z22 9s. (a) Replace N by its T-equivalent. (b) Use part (a) to nd input current i1 for vs cos 1000t (V).

using Problem 4.34. (b) It is seen from Problem 4.34 that a distribution whose moment generating function is (1 chi-square distribution with v1 v2 degrees of freedom. This must be the distribution of Z, by Theorem 3-10, page 77. By generalizing the above results, we obtain a proof of Theorem 4-4, page 115. 2t)

4.36. Let X be a normally distributed random variable having mean 0 and variance 1. Show that X2 is chi-square distributed with 1 degree of freedom.

(a) The network is reciprocal. Therefore, its T-equivalent exists. shown in the circuit of Fig. 13-22(b). Its elements are found from (6) and

We want to find the distribution of Y X2 given a standard normal distribution for X. Since the correspondence between X and Y is not one-one, we cannot apply Theorem 2-3 as it stands but must proceed as follows.

Fig. 13-22

0, it is clear that P(Y P(Y y)

0. For y y)

Za Z11 Z12 4s 3s s Zb Z22 Z21 9s 3s 6s Zc Z12 Z21 3s (b) We repeatedly combine the series and parallel elements of Fig. 13-22(b), with resistors being in k

Fig. 8-16.

0, we have P( !y X 2 !2p 30 !y)

as shown in the following. Zin s Vs =I1 s 3s 6 6s 12 3s 4 9s 18 (mA). or Zin j 3j 4 5 36:98 k

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