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where the integral is taken over the region 5 of the yz plane for which y x !z>v. We first fix z and integrate with respect to y from ` to x !z>v. Then we integrate with respect to z from 0 to ` . We therefore have F(x) !2p2v>2 (v>2) 3z 1

Fig. 9-2.

3 0 u

Fig. 13-25

F(x)

13.15 The h-parameters of a two-port network are given. Show that the network may be modeled by the network in Fig. 13-26 where h11 is an impedance, h12 is a voltage gain, h21 is a current gain, and h22 is an admittance.

w(v 1)>2e w dw d du u2 >v)(v 1)>2 0 (1

Fig. 13-26 Apply KVL around the input loop to get V1 h11 I1 h12 V2 Apply KCL at the output node to get I2 h21 I1 h22 V2 These results agree with the de nition of h-parameters given in (23).

` (1

4.43. The graph of Student s t distribution with 9 degrees of freedom is shown in Fig. 4-19. Find the value of t1 for which (a) the shaded area on the right (b) the total shaded area (c) the total unshaded area 0.05, 0.99, 0.01, 0.05,

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An interface speci cation is a speci cation of the external behavior of a module. It should be complete enough so that the calling module knows exactly what the called module will do under any circumstance. It should also be complete enough so that the implementer knows exactly what information must be provided. The interface speci cations in an OO model are often the signatures of the public methods and the semantics associated with the methods. Interfaces can also be speci ed as part of a formal speci cation of the behavior of the whole system. Interfaces can also be the invariants, preconditions, and post-conditions for a method.

13.16 Find the h-parameters of the circuit in Fig. 13-25.

Fig. 4-19

; h12 0; h21 0:4; h22 1=5 0:2

(a) If the shaded area on the right is 0.05, then the area to the left of t1 is (1 0.05) 0.95, and t1 represents the 95th percentile, t0.95. Referring to the table in Appendix D, proceed downward under the column headed v until entry 9 is reached. Then proceed right to the column headed t0.95. The result 1.83 is the required value of t. (b) If the total shaded area is 0.05, then the shaded area on the right is 0.025 by symmetry. Therefore, the area to the left of t1 is (1 0.025) 0.975, and t1 represents the 97.5th percentile, t0.975. From Appendix D, we find 2.26 as the required value of t. (c) If the total unshaded area is 0.99, then the total shaded area is (1 right is 0.01> 2 0.005. From the table we find t0.995 3.25. 0.99) 0.01, and the shaded area to the

(d) If the shaded area on the left is 0.01, then by symmetry the shaded area on the right is 0.01. From the table, t0.99 2.82. Therefore, the value of t for which the shaded area on the left is 0.01 is 2.82. (e) If the area to the left of t1 is 0.90, then t1 corresponds to the 90th percentile, t0.90, which from the table equals 1.38.

13.17 Find the h-parameters of the circuit in Fig. 13-25 from its Z-parameters and compare with results of Problem 13.16.

4.44. Find the values of t for which the area of the right-hand tail of the t distribution is 0.05 if the number of degrees of freedom v is equal to (a) 16, (b) 27, (c) 200.

Refer to Problem 13.13 for the values of the Z-parameters and DZZ . Use Table 13-1 for the conversion of the Z-parameters to the h-parameters of the circuit. Thus, h11 DZZ 20 4 5 Z22 h12 Z12 0 Z22 h21 Z21 2 0:4 5 Z22 h22 1 1 0:2 Z22 5

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