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For v greater than 30, we can use the fact that ( !2x 2 !2v 1) is very closely normally distributed with mean zero and variance one. Then if zp is the (100p)th percentile of the standardized normal distribution, we can write, to a high degree of approximation, !2x2 p from which x2 p

Fig. 13-23 (a) From application of KVL around the input and output loops we nd, respectively, V1 2I1 2I2 2 I1 I2 4I1 V2 3I2 2 I1 I2 2I1 5I2 The Z-parameters are Z11 4, Z12 0, Z21 2, and Z22 5. (b) The circuit of Fig. 13-23(b), with two resistors and a voltage source, has the same Z-parameters as the circuit of Fig. 13-23(a). This can be veri ed by applying KVL to its input and output loops.

1)2 !99)2 69.2, which agrees very well with the 124.0 (actual value 124.3).

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Design the library classes/data structures from the data items in the object model shown in Fig. 9-1 for the library problem (see Examples 8.1 and 2.6). The data design and the architectural phases have been combined in this example. The concentration in this example is on the loan and checkout functionality, with little regard for the other necessary tasks, such as administration, cataloging, assigning overdue fines, retiring books, and patron maintenance. The domain entity book is probably not going to continue into the design. It will be combined with copy into a class/data structure that stores all the information about a copy. It will probably use the ISBN and a copy number as the unique identifier. The patron information will be stored in a second data structure. Each record is probably identified by an unique patron ID number. The loan information may or may not be a separate data structure. If borrowing information needs to be saved beyond the return of the book, then it had better be a separate class/data

13.13 (a) Obtain the Y-parameters of the circuit in Fig. 13-23(a) from its Z-parameters. (b) Find an equivalent model which uses two positive-valued resistors and one dependent current source.

1)2 1)2

(1.64

(a) From Problem 13.12, Z11 4, Z12 0, Z21 2; Z22 5, and so DZZ Z11 Z22 Z12 Z21 20. Hence, Y11 Z22 5 1 DZZ 20 4 Y12 Z12 0 DZZ Y21 Z21 2 1 20 10 DZZ Y22 Z11 4 1 DZZ 20 5

!2(100)

(b) Figure 13-24, with two resistors and a current source, has the same Y-parameters as the circuit in Fig. 13-23(a). This can be veri ed by applying KCL to the input and output nodes.

(1.64

!199)2

13.14 Referring to the network of Fig. 13-23(b), convert the voltage source and its series resistor to its Norton equivalent and show that the resulting network is identical with that in Fig. 13-24.

structure. Otherwise, the patron ID can be part of the copy class/data structure along with the due date of the book. Note in Fig. 9-2 that many data items have been added that are more in the implementation/solution space than in the problem/domain space. It can be argued that ISBN is part of the problem space instead of the solution space, but many library systems do not allow normal users to search by ISBN.

Since Y is normally distributed with mean 0 and variance 1, its density function is 1 e y2>2 !2p Since Z is chi-square distributed with v degrees of freedom, its density function is (1) (2) 1 z(v>2) 1e 2v>2 (v>2)

The Norton equivalent current source is IN 2I1 =5 0:4I1 . But I1 V1 =4. Therefore, IN 0:4I1 0:1V1 . The 5-

Because Y and Z are independent, their joint density function is the product of (1) and (2), i.e., 1 z(v>2) !2p 2v>2 (v>2) for ` y `, z 0. Y> !Z>v is P(T x) P(Y x!Z>v)

The distribution function of T F(x)

resistor is then placed in parallel with IN . The circuit is shown in Fig. 13-25 which is the same as the circuit in Fig. 13-24.

1 z(v>2) 1 e !2p2v>2 (v>2) 6

Fig. 13-24

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