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(a) Using the results in Problems 2.8 and 2.9, we have f ( y u x) so that with x 2 f ( y u 2) (b) (4 y)>42 11>21 4 22 y f (x, y) f1(x) (2x y)>42 f1(x)

9R2 , C2 100 pF 9C1 , and v1 104 t V. the switch is closed. Ans: i 1:0001 mA Find i at 1 ms after

f (1 u 2)

Fig. 8-44

5 22

f (x, y) find (a) f ( y u x), (b) P(Y (a) For 0 x 1, f1(x)

Lead network. The circuit of Fig. 8-45(a) is called a lead network. (a) Find the di erential equation relating v2 to v1 . (b) Find the unit-step response of the network with R1 10 k

Potential Volume, V*

e4 0

, and C 1 mF. (c) Let v1 cos !t be the input and v2 A cos !t be the output of the network of Part (b). Find A and for ! at 1, 100, 331.6, 1100, and 105 , all in rad/s. At what ! is the phase at a maximum dv2 R1 R2 dv 1 1 v ; b v2 1 10e 1100t u t Ans: a v2 1 dt R1 R2 C dt R1 C 1 11 (c) ! A 1 0.091 0.58 100 0.128 39:88 331.6 0.3015 56.48 1100 0.71 39.88 105 1 0.58

3 1 2

dx).

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p Phase is maximum at ! 100 11 331:6 rad/s 8.49 Lag network. The circuit of Fig. 8-45(b) is called a lag network. (a) Find the di erential equation relating v2 to v1 . (b) Find the unit-step response of the network with R1 10 k

1 3 30 4

, and C 1 mF. (c) Let v1 cos !t be the input and v2 A cos !t be the output of the network of Part (b). Find A and for ! at 1, 90.9, 301.5, 1000, and 105 , all in rad/s. At what ! is the phase at a minimum dv dv 10 90:91t e Ans: a v2 R1 R2 C 2 v1 R2 C 1 ; b v2 1 u t 11 dt dt

3 4 0

f ( y u x)

Fig. 8-45

The potential volume is the minimal size of a solution to the problem, solved in any language. Halstead assumes that in the minimal implementation, there would only be two operators: the name of the function and a grouping operator. The minimal number of operands is 2 : V 2 log2 2 2 2

f (x, y) f1(x)

1 1 0.58

31>2

90.9 0.71 39.88

31> 2

9 16

301.5 0.3015 56.48

2.29. The joint density function of the random variables X and Y is given by f (x, y) e 8xy 0 0 x 1, 0 otherwise y x

Since we have the actual volume and the minimal volume, it is natural to take a ratio. Halstead divides the potential volume by the actual. This relates to how close the current implementation is to the minimal implementation as measured by the potential volume. The implementation level is unitless. L V =V The basic measures described so far are reasonable. Many of the ideas of operands and operators have been used in many other metric e orts. The remaining measures are given for historical interest and are not recommended as being useful or valid.

1000 0.128 39.88

Find (a) the marginal density of X, (b) the marginal density of Y, (c) the conditional density of X, (d) the conditional density of Y.

The region over which f (x, y) is different from zero is shown shaded in Fig. 2-17.

105 0.091 0.58

Fig. 2-17

p Phase is minimum at ! 1000= 11 301:5 rad/s 8.50 In the circuit of Fig. 8-46 nd the relationship between v2 and v1 for (a) k 103 , case nd its unit-step response; that is, v2 for v1 u t . 6 dv2 4 106 v2 4 107 v1 ; v2 10 1 e 4 10 t u t Ans: a dt 9 dv2 4 108 v2 4 109 v1 ; v2 10 1 e 4 10 t u t b dt (b) k 105 . In each

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