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input resistance of the ampli er reduces the overall gain very slightly, as seen by comparing Tables 5-4 and 5-5. The feedback has made the input resistance of the ampli er less e ective in changing the overall gain. Table 5-5 k 1 10 100 1000 1 v2 =vs 0:704 3:106 4:713 4:97 5:00

1 2 (5

Step I. Table 10-11 shows the array of grades and ranks. Note that the rank for the two grades of 75 is 1 6) 5.5, while the rank for the three grades of 84 is 3(11 12 13) 12. Step 2. Rewriting Table 10-10 in terms of ranks gives us Table 10-12. Check: R1 73, R2 158, and N N(N 2 1) N1 N2 (21)(22) 2 9 12 231 21; thus R1 R1 R2 R2 73 158 231 and

. Find v2 =v1 for k 1; 10; 100; 1000; 1. Repeat for Ri 1: Discuss the effects of Ri and k on the overall gain. Show that, for k 1 and Ri 6 0; the gain of the amplifier is independent of Ri and is equal to R2 =R1 :

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Table 10-11 53 66 70 73 75 75 79 81 82 83 84 84 84 85 86 87 88 90 91 92 95 1 2 3 4 5.5 7 8 9 10 12 14 15 16 17 18 19 20 21

Fig. 14-4

(a) Apply KCL to currents leaving node A to obtain v1 vs v1 v2 v1 0 1 5 Ri From the ampli er we get v2 kv1 or v1 v2 =k. rearranging terms we get Substituting for v1 in the KCL equation and

Table 10-12

Sum of Ranks Morning class Afternoon class 4 15 16 8 7 12 5.5 17 9 18 2 14 21 12 5.5 20 3 10 19 1 12 73 158

36

Step 3. U mU Therefore, N1N2 2 N1N2 (9)(12) 2 Z N1(N1 2 54 U 1) R1 (9)(12) N2 12 80 54 14.07 1.85 (9)(10) 2 1) 73 80 198

, c 1=11 which, substituted into (36), gives v2 5k vs 11 k For Ri 1 we get c 1=6 and so v2 5k vs 6 k 38 37

(9)(12)(22) 12

Since we wish to test the hypothesis H1 that the morning class performs worse than the afternoon class against the hypothesis H0 that there is no difference at the 0.05 level, a one-tailed test is needed. Referring to Fig. 10-2, which applies here, we have the decision rule: Accept H0 if z Reject H0 if z 1.645. 1.645.

Table 5-6 gives values of v2 =vs in (37) and (38) versus k. Note that (38) is identical with (32).

person d) self.birthfam.child_include(self) e) self.marriage->size = 1 f) self.marriage.wife.birthdate < self.birthdate a. This invariant says that the set of persons is the same as the set of all children or every person has his or her birth marriage listed. This is a reasonable invariant, and it is true if the data is complete. b. This invariant says that every child has his or her birth family listed and it matches the instance that points to the person as a child. This is reasonable and is always true. c. This says that every husband is older than his wife. This invariant can be stated, but it does not match reality. d. This states that the set of children (siblings) reachable from the birthfam includes the person. This is reasonable and is always true. e. This states that the set of marriages for a person is only one. It does not match reality. f. Either your own birthday (if female) or your spouse s (if male) is less than yours. Not reasonable. Not always true. 3. Write OCL constraints for a restaurant without a smoking section that seats customers in order of arrival. Class group Char* name Int number Int arrivalorder Class waitlist Group* list[MAX] Int listptr Void addtolist(group* newgroup) Group* seatnext()

Since the actual value of z 1.85 1.645, we reject H0 and conclude that the morning class performed worse than the afternoon class at the 0.05 level. This conclusion cannot be reached, however, for the 0.01 level (see Problem 10.33).

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