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0:31 2:38 4:51 4:95 5:00 Ri 1 0:71 3:12 4:72 4:97 5:00

10.9. Find U for the data of Table 10-13 by using (a) formula (2) of this chapter, (b) the counting method (as described in Remark 4 of this chapter).

(a) Arranging the data from both samples in an array in increasing order of magnitude and assigning ranks from 1 to 5 gives us Table 10-14. Replacing the data of Table 10-13 with the corresponding ranks gives us Table 10-15, from which the sums of the ranks are R1 5 and R2 10. Since N1 2 and N2 3, the value of U for sample 1 is U N1N2 N1(N1 2 1) R1 (2)(3) 2. (2)(3) 2 5 4

Comparing the two columns in Table 5-6 we see that the smaller Ri reduces the overall gain G . However, as the open-loop gain k increases, the e ect of Ri is diminished. As k becomes very large, v2 =v1 approaches 5 unless Ri 0.

Table 10-13

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in the circuit of Fig. 5-33. Replace the circuit to the left of node A including vs , R1 , and Ri by its Thevenin equivalent. Then use (5) to derive (36).

Table 10-14

Class restaurant Waitlist* waiting Void arrive(group* newgroup) Group* seat() Waitlist Self.Listptr = self.list->size Void waitlist::addtolist(group* newgroup) Pre: self.listptr < MAX Post: forall(i | list[i] = list[i-1]@pre) List[0] = newgroup Listptr = listptr@pre+1 Group* waitlist::seatnext() Pre: self.listprt > 0 Post: Result = list[listptr@pre] Self.listptr = self.listprt@pre - 1 Group* Restaurant::seat Pre: waiting.waitlist->size > 0 Post: waiting.waitlist->size = waiting. waitlist@pre->size -1 Result = waiting.seatnext() and forall (x : group | waiting.seatnext().arrivalorder <= x.arrivalorder)

10 1 14 2 17 3 22 4 25 5

: From (5), v2 1 b where Therefore, v2 5 1 Ri k Ri 5Ri k v v 6Ri 5 1 Ri k= 6Ri 5 1 Ri s 6Ri 5 Ri k s b RTh Ri RTh R2 6Ri 5 k v 1 bk Th and 1 b 5 1 Ri 6Ri 5 R i vs Rv i s R1 Ri 1 Ri R1 Ri Ri R1 Ri 1 Ri

Table 10-15

which is identical with (36).

(b) Let us replace the sample values in Table 10-14 with I or II, depending on whether the value belongs to sample 1 or 2. Then the first line of Table 10-14 becomes Data From this we see that Number of sample 1 values preceding first sample 2 value Number of sample 1 values preceding second sample 2 value Number of sample 1 values preceding third sample 2 value Total 1 1 2 4 I II II I II

Thus the value of U corresponding to the first sample is 4. Similarly, we have Number of sample 2 values preceding first sample 1 value Number of sample 2 values preceding second sample 1 value Total Thus the value of U corresponding to the second sample is 2. Note that since N1 2 and N2 3, these values satisfy U1 0 2 2

Find the output voltage of an op amp with A 105 and Vcc 10 V for v 0 and v sin t (V). Refer to Figs. 5-7 and 5-8.

N1N2; that is, 4

CHAP. 5]

(2)(3)

Z1 (see Operand) Z2 (see Operator) Z2 * (see Potential operand)

10.10. A population consists of the values 7, 12, and 15. Two samples are drawn without replacement from this population; sample 1, consisting of one value, and sample 2, consisting of two values. (Between them, the two samples exhaust the population.) (a) Find the sampling distribution of U. (b) Find the mean and variance of the distribution in part (a). (c) Verify the results found in part (b) by using formulas (3) of this chapter.

Because of high gain, saturation occurs quickly at jv2 j 105 jvd j 10 V We may ignore the linear interval and write v2 where vd v v sin t (V). 10 V 10 V vd > 0 vd < 0 or jvd j 10 4 V

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