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EXAMPLE 15.11 Find the transfer function V3 =Vs in the ideal op amp circuit of Fig. 15-11(a). The op amp is replaced by a voltage-dependent voltage source with a gain of 106 [see Fig. 15-11(b)]. The source le is

Table 5-6

118 126 127 135 136 144 145 153 154 162 163 171 172 180

CHAP. 15]

3 5 9 12 5 4 2 40

Fig. 15-11 Inverting Vs E1 R1 R2 .TF .END op amp circuit, Fig. 15-11 1 0 DC 12 3 0 0 2 1E6 1 2 1k 2 3 2k V(3) Vs

The histogram and frequency polygon for each of the two cases considered in Problem 5.28 are given in Figs. 5-7 and 5-8. Note that the centers of the bases of the rectangles are located at the class marks.

Fig. 5-7

The transfer function is written in the output le: NODE (1) VOLTAGE 12.0000 NODE (2) VOLTAGE 24.00E 06 NODE (3) VOLTAGE 24.0000

3,4,8

Fig. 5-8

SMALL-SIGNAL CHARACTERISTICS V(3)/Vs 2:000E + 00 INPUT RESISTANCE AT Vs 1.000E + 03 OUTPUT RESISTANCE AT V(3) 0.000E + 00

5.30. Five pennies were simultaneously tossed 1000 times and at each toss the number of heads was observed. The numbers of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in Table 5-7. Graph the data.

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The data can be shown graphically either as in Fig. 5-9 or Fig. 5-10. Figure 5-9 seems to be a more natural graph to use. One reason is that the number of heads cannot be 1.5 or 3.2. This graph is a form of bar graph where the bars have zero width, and it is sometimes called a rod graph. It is especially useful when the data are discrete. Figure 5-10 shows a histogram of the data. Note that the total area of the histogram is the total frequency 1000, as it should be.

.SUBCKT Statement A subcircuit is de ned by a set of statements beginning with .SUBCKT hnamei hexternal terminalsi

Table 5-7

and terminating with an .ENDS statement. Within a netlist we refer to a subcircuit by Xaa hnamei hnodesi

Number of Heads 0 1 2 3 4 5 TOTAL Number of Tosses (frequency) 38 144 342 287 164 25 1000

3,4,5

Fig. 5-9

EXAMPLE 15.12 Given the circuit of Fig. 15-12(a), nd Is , If , V2 , and V6 for Vs varying from 0.5 to 2 V in 0.5-V steps. Assume a practical op amp [Fig. 15-12(b)], with Rin 100 k

Fig. 5-10

, and an open loop gain of 105 . The source le employs the subcircuit named OPAMP of Fig. 15-12(b) whose description begins with .SUBCKT and ends with .ENDS. The X1 and X2 statements describe the two op amps by referring to the OPAMP subcircuit. Note the correspondence of node connections in the X1 and X2 statements with that of the external terminals speci ed in the .SUBCKT statement. The source le is Op amp circuit of Fig. 15-12 .SUBCKT OPAMP Rin 1 2 Cin 1 2 Rout 3 5 using .SUBCKT 1 2 3 4 10 E5 10 pF 10 k

Computation of mean, variance, and moments for samples 5.31. Find the arithmetic mean of the numbers 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.

There are six 5s, two 3s, two 6s, five 4s, two 2s, and three 8s. Then a fx n (6)(5) (2)(3) 6 (2)(6) 2 2 (5)(4) 5 2 (2)(2) 3 (3)(8) 96 20

5 4 1 0 1 2 2 3 3 4 4 5 5 6 6 2 0 3 0 5 Vs DC V(6)

5.32. Four groups of students, consisting of 15, 20, 10, and 18 individuals, reported mean weights of 162, 148, 153, and 140 lb, respectively. Find the mean weight of all the students.

10 E5

valid combinations of T and F will be used. If there are three conditions, there may be 23 8 subdomains (columns). Additional rows will be used for values of a, b, and c and for the expected output for each subdomain.

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