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.PROBE generates a data le *.DAT which can be plotted in post-analysis by evoking the Probe program. The syntax is :PROBE houtput variablesi

3,3,3

According to Theorem 5-6, x2 nS2 >s2 2S2 >10.8 has the chi-square distribution with 1 degree of freedom. From the table in Appendix E it follows that P(S2 s2) 1 P(x2 1.33) 0.25

EXAMPLE 15.8 Find the value of Vs in the circuit in Fig. 15-8 such that the power dissipated in the 1-k

Therefore, we would expect about 25% of the samples, or 6, to have variances greater than 7.2. (b) From Problem 5.18 we find by counting that there are actually 6 variances greater than 7.2, so that there is agreement.

resistor is zero. Use the .DC command to sweep Vs from 1 to 6 V in steps of 1 V and use .PRINT to show I Vs , V(1,2), and V(2).

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Let Y nS2 , n n 1. Then since the Xj are normally distributed with mean m and s2 s> !n # variance s2, we know (Problem 5.21) that X is normally distributed with mean m and variance s2 >n, so that y is normally distributed with mean 0 and variance 1. Also, from Theorem 5-6, page 158, or Problem 5.22, Z is chi square distributed with n n 1 degrees of freedom. Furthermore, it can be shown that Y and Z are independent. It follows from Theorem 4-6, page 116, that , Z T has the t distribution with n Y !Z>n # X S> !n m 1 # X

5.25. According to the table of Student s t distribution for 1 degree of freedom (Appendix D), we have P( 1.376 T 1.376) 0.60. Check whether this is confirmed by the results obtained in Problem 5.1.

Fig. 15-8

# From the values of X in (1) on page 155, and the values of S2 in Problem 5.18(a), we obtain the following # values for T (X m)>(S> !1): ` 7.0 1.0 0.33 0.11 7.0 ` 1.0 0.20 0.25 1.0 1.0 c 1.0 1.0 0.33 0.20 1.0 ` 2.33 0.11 0.25 1.0 2.33 `

0,4,0

The source le is DC sweep, Fig. 15-8 Vs 1 0 DC 1V Is 0 2 DC 1 mA R1 1 2 1k R2 0 2 2k .DC Vs 1 6 1 .PRINT DC I(Vs) V(1,2) V(2) .END The results in the output le are DC TRANSFER CURVES Vs I(Vs) 1:000E 00 3:333E 04 2:000E 00 1:333E 12 3:000E 00 3:333E 04 4:000E 00 6:667E 04 5:000E 00 1:000E 03 6:000E 00 1:333E 03 The answer is Vs 2 V. EXAMPLE 15.9 Write the source le for the circuit in Fig. 15-9(a) using commands .DC, .PLOT, and .PROBE to nd the I-V characteristic equation for I varying from 0 to 2 A at the terminal AB. First, we connect a dc current source Iadd at terminal AB, sweep its value from 0 to 2 A using the .DC command, and plot V versus I. Since the circuit is linear, two points are necessary and su cient. However, for clarity of the plot, ten steps are included in the source le as follows: V(1,2) 3:333E 01 1:333E 09 3:333E 01 6:667E 01 1:000E 00 1:333E 00 V(2) 1:333E 00 2:000E 00 2:667E 00 3:333E 00 4:000E 00 4:667E 00

There are actually 16 values for which 1.376 T 1.376 whereas we would expect (0.60) (25) 15. This is not too bad considering the small amount of data involved. This method of sampling was in fact the way Student originally obtained the t distribution.

Denote the samples of sizes m and n by X1, . . . , Xm and Y1, . . . ,Yn, respectively. Then the sample variances are given by S2 1 # # where X, Y are the sample means. 1 m a (Xj

CHAP. 15]

# X)2,

Fig. 15-9

S2 2

# Y)2

3,8,3

Terminal Characteristic, Fig. 15-9 Iadd 0 5 DC Is 0 4 DC Vs 3 2 DC R1 0 1 1 R2 1 2 2 R3 3 4 3 R4 4 5 2 .DC Iadd 0 2 .PLOT DC V(5) .PROBE .END The output is shown in Fig. 15-9(b).

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