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the terminal current is measured, the input admittance Y I1 =V1 1=Z represents the frequency response. For the two-port network of Fig. 12-1(b), the following frequency responses are de ned: Input impedance Zin j! V1 =I1 Input admittance Yin j! 1=Zin j! I1 =V1 Voltage transfer ratio Hv j! V2 =V1 Current transfer ratio Hi j! I2 =I1 Transfer impedances V2 =I1 and V1 =I2

If m1 and m2 denote population mean IQs of students from the two areas, we have to decide between the hypotheses H0: m1 m2, and there is essentially no difference between the groups H1: m1 2 m2, and there is a significant difference between the groups Under the hypothesis H0, T # X1 s 21>n1 # X2 1>n2 where s n1S2 n2S2 1 2 A n1 n2 2

EXAMPLE 12.1 Find the frequency response V2 =V1 for the two-port circuit shown in Fig. 12-2. Let YRC be the admittance of the parallel RC combination. Then, YRC 10 6 j! 1=1250. V2 =V1 is obtained by dividing V1 between ZRC and the 5-k resistor. H j! V2 ZRC 1 1 V1 ZRC 5000 1 5000YRC 5 1 10 3 j! 1 jHj p tan 1 10 3 ! 5 1 10 6 !2 2a 2b

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(a) On the basis of a two-tailed test at a 0.01 level of significance, we would reject H0 if T were outside the range t0.995 to t0.995, which, for n1 n2 2 16 14 2 28 degrees of freedom, is the range 2.76 to 2.76. Therefore, we cannot reject H0 at a 0.01 level of significance. (b) On the basis of a two-tailed rest at a 0.05 level of significance, we would reject H0 if T were outside the range t0.975 to t0.975, which for 28 degrees of freedom is the range 2.05 to 2.05. Therefore, we cannot reject H0 at a 0.05 level of significance. We conclude that there is no significant difference between the IQs of the two groups. (c) The P value is P(T 1.45) software, P 0.158. P(T 1.45). The table in Appendix D shows 0.1 P 0.2. By computer

7.19. At an agricultural station it was desired to test the effect of a given fertilizer on wheat production. To accomplish this, 24 plots of land having equal areas were chosen; half of these were treated with the fertilizer and the other half were untreated (control group). Otherwise the conditions were the same. The mean yield of wheat on the untreated plots was 4.8 bushels with a standard deviation of 0.40 bushels, while the mean yield on the treated plots was 5.1 bushels with a standard deviation of 0.36 bushels. Can we conclude that there is a significant improvement in wheat production because of the fertilizer if a significance level of (a) 1%, (b) 5% is used (c) What is the P value of the test

Fig. 12-2

If m1 and m2 denote population mean yields of wheat on treated and untreated land, respectively, we have to decide between the hypotheses H0: m1 H1: m1 Under the hypothesis H0, T Then s 12(0.40)2 12(0.36)2 12 12 2 A 0.397 and T 5.1 4.8 0.397 !1>12 1>12 1.85 # X1 s !1>n1 # X2 1>n2 where s n1S2 n2S2 1 2 A n1 n2 2 m2, and the difference is due to chance m2, and the fertilizer improves the yield

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Alternative solution: First we nd the Thevenin equivalent of the resistive part of the circuit, VTh V1 =5 and RTh 1 k , and then divide VTh between RTh and the 1-mF capacitor to obtain (2a).

We have M(t) Letting (x )> E(etX )

` ut 3 `e svt (v2>2)

A resistive voltage divider under a no-load condition is shown in Fig. 12-3, with the standard twoport voltages and currents. The voltage transfer function and input impedance are Hv1 ! R2 R1 R2 Hz1 ! R1 R2

1 22p

The 1 in subscripts indicates no-load conditions. Both Hv1 and Hz1 are real constants, independent of frequency, since no reactive elements are present. If the network contains either an inductance or a capacitance, then Hv1 and Hz1 will be complex and will vary with frequency. If jHv1 j decreases as

(s2t2>2)

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