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Figure 10-1 is a graph of the binomial distribution (and a normal approximation to it) that gives the probabilities of x heads in 12 tosses of a fair coin, where x 0, 1, 2, c, 12. From 4 the probability of x heads is Pr5x6 whereby Pr{0} 0.00024, Pr{l} a 12 1 x 1 12 ba b a b 2 2 x

Fig. 5-29

0.00293, Pr{2}

0.01611, and Pr{3}

A frequency-selective ampli er whose gain decreases from a nite value to zero as the frequency of the sinusoidal input increases from dc to in nity is called a low-pass lter. The plot of gain versus frequency is called a frequency response. An easy technique for nding the frequency response of lters will be developed in 13. The leaky integrator of Fig. 5-24 is a low-pass lter, as illustrated in the following example.

Fig. 10-1

EXAMPLE 5.22 In Example 5.18 let v1 sin ! t. Find jv2 j for ! 0; 10; 100; 103 ; 104 , and 105 rad/s. By repeating the procedure of Example 5.18, the frequency response is found and given in Table 5-1. response amplitude decreases with frequency. The circuit is a low-pass lter.

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Since H1 is the hypothesis that there is a difference between the machines, rather than the hypothesis that machine I is better than machine II, we use a two-tailed test. For the 0.05 significance level, each tail has the associated probability 1(0.05) 0.025. We now add the probabilities in the left-hand tail until the sum exceeds 2 0.025. Thus Pr{0, 1, or 2 heads} Pr{0, 1, 2, or 3 heads} 0.00024 0.00293 0.01611 0.01928 0.07299

2. Given the object model shown in Fig. 14-3, explain each OCL statement. What does it specify Is the OCL invariant reasonable Is it always true

Since 0.025 is greater than 0.01928 but less than 0.07299, we can reject hypothesis H0 if the number of heads is 2 or less (or, by symmetry, if the number of heads is 10 or more); however, the number of heads [the signs in sequence (1) of this chapter] is 3. Thus we cannot reject H0 at the 0.05 level and must conclude that there is no difference between the machines at this level.

Table 5-1. Frequency Response of the Low-pass Filter !, rad/s f , Hz jv2 =v1 j 0 0 1 10 1.59 1 100 15.9 0.995 103 159 0.707 104 1:59 103 0.1 105 15:9 103 0.01

For a normal approximation to the binomial distribution, we use the fact that the z score corresponding to the number of heads is Z X s m X Np !Npq .

The circuit of Fig. 5-30 compares the voltage v1 with a reference level vo . Since the open-loop gain is very large, the op amp output v2 is either at Vcc (if v1 > vo ) or at Vcc (if v1 < vo ). This is shown by v2 Vcc sgn v1 vo where sgn stands for sign of. For vo 0, we have Vcc v1 > 0 v2 Vcc sgn v1 Vcc v1 < 0

Because the variable X for the binomial distribution is discrete while that for a normal distribution is continuous, we make a correction for continuity (for example, 3 heads are really a value between 2.5 and 3.5 heads). This amounts to decreasing X by 0.5 if X Np and to increasing X by 0.5 if X Np. Now N 12, m Np (12)(0.5) 6, and s !(12)(0.5)(0.5) 1.73, so that !Npq z (3 0.5) 1.73 6 1.45

Since this is greater than 1.96 (the value of z for which the area in the left-hand tail is 0.025), we arrive at the same conclusion in Problem 10.1. Note that Pr5Z 1.456 0.0735, which agrees very well with the Pr5X 3 heads6 0.07299 of Problem 10.1.

Fig. 5-30 EXAMPLE 5.23 In Fig. 5-30, let Vcc 5 V, vo 0, and v1 sin !t. Find v2 . For 0 < t < =!, v1 sin !t > 0 For =! < t < 2 =!, v1 sin !t < 0 v2 5 V One cycle of v2 is v2 5 V

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