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CHAPTER 9 Impedance Transformation
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CASE of the low-pass RC circuit of Fig. 190, upon setting !=!1 1 in eq. (321), we have that dB 10 log 2 3 decibels hence the half-power frequency for Fig. 190 is at !=!1 1, that is, for ! !1 1=RC (making use of eq. (317)). Problem 165 Suppose, in Fig. 190, that C 0:05 mF. Find the required value of R if the halfpower frequency is to be 7.2 kHz. Problem 166 In problem 165, at what frequency, in kHz, will the power gain be 6 decibels So far we ve concentrated our attention on the amplitude response of Fig. 190 (given by eq. (319)). Let us now complete our work with an examination of the PHASE response of the network. To do this, we can begin with the basic relationship " " Vo GVi where Vi is the reference input rms voltage. Thus, by eq. (318), " Vo Vi Vi 1 j !=!1 1 jh 323 322
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where, for convenience, we ve temporarily set (!=!1 h. Now let s rationalize the last fraction; that is, let us multiply the numerator and denominator by the conjugate of the denominator, (1 jh),* so that eq. (323) becomes ! 1 h "o j 324 V Vi A jB Vi 1 h2 1 h2 where A 1= 1 h2 and B h= 1 h2 Equation (324) shows that component A is IN PHASE with the reference vector Vi , while component B LEADS Vi by 90 degrees, the vector sum of A and B being IN PHASE with the " output voltage vector Vo , as shown in Fig. 192, in which  (phi or fee ) is the PHASE " ANGLE (phase shift) between the output voltage Vo and the input reference vector Vi .
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Fig. 192. Here, tan / opp=adj B=A; thus, / arctan (B/A).
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Fig. 193. Here, tan / opp=adj h; thus, / arctan ( h); that is, / arctan ( x=x1 ).
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" * Doing this will show, separately, the real and imaginary components of Vo .
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CHAPTER 9 Impedance Transformation
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If, now, in Fig. 192 we replace A and B with their values de ned in connection with eq. " (324) then Fig. 192 becomes Fig. 193, which shows that the output voltage vector Vo LAGS the input reference voltage by the angular amount of  arctan !=!1 arctan !=!1 325
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Your calculator, applied to the above equation, should produce the following table of values.
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!=!1 0.01 0.02 0.04 0.06 0.08 0.10 8 0.6 1.2 2.3 3.4 4.6 5.7 !=!1 0.2 0.4 0.6 0.8 1.0 2.0 8 11.3 21.8 31.0 38.7 45.0 63.4 !=!1 4 6 8 10 50 100 8 76.0 80.5 82.9 84.3 88.9 89.4
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Inspection of the table brings out the following facts concerning the PHASE-SHIFT characteristics of the basic low-pass network of Fig. 190. 1. The phase angle  is negative, meaning that the output voltage LAGS the input voltage. For this reason, Fig. 190 is called a lag network in control system terminology. It has been shown that, if a network is to produce no phase distortion,  must be proportional to !; that is, the ratio of  to ! must be constant, =! k (note 20 in the Appendix). The above table of values shows that this requirement is very nearly satis ed, in the case of Fig. 190, for low frequencies. To see that this is true, let us set !1 1 and, using the above table of values, construct the following table, where  is in degrees and the ratio =! is rounded o to the nearest whole number.
0.6 1.2 2.3 3.4 4.6 5.7 11.3 21.8 ! 0.01 0.02 0.04 0.06 0.08 0.10 0.20 0.40 =! 60 60 58 57 58 57 57 55  31.0 38.7 45.0 63.4 76.0 80.5 84.3 88.9 ! 0.60 0.80 1.00 2.00 4.00 6.00 10.00 50.00 =! 52 48 45 32 19 13 8 2
The table plainly shows that, for practical purposes, the ratio =! is constant for all low frequencies, up to ! 0:04; that is, since we re using !1 1 here, up to the value of !=!1 0:04. Actually, for most practical purposes, we can say that =! is constant up to the value !=!1 1; that is, to ! !1 , where !1 is the half-power frequency. Thus our CONCLUSIONS regarding Fig. 190 can be summarized as follows. Figure 190 is a low-pass network having, for practical purposes, no amplitude or phase discrimination for frequencies LESS than the half-power frequency, but increasing amounts of such discrimination as the frequency increases beyond the half-power frequency.