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CHAPTER 9 Impedance Transformation
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Problem 167 " In Fig. 190, given that R 10,000 ohms and C 0:25 mF, if Vi Vi 10 volts rms, "o if (a) the frequency is 30 Hz, nd the amplitude and phase of the output voltage V (b) the frequency is increased to 300 Hz. Problem 168 In the above problem (same values of R and C), suppose an ideal square wave of voltage, having a steady frequency of 30 square waves per second, is applied to the input. Explain why the output voltage waveform will not also be an ideal square wave.
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CASE II: BASIC HIGH-PASS RC NETWORK
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This is a series RC circuit in which the output voltage appears across the resistor R, as shown in Fig. 194.
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Fig. 194
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In the gure, it is given that the AMPLITUDE of the reference input signal, " Vi Vi =08 Vi , is to remain constant, while the FREQUENCY is allowed to increase from a very low value to progressively higher values. From the gure, note that as the frequency of Vi increases the reactance of C decreases, and thus as the FREQUENCY INCREASES the OUTPUT VOLTAGE ACROSS R INCREASES; in other words, the higher the frequency, the higher is the output voltage. Thus the circuit of Fig. 194 discriminates against the lower frequencies and is therefore a high-pass type of network. The algebra for Fig. 194 parallels that for Fig. 190, as follows. First, by Ohm s law, " " Vo RI RVi R jXC
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Now set jXC 1=j!C; doing this, then multiplying the numerator and denominator by j!C, we have " j!RC Vo " G Vi 1 j!RC 326
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which corresponds to eq. (316). Now, again setting RC 1=!1 , as in eq. (317), the last equation becomes " G where h !=!1 . j !=!1 jh 1 j !=!1 1 jh 327
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CHAPTER 9 Impedance Transformation
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Equation (327) contains all the information concerning both the amplitude and phase response of the network of Fig. 194. We ll rst investigate the amplitude response, as follows. The AMPLITUDE response is determined by the MAGNITUDE of eq. (327); thus, recalling that if A and B are complex numbers, jA=Bj jAj=jBj, we have that h " jGj p h 1 h2 1=2 328 1 h2 which corresponds to eq. (319). Now, to express the amplitude response in decibels, let us, in the manner of eq. (320), write that dB 20 log h 1 h2 1=2 Now make use of the fact that log XY log X log Y,* and that log X n n log X.{ Doing this, the last equation becomes dB 20 log !=!1 10 log 1 !=!1 2 329 " which expresses the ratio of Vo =Vi in Fig. 194 in terms of decibels. Using your calculator, you can quickly verify that the following table of values is correct for eq. (329).
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!=!1 0.02 0.04 0.06 0.08 0.10 dB gain 34.0 28.0 24.5 22.0 20.0 !=!1 0.2 0.4 0.6 0.8 1.0 dB gain 14.2 8.6 5.8 4.1 3.0 !=!1 2.0 4.0 6.0 8.0 10.0 dB gain 0.97 0.26 0.12 0.07 0.04
A plot of the above data gives Fig. 195, the curve of dB gain versus frequency ratio.
Fig. 195
* See note 22 in Appendix. { See eq. (9-A) in note 19 in Appendix.
CHAPTER 9 Impedance Transformation
Figure 195 veri es that the circuit of Fig. 194 is a high-pass network because the higher the frequency, the less is the attenuation of the input signal as it appears at the output terminals. In the discussion following Fig. 191 the term half-power frequency was introduced as being the frequency at which the power gain of a network is down 3 decibels from its maximum or reference value. Note, now, that setting ! !1 in eq. (329) gives the value dB 20 log 1 10 log 2 3 decibels showing that the value of ! at the half-power frequency in Fig. 194 is ! !1 1=RC (by eq. (317)). Problem 169 Suppose, in Fig. 194, that R 1200 ohms. Find the required value of C if the halfpower frequency is to be 2.2 kHz. Problem 170 (a) In problem 169, at what frequency, in hertz, will the power gain be 6 decibels (Answer: 1274 Hz) (b) At what frequency will the power gain be 2 decibels (Answer: 2877 Hz) Next, to investigate the PHASE RESPONSE of Fig. 194 we ll basically repeat the " " procedure for Fig. 190, as follows. First, by eq. (322), Vo GVi , which, after rationalizing " the value of G given by eq. (327), becomes " # h2 h "o V Vi A jB Vi j 330 1 h2 1 h2 " showing that Vo is the vector sum of the voltage drops AVi and jBVi , as shown in Fig. 196.
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