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CHAPTER 9 Impedance Transformation
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Problem 181 Suppose, in Fig. 208, that the inductance of each coil is 800 microhenrys and that C 0:04 microfarads. What value of RL should be used At what frequency will the characteristic impedance be equal to zero What value of impedance will the generator see at the frequency at which Z0 0 (d) Why doesn t the generator see zero impedance in part (c) r 1 C Thus, since , we have that RL L r r p !L C L2 C ! LC !L ! RL L L hence eq. (348) becomes " 1 V2 p 2 LC=2 j! LC 1 !2 LC=4 V1 1 ! which has the form " 1 V2 V1 A jB In the above the variable is frequency, ! radians/second, in which ! denotes ANY FREQUENCY in general. It will be helpful, however, in understanding what eq. (351) says, if the equation is written in terms of the ratio of ! to some PARTICULAR VALUE of ! that we ll denote by !0 . In doing this, !0 can be any particular value we choose, but in the case of eq. (351) one convenient way would be to set !2 0 that is, LC 4 !2 0 352 4 LC 351 (a) (b) (c)
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and if you now substitute the above value of LC into eq. (351) you should nd that " 1 V2 V1 1 2 !=!0 2 j2 !=!0 1 !=!0 2 353
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" so that the ratio 2 =V1 is now expressed in terms of the value of ! relative to the xed p V value !0 2= LC , and we say that the equation is normalized relative to !0 . The advantage of doing this is that now we can investigate eq. (351) in general terms without having to carry along the values of L and C. As a further step let us make the substitution h !=!0 thus eq. (352) becomes " 1 1 V2 V1 1 2h2 j2h 1 h2 A jB 355 354
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CHAPTER 9 Impedance Transformation
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Now let it be given that we wish to investigate only the manner in which the magnitude " of V2 =V1 varies with h; in which case you can verify that eq. (355) becomes   "  V2  1 1   p p 356 V  2 B2 A 1 4 h6 h4 1 thus,   "  V2    1 4 h6 h4 1=2 V  1
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357
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which in terms of decibels becomes (see eqs. (319), (320), and (321)) dB 10 log 1 4 h6 h4 358
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Now, using your calculator, you can verify that the following table of values is correct for eq. (358), in which we ve rounded o dB values to two decimal places.
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h 0.1 0.2 0.4 0.6 0.8 0.9 dB 0.00 0.03 0.39 1.75 3.87 3.00 h 1.0 1.2 1.4 1.6 1.8 2.0 2.2 dB 0.00 6.67 11.97 16.22 19.73 22.86 25.57
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A plot of the above results on semi-log paper is given in Fig. 209, with a brief discussion following.
Fig. 209
Thus the gain of the low-pass network of Fig. 208 is practically constant from h 0 to h 0:3; then, as h increases, the gain rises to a maximum value of 3.87 dB for h 0:8, after which the gain rapidly decreases as h increases in value. The rise in gain from h 0:3 to
CHAPTER 9 Impedance Transformation
h 1:0 is due to a resonant condition that comes into play between L and C, the e ect peaking, very approximately, at h 0:8. Then, as h increases in value beyond h 1:0, the condition of resonance is lost and the gain begins to rapidly decrease. It is thus the presence of both inductance and capacitance, L and C, having the possibility of resonance, that causes the curve of Fig. 209 to be so much improved over the simple curve of Fig. 191. As inspection of Fig. 209 shows that the gain falls o rapidly for values of h greater than 1, and for that reason h 1 is sometimes taken to be the cut-o condition for Fig. 208. Since h !=!0 , it follows that h 1 when ! !0 ; thus the cut-o frequency, !c , can be taken to be equal to !0 and hence, by eq. (352), 2 !c p LC 359
Lastly, we should mention that the term constant-k is applied to Fig. 208 because the " " values of Z1 and Z2 are such that their product has a constant value, independent of frequency; thus " " Z1 Z2 j!L 1=j!C L=C Problem 182 In Fig. 208, suppose C 0:015 mF. If a cut-o frequency of 100,000 Hz is desired, what must be the inductance of each coil (Answer: 337.7 mH) Problem 183 " Using eq. (355), show that the phase shift of V2 with respect to V1 is equal to  arctan 2h 1 h2 = 1 2h2
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