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CHAPTER 9 Impedance Transformation
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Problem 175 If the open-circuit and short-circuit measurements on a certain symmetrical T network give magnitudes of 15 and 25 ohms and angular displacements of 35 and 75 degrees, nd the characteristic impedance of the network. " In the foregoing discussion we should remember that the value of an impedance Z (assuming given values of R, L, and C) depends upon the value of the frequency !. Thus " the value of Z0 given by eq. (340) depends upon the value of !, and is di erent for each value of !. The di erence between Figs. 202 and 203 is that Fig. 203 is for the particular case for which q " "2 " " " ZL Z0 Z1 Z2 Z1 =4 " " which can be exactly true, in general, for only one frequency, because ZL and Z0 will be represented by di erent mathematical equations. This e ect will be considered later on in our treatment of the constant-k lter. Problem 176 In Fig. 206, the values are in henrys, farads, and ohms. Find the frequency, in rad/ (Answer: 173.21 rad/sec) sec, at which Zin 2 ohms.
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Let us continue with the symmetrical T of Fig. 203, which depicts the particular con" " " dition in which Zin ZL Z0 . As we know, eq. (340) applies to this condition and, in addition to eq. (340), several other useful relationships exist. As an aid in nding these relations let us begin by redrawing Fig. 203 as Fig. 207.
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Fig. 207
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By Kirchho s voltage law, the sum of the voltage drops AROUND ANY CLOSED PATH IN A NETWORK is equal to the sum of the generator voltages in that path. Thus,
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CHAPTER 9 Impedance Transformation
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in Fig. 207, around the closed path abcda we have that " " " " " " " V1 Z1 I1 =2 Z1 I2 =2 Z0 I2 " " " or, since V1 I1 Z0 ; " " " " " " " " Z0 I1 Z1 I1 =2 Z1 I2 =2 Z0 I2 from which we have the important relationship " Z1 " " I1 Z 0 2 " " I2 Z " Z0 1 2 342
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Problem 177 " " " Find the ratio of I1 to I2 in Fig. 206 for the frequency at which Zin 2 ohms. (Answer: 0:500 j0:866 As you can see from the above, eq. (342) was derived for the particular frequency for " " " which Zin ZL Z0 . For any frequency in general, eq. (342) becomes " " " I1 ZL Z1 =2 " " " Zin Z1 =2 I2 " " in which ZL will be a given load impedance and Zin can be found by eq. (338). Problem 178 Derive eq. (343), using the same procedure as in deriving eq. (342). Problem 179 " " In Fig. 206, nd I1 =I2 for ! 200 rad/sec. 343
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Problem 180 If the T network in problem 174 is terminated in its characteristic impedance, nd " " the magnitude and phase angle of I1 relative to I2 . It s important to note, here, that the equation for the current ratio can also be written " in the following form, in which ZL denotes ANY VALUE of load impedance: " " " I1 Z1 ZL 344 * " 1 2Z 2 Z 2 " " I2 " " Or, setting ZL Z0 , then making use of eq. (340), and noting that r s " " "2 Z2 Z1 1 Z1 " Z1 Z2 1 "2 " " 4 Z2 4Z2 Z2 " " eq. (344) becomes, for the special case in which ZL Z0 , s " " " " !2 I1 Z1 Z1 Z1 " " " 1 2Z2 Z2 2Z2 " I2
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