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CHAPTER 9 Impedance Transformation
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following results, as you should verify: Switch #2 open closed open open " " " Z Z Z C " Z1O " A B "B ZC " ZA Z " " Z Z " Z1S " A B " ZA ZB " " " Z Z ZB " Z2O " C A " " ZA ZB ZC " " Z Z " Z2S " B C " ZB ZC
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290 291 292 293
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" " Now, as an exercise in algebraic manipulation, let s verify that the values of ZA and ZB given by eqs. (285) and (286) are correct; one way to do this is as follows. First, by eq. (291), " " Z Z " ZB " A 1S " ZA Z1S Next, using the relationships found in the above switching table, we have that " " " Z Z Z " " Z1S Z2O " A "B C " ZA ZB ZC 295 294
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Next, again making use of the relationships in the switching table, you can verify that " " " Z2O Z1O Z1S "2 "2 ZA ZC " " " ZA ZB ZC 2 296
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Now invert both sides of the last equation, then take the square root of both sides, then multiply, respectively, the left-hand and right-hand sides of the result by the left-hand and right-hand sides of eq. (295); doing this, you should nd that " " " Z1S Z2O Z0 " p ZB 00 " " " " Z Z2O Z1O Z1S thus verifying that eq. (286) is correct. Now, in the above, substitute the right-hand side of " " eq. (294), in place of ZB , then solve for ZA to verify that eq. (285) is also correct. " "A and ZB (from eqs. (285) and (286)) can be substituted into, for Next, the values of Z " example, eq. (292), the result then being solved for the value of ZC , which will prove that eq. (287) is correct. The above results can be summarized in the statement that any linear, bilateral network, containing no internal generators, can be represented, at a single frequency, by a T or pi network. Problem 155 The network in Fig. 177 is composed of pure resistances having values in ohms, as shown. Find the equivalent T network. Is the answer valid at all frequencies
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CHAPTER 9 Impedance Transformation
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Fig. 177
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Problem 156 In Fig. 178, it is given that C 2 F and L 150 H, the resistance values being in ohms, as shown. Draw the diagram of the equivalent T network, showing the required values of inductance, capacitance, and resistance, for operation at 105 radians/second.
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Fig. 178
Problem 157 Draw the T network equivalent of the purely resistive bridge-type network shown in Fig. 179. Resistance values are in ohms.
Fig. 179
Problem 158 Suppose the following measurements are made on a certain network at a particular frequency of interest: " Z1O 18 j12 ohms; " Z1S 8 j18 ohms; " Z2O 20 j12 ohms:
CHAPTER 9 Impedance Transformation
Find the values of the equivalent pi representation of the network, at the frequency of interest.
Conversion of Pi to T and T to Pi
In practical work it s sometimes helpful to convert a given pi network into an equivalent T network, or to convert a given T network into an equivalent pi network. This can be done as follows, in which we ll continue to use the standard notation of Figs. 174 and 175. " " " " In section 9.2 we de ned the quantities Z1O , Z1S , Z2O , and Z2S as being the values of external measurements made at the input and output terminals of a network. It follows that if two networks are to be equivalent, then the values of these external measurements must be the same for both networks. Algebraically, this means that the right-hand side of eq. (278) must be equal to the right-hand side of eq. (290), the right-hand side of eq. (279) must be equal to the right-hand side of eq. (291), the right-hand side of eq. (280) must be equal to the right-hand side of eq. (292), and likewise for eqs. (281) and (293); thus the following system of equations must be satis ed: " " " Z Z Z C " " Z1 Z3 " A B "B ZC " ZA Z " " " " Z Z Z Z " Z1 " 2 3 " A B "3 ZA ZB " Z2 Z " " " Z Z Z C " " Z2 Z3 " A " B " ZA ZB ZC " " " " Z Z Z Z " Z2 " 1 3 " B C " " Z1 Z3 ZB ZC 297 298 299 300
Equations (297) through (300) express the relationships that must always exist between two equivalent T and pi networks. Making use of these relationships, we can derive equations that will allow us to convert from one type of network to the other. " " " Suppose, for example, that ZA , ZB , and ZC are known, and we wish to nd the equations for calculating the equivalent T network. After several false starts, we nd that the following procedure will work. First, subtract eq. (298) from eq. (297) to get " " " " " " " Z Z Z Z ZC ZA ZB " Z3 " 2 3 " A B "3 ZA ZB ZC ZA ZB " " " " Z2 Z or, after putting the left side over its common denominator and the right side over its common denominator, we have "2 "2 " Z3 ZA ZC " " " " " " " Z2 Z3 ZA ZB ZC ZA ZB We ve so far made use of eqs. (297) and (298); we can now make use of eq. (299), as " " " " follows. Multiply both sides of the last equation by Z2 Z3 , then replace Z2 Z3 by " the right-hand side of eq. (299). Doing this, then solving for Z3 , you should nd that " " ZA ZC " Z3 " " " ZA ZB ZC 301
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