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Home Impedancematching Networks Type I Problem  
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Type I ProblemAuthor: Edmund A. Laport
Problem^{1}: Design a network which will match a 500 ohms resistance load to a 100ohm source with a phase shift of 30 degrees through the network.
For vector circuit calculations in general, the first step is to set up the original problem in vector form, using convenient scales for potentials and currents and taking the power derived therefrom into consideration. Draw a potential vector and a current vector in phase in a reference direction representing, according to some convenient scale, the 500ohm load (for example, V_{0} = 500 volts and I_{0} = 1 ampere). According to this arbitrarily chosen scale for the vectors, the power W_{0} represented would be 500 watts. If the impedancematching network is to have pure reactances, the power input to the network must equal the power represented in the load. The next step is to draw a potential vector and a current vector, in phase, according to the same scale of vectors, which will have a ratio representing the input resistance, a product which gives a power equal to that in the load, and a direction which gives the specified phase shift. Since W_{0} = 500 watts, and Figure 5.1 illustrates these first two steps for the problem outlined. This figure represents the problem written vectorially.
We now have a choice of a T or a π solution. Let us choose a T network first and label all the potentials and currents in the load, network, and input in a manner which will satisfy Kirchhoff's laws (see Fig. 5.2). At this point we know V_{0} and I_{0}, V_{in} and I_{in}, R_{0} and R_{in}. We represent the network elements as X's because we do not know what is required yet. We also know, from elementary alternatingcurrent theory, that
Similarly, we know that for circuit elements which are purely reactive, V_{1} is perpendicular to I_{0}, V_{3} is perpendicular to I_{in}, and V_{2} is perpendicular to I_{1}. These facts give us enough information to complete the vector diagram for the entire network. This vector diagram has all the information necessary to reveal the nature and magnitude of X_{1}, X_{2}, and X_{3}. There are only three currents in the circuit, and at the outset of the problem we already know two of them. The unknown current I_{1} can be drawn immediately by connecting the tips of I_{0} and Iin. The direction of I_{1} is that which, when added to I_{0}, gives a vector sum Iin. Thus, I_{1} is directed toward Iin. Its magnitude is determined by its length according to the scale for the current vectors. The intersection of perpendiculars through V_{0} and V_{in} locates V_{2}, as shown in Fig. 5.3. The directions are found from considering that V_{0} + V_{1} = V_{2} vectorially and V_{in} = V_{2} + V_{3} vectorially. From the scales for the diagram
where V_{1} lags I_{0} by 90 degrees,
where V_{2} lags I_{1} by 90 degrees, and
where V_{3} leads I_{in} by 90 degrees. The network becomes that shown in Fig. 5.4. A check on the accuracy of construction is offered by the angle between I_{1} and V_{2} vectors, which should be 90 degrees. The problem specified a phase difference between V_{0} and V_{in}. A different solution results from each different value of phase shift through the network. If phase shift is immaterial, we can find a solution where X_{1} = 0 and the circuit is simplified to two reactive elements. This is the case where V_{3} passes through V_{0} and V_{1} vanishes; also V_{2} = V_{0}. The phase shift θ becomes cos1 (Vin/V_{0}). For values of φ greater than the value where X_{1} = 0, the sign of X_{1} changes from j to +j.^{2}
When φ is positive instead of negative, the signs of all the elements reverse likewise. Taking now the πnetwork solution for the same problem, the circuit becomes that shown in Fig. 5.5. In this case, we know V_{0}, I_{0}, V_{in}, I_{in} and φ. We also know X_{1}, X_{2} and X_{3} to be pure reactances. Therefore
I_{1} must be perpendicular to V_{0}, I_{2} must be perpendicular to V_{1}, and I_{3} must be perpendicular to V_{in}. Since we know V_{0} and V_{in}, we can draw V_{1} immediately, as shown in Fig. 5.6. From the scales of the diagram
The network becomes that shown in Fig. 5.7. If the value of φ is immaterial, there is a value at which I_{3} = 0 and X_{3} = . This particular solution simplifies the circuit to two reactive elements. For values of φ greater than this particular value, the sign of X_{3} reverses.
To demonstrate the solution for the twoelement circuit to obtain this transformation, we set up input and output vectors but do not specify a phase shift. Instead we allow the input vectors to assume a position which, in the T case, causes V_{3} to pass directly through V_{0} and, in the π case, causes I_{1} to pass through Iin. For example in the π case we drawthe loci of V_{in} and I_{in} for variable φ, as shown in Fig. 5.8. Then we erect I_{1} perpendicular to V_{0} through I_{0}. Where this cuts the locus of I_{in} locates the vector Iin.
In the case of the T, X_{1} becomes 0, and in the case of the π, X_{3} becomes . Thus, the T and π solutions merge into a common solution at this value of φ where two reactive elements suffice to solve the problem, as shown in Fig. 5.9.
Where φ is larger, for the T case, the diagram becomes that shown in
Fig. 5.10. The circuit becomes that shown in Fig. 5.11. Complete curves of reactances for both π and T networks for this transformation as a function of φ are reproduced in Figs. 5.12 and 5.13 to show the nature of the variations. Any impedancematching problem can be reduced to a type I problem by correcting the power factor of the load or of the generator, when they are other than resistive. This can always be done in either of two ways by adding series reactance or by adding parallel reactance.
The choice depends upon the relative physical convenience of the ohmic values obtained, whether the desired phase difference is between the currents or the potentials, and the cost or convenience of obtaining the proper reactive components. If a T network is to be used, series powerfactor correction of the load impedance or the generator impedance has the advantage that the powerfactorcorrecting reactors can be combined in value with those found to be necessary from the design synthesis, so that fewer components are ultimately required  three instead of four or five. In the same sense, if a π network is chosen, the parallel method of powerfactor correction permits the use of three reactive components instead of four or five by combining the powerfactorcorrecting reactances with those derived from the network synthesis.
This is virtually what is done in the type II problems. In general, it is always wise to employ network designs requiring the least possible energy storage  that is, the least number of total voltamperes in all the reactive components. This not only economizes the ratings of the individual components but minimizes the selectivity of the network.This has special importance whenever the bandwidth to be transmitted exceeds 1 percent. The various individual reactances for a network can be made up of lumped capacitance or inductance; or equivalent values of transmissionline sections may be used at frequencies where line sections are preferable.


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