qr code vb.net library Notes Regarding the Interconnection Formulas in .NET

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Notes Regarding the Interconnection Formulas
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In the preceding work dealing with the interconnection of two-ports, we ve assumed that each two-port operates in the normal balanced mode, meaning that both input terminals carry the same current I1 and both output terminals carry the same current I2 , as shown in Fig. 289. If we have a situation in which this condition is not true, then we do not
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CHAPTER 11 Matrix Algebra. Networks
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Fig. 289
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have the normal two-port network de ned by Fig. 277, and the equations we ve developed for interconnected two-ports are not valid. To understand how such current imbalances can occur, consider the series connection of two two-port networks, a and b, shown in Fig. 290, in which we ll concentrate our attention on the input and output currents of network a.
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Fig. 290
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In Fig. 290, depending upon the type of networks that blocks a and b represent, it is possible that an unwanted loop current I3 can ow in the interconnection, as shown. If this happens, inspection of the gure shows that, in general, the current at terminal 1 will not be equal to the current at terminal 1 0 . In such a case, network a is no longer a normal balanced two-port, and the two-port equations previously derived will not be valid. Therefore, the condition that the previously derived formulas for series-connected twoports be valid is that no circulating current can exist between the networks, that is, that I3 0 in Fig. 290. Fortunately, a simple test can be applied to determine whether the basic equation (534), found in section 11.9, is valid for a given series connection of two-ports. The test setup is shown in Figs. 291 and 292. The test is carried out as follows. The rst step, shown in Fig. 291, is to connect the inputs of the two networks in series, leaving the outputs open-circuited, as shown. Now imagine a signal voltage Vs to be applied to the series-connected inputs, as shown. Then eq. (534) is valid if, and only if, the voltage V is zero. The second step in the test is to apply the test signal Vs to the series-connected outputs with the inputs open-circuited, as shown in Fig. 292. Again, eq. (534) is valid only if V 0. If the foregoing test shows that the two given two-ports will not satisfy the requirement that V 0 for the series connection, they can sometimes be put into a di erent but equivalent form, for which V will be zero.
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Matrix Algebra. Networks
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Fig. 291
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Fig. 292
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As an example of this, consider rst Fig. 293, which shows the test setup of Fig. 291 for a proposed series connection of two particular two-ports. In Fig. 293, note that V will not be equal to zero, because of the voltage drop across R, and hence Eq. (534) will not apply if the two networks are connected in series. If, however, we convert the top network into its T equivalent (section 9.2), then Fig. 293 becomes Fig. 294, for which V does equal zero. Hence, if Fig. 293 is put into the equivalent form of Fig. 294, then eq. (534) will be valid for the series connection of the two networks.
Fig. 293
Fig. 294
It s also possible that an unwanted loop current can ow in the parallel connection of Fig. 285. The test setup to determine whether eq. (536) is valid for a parallel connection of two two-port networks is shown in Figs. 295 and 296. To apply the test we begin with Fig. 295, in which a signal voltage Vs is applied to the parallel inputs, with the two outputs, previously connected in parallel, now disconnected and with each short-circuited, as shown. The voltage V is now calculated or measured. The operation is then repeated in the reverse direction, as shown in Fig. 296. Only if V 0 for both test conditions is eq. (536) valid for the parallel connection of the given two-ports. If V does not equal zero in a given case, it may be possible to transform one, or both, of the two-ports into an equivalent network for which V 0, in a manner such as was done for the series connection of Fig. 293. In the preceding discussions we ve found that if a test setup shows V not equal to zero, we must then try to change one or both of the networks into an equivalent form for which
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