110_1_from Su_TwoPorts_chapter

110_1_from Su_TwoPorts_chapter - 532 F UNDAMENTALS O F...

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533 532 FUNDAMENTALS OF CIRCUIT ANALYSIS The arrangement in Fig. 10.24 assumes an ideal voltage source at port 1. A more practical situation is that in Fig. 1O.25-the doubly terminated two-port. E. and Z. is the Thevenin's equivalent of that part of the network looking to the left from port 1. ZL represents the equivalent impedance seen by the two-port from port 2 to its right. Suppose [FJ of N is given and we are interested in the voltage gain E2/ E •. Then we have El = AE2 - BI2 (10.77) II = CE2 - DI2 (10.78) Zs II 12 + + ZL El N E% Figure 10.25: A doubly terminated two-port. We also have E2 E2 = -I2ZL or 12 = - ZL (10.79) II = E. - El El = E. - liZ. or (10.80) Z. Substitute Eq. (10.77) into Eq. (10.80) and equate it to Eq. (10.78). We get E. - (A2 - BI 2) = CE 2 _ DI% II = Z. (10.81) Solving for 12 and using Eq. (10.79), we get 12 = AE2 + CE2 Z. - E. = _ E2 (10.82) B+DZ. ZL TWO-PORT NETWORKS Eq. (10.82) involves only E. and E 2• Hence, E2 ZL (10.83) E. - AZL+B+CZLZ.+DZ. is easily obtained. \ \ EXERCISE \1.0.5.1 Obtain Y21 = 12/E. for the arrangement in Fig. 10.25 in terms of the two-port open-circuit impedance parameters. Ans.
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This note was uploaded on 03/28/2011 for the course EE EE 110 taught by Professor Gupta during the Winter '09 term at UCLA.

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110_1_from Su_TwoPorts_chapter - 532 F UNDAMENTALS O F...

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