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Now nd the values of A, B, C, D, by making use of, from problem 254, the matrix equation, # ! " 2 A B d h h11 h22 dhh11 h11 C D dhh22 h22 h11 h22 1 256. As noted in the solution to problem 255, the network matrix in the answer to problem 254 is of the form ! A B C D
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the inverse of which is (see problem 232 if you wish) ! ! A B 1 D B 1 AD BC C A C D Now substitute, into the last expression, the values of A, B, C, D, found from inspection of the matrix equality that appears in the solution to problem 255. Doing this gives, after some simpli cation, the answer ! 1 h11 h22 1 dh h11 1 1 h 2 dh h11 h22 2 1 dh h22 d h h11 h22 257. Note that eq. (545) is of the form ! ! ! ! A B I1 AI1 BI2 V1 CI1 DI2 V2 C D I2 which, by the de nition of equal matrices, shows that AI1 BI2 V1 CI1 DI2 V2 Thus    V1 B     V D  DV BV 1 2 2 I1 AD BC AD BC
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which, upon substituting in the values of A, B, C, and D, gives the required answer. 258. In eq. (545) we must rst write the transistor z-parameters (Z11 ; Z12 ; Z21 ; Z22 ) in terms of h-parameters, which is most easily done by making use of the conversion chart in section 11.8. Doing this, eq. (545) becomes 2    3 dh h12 ! 6 ! Z Z 7 V1 h 7 I1 6 h22   22 7 6 ; answer: 5 I2 4 h21 1 V2 Z Z h22 h22 259. No, because Z in eq. (543) is a singular matrix. (See discussion just prior to eq. (484) in section 11.3.)
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260. First write eq. (548) in the inverse form ! 1 ! ! V1 Vo 1 Z0 Io I1 1=Z 1 Z 0 =Z in which (as in the solution to problem 256) A 1, B Z 0 , C 1=Z and D 1 Z 0 =Z ; thus, using the special formula noted in the solution to problem 256, we nd that ! ! ! Vo 1 Z 0 =Z Z 0 V1 ; answer: Io I1 1=Z 1 261. We can make use of eqs. (506) and (507) in section 11.6 as follows. First, from Fig. 307, note that V2 IL ZL and thus eqs. (506) and (507) become V1 h11 I1 h12 ZL IL 0 h21 I1 1 h22 ZL IL We must now solve the foregoing two simultaneous equations for IL . This can be done by using either determinants or the method of elimination. Thus, if you multiply the rst equation by h21 and the second equation by h11 , then add the two equations together, you should nd that IL h21 V1 h12 h21 ZL h11 1 h22 ZL and I2 IL
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which, upon substituting in the given h-values, ZL 150 ohms, V1 12 volts, should give you the required answer. 262. Since V2 RL IL and I2 IL , eq. (514) becomes ! ! ! a11 a12 RL IL V1 I1 a21 a22 IL thus V1 I1 ! a11 RL I L a12 IL a21 RL I L a22 IL !
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hence, by the de nition of equal matrices (section 11.1), V1 a11 RL a12 IL thus IL V1 a11 RL a12
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We must now make use of the conversion chart in section 11.8 to express the required a-parameters in terms of the given h-parameters. Carefully doing this, you should nd that a11 0:14 and a12 20, and now, substituting these a values into the last equation above, along with V1 12 and RL 150, you should nd that IL 0:2927 amperes, as before.
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