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263. Here we have a cascade connection of three two-ports in the manner of Fig. 288 and eq. (541) in section 11.9, in which, for Fig. 308, in terms of a coe cients, " # a11 a12 ; for first transistor a1 a21 a22 " # 1 0 a2 ; shunt impedance of Z R ohms in terms of a coefficients* 1=R 1 " # a11 a12 a3 ; for second transistor same values as for first transistor a21 a22 Thus we now have, for Fig. 308, in the manner of eq. (540) I7 IL , V7 RL IL , ! ! ! ! ! V1 1 0 a11 a12 RL IL a11 a12 I1 a21 a22 1=R 1 a21 a22 IL We must now take the rst matrix times the second matrix, then take that result times the third matrix, then that result times the fourth matrix. Upon doing this, carefully following the rule for matrix multiplication from section 11.2, you should nd that the result is the matrix expression ! ! a11 a12 =R a11 a12 a21 RL IL a11 a12 =R a12 a12 a22 IL V1 I1 a21 a22 =R a11 a21 a22 RL IL a21 a22 =R a12 a2 IL 22 Now, using the same procedure as in the solution to problem 262 (noting that RL IL VL , you can verify that VL V1 RL a11 a12 =R a11 a12 a21 RL a11 a22 a12 =R a12
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We must now turn to the conversion chart in section 11.8 to nd the values of the above a-parameters in terms of the given h-parameters. To do this we note that, from the fth row of the chart, a11 dh=h21 0:0085 a21 h22 =h21 0:000 012 5 a12 h11 =h21 25 a22 1=h21 0:025
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Now carefully substitute these values (including V1 0:001, R 500, RL 900, into the above formula for VL . Doing this, you should nd that VL 0:9 0:319 571 2:816 275 0:3196 volts approx:; answer: In regard to the above problem it should be noted that the total phase shift produced by two CE stages in cascade (resistive loads) is equal to 1808 1808 3608, which e ectively amounts to zero degrees of phase shift between input and output signals; thus we can disregard the 1808 of phase shift produced in each stage.
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* Upon applying the conversion chart of section 11.8 to eq. (543).
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264. By the rules for matrix multiplication (and dispensing with the overscore notation) we have 32 3 2 3 2 A A1 1 1 1 76 7 6 7 6 2 4 a a 1 54 A2 5 4 B 5 C a2 a 1 A0 265. The basic problem here is to nd the inverse of the (3 3) matrix, as indicated. This can be done by following the procedure summarized in connection of eq. (482) in section 11.3. In the present case you should nd, for the rst step, noting that a4 a3 a a, that D 3a a 1 Next, you can check that the cofactor form of the (3 3) matrix is (again noting that a4 a) 3 2 2 a a a a2 a2 a 7 6 6 1 a 1 a2 a a2 7 5 4 1 a2 2 6 6 4 1 a a a 1 a 1 a2 a a a 1 a 1 a 1 a a 1 3
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Now transpose the above matrix, then multiply by 1=D; doing this, the statement of problem 265 becomes 2 32 3 2 3 A1 A a 1 a 1 1 6 76 7 6 7 1 A2 5 4 a a 1 54 B 5 4 3a C a a a A0 Now, on the right-hand side of the above equation, take the product of the (3 3) matrix and the (3 1) matrix, then multiply the result by 1/3a. You now have the equality of two (3 1) matrices and thus, in accordance with the law of equal matrices, you have found that 1 aA B a 1 C 3a 1 A2 aA a 1 B C 3a 1 A0 aA aB aC 3a A1
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