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1. (a) A (9, 2), B (2, 4), and C9(5, 7); (b) A ( 4, 2), B (3, 4), and C (0, 7); (c) A ( 5, 1), B (2, 1), and C ( 1, 4); (d) A ( 2, 14), B ( 4, 7), and C ( 7, 10); (e) A (12, 4), B ( 2, 8), and C (4, 14); (f) A ( 14, 6), B (7, 12), and C ( 2, 21); (g) A (0, 1), B ( 2, 6), and C ( 5, 3) 2. (a) A (6, 1), B (7, 2), and C (8, 1); (b) A (1, 5), B (2, 2), and C (3, 5); (c) A ( 2, 1), B ( 1, 4), and C (0, 1)

For the parallel network shown in Fig. 12-54, (a) nd the value of R for resonance; (b) convert the RC branch to a parallel equivalent. Ans: a 6:0 ; b Rp 6:67 ; XCp 20 For the network of Fig. 12-55(a), nd R for resonance. Obtain the values of R 0 , XL , and XC in the parallel equivalent of Fig. 12-55(b). Ans: R 12:25 ; R 0 7:75 ; XL 25 ; XC 25

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3. (a) translation to the right 5 spaces, P(x, y) A P (x 5, y); (b) translation down 5 spaces and to the right 3, P(x, y) A P (x 3, y 5); (c) translation down 3 spaces and to the left 5, P(x, y) A P (x 5, y 3) 4. (a) P(x, y) A P (x, y 5); (b) P(x, y) A P (x 6, y); (c) P(x, y) A P (x 7, y (d) P(x, y) A P (x 8, y 2); (e) P(x, y) A P (x 1, y 4) 3);

where KF V2 =V1 is obviously the forward voltage-gain ratio of the ampli er. In a similar manner, Yo and the use of (8.45) in (8.47) gives us I V V2 Yo 2 1 YF V2 V2 Y2 KR 1 YF Y2 1 KR YF 8:48 Io I2 IF V2 V2 8:47

Fig. 12-55 12.31 Branch 1 of a two-branch parallel circuit has an impedance Z1 8 j6 at ! 5000 rad/s. Branch 2 contains R 8:34 in series with a variable capacitance C. (a) Find C for resonance. (b) Sketch the admittance locus diagram. Ans: a 24 mF b See Fig. 12-56

5. (a) A ( 1, 3), B ( 5, 3), C ( 4, 1), and D ( 2, 1); (b) A0(1, 5), B0 (5, 5), C0(4, 3), and D0(2, 3); (c) A-(15, 3), B-(11, 3), C-(12, 1), and D-(14, 1) 6. (a) reflection across the line x 2, P(x, y) A P (4 x, y); (b) reflection across y P(x, y) A P0(x, 6 y); (c) reflection across x 5, P(x, y) A P-( 10 x, y) 7. (a) P(x, y) A P (x, 10 y); (b) P(x, y) A P ( 4 x, y); (c) P(x, y) A P (x, 2 y); (d) P(x, y) A P (5 x, y) 8. (a), (b), (e), and ( f ) 9. (a) A (2, 1), B (2, 4), C (1, 5), and D (1, 2); (b) A0( 1, 2), B0( 4, D0( 2, 1); (c) A-( 2, 1), B-( 2, 4), C-( 1, 5), and D-( 1, 2) 2), C0( 5, 1), and 3,

where KR V1 =V2 is the reverse voltage-gain ratio of the ampli er. Equations (8.46) and (8.48) suggest that the feedback admittance YF can be replaced with two shuntconnected admittances as shown in Fig. 8-12(b). When this two-port network is used to model an ampli er, the voltage gain KF usually turns out to have a large negative value, so that 1 KF YF % jKF j YF . Hence, a small feedback capacitance appears as a large shunt capacitance (called the Miller capacitance). On the other hand, KR is typically small so that 1 KR YF % YF .

Fig. 8-22

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