EECE 290 Homework, Spring 2008

18.29 The linear transformer in the circuit shown in Fig. P18.29 has a coefficient of coupling of 0.75. The transformer is driven by a sinusoidal voltage source whose internal voltage is vg = 260 cos 4000t V. The internal impedance of the source is 25 + j0 Q. a) Find the frequency-domain a parameters of the linear transformer. b) Use the a parameters to derive the Thévenin equivalent circuit with respect to the termi- nals of the load. c). Derive the steady-state time-domain ex- pression for v2. Figure P18.29 a 4 we ANN @+ S' | @ -~------4 _ LG an ioe | + 2 | a ° | AM | C(x 1 oO | e © ray e+ xs | @ C va} N P 18.29 [a] For 5 = 0: ty _ jisan wnnn 22 — > 502 pay 400 <— 4? AMY ' WW ® ats * 35022 ¢ 48000 - f ; ® Yi _ -2Y 4 = —j150l, = —j150 50+ 750 1+fl Y 1+jl -l+jl aya its _ TV lmao = 33 3 i 1 j “1 3 In-o 150 150 For Vo = 0: qT, 41502 I, — SY s0n ew 4002 SC — eo Wy fA + | | Vi ; c 45022 ¢ je8000 ~ © ® 0 = —7150/; + (400 + 7800) I> 50+ 750 —j150 —j150 400 + j800 = 2500(1 + j24) 50+ 750M) Ny = = j150Vy —j150 0 1 ke. _7150V; >A 2500(1 + 724) -V —50 = = ——(24—j1)Q a2 lb \vsco 3 ( jl) j150L; = (400 + 7800), I; 8 . p= =—-(2 1 a22 b lyeco 3 | ji) Ve 260/0° ( aitanZ, (—1+j1)/3+ 925/150 —2+4j2+j1 —24 73 = 120(—2 — 73) = 432.67/— 123.69° V Zo = U2 t 422% _ [-(50/3)(24 — 71)] + [(-8/3)(2 — 71) (25)] Th [b] Vn = Qy1 + An Zy [(-1 + 1)/3] + [(7/150)(25)] _ —100(24 — j1) — 16(2 — j1)(25) | —3200 + j500 —24 924+ 71 — =24 53 = 607.69 + 661.542 Wy YN -& 607.699 4661.54 . + 432 .614123.69 v C) y 2 10009 t 1000 Vo = 2.67/— 123.69°) = 248.88/— 146.06° [el Ve = se57 a9 4 jOOL Sd (132-67/— 123.69") = 248.88/ — 146.06 U2(t) = 248.88 cos(4000¢ — 146.06°) V 18.31. The b parameters for the two-port circuit in Fig. | P18.31 are 1 . br =1+ jz; big = —-14 j4Q; 1 . bu = 3S; bag =14 jl. The load impedance Z, is adjusted for maximum average power transfer to Z,. The ideal volta source is generating a sinusoidal voltage of Vg = 90cos 80001 V. a) Find the rms value of V3. b) Find the average power delivered to Z,. c) What percentage of the average power developed by the ideal voltage source is delivered by Z,? Figure P18.31 60, | wn ~e e—t—_——_ +! [+ + by bi | v7) Vil {Ms Zr = by bx | — i - ° oT oe J biiZy + die bo1Zq + bo» bu Z,y =6 + 72; boi Zag =2 P 18.31 [a] Zth = 6+j2-1+j4 54+ 96 Zr, = = —=2.14+ 71.39 me O14 il 3471 rd Z, = Zi, =2.1-j1.32 Vo AbzZy, Vy - bie + biiZy + bao Zr, + be Zy LZ, 34+ jl Ab= a + jl) —(-1+ j4)(1/3) =1 by Zy = 6+ j2 boZy, = (1+ jl)(2.1 — 71.3) = 3.4 + j0.8 ‘ bx ZZ, = 4.2—- j2.6 Vo _ 2.1—j1.3 21-513 V, —14+j4+64+j2+3.44+70.8+4.2—92.6 12.6 + 394.2 Vy = 242 I18 (990°) = 10.71 — [12.86 = 16.74/— 50.19° V 2 72.64 74.2 i nes . 16.74 Vo(rms) = 5 = 11.84V —(10.71 — 712.86) - I= =6. STA [b] J 21-713 6.78 /161.57 6.78 In(Qgms) = —= = 4.79A a( ) Ja P = (4.79)?(2.1) = 48.21 W In —Ab [e] = = —~ by + ba Zy -1 _ =3 (34+ 91)/3+ (21—J1.3)/3 5.1— 70.3 I 1—j0. 1 ET 03 4 jo Ts —3 Ty = (-1.7 + j0.1)(6.78/161.57°) = 11.54/— 21.80°A 1 P, (dev) = 5 (11.54)(90) cos(21.80°) = 482.15 W 48.21 GH _ = 482.15 (100) 10% Figure P18,32 ___ >) a | Z |» e § | + | hy 1 hyo | Ve C) | [Vo ZL = | hy; hy | _— | l _— eae CK XX _/ 18.33 For the terminated two-port amplifier circuit in Fig. P18.32, find a) the value of Z, for maximum average power transfer to Z;, b) the maximum average power delivered to ZL c) the average power developed by the ideal voltage source when maximum power is de- livered to Z,. Zq + Ai P 18.33 [a] ZTh = hooZ, + Ah g Ah = (500)(50 x 107°) — 50 x 1073 = —25 x 1073 hogZy = 50 x 10-°(1500) = 75 x 1073 7 2000 75 x 10-3 — 25 x 10-3 Zy, = Zn, = 40+ jOKQ 27h = 40,000 + j02