IRWIN 9e 12_65

IRWIN 9e 12_65 - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 91E 1 12.65 The auternatie gain centre] eircuit in Fig. [2.65 is used to hntit the transcenttuetance. rafts". [a] Find an expression for “at. irt terms at arm. R5. anti EL. [b] Express the asymptetic transeenductanee. iofa-m. in terms 01' R5 anti RL at em = I] and as an approach- es infinity. Given R1 anti R5 values in the circuit diagram. what are the vahies cut the asymptotic transcenduetanee'? [c] 1|What are the ee-nsequences cut your results in {h}? [d] 11' am must he an more than if“ for preper opera- tion. what is the rnininturn transeenductanee tear the t'unctiena] eircuit'.J 15cc 4 st rnicn Rt: —D4"." Figure P12J55 SOLUTION: Chapter 12: Variable — Frequency Network Performance Problem 12. 65 Irwin, Basic Engineering Circuit Analysis, 9/E \10: Vin 97"). RI. : 201mg;l Vim RL Va: 2015, VD, KL Lo, : 9m, (Li—VD) iOi : QOIAGL, (H'VO) r0, : 20( ii.) (w'vo) R63 v0: 70w” RL (ELM-V0) KC» V0: @400 Vin flg- ISDDVLnVofll; no, '20: VD: Vinyvo [000) locoo [0000 \Io: SHovLm— isoanVO VO< H'leoul‘n) :qu VL‘M v0: Shim“,1 I HISQVQ » Problem 12.65 Chapter 12: Variable — Frequency Network Performance Irwin, Basic Engineering Circuit Analysis. 9/E . 3 (b) 61m: 10 VLV» 67m?» $2.0m ( ML 3 Vin BL Gim: i—Héovm (awn—)0 ,Gimwao'wsfim—fi éqoo Km 0» *44, vaieup 6? Vin thheQJM,'tAQ VOJAM— at“ Gm dfiCMaJZA' . Jo in IZin *0 LL. R... COL) QM —_) ii'iso(u) G’m "5 WWI/LS 7-: minimum 62m~ Chapter 12: Variable — Frequency Network Performance Problem 12. 65 ...
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IRWIN 9e 12_65 - Irwin, Basic Engineering Circuit Analysis,...

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