Question

The transfer function of an amplifier is given by A = A A = {1+3) where Ao and wo are positive constants and Ao is 46dB. What is the absolution value of the gain (in dB) forw = wo? none of these 43 dB 40 dB 39 dB 46 dB

Answer 1

Given, Transfer function. A to 6. Ao in dB = 46dB 20 log to = 46 → con all places to Gain & IAI= Gain = Ao Gain in dB 10.) Grain in de -20 log ( magne) =20log Ao - 2017 (146.)) Gain in dB at w=wo = 2000g Ao - zolog (sterk)). - 464 - 20 log2 => 46–20(0.301) => 466 46 -6.02 Gain in dB at w=wo - 40dB ophon (C). 1 (fromegn 20log A=46)

Given. T VOD Robim Vout od M2 ID Vin o man divided between Rp The small signal dran Current of My smilin is and the impedance seen into source of M2. Thus, current through ma is ID = (2m, Vin) (Rp) ( current divider formula ) In small I= Gm, Vin Rp. 9m2 +9m2 Rp de signal model pozo. Output voltage yout - - I. . Rp. Vout= - Im, Vin Rp. 2m2.Ro.. 1+3m. Rp Vout - gm, Rp. Roe ama + Rp Gwen, Rp >> Transconductance me. So Amat Rp vout - - 9m, RO RD = -2m, RD. RP Ophon (A)
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