Question

VA & UB It + R с + + V) R UL 361 Fig. 1 A Passive Bandpass RC/CR Filter

Part 8 in computing |H|Max is what I would like to ask answered

Answer 1

R M Vhe. Vsc. B A a 1, U2 c Ysc Iy Ji TT Apply KCL at node A, I, = I₂ + I3 V,Cs) - Van R Va Ysc + Va - V₂ Yse. 11.08) R Va + Vx SC + Un SC - VASC. 1,05) R Va [ť + 2567 - vse Эсу) Apply kle at node B, I, I4 U2 - V-V2 YSC Va SC Vg Vas vo(483 +(2) put (2) in (1) 11,05) V, (S) Va Litresc] – 4,9256 ve s ' [1+ Rese] – vcs esc (1+rSc)(1+225c) - Rse VI . V₂ RSC CS Scanned with CGMScanner

Vi= 1₂ It RSC + ars(+an's c² 2 2²s²c? _ R² s² (² RSC sls RS C R?s?c? + 3.125+). Rillide by 2²2, Me s cvec) 403) ui $2 +5 (318c) + öze? H(w). jW/RC -w² + jw + RZCZ RC co H(jw) = RO (ee-w) + 3jw 12C resonat frequency, 1 w? - R²(2 R = 15k2 c=lonf RO. - fo añRO - f 10610.33 H₃ 3 -9 21 x 15X10 X 10X10 10.61 kih CS Scanned with CamScanner

will resonant frequeny on Manimun amplitude ouun. at 10.61 kHz Ito { given value is fo=1061 H3., This Value for either R = 150k, OY co 100 OF Momimun Magnitude is at to, litewol دينا 3 RO "Chan-w) + 3 in Rc Wallace W/RC 3W/Rc. - I mon. 3 is Manimu 13 Amplitude in dB, IH) = 20 log (13) for two wa the 1H1 -9.5424 dB for given to 1061, jf 140 H 01 0.0966 j 1061 10610 110611 11061 +13 110610) 10610) 2 | 4+1 3 ( 10.99+0:53 CS Scanned with CamScanner