Question

VO D IO 2. In the above circuit, the initial conditions are A. Determine the transfer function, H(s) VI(S) k, b and c are constants. B. Determine the type of filter impleme C. Determine the values for R.C, R, and R. that Cons are all zero and the op-amp is ideal. (1.6 pts) in the form of H(s) = k zubte, where W, = 25000 rad/s, a bandwidth of R - 50000 requirement that ALL resistors must have a res emented by the circuit. and R, that will yield a center frequency of 01% = 50000 rad/s, and \H (jw.)] = 35, under the must have a resistance greater than or equal to 1ks.

Answer 1

cc w16) V (S) on4 -(2) V+ = Vils) - V- ..Abbly voltage division rule "Va(s) = Ra v. 15) - C Rotro On output side __Vo CG) = Ysc Vals) = 1 _Vulsel - R+ Ysc 1 +SRC A bly voltuge divison in input side Vols) = (1+SRC)V. (S) --- VAS) = R V;(s)_ O - R+Yac put o in Ra Vuls) = SCR Vi (9) Rat Ro 1+SRC y (9) Ba_(I+SRC) = SRCV: (5) Ratho ITSRC - VIS) = SRC Ratral V:15) 1+SRC) ? | Ral __e fra +RBL SRG Re 2(RC)? +1+25RC LIS) = V015) =_lratrol 5 __ _vils) Lara JRC [s2+25+17 Lambe _ LRC (RC? _H(S) = K b § __52+bste k=Patro b=2 c l 2Ra RĆ (RC)2

10 H19) = Rat Rb). S Pia RC [52-+ 2 y +60137 Low THtw) Pot sto HIS) = o Low Put S=00 H S) = Rat RW.191 2 Ra IRC) 82(1+2 + 1 Put sro RCS (RCS) HIS) = 0 LAU&= 0 (Wed)_H(S) = __ Low At SE 2 (W=0) H.(S) = 0 Hogh L Band pass filter ---High Tlow © -S=0- Centre frequency Ub = 26000 (S=0 Band width B = 50000 TH(jwo) = 36 Résistance > 1 kr |qemeral funchenoy Bambass_filter HS) = K b'S - 52+BS+ W? H15)= Thatrol 25 Azra IRC5+35+)27 B = 50.000 wo² = 16 2 = 50000 W, BI H(jw) = Rat Ro 2 ja 26000 = 1 (2Ra , RC (jw2 + 1 -w2) FRC (RC) RC RC = 35

- Juljuste pour le che non Wo =35 Hljwoll= (Patrol 2 | 2 Ral Re Oo Pot Wo = 25000 = 25000 150 0 0 = sooot : 35 = Rat Rb = 70 Ratko (50000) (25000) 2 Ra (25000) (50.000) RatRb = 35 2 Ra Rat RG=70ra = Ra - 1 Rb 69 Ra 69 Ra= Rb

Raai 69 ਉਧ ਨ - 69 ooo Ras1ooon Ra=1k2 69 000 1 R₂ = 69 KSZ

+= 25000 Ro Все т е Тоо - Чx 105 25000 2500000 RC= Onu x 10-6 RC = (0.4 X 10-4) 1000_ Tooo Lot = 10.4 X 109) 10oo R = (2000) (0.9 X1009) R=2000 C= 0:2 X10-9 F IR = 2 ks C=0.2nF TC = 0.2 nanofarad 0
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