Question

(25 credits) For circuit shown in the figure below do the following: Rs Ro Rs с, R, Rj Ri 100 k Rs= R,-10 k C3-100 nF R2 = 10k R,-10M C-10 nF Simplify your expressions as much as possible. First solve using variables and substitute values at the end (a) (2 credits) What type of configuration Operational Amplifier 1 shows? You answer:

Answer 1

w Just draw the part of the circuit highlighting op-cump 1. - W WW vollage at the and a ve terminale is same. So node voltage of node * is zero. Apply KCL at node o bed-back vollage Vy is negative (66) antiphase with respect to input vor so it is in inverting conbiguation. (b) Dow the part of córcut showing op-amp 2

Here voltage at ave to allage at eve terminal is given as Vz nely KCL at node of Vz Ra ka ve = Vz - Yout Herre Vaut í tve (or) in phase with V I so it is in Non-investing configuration. © Impedance of Capacitor cis z = Joc=56 | Now drawing the circuit using impedances. Vx RM Rs EMMA wwny trovout

In som epution 0 of solution of ca) cubatibak given Ps, F10M.0. , Puy-IMO :) Daew the part of cacut to show op-amp 2 Because of visitial ground concept voltage at the terminal becomes Vzo Apply KCL at this que terminalo ng Va de

cabilitate this vz En er @ from solution (6) substitute giver seekstor values Kaut are ] [ + te dy = (%) (2) vy (4) Dolow the part of the circút showing RC stages Zc2 w op-empt) let va, Vb be the node voltages

>Apply KeL at a RR, Vaz te, v[ hit has + 7,7 Eva [zai ( hitta) + 1] zva [ , (* * *s) + ] substitute given values av ( )] Valle 61 10 -Apply KCL of node of va Va-Vy 21 zci V = 7, x[*E, *] =x[zolt teta) .

Tr(st?) » = А * - Ст? | = fЯ е «ро ғ 12, 44, 4 Тус. | а) t. 1111° 1 1 (10) "д - 1 р (сэн 4 ) | pun wungl 19 7poste

shokitaka given value substitada ego , @ 1 11 ovo 15 ti 2.431X109 + 2,2 x 1,8 +2.431 1.2.1 x 10 2- 52 - w [13*+1] (04) w = ve[ures et mange sangue e 1 10" 414 – mon -0.17 ve Surgery used to samo-> 3.32 / X104 + 3.42X108 f 2 Vaidot s’ +(3.321X1952 +3.42X10)s +0.1X102|| 53