Question

Experimental methodology

Problem 1 (25 Pts) Consider the OP amp circuit shown below with R = 100kN and C = 1(10)-SP: Part a) 10 pts Find the complex transfer functions for the circuit in terms of frequency f (Hz). Part b) 5 pts Compute the gain of sat as a function of frequency f (Hz). Part e) 10 pts Compute the corresponding gains at 100, 1000, 10000 Hz. R 3R 5R protein Vi - Vo

Answer 1

Solution R 3R C SR Mum Vi vo + loso R= 100 kr cz is This amplifier. inverting a op - amp a feedback impedance with Grain of a inverting Op - amp lp impedance zi . is If and zf N6 --> Zi Vi a Complex Transfer function of system have obtain the Transfer function To Convert the to in No S- domain for frequency domain but Sa ja jaaf of sa

R ulu 3R mu Isc Vils) SR Vols) impedance of feed back path (af) If (RII Śc ) + 3R RX 5c + 3R Rt se 3R b + R RSCHI 4+ 3RSC R j n It RSC input impedance (2) V Zi SR Vo () - output voltage elp voltage Virs)

for inverting op- amp amplifier Vo(s) 2f Ii Vils) Vols) R 이무 4 + 3 RSC it RSC Vi (5) SR 3 RSC H(S) Vols) Vics) $(". It RSC This is the Trawsfer function in s-domain for frequency put sz jw = janf -domain Volf) T.F. = 4 + 3 RC (f) 1 + j Rc (anf) Vilf) put Ra 100 kr 12 105 N los f Volf) Tf. !) دیا ہے ut j 3 xlox105(227) It j 105X16S ( 204 ) I S Vilf) Volf) Vilf) Tof. L 5 4+ jant 1 + j2NF G This is the transfer function of system in term of f.

b) Grain in terms of frequency (f) vi Gain of a olp voltage system is magnitude of ratio of elp voltage. to in eg. ☺ we have already voltage ratio Vo in terms of f. vi Vo Cain (Av) $15 1 5 44.jgnt It ja af vi Av. = 1 L دیا۔ 542 +GAF? JR + (20f) 2 vo L Av 1) Vi th 16 + 3662 JH 402f² 2 This of the system as a function is the gain frequency (f) of Gain at f 100, 1000, 10000 H2 at fa 100 H2 AV put f 2100 in 2 eq.

Av, 12 at 16+ 3612 (100) It u A2 (100) 2 1884.9598 ut 0.600 628.3193 Avi 0.6 Gain în desibal 0.6 Av, 20 logi dB ANI 4.437 dB → at f 1000 H2 frequency is very is very high f - 1000 H f² (1000? 164 36A2f² Ava ما) - :) - + YaLfL Li 36-4f2 >>> 16 & un2f² os. J 3672.f2 Av, a 히 :) 4R2f2

Ave a at (3) | Av. 0.6 also 10000 Hz similarly foo fa Grain Ava 0.6 AN, I AV2 = AV3 Ansi 4.487 dB - AV - Av₂ = AV so frequency is very high ather tem în as a compare tramfer function will be negligible to frequency that's why gain Constant at gain is almost 0.6 of a system To obtain transfer function laplase transform. م لرم use ** Gain of is always UP the ratio of magnitude voltages. olp &