Question

3 repeat problem 2 witht he circuit to the right. c part

iii,iv,v. and d

v.0) 2. (20 pts) Consider this circuit configuration. a. [10 pts) Find the voltage gain of the circuit, A, =V./V. b. [04 pts) What is the gain for the following frequencies? i. W = 0 rad/s (A DC signal) +0 ii. 6 (l.e., Frequency approaches infinity) c. [05 pts] Using R = 1 kn and C = 1 uF, find v.t) if: 1. v(t) = 1 V (ADC signal) ii. vC) = 1 cos(1000) v ill. v(t) = 1 cos(1,000) iv. vi(t) = 1 cos(10,000) V V. v.(t) = 1 cos(100,0001) V d. (01 pt) What does the circuit appear to be doing? 3. [10 pts) Repeat Problem 2 using the circult shown to the right. (Each part is half the point value as the corresponding part from Problem 2). v. +0 OX RS v. )

Answer 1

. ANSWER 2 The opamp eirmit is a voltage follower (unity - gain cirenit). So, the output is same as input. However, in the input side, there is a low pass RC ciruit which rejects the high frequency, cSignals do with CsCamScanner

1. Now, ent off frequency : Fe = 1 rad/s. . RO gc@lv. Icos (100+) v lui Icos (tooot) v. @ This circuit acts as a low pass voltage follower. RC circuit:: CANSWER). In this case, it exactly the opposite 30 les (coust) V. (6) Aces ( 100014) • I cos(100000 t) v @ This cirenit acts as a high pass voltage follower Rc vinuit. (answer) X cs Scanned with CamScanner

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