# 1 now looking at the upper and upper left sides of

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(1) Now looking at the upper, and upper left, sides of the above diagram we note that the current pass- ing through , the internal op-amp resistance, is the same current that passes through the node marked and through and . Moreover, this current, , is (nearly)zero according to Eqn. (1). Note that this argument is consistent with the fact that and further means that the voltage is the same as . We therefore simplify the above diagram as follows: Next we note that voltage divider relationship holds in relation to and : . (2) Problem 2: The data set for the problem, given below show a gradual drop in amplitude of the output voltage The graph of the 20log(Vout) versus log(omega) is shown below. As it is seen we have a corner fre- quency of approximately rad/sec. This corresponds to . This is also evi- dent from the data where the gain drops to 0.71 at 1000 Hz. This means that , (3) or approximately 16 kOhm. v + v - i 0 i R i v i R 1 R 2 i v + v - V A v i v o R 3 R f v i i f v i v o v o R 3 R f + R 3 ----------------- v i = 25.00 20.00 15.00 10.00 5.00 0.00 1.80 2.10 2.50 2.80 3.10 3.50 3.80 4.10 4.50 4.80 5.10 20*log(Gain) 20*log(Gain) Corner Frequency ω c log 3.8 = ω 10 3.8 6280 = = 1000Hz R 1 6280 C --------------- 1 6280 0.01 10 6 ( ) ( ) ------------------------------------------- 15923 = = =
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