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Solution to Exercise 55: Since this is an ideal op-amp, we know that: Rf Vout =− Vin Rs (1) This tells us that the voltage at output of the op-amp is: −Vs Rf Rs = −4mVp−p (2) 12K 1.2K = −40mVp−p (3) (4) Now we have to deal with the potentiometer. A potentiometer is essentially a voltage divider: R1 Vin R2 Vout We are told that the wiper of the potentiometer is one quarter of the way up from the ground connection; in other words: R1 = 3R2 (5) So the voltage divider (alone) gives us Vout = Vin = Vin R2 R1 + R2 R2 3R2 + R2 = 1 Vin 4 (6) (7) (8) Now we can combine equation (4) with equation (8) to get our answer, the level of the output voltage: 1 Vout = 1 (−40mVp−p ) 4 = −10mVp−p (9) (10) An ideal op-amp doesn't change the frequency or the shape, so the output wave is still a sine wave at 440 Hz, but due to the negative sign it is inverted i.e. when the input sine wave rises, the output sine wave falls. 2