HW9_Soln - Escuti/Brickley Fall 2008 ECE200 Fall 2008...

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Escuti/Brickley Fall 2008 ECE200 – Fall 2008 – Homework 9 Operational Amplifiers, Filters Page 1 of 10 PROBLEM 1 The circuit shown below is used to divide an input signal between the left and right channels of a stereo system. a) Derive equations for the signals applied to the left and right channel audio speakers as a function of x. The two resistors, ( ) Ω k x 10 and ( )( ) Ω k x 10 1 represent a potentiometer with 0 < x < 1. (10 pts) b) If () ( ) ( ) t mV t v in 1000 2 cos 100 π = is applied as the test signal and x = 0.2, find the power delivered to the left and right channels. (5 pts) Solution: We note that the voltage at Node A is identical to the input voltage since the first stage is a unity gain amplifier (voltage follower). It serves as a buffer between the input source and the rest of the circuit. Inspecting the circuit carefully, we note that the signal at node A is passed through a voltage divider (on each channel) before it is applied to the non-inverting input of either op-amp. Therefore, by voltage division, A B v K x K K x v × Ω + Ω Ω = ) 10 ( 1 ) 10 ( Since ) ( t v v in A = , () ) ( 10 1 10 ) ( 10 1 10 t v x x t v x K K x K v in in B + = + =
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Escuti/Brickley Fall 2008 ECE200 – Fall 2008 – Homework 9 Operational Amplifiers, Filters Page 2 of 10 Similarly, () ( ) ( ) A C v K x K K x v × Ω + Ω Ω = 10 1 1 10 1 ) ( 1 10 1 1 10 ) ( 1 10 1 1 10 t v x x t v x K K x K v in in C + = + = The two op-amps form two independent non-inverting amplifier stages with the same voltage gain, 11 1 10 1 1 = + = + = K K R R A n f v This yields the following output voltages: ) ( 11 10 1 10 t v x x v in LEFT × × + = ) ( 11 ) 1 ( 10 1 ) 1 ( 10 t v x x v in RIGHT × × + = b) ( ) ( ) t V t mV
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This note was uploaded on 03/10/2009 for the course ECE 200 taught by Professor Escuti during the Fall '07 term at N.C. State.

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HW9_Soln - Escuti/Brickley Fall 2008 ECE200 Fall 2008...

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