Quiz2Solutions

# Quiz2Solutions - Solutions to MAE140 Linear Circuits Quiz 2...

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Unformatted text preview: Solutions to MAE140 Linear Circuits Quiz # 2 --November 12, 2002 Question 1 1. Obvious labeling of nodal voltages. 2. Node A: G1 (vin - v A ) + G1 (v B - v A ) = 0, iN = 0 v B = 2v A - vin iN = 0, iout = GL vC Node C: iout + G3 (vC - v E ) = 0 Node D: iP = 0, G2v D + G2 (v D - v E ) = 0. 3. OpAmp voltage relationships: v A = v D v B = vC From Node C, iout = G3 (v E - vC ) = G3 (v E - v B ) Now substitute from Node D: 2v D = v E and v A = v D to yield iout = G3 (2v A - v B ) Next use the Node A relation: v B = 2v A - vin to yield iout = G3vin . 1 This is theiout = gvin relationship with g = G3 = . R3 Question 2 (a) For non-inverting amplifier, the input current iP = for all voltage. Thus 0 Rin = W . For the voltage follower, the same argument holds, so Rin = W . For the inverting amplifier/summer, the OpAmp voltage relationships force v v v N = 0 . So i1 = 1 and i2 = 2 . Thus Rin1 = R1 and Rin 2 = R2 . R1 R2 R4 v 2 . So we have the For the differential amplifier, we have v N = v P = R3 + R4 following currents entering the circuit ^ 1 1 R4 i2 = v 2 and i1 = v1 v 2 ~. R3 + R4 R1 R3 + R4 Thus, Rin 2 = R3 + R4 and Rin1 depends on the other voltage input. (b) Note that we can use the differential amplifier to achieve infinite input impedance to the non-inverted input by choosing R4 = . There are many possible answers to this design. My version is as follows. v1 + R1 v2 + R2 R5 R4 + - - R3 v3 vO R6 R7 Here the resultant function is R R R + R1 R7 R4 + R5 vO = - 5 - 2 v1 + 2 v 2 + v 3 R4 R1 R1 R6 + R7 R4 To match the input impedance requirements we need to ensure that R6 + R7 = 1000W . The other impedance requirements are automatically satisfied by the circuit design. (c) I think that three is as low as one can go. It is not a nice design though. It would be better to take more OpAmps and remain simpler. Question 3 (a) dvC (t) dvC (t) and 0 from the dt dt description of steady state. So iC (t) 0. Similarly, using di (t) di (t) v L (t) = L L and L 0, v L (t) 0. Thus at steady state, dt dt capacitors look like open circuits (i=0) and inductors look like short circuits (v=0). (ii) For the circuit shown, at steady state the inductor L2 shorts out the resistor R2 capacitor C2. Similarly, inductor L1 is replaced by a and short circuit and capacitor C1 is open circuit in parallel with resistor R1. Consequently, the steady state current flowing around the loop is 5V across R1=1KW or 5mA clockwise. So the steady-state circuit variables are: v R1 = 5V, iR1 = 5mA, vC1 = 5V, iC1 = 0mA v R2 = 0V, iR2 = 0A, vC 2 = 0V, iC 2 = 0A v L2 = 0V, iL2 = 5mA, v L1 = 0V, iL1 = 5mA R s vi (s) = v and (b) For the RC circuit, using voltage division vO1 (s) = 1 1 i R+ s+ sC RC 1 1 vO2 (s) = sC vi (s) = RC vi (s) . 1 1 R+ s+ sC RC R s v (s) and vO2 (s) = L vi (s). For the RL circuit we have similarly vO1 (s) = R i R s+ s+ L L 1 R By choosing = , we can realize exactly the same s-domain voltage relationships RC L using either approach. It is easy to produce small capacitances (1pF to 100mF) and to produce small inductances (1mH to 100mH). The choice of circuit depends on the ease of producing the required time constant. (i) For a capacitor we have ic (t) = C ...
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