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# final-sol - MAE140 - Linear Circuits - Fall 09 Final,...

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Unformatted text preview: MAE140 - Linear Circuits - Fall 09 Final, December 7 Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a hand calculator with no communication capabilities (ii) You have 170 minutes (iii) Do not forget to write your name, student number, and instructor (iv) On the questions for which we have given the answers, please provide detailed derivations. L L R L R L/2 A B (a) Question 1, Part I R C L D C i(0) L v(0) C (b) Question 1, Part II Figure 1: Circuit for Question 1. 1. Equivalent Circuits Part I: [5 points] Assuming zero initial conditions, find the impedance equivalent to the circuit in Figure 1(a) as seen from terminals A and B. The answer should be given as a ratio of two poly- nomials. Part II: [5 points] Assuming that the initial conditions of the inductor and capacitor are as indicated in the diagram, redraw the circuit shown in Figure 1(b) in the s-domain. Then use source trans- formations to find the s-domain Norton equivalent of this circuit as seen from terminals C and D. ( Hint: Use an equivalent model for the inductor in which the initial condition appears as a cur- rent source, and an equivalent model for the capacitor in which the initial condition appears as a voltage source) Solution: Part I: Since all initial conditions are zero, it is easy to trans- form the circuit to the s-domain. [1 point] sL sL R sL R sL/2 A B Now, we combine the two impedances (inductors) in parallel to arrive to the circuit on the right. [1 point] R sL R sL/2 A B sL/2 And now the two impedances in series to arrive to the circuit on the right. [1 point] R sL R sL A B Page 2 Now again the two impedances in parallel to arrive to the circuit on the right. [1 point] sL R A B RLs R+sL Finally, we arrive at the equivalent impedance by merging together all the impedances in series [1 point] A B R +s L+ 3RLs R+sL 2 2 2 Part II: We begin by transforming the circuit to the s-domain taking good care of the initial con- ditions using the hint. [1 point for capacitor transformation; 1 point for inductor transformation] R sL D C sC 1 v(0) C s i(0) L s Next, we combine the resistor and the ca- pacitor in series [1 point] sL D C v(0) C s i(0) L s sC 1 R+ Page 3 Now, we do a source transformation with the voltage source and the impedance in se- ries [1 point] sL D C Cv(0) C sRC + 1 i(0) L s sC 1 R+ Combining everything together, we get the Norton equivalent [1 point] D C sC 1 R+sL+ Cv(0) C sRC + 1 i(0) L s + Page 4 i(t) L 1 2 3 A B C i(0) L R 2 v x v x 2 R 2 R Figure 2: Nodal and Mesh Analysis Circuit 2. Nodal and Mesh Analysis Part I: [5 points] Formulate node-voltage equations in the s-domain for the circuit in Figure 2. Use[5 points] Formulate node-voltage equations in the s-domain for the circuit in Figure 2....
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## This note was uploaded on 01/05/2011 for the course MAE MAE 131A taught by Professor Vladolubarda during the Spring '10 term at UCSD.

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final-sol - MAE140 - Linear Circuits - Fall 09 Final,...

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