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sol_exam2 - CAAM 335 MATRIX ANALYSIS Examination 2 Posted...

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CAAM 335 · MATRIX ANALYSIS Examination 2 Posted Friday, 21 March 2008. Due no later than 5pm on Tuesday, 25 March 2008. Instructions: 1. Time limit: 3 uninterrupted hours . 2. There are five problems worth a total of 105 points. (Any points above 100 will be regarded as bonus credit.) Please do not look at the questions until you begin the exam. 3. You may not use any outside resources, such as books, notes, problem sets, friends, calculators, or MATLAB. 4. Please answer the questions thoroughly and justify all your answers. Show all your work to maximize partial credit. 5. Print your name on the line below: 6. Time started: Time completed: 7. Indicate that this is your own individual effort in compliance with the instructions above and the honor system by writing out in full and signing the traditional pledge on the lines below. 8. Staple this page to the front of your exam.
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CAAM 335 · MATRIX ANALYSIS 1. [40 points: (a)=8 points; (b)=4 points; (c)=10 points; (d)=8 points; (e)=10 points] Consider the circuit below. Each capacitance is 1 Farad, the horizontal resistor is 1 Ohm and each vertical resistor is 1/3 Ohm. y 1 ? y 2 ? y 4 ? y 5 ? y 3 - x 1 x 2 (a) Carefully express Kirchhoff’s Current Law at each node, and write the resulting two equations as a system x 0 ( t ) = Bx ( t ) , where B = ± - 4 1 1 - 4 ² . (1) Hints: Recall that capacitors give y j ( t ) = Ce 0 j ( t ), where e j denotes the potential drop across the capacitor. You can solve this problem directly from Kirchhoff’s Current Law without the formality of the Strang Quartet. If you cannot complete the problem, use the B matrix in (1) andproceed with the rest of the question. (b) Construct the resolvent, ( sI - B ) - 1 , by making use of the fact that ± a b b a ² - 1 = 1 a 2 - b 2 ± a - b - b a ² . (c) Identify the two poles of the resolvent, λ 1 and λ 2 , and expand ( sI - B ) - 1 in partial fractions, i.e., ( sI - B ) - 1 = 1 s - λ 1 P 1 + 1 s - λ 2 P 2 . Give clear numerical values to λ 1 and λ 2 and the two 2-by-2 matrices P 1 and P 2 . (d) Show that P 1 and P 2 are both projections. Compute and draw bases for the null and column spaces of both P 1 and P 2 . (e) Now suppose that x (0) = [ v 1 v 2 ] T Volts and evaluate x ( t ) = L - 1 { ( sI - B ) - 1 x (0) } = 1 2 πi Z C exp( st )( sI - B ) - 1 x (0) ds, where C is a closed curve with both λ 1 and λ 2 in its interior. Your answer should be x ( t ) = exp( - 3 t ) ( v 1 + v 2 ) 2 ± 1 1 ² + exp( - 5 t ) ( v 1 - v 2 ) 2 ± 1 - 1 ² . (2) 1
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CAAM 335 · MATRIX ANALYSIS Solution: (a) Let R h and R v denote the horizontal resistor and vertical resistor, respectively. One can follow the Strang Quartet to solve this problem, but let us approach this using Kirchhoff’s Current Law directly. At node 1,
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This note was uploaded on 04/07/2008 for the course ELEC 335 taught by Professor Cavallaro during the Spring '08 term at Rice.

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sol_exam2 - CAAM 335 MATRIX ANALYSIS Examination 2 Posted...

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