solutions_test2_s07_s001

solutions_test2_s07_s001 - ECE 220, Section 001, Test 2...

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Unformatted text preview: ECE 220, Section 001, Test 2 Solutions 22 March 2007 Problem 1. (20 points) The following equation describes the relationship between the input voltage phasor (i.e., V in ( )) and the output voltage phasor (i.e., V o ( )) in a circuit: V o ( ) = e j 1- 2 . 5 + j 2 V in ( ) In the above expression, is the frequency of the input signal v in ( t ). Determine the output voltage v o ( t ), as a function of time, when the input voltage is the signal v in ( t ) =- 2sin(2 t- / 4) v o ( t ) = 0 . 25 cos(2 t + 2 + / 2) Solution: We need to calculate the phasor of the input signal first. Using the identity- sin( a ) = cos( a + / 2) we can rewrite the input signal as: v in ( t ) = 2cos(2 t- / 4 + / 2) = 2cos(2 t + / 4) The input voltage phasor, V in ( ), is then given by V in ( ) = 2 e j/ 4 , where, of course, = 2 . The output voltage phasor, V o (2), is then given by V o (2) = e j 2 1- 2 . 5 2 + j 2 2 V in (2) 1 = e 2 j 1- 5 + j 4 V in (2) = e 2 j- 4 + j 4 2 e j/ 4 = 2 e j (2+ / 4)- 4 + 4 j = 2 q (- 4) 2 + 4 2 e j (2+ / 4) e j arctan(4 /- 4) = 2 4 2 e j (2+ / 4) e- j/ 4 = 0 . 25 e j (2+ / 2) (1) Therefore, in the time domain, the output voltage v o ( t ) is given by v o ( t ) = 0 . 25 cos(2 t + 2 + / 2) 2 Problem 2. (15 points) Consider the differential equation dv ( t ) dt + 10 v ( t ) = e- 2 t u ( t ) (2) Let v (0) = 1. Find the particular solution v p ( t ), for t 0. Show all steps involved....
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This note was uploaded on 08/24/2009 for the course ECE 220 taught by Professor Nilson during the Summer '08 term at N.C. State.

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solutions_test2_s07_s001 - ECE 220, Section 001, Test 2...

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