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solutions_test2_s07_s001

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

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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 ( ω )) and the output voltage phasor (i.e., ˜ V o ( ω )) in a circuit: ˜ V o ( ω ) = e 1 - 2 . 5 ω + j 2 ω ˜ V in ( ω ) In the above expression, ω 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 ) = - 2 sin(2 t - π/ 4) v o ( t ) = 0 . 25 cos(2 t + 2 + π/ 2) Solution: We need to calculate the phasor of the input signal first. Using the identity - sin( a ) = cos( a + π/ 2) we can rewrite the input signal as: v in ( t ) = 2 cos(2 t - π/ 4 + π/ 2) = 2 cos(2 t + π/ 4) The input voltage phasor, ˜ V in ( ω ), is then given by ˜ V in ( ω ) = 2 e jπ/ 4 , where, of course, ω = 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

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= e 2 j 1 - 5 + j 4 ˜ V in (2) = e 2 j - 4 + j 4 2 e jπ/ 4 = 2 e j (2+ π/ 4) - 4 + 4 j = 2 q ( - 4) 2 + 4 2 e j (2+ π/ 4) e j arctan (4 / - 4) = 2 4 2 e j (2+ π/ 4) e - jπ/ 4 = 0 . 25 e j (2+ π/ 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 + π/ 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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solutions_test2_s07_s001 - ECE 220 Section 001 Test 2...

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