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solutions_test3_section1_f04

# solutions_test3_section1_f04 - ECE 220 section 1 Test 3...

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ECE 220, section 1 Test 3 Solutions 18 November 2004 Problem 1. (10 points) Consider the differential equation 1: d 3 v ( t ) dt 3 + 2 d 2 v ( t ) dt 2 + 3 dv ( t ) dt + 4 v ( t ) = cos(2 π 100 t ) u ( t ) . (1) Let v (0) = 0 , ˙ v (0) = 0 , ¨ v (0) = 0. Determine the total solution in the Laplace domain; in other words, find V ( s ). No need to determine v ( t ). Solution Take Laplace Transforms (LT) of both sides of Equation 1. We have L " d 3 v ( t ) dt 3 + 2 d 2 v ( t ) dt 2 + 3 dv ( t ) dt + 4 v ( t ) # = L [cos(2 π 100 t ) u ( t )] h s 3 V ( s ) i + 2 h s 2 V ( s ) i + 3 [ sV ( s )] + 4 V ( s ) = s s 2 + (2 π 100) 2 (2) h s 3 + 2 s 2 + 3 s + 4 i V ( s ) = s s 2 + 40000 π 2 V ( s ) = s ( s 3 + 2 s 2 + 3 s + 4)( s 2 + 40000 π 2 ) In deriving Equation 2 we used the following facts: 1. The linearity property of the LT 2. The differentiation property of the LT: L " d 3 v ( t ) dt 3 # = s 3 V ( s ) - s 2 v (0) - s ˙ v (0) - ¨ v (0) L " d 2 v ( t ) dt 2 # = s 2 V ( s ) - sv (0) - ˙ v (0) L " dv ( t ) dt # = sV ( s ) - v (0) 3. The LT of the cos function is (see the tables in the test fact sheet) L [cos(2 π 100 t ) u ( t )] = s s 2 + (2 π 100) 2 1

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So, finally, V ( s ) = s ( s 3 +2 s 2 +3 s +4)( s 2 +40000 π 2 ) 2
Problem 2. (15 points) Consider the Laplace transform in equation 3: V ( s ) = s ( s + 1)( s 2 + 1) (3) Determine the corresponding signal v ( t ). Solution The roots of ( s 2 + 1) are ± j . The root of s + 1 is - 1. Use Partial Fraction Expansion to rewrite Equation 3 as V ( s ) = K 1 s + 1 + K 2 s - j + K * 2 s + j (4) In Equation 4 we used the fact that the coefficients K 2 and K * 2 are complex conjugates of each other. Now K 1 = ( s + 1) V ( s ) | s = - 1 = ( s + 1) s ( s + 1)( s 2 + 1) fl fl fl fl fl s = - 1 = s ( s 2 + 1) fl fl fl fl fl s = - 1 = - 1 (( - 1) 2 + 1) = - 0 . 5 (5) K 2 = ( s - j ) V ( s ) | s = j = ( s - j ) s ( s + 1)( s 2 + 1) fl fl fl fl fl s = j = s ( s + 1)( s + j ) fl fl fl fl fl s = j = j ( j + 1)( j + j ) = j - 2 + 2 j = 1 4 + 4 e j ( π/ 2 - 3 π/ 4) = 1 2 2 e - jπ/ 4 (6) Substitute Equations 5 and 6 into Equation 4. We have V ( s ) = - 0 . 5 1 s + 1 + 1 2 2 e - jπ/ 4 1 s - j + 1 2 2 e jπ/ 4 1 s + j and, in the time domain (see the table in the fact sheet): v ( t ) = - 0 . 5 e - t u ( t ) + 1 2 2 e - jπ/ 4 e jt u ( t ) + 1 2 2 e jπ/ 4 e - jt u ( t ) = - 0 . 5 e - t u ( t ) + 1

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solutions_test3_section1_f04 - ECE 220 section 1 Test 3...

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