solutions_test3_section1_f04

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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(2 100 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(2 100 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 + (2 100) 2 (2) h s 3 + 2 s 2 + 3 s + 4 i V ( s ) = s s 2 + 40000 2 V ( s ) = s ( s 3 + 2 s 2 + 3 s + 4)( s 2 + 40000 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(2 100 t ) u ( t )] = s s 2 + (2 100) 2 1 So, finally, V ( s ) = s ( s 3 +2 s 2 +3 s +4)( s 2 +40000 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) =- . 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 ( /...
View Full Document

This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.

Page1 / 12

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online