Solution_to_problem_3.6 - 3.6 1 Solution to problem 3.6 a)...

Info iconThis preview shows pages 1–2. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.6 1 Solution to problem 3.6 a) ( ) ( ) ( )( )( ) 4 3 2 1 + + + + = s s s s s s X We apply partial fraction expansion: ( ) ( ) ( )( )( ) 4 3 2 4 3 2 1 3 2 1 + + + + + = + + + + = s s s s s s s s s X We use Heaviside expansion to find values for 1 , 2 and 3 Multiplication by (s+2) and putting s equal to -2 gives: ( ) ( )( ) 2 1 4 3 1- = + + + = s s s s s 1 2 2 1 = = Multiplication by (s+3) and putting s equal to -3 gives: ( ) ( )( ) 3 2 4 2 1- = + + + = s s s s s 6 1 6 2- =- = Multiplication by (s+4) and putting s equal to -4 gives: ( ) ( )( ) 4 3 3 2 1- = + + + = s s s s s 6 2 12 3 = = We can thus write: ( ) 4 6 3 6 2 1 + + +- + = s s s s X By taking the inverse Laplace transform we obtain the time domain solution: ( ) t t t e e e t x - - - + - = 4 3 2 6 6 b) ( ) ( )( ) ( ) 4 3 2 1 2 + + + + = s s s s s X One term in the denominator is of the form: 1 2 d s d s + + with (d 1 ) 2 /4<d The system thus has complex roots, and can be rewritten as:...
View Full Document

This note was uploaded on 11/20/2009 for the course CHME DTU-abroad taught by Professor Rafiqulgani during the Spring '09 term at Rensselaer Polytechnic Institute.

Page1 / 3

Solution_to_problem_3.6 - 3.6 1 Solution to problem 3.6 a)...

This preview shows document pages 1 - 2. Sign up to view the full document.

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