{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution_to_problem_3.6

# Solution_to_problem_3.6 - 3.6– 1 Solution to problem 3.6...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
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 α α α We use Heaviside expansion to find values for α 1 , α 2 and α 3 Multiplication by (s+2) and putting s equal to -2 gives: ( ) ( )( ) 2 1 4 3 1- = + + + = s s s s s α 1 2 2 1 = = α Multiplication by (s+3) and putting s equal to -3 gives: ( ) ( )( ) 3 2 4 2 1- = + + + = s s s s s α 6 1 6 2- =- = α Multiplication by (s+4) and putting s equal to -4 gives: ( ) ( )( ) 4 3 3 2 1- = + + + = s s s s s α 6 2 12 3 = = α 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

{[ snackBarMessage ]}

### Page1 / 3

Solution_to_problem_3.6 - 3.6– 1 Solution to problem 3.6...

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

View Full Document
Ask a homework question - tutors are online