s251ex2Laplace(fa03) - MATH 251 Fall 2003 Exam II Laplace...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 251 Fall 2003 Exam II Laplace Transforms Make-Up Exam Solutions 6. (14 pts) Write the following function in terms of unit step functions, and ±nd its Laplace trans- form. g ( t ) = t 2 + 1 0 t < 1 e - 3 t + 1 1 t < 2 1 t 2 g ( t ) = (1 - u 1 ( t ))( t 2 + 1) + ( u 1 ( t ) - u 2 ( t ))( e - 3 t + 1) + u 2 ( t ) = t 2 + 1 + u 1 ( t )( e - 3 t + 1 - t 2 - 1) + u 2 ( t )(1 - e - 3 t - 1) = t 2 + 1 + u 1 ( t )( e - 3 t - t 2 ) - u 2 ( t ) e - 3 t G ( s ) = L{ g ( t ) } = 2 s 3 + 1 s + e - s L{ e - 3( t +1) - ( t + 1) 2 } - e - 2 s L{ e - 3( t +2) } = 2 s 3 + 1 s + e - s L{ e - 3 e - 3 t - t 2 - 2 t - 1 } - e - 2 s L{ e - 6 e - 3 t } = 2 s 3 + 1 s + e - s [ e - 3 s + 3 - 2 s 3 - 2 s 2 - 1 s ] - e - 2 s e - 6 s + 3
Background image of page 1

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

View Full DocumentRight Arrow Icon
8. (12 pts) Find the inverse Laplace transform of: F ( s ) = s 2 - 4 s 3 + 6 s 2 + 9 s First simplify F ( s ) using partial fractions: F ( s ) = s 2 - 4 s ( s 2 + 6 s + 9) = s 2 - 4 s ( s + 3) 2 = a s + b s + 3 + c ( s + 3) 2 = a ( s + 3) 2 + bs ( s + 3) + cs s ( s + 3) 2 . Therefore,
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.

Page1 / 3

s251ex2Laplace(fa03) - MATH 251 Fall 2003 Exam II Laplace...

This preview shows document pages 1 - 3. Sign up to view the full document.

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