test1_07_solutions

test1_07_solutions - MATH 3705 B Test 1 January 2007...

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

View Full Document Right Arrow Icon
MATH 3705* B Test 1 January 2007 SOLUTIONS Questions 1-5 are multiple choice. Circle the correct answer. Only the answer will be marked. [12 marks] 1. [2] L{ e 2 t cos(3 t ) } = ( a ) s +2 ( s +2) 2 +9 ( b ) s s 2 ( c ) s 2 ( s 2) 2 ( d ) None of the above 2. [2] L{ t sin(3 t ) } = ( a ) 6 s ( s 2 +9) 2 ( b ) s 2 9 ( s 2 2 ( c ) 9 s 2 ( s 2 2 ( d ) None of the above 3. [2] L{ u ( t 1) e 2 t } = ( a ) e s s 2 ( b ) e 1 s s ( c ) e 2 s s 2 ( d ) None of the above 4. [3] L 1 F 2 s s 2 25 k = ( a ) e 5 t e 5 t ( b ) e 5 t + e 5 t ( c ) cos(5 t )( d ) None of the above 5. [3] L 1 F se 3 s s 2 +4 k = ( a ) u ( t 3) cos(2 t b ) u ( t +3)cos[2( t +3)] ( c ) u ( t 3) cos[2( t 3)] ( d ) None of the above Answers: 1.(a), 2.(a), 3.(c), 4.(b), 5.(c) .
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 6. [4 marks] Find L{ f ( t ) } ,where f ( t )= F t ,i f 0 & t< 2; e 5 t f t 2 . Solution: Rewrite f ( t ) using the unit step function u ( t a ): f ( t )=[ u ( t ) u ( t 2)] t + u ( t 2) e 5 t = u ( t ) t + u ( t 2)( t + e 5 t ) . Find the Laplace transform of each term of the expression above, using the Second Shifting theorem with a =0and a = 2 respectively: L{ u ( t ) t } = e 0 t L{ t } = 1 s 2 , L{ u ( t
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 07/16/2009 for the course MATH 3705 taught by Professor Jaberabdualrahman during the Winter '08 term at Carleton CA.

Page1 / 3

test1_07_solutions - MATH 3705 B Test 1 January 2007...

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