Test 1 Solution

# Test 1 Solution - MATH3705 A - Test 1 Name and Student...

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

Name and Student Number: Total points: 20. No partial marks for Questions 1-4. Closed book! Formula Sheet and Non-programmer calculators are allowed! 1. L { t sin 3 t } = [2] (a) 3 s ( s 2 +3) 2 (b) 3 s s 2 +9 (c) 2 sin s ( s 2 +9) 2 (d) - 2 e s ( s 2 +3) 2 (e) 6 s ( s 2 +9) 2 Solution: (e) L { sin 3 t } = 3 s 2 +9 , L { t sin 3 t } = - ( 3 s 2 +9 ) = 6 s ( s 2 +9) 2 . 2. Find L { f ( t ) } , where f ( t ) = 2 t 3 - e 3 t cos(4 t ) . [2] (a) 12 s 4 - ( s +3) ( s +3) 2 +4 (b) 6 s 4 - ( s 3) ( s 3) 2 +16 (c) 12 s 4 - ( s 3) ( s 3) 2 +16 (d) 6 s 4 - ( s +3) ( s +3) 2 +16 (e) 12 s 4 - ( s +3) ( s +3) 2 +16 Solution: (e) By linearity of LT and the First Shift Theorem, we have F ( s ) = 2(3!) s 4 - ( s + 3) ( s + 3) 2 + 16 . 3. Let f ( t ) be 2-periodic for t 0, and f ( t ) = { 0 , 0 t < 1; t, 1 t < 2. Find L { f ( t ) } . [2] (a) (1+ s ) e - s (1+2 s ) e - 2 s s 2 (1 e - 2 s ) (b) se - s (1+2 s ) e - 2 s s 2 (1 e - 2 s ) (c) (1+ s ) e

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.

## This note was uploaded on 07/11/2011 for the course ECE 45 taught by Professor Lee during the Spring '08 term at Alfred University.

### Page1 / 3

Test 1 Solution - MATH3705 A - Test 1 Name and Student...

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

View Full Document
Ask a homework question - tutors are online