Test 1 Solution

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

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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
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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.

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

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