SampleTest2v2

# SampleTest2v2 - Theorem to find the convolution of sin t...

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MATH3705 Solution to Test 2 Summer 2006 1 SOLUTION TO TEST 2 MATH3705, Summer 2006 Total = 20 1. (4 marks) Find the Laplace transformation of the function f ( t ) = u ( t - 1) e - 2 t cos ( t - 1). Solution . L { f } = [ { u ( t - 1)cos ( t - 1)}] s +2 = [ e s s / ( s 2 + 1)] s +2 = e ( s +2) ( s + 2) / [( s + 2) 2 + 1]. 2. (4 marks) Find the inverse Laplace transformation of F ( s ) = 2 2 2 s se s s - + + . Solution . F ( s ) = e s ( F 1 ( s ) - F 2 ( s )), where F 1 ( s ) = 2 1 ( 1) 1 s s + + + , F 2 ( s ) = 2 1 ( 1) 1 s + + . - 1 { F 1 } = e t cos t , - 1 { F 2 } = e t sin t . Hence, - 1 { F } = u ( t - 1) e ( t - 1) (cos ( t - 1) - sin ( t - 1)). 3. (3 marks) Express the following function by a single expression using the unit step function: f ( t ) = 0 1 1 1 t t t ≤ ≤ . Solution . f ( t ) = [ u ( t ) - u ( t - 1)] t + u ( t - 1) = u ( t ) t - u ( t - 1)( t - 1). 4. (3 marks) Use the formula - 1 { s / ( s 2 + 1) 2 } = ( t sin t ) / 2, and the Convolution

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Unformatted text preview: Theorem to find the convolution of sin t and cos t . Solution . {sin t * cos t } = {sin t } {cos t } = s / ( s 2 + 1) 2 . Hence, sin t * cos t = -1 { s / ( s 2 + 1)} = ( t sin t ) / 2. 5. (6 marks) Solve the differential equation y' + y = 1 + δ ( t-1), y (0) = y' (0) = 0. Solution . sY + Y = 1 / s + e s . MATH3705 Solution to Test 2 Summer 2006 2 Y = 1 / [ s ( s + 1)] + e-s / ( s + 1). L-1 {1 / [ s ( s + 1)]} = -1 {1 / s -1 / ( s + 1)]} = t-e t . -1 { e s / ( s + 1)} = u ( t-1) e ( t-1) . y = t-e-t + u ( t-1) e ( t-1) ....
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## 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.

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SampleTest2v2 - Theorem to find the convolution of sin t...

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