Tutorial 2

Tutorial 2 - 3 s 3 2 16 5 Find L f t where f t = t ≤ t< 2 e 2 t sin 3 t-2 t ≥ 2 Solution We write f t as f t = u t-u t-2 t u t-2 e 2 t sin 3

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MATH3705 Tutorial 2 1. Calculate the Laplace transform of f ( t ) = sinh( at ) = e at - e at 2 Solution: L (sinh( at )) = L ( e at - e at 2 ) = 1 2 L ( e at ) - 1 2 L ( e at ) = 1 2 1 s - a - 1 2 1 s + a = 1 2 s + a - ( s - a ) s 2 - a 2 = a s 2 - a 2 2. Using the shift theorem ﬁnd the Laplace transform of f ( t ) = e 2 t t 2 Solution: Recall the ﬁrst shift theorem says L ± e at f ( t ) ² = F ( s - a ) where L ( f ) = F ( s ). Now, we know that L ± t 2 ² = 2! s 3 = 2 s 3 so, by the shift theorem L ± e 2 t t 2 ² = 2 ( s - 2) 3 3. Using the Laplace transform solve the diﬀerential equation f ′′ + f - 6 f = e 3 t with boundary conditions f (0) = f (0) = 0. Solution: The subsidiary equation is s 2 F + sF - 6 F = 1 s + 3 or F = 1 ( s + 3) 2 ( s - 2) As before, we do partial fractions 1 ( s + 3) 2 ( s - 2) = A s + 3 + B ( s + 3) 2 + C s - 2 1 = A ( s + 3)( s - 2) + B ( s - 2) + C ( s + 3) 2 1

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s = - 3 gives B = - 1 / 5 and s = 2 gives C = 1 / 25. Putting in s = 1 we ﬁnd 1 = - 4 A + 1 5 + 16 25 and so A = - 1 / 25. Putting all this together says that f = - 1 25 e 3 t - t 5 e 3 t + 1 25 e 2 t 4. Find L { f ( t ) } , where f ( t ) = t + 2 t 3 - e 3 t cos(4 t ) . Solution: By the First Shift Theorem, we have L { e 3 t cos(4 t ) } = [ L { cos(4 t ) } ] | t +3 = [ L s s 2 + 16 ] | s +3 = ( s + 3) ( s + 3) 2 + 16 . By linearity of LT, we have F ( s ) = π 2 s 3 / 2 + 12 s 4 - ( s
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Unformatted text preview: + 3) ( s + 3) 2 + 16 . 5. Find L { f ( t ) } , where f ( t ) = { t, ≤ t < 2; e 2 t sin 3( t-2) , t ≥ 2. Solution: We write f ( t ) as f ( t ) = [ u ( t )-u ( t-2)] t + u ( t-2) e 2 t sin 3( t-2) . Then F ( s ) = 1 s 2 + e − 2 s [ 3 e 4 ( s-2) 2 + 9-1 s 2-2 s ] . 6. Find L − 1 ± e-3 s s ( s 2 +2 s +2) } . Solution: Let G ( s ) = 1 s ( s 2 +2 s +2) . By partial fraction, we have G ( s ) = 1 2 s-1 2 s + 1 s 2 + 2 s + 2 = 1 2 ( 1 s )-1 2 ( s + 1 ( s + 1) 2 + 1 )-1 2 ( 1 ( s + 1) 2 + 1 ) . Thus g ( t ) = 1 2-1 2 e − t cos( t )-1 2 e − t sin( t ) . By the Second Shift Theorem, f ( t ) = u ( t-3) [ 1 2-1 2 e − ( t − 3) cos( t-3)-1 2 e − ( t − 3) sin( t-3) ] . 2...
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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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Tutorial 2 - 3 s 3 2 16 5 Find L f t where f t = t ≤ t< 2 e 2 t sin 3 t-2 t ≥ 2 Solution We write f t as f t = u t-u t-2 t u t-2 e 2 t sin 3

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