BME211HW9 - Homework#9 Solutions Ex 14.6-1 Taking Laplace Transform of the differential equation s 2 F s s f 0 f 0 5 F s f 0 6 F s = s 10 s 3 Using

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Unformatted text preview: Homework #9 Solutions Ex. 14.6-1 Taking Laplace Transform of the differential equation: s 2 F ( s ) - s f ( 0 ) - f ' ( 0 ) + 5 F ( s ) - f ( 0 ) 6 F ( s ) = s + 10 s +3 Using the given initial conditions ( s 2 + 5s + 6) F ( s ) = 10 2 s 2 +16 s + 40 + 2 s + 10 = s +3 s +3 F ( s) = 2 s 2 +16 s + 40 A B C = + + 2 ( s + 3 ) ( s + 2 ) ( s + 3 ) ( s + 3) ( s + 3) ( s + 2 ) where A = - 10, B = - 14, and C = 16 . Then F (s) = ( s +3) -10 2 + -14 16 + s +3 s + 2 f (t ) = - 10te -3t - 14e -3t + 16e -2t for t 0 P14.2-1 L f1 ( t ) A F1 ( s ) A = f1 ( t ) = cos ( t ) As L s F ( s ) = s2 + 2 F1 ( s ) = 2 2 L s + P14.2-2 L tn = n! s n +1 F ( s ) = L t1 = 1 1! = 2 1+1 s s P14.2-3 Linearity: L 1 f1 ( t ) + a2 f 2 ( t ) a1 F1 ( s ) + a2 F2 ( s ) a = Here a1 = a2 = 1 L 1 ( t ) L -3t f = e = L 2 ( t ) L [ t ] = f = so F ( s ) = 1 1 + 2 s +3 s 1 = F1 ( s ) s+3 1 = F2 ( s ) s2 P14.3-1 ( 1-e A Ae - sT F ( s ) = AL ( t ) AL ( t -T ) - u - u = =A s s s P14.3-7 f ( t ) = A ( t ) - u ( t -T ) u - sT ) 3 e - st F ( s ) = f (t ) e dt = e dt = 3 0 0 -s - st 2 - st 2 = 0 3(1-e -2 s ) s P14.4-3 F (s) = ( s +1) ( s - 2 ) 2 5 s -1 = A B C + + 2 s +1 ( s +1) s -2 5 s -1 2 s =2 where Then B= A= 5 s -1 s -2 = 2 and C = s =-1 ( s +1) =1 = -1 s =-1 d -9 2 ( s +1) F ( s ) s =-1 = 2 ds ( s -2) Finally F (s) = -1 2 1 + + 2 s +1 ( s +1) s -2 f ( t ) = e - t + 2 t e - t + e 2t ( t ) - u P14.4-6 F ( s) = where A = sF ( s ) and s =0 2( s+3) A B C = + + s( s+1) ( s + 2 ) s s +1 s + 2 = 3, B = ( s +1) F ( s ) 2( s +3) s ( s +1) = s =-1 2( s + 3 ) s( s + 2 ) = -4 = 2( s +3) ( s +1) ( s + 2 ) s =0 s =-1 ( s+2) F ( s ) Finally s = -2 = s =-2 = C =1 3 -4 1 F ( s) = + + s s +1 s + 2 f ( t ) = ( 3 - 4e - t + e -2t ) u ( t ) P14.4-9 (a) where A = sF ( s ) |s =0 = F ( s) = s( s +1) s 2 -5 2 = A B C + + s s +1 ( s +1) 2 -5 1-5 2 = -5 and C = ( s +1) F ( s ) |s =-1 = =4 1 -1 2 Multiply both sides by s ( s + 1) s 2 - 5 = -5 ( s +1) + Bs ( s +1) + 4 s B = 6 2 Then F ( s) = Finally -5 6 4 + + s s +1 ( s +1) 2 f ( t ) = ( -5 + 6 e -t + 4 t e - t ) , t 0 (b) where A= and F ( s) = 4s 2 ( s +3) 3 = A B C + + 2 ( s + 3 ) ( s + 3) ( s + 3 ) 3 1 d2 d 3 3 ( s +3) F ( s ) 3 = 4, B = s +3) F ( s ) 3 = -24 ( 2 s =- s =- 2 ds ds C = ( s +3) F ( s ) s =-3 = 36 3 Then F ( s) = Finally 4 -24 36 + + 2 ( s + 3 ) ( s + 3) ( s + 3 ) 3 t 0 f ( t ) = ( 4 - 24 t + 18t 2 ) e -3t , P14.6-7 KCL: v1 5 + i = 7 e -6 t KVL: 4 di di + 3 i - v1 = 0 v1 = 4 + 3 i dt dt di 35 + 2 i = e -6 t dt 4 Then di 4 + 3i dt + i = 7 e -6 t 5 Taking the Laplace transform of the differential equation: s I ( s) - i (0) + 2 I ( s) = 35 1 35 1 I (s) = 4 s +6 4 ( s + 2)( s + 6) Where we have used i (0) = 0 . Next, we perform partial fraction expansion. 1 A B 1 = + where A = ( s + 2) ( s + 6) s + 2 s + 6 s +6 Then I (s) = 35 35 35 1 35 1 - i (t ) = e -2t - e -6t A, t 0 16 s + 2 16 s + 6 16 16 = s =-2 1 1 1 and B = =- 4 s + 2 s = -6 4 ...
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This note was uploaded on 03/07/2008 for the course BME 211 taught by Professor Cain during the Winter '07 term at University of Michigan.

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