UNIVERSITY OF INDONESIA FACULTY OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING

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1 Engineering Mathematics (MCS-1007) UNIVERSITY OF INDONESIA FACULTY OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING ENGINEERING MATHEMATICS (MCS-1007) 1. Course Name/Units : Engineering Mathematics/4. Department/Semester : Mechanical Engineering /3 3. Program/Period : S1/ First semester Date/Time/Room : I. Tuesday : (K105) : II. Thursday : (K06) 5. Instructor : 6. Topics : MCS : Inverse Laplace Transform Inverse Laplace Transform Definition As discussed before, the Laplace Transform can be used to solve differential equations. However, before it can be applied, we must learn the inverse Laplace Transform. So far, we have been given functions of t and found their Laplace Transforms. Now, we consider the problem of finding a function f(t), when the Laplace Transform, F(s), has been given. -1 If L f ( t) F( s) we write f ( t) F( s) L and call L -1 the invers Laplace Transform operator. Clearly the Laplace Transforms Tables previously given and the operation rules of Laplace Transforms will help us to do this. However, the following hints of Laplace Transform forms and table of inverse Laplace Transform give other convenient shortcuts. Form 1 Form Form 3 Form 4 Form 5 Form 6 Form 7 Form 8 Form 9 Form 10 Form 11 Form 1 MCS Inverse Laplace Transform - 1

2 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 1 MCS Inverse Laplace Transform -

3 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form Inverse Laplace Transform of Function of Form 3 Inverse Laplace Transform of Function of Form 4 MCS Inverse Laplace Transform - 3

4 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 5 Inverse Laplace Transform of Function of Form 6 MCS Inverse Laplace Transform - 4

5 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 7 MCS Inverse Laplace Transform - 5

6 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 8 Inverse Laplace Transform of Function of Form 9 MCS Inverse Laplace Transform - 6

7 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 10 Inverse Laplace Transform of Function of Form 11 MCS Inverse Laplace Transform - 7

8 Engineering Mathematics (MCS-1007) Inverse Laplace Transform of Function of Form 1 Example: Find the Inverse Laplace Transform of the following functions: s 10 (a) (c) (e) 4 s 1 s s s 1 15 (b) (d) s 1 s 1 9 s 3 6s 13 Solution: (a) Reffering directly to the table -1 s f ( t) L cost s 1 (b) Using Linearity property of the inverse transform and then reffered to the table -1 s 1-1 s -1 1 f ( t) L L L cost sint s 1 s 1 s 1 (c) The function is written to match exactly the standard forms given in the table, with possibly a constan factor being present. Often the denominator needs to be written in standard form 3 t 3 t t e 5t e f ( t) L 10L s s 3! t (d) f ( t) L 5L 5e sinh3t s 1 9 s s 3-1 s 3-1 s 6 3 (e) f ( t) L L L s 6s 13 s 3 s 3 s 3 3t -1 s t 3e sint L L e cost s 3 s 3 MCS Inverse Laplace Transform - 8

9 Engineering Mathematics (MCS-1007) Some Useful Technique 1. Using Partial Fractions to Find the Inverse Laplace Transform The Inverse Laplace Transform of a fraction is often best found by expressing it as its partial fractions, and finding the inverse transform of these partial fractions. Example: Find the Inverse Laplace Transform of the following functions: 4s 1 6s 8 3s 11s 14 (a) (b) (c) 3 s s s 3s s s 11s 5 Solution: (a) We express 4s 1 4s 1 4s 1 A B s s s s 1 s s 1 s s as its partial fractions: Using the technique from basic calculus we find that A = 1, B = 3 and hence -1 4s t f ( t) L L L L 1 3e s s s s 1 s s 1 (b) Using the same technique of partial fraction we can express: 6s 8 4, hence s 3s s 1 s -1 6s t f ( t) L L L L e 4e s 3s s 1 s s 1 s (c) Using the same technique of partial fraction we can express: 3s 11s 14 s 3, hence 3 s s 11s 5 s 4 s 6s s 11s 14-1 s 3 f ( t) L L 3 s s 11s 5 s 4 s 6s s 3 4t 3t L L e e cost s 4 s 6s 13 t. Using Complex Numbers to the Inverse Laplace Transform Essentially the method is using partial fractions, but where all the factors in the denominator are linear that is, there is no quadratic factors. When complex numbers are allowed, the unknown constants are evaluated using particular values of s or equating coefficients. Example: s 3 Find the Inverse Laplace Transform of the following functions s 6s 13 Solution: We first factorized the denominator. To do this we solve s + 6s + 13 = 0 using the usual formula of roots of quadratic equations. Hence, MCS Inverse Laplace Transform - 9

10 Engineering Mathematics (MCS-1007) 6 6 (4)(1)(13) i s 3 i (1) It then follows that the denominator can be factorized as (s a)(s b) where a = -3 +i and = -3 i. Then using the partial fractions method b s 3 s 3 A B s 6s 13 s a s b s a s b The unknown constants A and B can now be found as follows. s 3 A B As b Bs a s a s b s a s b s a s b Substitute s = a then a 3 A a b B a a A a b since a 3 i and b 3 i then: 1 3 i 3 A3 i 3 i i A(4 i) A Equating the coefficients of s 1 A B B 1 So, s s 6s 13 s a s b The Inverse Laplace Transform can now be found -1 s f ( t) L L L s 6s 13 s a s b s a s b 1 at bt 1 3it 3i t 1 3t it it e e e e e e e Using the theorem of complex number: t it it 3 t ( ) cos sin cos sin 3 t f t e e e e t i t t i t e cos t MCS Inverse Laplace Transform - 10