6 Inverse Laplace Transforms PDF

Title	6 Inverse Laplace Transforms
Course	Engineering Mathematics
Institution	Technological University Dublin
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Summary

Inverse Laplace Transforms. This process is the reverse of finding the Laplace Transform of a function, f(t). Given the Laplace Transform F(s) , find the function f (t) , to which it belongs....

Description

DT022/2, DT024/2

Engineering Mathematics III

6 Inverse Laplace Transforms

This process is the reverse of finding the Laplace Transform of a function, f ( t) . Given the Laplace Transform F( s) , find the function f (t) , to which it belongs. L  f (t )  F ( s )



L1 F ( s )  f (t )



 k  L1  2  sin kt 2 s k 

For example,

Lsin kt 

k s  k2 2

Example 1

 3  L1  2   sin 3t s  9 

Example 2

 s  cos 4 L1  2 t  s  16 

Example 3

  1  2t L1  e s  2 

Example 4

 s  cosh 5 L1  2 t  s  25 

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Example 5

6s     t sin 3 t L1 2 2  ( s  9)  Example 6

 5  L1  2   sinh 5t s  25 

Example 7

3  1  L1    3L1    3 1  3 s  s 

Further Exercises

Determine the inverse Laplace transform of the following functions, with reference to the tables supplied:

1.

 s  L1  2   s 16 

2.

 6  L1  2   s  36 

3.

 7  L1   s

4.

 1  L1   s 4

5.

  3  L1    s  5

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6.

10 s    L1  2 2  ( s  25) 

7.

  3 L1   2 ( s  4)  9 

8.

 s2  36  L1  2 2  ( s  36) 

Answers to all of the exercises above can be found by reading the table of Laplace transforms in reverse.

We will now look at additional examples where further manipulations have to be carried before we can find the inverse transform.

Partial Fractions

What happens if we want to find  3s 1  L1  2  s  s6

We don’t have any standard transform for this expression. We can however separate the expression into partial fractions as follows:

3s  1 1 2       s s 6 s 2 s 3 2



 3s  1  1  1  1  2  L1  2  L   L   s  s  6  s  2  s  3 

 e 2 t  2e 3t

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Rules for finding partial fractions

1.

Factorise the denominator into its prime factors.

Example

2.

1 1 becomes ( s  1)( s  2) s  3s  2 2

A linear factor, e.g., ( s  a) gives a partial fraction

A where A is a sa

constant to be determined.

Example

3.

A repeated factor (s  a )2 gives 

Example

4.

A B 1   (s  1)(s  2) s  1 s  2

A B 1 1 giving  goes to  s  2 (s  2)2 ( s  2)( s  2) s  4s  4 2

Similarly, (s  a )3 gives 

Example

1 1 giving goes to ( s  2)( s  2)( s  2) ( s  2)( s  4 s  4) A B C   2 s  2 (s  2) ( s  2)3

A quadratic factor

Example

A B C   2 s  a ( s  a) ( s  a)3

2



5.

A B  s  a ( s  a)2

1 Ps  Q gives 2 s  ps  q s  ps  q 2

15 s 2  s  2 A Bs  C  2 goes to 2 (s  5)(3s  4s  2) s  5 3s  4 s  2

6 Inverse Laplace Transforms

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Example 8

s  20    Find L 1  2   s 6 s8 

Solution



s  20 s  20 A B    s  6s  8 (s  2)(s  4) s  2 s  4 2

s  20 A( s  4)  B( s  2)  ( s  2)( s  4) ( s  2)( s  4)

Multiplying both sides by the denominator gives 

s  20  A(s  4)  B (s  2)

Substituting s  4 into both sides of the equation gives 4  20   16  A(0)  B (2)



B 8



A  9

Substituting s  2 gives 2  20   18  A( 2)  B(0)

So, returning to the original problem, we get 8  1   9   1  8   s  20  1   9 L1  2   L   L   L   s  6s  8  s  2 s 4  s 2  s  4  1  4t 2t 1  1   9L1    8L    8e  9e  s 2   s 4 

Exercise 9

s  19    2 t 5t Show that L 1  2   3e  2e  s  3 s  10  6 Inverse Laplace Transforms

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Exercise 10

 5s  1  4t 3 t Show that L1  2   3e  2e s s 12    

Exercise 11

 9 s 8  2t Show that L1  2   4 5 e s s 2   

Example 9

 s 2  15s  41  Determine L1  2  ( s  2)( s  3) 

Solution



s 2  15 s  41 A B C    2 ( s  2)( s  3) s  2 s  3 ( s  3) 2 s 2  15 s  41 A( s  3) 2  B( s  2)( s  3)  C ( s  2)  ( s  2)( s  3) 2 ( s  2)( s  3) 2

Multiplying both sides by the denominator gives s2  15s  41  A( s  3)2  B( s  2)( s  3)  C ( s  2)

We will choose suitable values of s to give us values of A, B and C Substituting s  3 into both sides of the equation gives 5  A(0)2  B (5)(0)  C (5)



C 1

Similarly, Substituting s  2 gives

6 Inverse Laplace Transforms

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75  A(5)  B (0)( 5)  C (0)



2

A3

Finally substituting s  0 into both sides of the equation gives 41  A(3)2  B(2)( 3)  C (2)

41  9 A  6B  2C

Since A  3 and C  1 , B can be calculated as follows: 41  9(3)  6 B  2(1)

So

41  27  2   6B



B  2

 s2  15 s 41  1  3 1  2 L1  L    2  2  ( s  2)( s  3)   s  2 s  3 ( s  3) 

3e 2t  2e3t  te3t

Exercise 12

 4s2  5s  6  t Show that L1    3e  cos 2t  3sin 2t 2 s s ( 1)( 4)    

Even Further Exercises

Determine the inverse Laplace transform of the following functions:

1.

1 L1   s

2.

1  4 L   s

3.

7 L1   s

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4.

 s  L1  2   s  25 

5.

 s  L1  2  s  16 

6.

 3  L1  2  s  9 

7.

 20  L1  2   s  16 

8.

 16s  L1  2 2   ( s  64) 

9.

 10s  L1  2   ( s  25)2 

10.

 (s 2  49) L 1  2 2 ( s  49)

11.

  s 3 L1   2 ( s  3)  25 

12.

 s 4  L1  2   s  8 s  25 

13.

 s 2  L1  2   s  4 s  20 

14.

  5 L1  2 2  ( s  3)  5 

15.

4   L1  2   s  2 s  17 

  

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16.

12   L1  2   s  4 s  13 

17.

15   L1  2   s  8 s  25 

18.

6   L1  2   s  10 s  29 

19.

3   L1  2   s  4 s  13 

20.

  s2  25 L 1  4  2  s  50 s  625 

21.

 5s  1  L1  2  s  s  12 

22.

 3s  1  L1  2   s  s 6 

23.

 s  11  L1  2   s  s 2 

24.

 9s  8  L1  2  s  2 s 

25.

 4 s 1  L1  2   s  7 s 6 

26.

 4s  34  L1  2   s  2 s 8 

27.

 3s  1  L1  2   s  3 s 2 

28.

1  5s  4  L  2  s  2 s 

29.

 2 s 8  L1  2   s  5 s 6 

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Answers to Even Further Exercises

1.

 1 L 1   1 s 

2.

 4  1 L1    4.L1    4  1  4 s   s

3.

7  1  L1    7.L1    7 s  s 

4.

s  s  1  L1  2  L  2 2  s  25  s 5

   cos5 t 

5.

 s   s  L 1  2 L1  2  2  s  16  s  4

   cos 4t 

6.

3   3  1   sin 3 t L1  2  L  2 2  s 9 s 3 

7.

 20   1  5 4 L1  2  L  2 2  s  16  s  4

8.

 16s  L1  2   t sin 8t  (s  64)2 

9.

 10s  L1  2  t sin 5t 2   (s  25) 

10.

2  1  (s  49) L  2 2 (s  49)

11.

  s 3 3t L1   e cos 5t 2 (s  3)  25 

   5sin 4 t 

   t cos 7 t 

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12.

 s 4  4t L1  2  e cos 3 t s  8 s  25 

13.

 s 2 L1  2 s  4 s  20

14.

  5 L1   e3 t sin 5t 2 2  s ( 3) 5    

15.

4  L1  2 s  2 s  17

 t 4  1    L  ( s 1) 2 4 2   e sin 4t     

16.

12  L1  2 s  4 s  13

 4 3  1  2t   L ( s 2) 2 3 2   4e sin 3t     

17.

 5 3  15   L1  2  L1   5 e4 t sin 3 t  2 2   s  8 s 25 ( s 4)  3 

18.

6  L1  2 s  10 s  29

19.

3  L1  2 s  4 s  13

20.

   1  s2  25  s2  25 L 1  4 L  2    t cos 5 t 2 2 ( s  25)  s  50 s  625 

21.

 5s  1 L1  2 s  s  12

 2t   e cos 4t

 3 2  1  5 t   L  ( s 5)2 22   3e sin 2 t     

 2t 3  1    L ( s 2) 2 3 2   e sin 3t     

5s  1  B   1  1  A   L  ( s 3)( s 4)   L  s 3  s 4       5s  1  A(s  4)  B( s  3)

Solve or A and B: s  4 :

21  B ( 7)

 B 3

s  3:

14  A(7)

 A 2

3 1 2  L  s  3 s  4

 3t  4t   2 e  3e 

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Engineering Mathematics III

B   3s  1  1  A L1  2 L    s s 2  3  s  s  6   3s  1  A(s  3)  B ( s  2)

Solve or A and B: s  2 :

5  A(5) 

A 1

s  3:

10  B(5)

B2



2   2t  3s  1   1  1 L 1  2 L    e  2 e 3t  2 s s  3  s  s  6  

23.

 s  11 L1  2 s  s  2

24.

4 5  1  9s  8  2t  L1   L  2    4  5e s  2 s  s s  2 

25.

 4s1 L1  2 s  7 s  6

1   1  5 s  1  1  5 6 t t   L  ( s  6)( s 1)   L  s  6  s 1   5 e  e     

26.

 4s  34 L1  2 s  2 s  8

B   1  4 s  34  1  A   L  ( s 2)( s 4)   L  s 2  s 4          

  1  4  3   t  2t  L s  1 s  2  4e 3e   

4s  34  A(s  4)  B ( s  2)

Solve or A and B: s  4 s2

18  A(0)  B (6)  B  3 42  A(6)  B (0)  A  7

3  7 L 1   s  2 s  4

27.

 3s 1 L1  2 s  3 s  2

  4t 2t  7 e  3e 

B   1  3s 1  1  A  L  ( s  1)( s  2)   L  s 1  s  2      

Solve or A and B: s  2

3s  1  A(s  2)  B (s  1)

5  A(0)  B ( 1)  B  5

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2  A(1)  B (0) 

A 2

 2  5    t   2t 5e L 1   2e s  1 s  2 

28.

B   5s  4  1  5 s  4  1  A L1  2   L  s( s  2)   L  s  s  2  s  2 s      5 s  4  A( s  2)  Bs

Solve or A and B: s2

6  A (0) 2B

s  0:

 4   2A

 B 3

 A2

3  2 2t L 1     2  3e s s  2 

29.

 2 s 8 L1  2 s  5 s  6

 1  A B   1  2 s  8   L  ( s 3)( s 2)   L  s 3  s 2           2s  8  A(s  2)  B( s  3)

Solve or A and B: s  2:

4  (1)B  B  4

s  3:

 2  A (1)  A  2

4  2 L 1   s  3 s  2

 2t 3t   4 e  2 e

6 Inverse Laplace Transforms

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2020 - 2021

DT022/2, DT024/2

Engineering Mathematics III

The cover up rule

This method only works when the denominator has non repeated linear factors. Example

F (s ) 

9s  8 s 2  2s

has partial fractions of the form

A B  s s 2

Using the ‘cover up’ rule, the constant A, the coefficient of

1 is found by temporarily s

covering up the factor s in the denominator of F ( s ) and finding the limiting value of what remains when s (the factor covered up) tends to zero.

So

 9 s 8  A  lim   4 s 0  s 2 

Similarly, B is obtained by covering up the factor ( s 2) . In this case, (s  2) goes to zero if s  2 .

Thus

 9 s 8  B  lim   5 s 2  s 



F (s ) 

9 s 8 4 5   2 s  2s s s  2

6 Inverse Laplace Transforms

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2020 - 2021...