Inverse Laplace Transform

If { ( )} = ( ) then ( )is called the inverse Laplace transform of f(s) and is denoted byL f s .

i. e., if { ( )} = ( ) then ( ) = Standard form:

1. L = 12. L = e3. L { } = !4. L = e !5. L { } =6. L = cosat7. L = 8. L = coshat9. L = e10. L = e cosbt

Examples:

1. Find L { + − }L { + − } = L + L − L= e + e − e

2. Find L { + }= L ss + 6 + 2L 1s + 6 + 4L ss + 5 − L { 1s + 5 }= 6 + 2/6 6 + 4 5 – 1/5 5

3. Find L {( ) }= L = L + 8L + 6L + 12L= ! + 8 ! + 6 ! + 12 ! = ! + + t +

4. Find L { ( ) }= 3/2L = 3/2[L − 2L + L ]= 3/2[1– ! + !] = 3/2[1–t + ]

5. Find L { }= L − + = L − 3L + 4L { }= 1– 3. + 2

6. Find L { }= L = L = L + 4L { }= 3 + 4/3 3

7. Find L { }}= L = L − 3L { }

= 2 − 32 28. Find {( ) }

= + 2 + 12 + 6 = { 1 + 8 + 12 + 6 }= 2! + 85! + 12 4! + 63! = 2 + 15 + 2 +

9. Find { }= { 2 − 1 + 2 + 5 }= 2 + 2 − 5 + 2 + 5 = 2 + 2 + 2 + 5 − 5 + 2 + 5

= 2 cos 5 − sin 5 = (2cos 5 − sin 5 )10. Find { }

= 1 − 1 + 2 = sin 2211. Find { }

+ 2 + 3 = + 2 − 2 + 2 + 3 = + 2 + 2 + 3 − 2 1 + 2 + 3 = cos 3 − 23 sin 3 = (cos 3 − 23 sin 3 )

12. Find { }= { + 5 − 3 + 2 }

− 3 + 8 − 3 + 2 = − 3 − 3 + 2 + 8 1 − 3 + 2 = cos 2 + 82 sin 2 = (cos 2 + 4 sin 2 )

13. Find { }= { 3 + 7 − 1 − 2 }= 3 − 1 + 10 − 1 − 2 = 3 − 1 − 1 − 2 + 10 1 − 1 − 2 = 3 cosh 2 + 102 sinh 2

14. Find { }= + 1( + 3) = + 3 − 3 + 1( + 3)

= 1( + 3) − 2( + 3) = − 2 = (1 − 2 )15. FindL ( )( )

s + s − 2s(s + 3)(s − 2) = As + Bs + 3 + Cs − 2s + s − 2 = A s + 3 s − 2 + Bs s − 2 + Cs s + 3puts = 0, − 2 = − 6A ∴ A = 1 3⁄s = 2, 4 = 10C ∴ C = 2 5⁄s = − 3,4 = 15B ∴ B = 4 15⁄∴ s + s − 2s(s + 3)(s − 2) = 13 ∙ 1s + 415 . 1s + 3 + 25 . 1s − 2L s + s − 2s(s + 3)(s − 2) = 13 L 1s + 415 L 1s + 3 + 25 L 1s − 2

= . 1 + . e + . e16. Find L ( )( )( )

1s(s + 1)(s + 2)(s + 3) = As + Bs + 1 + Cs + 2 + Ds + 3

1 = A s + 1 s + 2 s + 3 + Bs s + 2 s + 3+ Cs s + 1 s + 3 + s(s + 1) s + 2

puts = 0,1 = 6A ∴ A = 1 6⁄s = − 1, 1 = B(− 1)(1)(2) ∴ B = − 1 2⁄s = − 2,1 = C(− 2)(− 1)(1) ∴ C = 1 2⁄s = − 3,1 = D(− 3)(− 2)(− 1) ∴ C = − 1 6⁄ 1s(s + 1)(s + 2)(s + 3) = 1/6s − 12s + 1 + 12s + 2 − 1/6s + 3L 1s(s + 1)(s + 2)(s + 3)

= 16 L 1s − 12 L 1s + 1 + 12 L 1s + 2 − 16 L 1s + 3= . 1 − . e + . e − . e

17. FindL 2s − 6s + 5(s − 1) s − 2 (s − 3) = A(s − 1) + B(s − 2) + C(s − 3)

2s − 6s + 5 = A s − 2 s − 3 + B(s − 1) s − 3 + C(s − 1) s − 2puts = 1, 1 = 2A ∴ A = 1 2⁄s = 2,1 = − B ∴ B = − 1

s = 3,5 = 2C ∴ C = 5 2⁄∴ 2s − 6s + 5(s − 1) s − 2 (s − 3) = 12 ∙ 1(s − 1) − 1s − 2 + 52 . 1s − 3L s + s − 2s(s + 3)(s − 2) = 12 L 1(s − 1) − L 1s − 2 + 52 L 1s − 3

= . e − e + . e

Inverse transform of logarithmic functions

and Inverse functions

= = − ′− ′ =

1) Find {log( )}Let = log( )

= log + − ( + ) = 1+ − 1+− = − 1+ + 1+

− = 1+ − 1+ = −

= 2) Find {log(1 − )}

Let = log 1 − = log = log − − log

′ = 2− − 2− = 2 − 2− − = {2} − { 2− }

= 2 − 2 cosh = (1 − cosh )

3) Find {log( ( ))}Let = log( ( ))

= log + 1 − − log( + 1)

= 2+ 1 − 1 − 1+ 1− = 1 + 1+ 1 − 2+ 1− = 1 + 1+ 1 − 2+ 1

= 1 + − 2= 1 + − 2

4) Find { }Let =

= − 11 + ( ) 1− = + {− } = { + } =

=

Convolution Theorem

Statement: If and then

Proof: Let ( ) = −Taking Laplace transform both sides, we get

= { − }= − = −

= − (By changing the order of integration)

− = ℎ = , = , = 0 = ∞, = ∞= ( )( ) = ( ) = −

Examples:

1. Using convolution theorem evaluate { ( )}Solution: Let = ( ) and = Therefore = = = ( )

= = = ( )( ) = −

( ) . = sin = sin= [ { }( ) ]0= [ − − − 1{− 0 − 1}]= [− − + ]

2. Using convolution theorem evaluate { ( )}Solution: Let = ( ) and = Therefore = = = ( ) = = 1 = ( )( ) = −

( ) . = 1 = sin= 1 [− cos ]0 = − [cos − 1]

= [1 − cos ]3. Using convolution theorem evaluate { ( )}

Solution: Let = ( ) and = Therefore = = = ( ) = = = ( )( ) = −

. = ( − )

= [ − − − 1 ]0= 0 − + − 1= + ( − 1)

4. Using convolution theorem evaluate {( ) }Solution: Let = and = Therefore = = cos = ( ) = = = ( )( ) = −

. = cos ( ) = sin + − 2= [sin 0 − { ( )}0]= [sin − 0 + {cos − }]= 2 sin

5. Using convolution theorem evaluate {( ) }Solution: Let = and = Therefore = = cos = ( ) = = cos = ( )

( ) = − . = cos ( − )

= [cos at − 2au + cos ]= [ + cos ]0= [− {− sin − } + cos { − 0}]= +

= [atcosat + sinat]6. Using convolution theorem, evaluate { ( )}

Let f s = ,g(s) = {f(s)} = { } = cosat = f(t){g(s)} = { } = cosbt = g(t)We know that{f(s)g(s)} = − { } = cos cos ( − ) = [cos{( + ) − bt} + cos{(a − ) + }] = + 0= [ + ]= [ + ]= ( ≠ )