Find the inverse Laplace transform of Fs 4s 4s2 4s 20 ft

Find the inverse Laplace transform of F(s) = -4s + 4/s^2 + 4s + 20 f(t) =

Solution

Given that

F(s) = ( -4s + 4 ) / ( s² + 4s + 20 )

Rewrite F(s) as follows

     F(s) = ( -4s - 8 + 12 ) / ( s² + 4s + 16 + 4 )

      F(s) = [ -4(s+2) + 12 ] / [ ( s+2)² + 16 ]

     F(s) = -4(s+2) / [ ( s+2)² + 16 ] + 12 / [ ( s+2)² + 16 ]

Take inverse laplace transform of F(s)

L^-1 { F(s) } = L^-1 { -4(s+2) / [ ( s+2)² + 16 ] + 12 / [ ( s+2)² + 16 ] }

L^-1 { F(s) } = -4 L^-1 { (s+2) / [ ( s+2)² + 16 } + L^-1 { 12 / [ ( s+2)² + 16 ] }.................................1

Compute

L^-1 { (s+2) / [ ( s+2)² + 16 }

Apply inverse laplace transform rule , if L^-1{ F(s) } = f(t) then L^-1( F(s-a) ) = e^atf(t)

Hence,

        L^-1 { (s+2) / [ ( s+2)² + 16 } = e^-2t .L^-1 { s/(s² + 16) }

                                                   = e^-2t . L^-1 { s/(s² + 42) }

                                                   = e^-2t.cos( 4t )                            [ since, L^-1( s/(s²+a²) ) = cos( at ) ]

L^-1 { 12 / [ ( s+2)² + 16 ] } = e^-2t.L^-1 { 12 / (s² + 16 ) }

                                        = e^-2t.L^-1 { 3.4 / (s² + 4² ) }

                                        = e^-2t.3.L^-1 { 4 / (s² + 4² ) }

                                        = 3e^-2t sin(4t)          [since, since, L^-1( a/(s²+a²) ) = sin( at ) ]

From equation 1

f(t) = L^-1 { F(s) } = -4 L^-1 { (s+2) / [ ( s+2)² + 16 } + L^-1 { 12 / [ ( s+2)² + 16 ] }

                          = -4e^-2t.cos( 4t ) + 3e^-2t sin( 4t )

Therefore,

         f(t) = -4e^-2tcos( 4t ) + 3e^-2t sin( 4t )