How to compute the inverse Laplace transformation of (s^2)/((s^2+1)^2)

metal1fc 2022-09-12 Answered
How to compute the inverse Laplace transformation of s 2 ( s 2 + 1 ) 2
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Illuddybopylu
Answered 2022-09-13 Author has 17 answers
F ( s ) = s 2 ( s 2 + 1 ) 2 = s 2 + 1 ( s 2 + 1 ) 2 1 ( s 2 + 1 ) 2
F ( s ) = 1 s 2 + 1 + 1 2 s d d s 1 ( s 2 + 1 ) .
Apply inverse Laplace Transform:
f ( t ) = sin t 1 2 0 t 1 × τ sin τ   d τ
f ( t ) = 1 2 ( sin t + t cos t )
Not exactly what you’re looking for?
Ask My Question
Skye Vazquez
Answered 2022-09-14 Author has 4 answers
Consider the Laplace transforms of sin ( a t ) and t cos ( a t ) which are
sin ( a t ) a s 2 + a 2 = a s 2 + a 3 ( s 2 + a 2 ) 2 t cos ( a t ) s 2 a 2 ( s 2 + a 2 ) 2 .
Now one can notice that
f ¯ ( s ) g ¯ ( s ) 0 t f ( t u ) g ( u ) d u
A second method is to use the convolution theorem,
f¯(s)g¯(s)≑∫t0f(t−u)g(u)du
where f ¯ ( s ) is the transform of f ( t ). Since
cos ( a t ) s s 2 + a 2
then
s 2 ( s 2 + a 2 ) 2 0 t cos ( a u ) cos ( a t a u ) d u [ 2 a t cos ( a t ) sin ( a ( t 2 u ) ) 4 a ] 0 t 1 2 a ( sin ( a t ) + a t cos ( a t ) ) .
Modified form s 3 ( s 2 + 1 ) 2
By using :
cos ( a t ) a t sin ( a t ) s ( s 2 a 2 ) ( s 2 + a 2 ) 2 cos ( a t ) + a t sin ( a t ) s ( s 2 + 3 a 2 ) ( s 2 + a 2 ) 2
then it can be shown that
1 2 ( 2 cos ( a t ) a t sin ( a t ) ) s 3 ( s 2 + a 2 ) 2 .
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more