Let F ( s ) = L { f ( t ) }After a Laplace transform of f(t) the expression is : F ( s ) = 2 s − 4 s 2 − 2 s + 2 which equals to F ( s ) = 2 ⋅ ( s − 1 ( s − 1 ) 2 + 1 − 1 ( s − 1 ) 2 + 1 ) let x = s − 1Therefore F ( x ) = x x 2 + 1 − 1 x 2 + 1 My question is, if I use Inverse Laplace transformation using the table do I get cos ⁡ ( t ) − sin ⁡ ( t ). Can I go back to the t domain with the x domain normally ?

Question

Let  F  (  s  )  =  L  {  f  (  t  )  }After a Laplace transform of f(t) the expression is :  F  (  s  )  =            2      s      −      4                      s        2            −      2      s      +      2      which equals to  F  (  s  )  =  2  ⋅      (                  s        −        1                    (        s        −        1                  )          2                +        1              −          1              (        s        −        1                  )          2                +        1              )  let  x  =  s  −  1Therefore  F  (  x  )  =      x                  x        2            +      1        −      1                  x        2            +      1      My question is, if I use Inverse Laplace transformation using the table do I get  cos  ⁡  (  t  )  −  sin  ⁡  (  t  ). Can I go back to the t domain with the x domain normally ?

Phoenix Morse · Accepted Answer

L              −      1        {  F  (  s  −  α  )  }  =      e          α      t                  L              −      1        {  F  (  s  )  }            L              −      1        {  F  (  s  −  1  )  }  =      e          t                  L              −      1        {  F  (  s  )  }            L              −      1            {    2          (                        s          −          1                          (          s          −          1                      )            2                    +          1                    −              1                  (          s          −          1                      )            2                    +          1                    )        }    =  2      e          t        (  cos  ⁡  t  −  sin  ⁡  t  )  .Where does the &quot;frequency shift&quot; property of the Laplace transform come from?:      L    {      e          α      t        f  (  t  )  }  =      ∫    0    ∞        e          −      s      t            e          α      t        f  (  t  )    d  t  =      ∫    0    ∞        e          −      (      s      −      α      )      t        f  (  t  )    d  t  =  F  (  s  −  α  )  .

batejavizb · Accepted Answer

F  (  s  )  =  2  ⋅      (                  s        −        1                    (        s        −        1                  )          2                +        1              −          1              (        s        −        1                  )          2                +        1              )  If you change the variable and substitute x=s−1 you have to substitute s on both sides:  F  (  x  +  1  )  =  2            (            x                  x        2            +      1        −      1                  x        2            +      1                  )                  L              −        1              {  F  (  x  +  1  )  }  =  2            L              −        1                  {                  (                    x                        x          2                +        1              −          1                        x          2                +        1                            )              }        e          −      t        f  (  t  )  =  2  (  cos  ⁡  (  t  )  −  sin  ⁡  (  t  )  )Therefore:  f  (  t  )  =  2      e          t        (  cos  ⁡  (  t  )  −  sin  ⁡  (  t  )  )So this line is not correct. Because your substitution on LHS is not correct s=x. When on RHS you substitute s=x−1.  F  (  s  )  =  2  ⋅      (                  s        −        1                    (        s        −        1                  )          2                +        1              −          1              (        s        −        1                  )          2                +        1              )                                F          (                      x                    )                ⏟                    x      =      s        =                                          x                                          x                2                            +              1                                −                      1                                          x                2                            +              1                                      ⏟                    x      =      s      −      1      It should be:                              F          (                      x            +            1                    )                ⏟                    x      =      s      −      1        =                                          x                                          x                2                            +              1                                −                      1                                          x                2                            +              1                                      ⏟                    x      =      s      −      1

If I use Inverse Laplace transformation using the table do I get cos(t)−sin(t). can I go back to the t domain with the x domain normally ?

Answered question

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