How do you evaluate the inverse transform below using convolution ? L − 1 [ s ( s 2 + a 2 ) 2 ] I tried L − 1 [ s ( s 2 + a 2 ) 2 ] ( t ) = ∫ 0 t sin ⁡ t ⋅ cos ⁡ ( a t − a τ ) d τ = sin ⁡ t ⋅ ∫ 0 t cos ⁡ ( a t − a τ ) d τ = sin ⁡ t ⋅ [ − 1 a sin ⁡ ( a t − a τ ) ] 0 t = sin ⁡ t ⋅ [ 0 − ( − 1 a sin ⁡ ( a t ) ) ] = 1 a sin 2 ⁡ ( a t ) What have I done wrong?

Question

How do you evaluate the inverse transform below using convolution ?             L              −        1                  [                  s                  (                      s            2                    +                      a            2                                )            2                                ]  I tried                                          L                          −              1                                                [                                    s                              (                                  s                  2                                +                                  a                  2                                                  )                  2                                                              ]                (        t        )                            =                  ∫          0          t                sin        ⁡        t        ⋅        cos        ⁡        (        a        t        −        a        τ        )                d        τ        =        sin        ⁡        t        ⋅                  ∫          0          t                cos        ⁡        (        a        t        −        a        τ        )                d        τ                                          =        sin        ⁡        t        ⋅                              [            −                          1              a                        sin            ⁡            (            a            t            −            a            τ            )            ]                    0          t                =        sin        ⁡        t        ⋅                  [          0          −                      (            −                          1              a                        sin            ⁡            (            a            t            )            )                    ]                                                  =                  1          a                          sin          2                ⁡        (        a        t        )            What have I done wrong?

siriceboynu1 · Accepted Answer

Instead of            L              −      1            (                  s        2                    (                  s          2                +                  a          2                          )          2                      )    (  t  )  =      ∫    0    t    sin  ⁡  t  ⋅  cos  ⁡  (  a  t  −  a  τ  )    d  τyou should have            L              −      1            (                  s        2                    (                  s          2                +                  a          2                          )          2                      )    (  t  )  =      1    a        ∫    0    t    sin  ⁡  (  a  τ  )  ⋅  cos  ⁡  (  a  t  −  a  τ  )    d  τbecause the convolution of functions f,g is defined as  (  f  ∗  g  )  (  t  )  =      ∫                  R              g  (  τ  )  ⋅  f  (  t  −  τ  )    d  τwhere   f  (  t  )  :=  cos  ⁡  (  a  ⋅  t  )  ⋅      1          [      0      ,      ∞      )        (  t  ),   g  (  t  )  :=      1    a    ⋅  sin  ⁡  (  a  ⋅  t  )  ⋅      1          [      0      ,      ∞      )        (  t  )

Keshawn Moran · Accepted Answer

Assume someone tells you to derivative       1          (              s        2            +              a        2            )       respect to s wherein a is a constant. You certainly will reply            −      2      s              (              s        2            +              a        2                    )        2                        (              1                  (                      s            2                    +                      a            2                    )                    )        ′    =            −      2      s              (              s        2            +              a        2                    )        2            or            (                        −          1                          2          (                      s            2                    +                      a            2                    )                    )        ′    =      s          (              s        2            +              a        2                    )        2            or      1    2    ×  (  −  1      )    1              (              1                  (                      s            2                    +                      a            2                    )                    )        ′    =      s          (              s        2            +              a        2                    )        2            But surely know that      L    (      t    1    f  (  t  )  )  =  (  −  1      )    1        F    ′    (  s  )Now can you find the proper f(t) form tow last equalities when you know that      L        (          1      a        sin    ⁡    (    a    t    )    )    =      1                  s        2            +              a        2            This another approach besides to yours and first answer.

How do you evaluate the inverse transform below using convolution ? ccL^(−1)[(s)/((s^2+a^2)^2)]

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