How do I compute the following transform? s − 1 2 s 2 + s + 6 I've gotten this far: 1 2 ⋅ s − 1 ( s + 1 4 ) 2 + 47 16

Question

How do I compute the following transform?            s      −      1              2              s        2            +      s      +      6      I&#039;ve gotten this far:      1    2    ⋅            s      −      1                                (          s          +                      1            4                    )                2            +              47        16

Jeremy Mayo · Accepted Answer

You can now use:            L        s        (                  e                    −        λ        t                            e                    i        t        ω              )    =      ∫    0    ∞              e              −      s      t                  e              −      λ      t                  e              i      t      ω            d    t  =      1          s      +      λ      −      i      ω      valid as long as   s  +  λ  &amp;gt;  0, and   ω  ∈      R  . From here, taking real and imaginary parts you conclude:            L        s        (                  e                    −        λ        t              cos    ⁡          (      ω      t      )        )    =            s      +      λ              (      s      +      λ              )        2            +              ω        2              ,              L        s        (                  e                    −        λ        t              sin    ⁡          (      ω      t      )        )    =      ω          (      s      +      λ              )        2            +              ω        2            Now you the decomposition you obtained and read off the coefficients, keeping in mind that the inverse Laplace transform has the form             e              −      λ      t            (    α    cos    ⁡    (    ω    t    )    +    β    sin    ⁡    (    ω    t    )    )   for some   λ  ,  ω  ,  α and   β

Parker Pitts · Accepted Answer

F  (  s  )  =            s      −      1              2              s        2            +      s      +      6        =            s      −      1              2      (              s        2            +              s        2            +      3      )        a      x    2    +  b  x  =  a      [    (    x    +          b              2        a                    )      2        −    (          b              2        a                    )      2        ]  So:  2  (      s    2    +      s    2    +  3  )  =  2      [    (    s    +          1      4              )      2        −    (          1      4              )      2        +    3    ]    =  2      [    (    s    +          1      4              )      2        +    (          47      15        )    ]  therefore:  F  (  s  )  =  (      1    2    )            s      −      1                        +                      5            4                    −                      5            4                                      (      s      +              1        4                    )        2            +      (              47        15            )        =  (      1    2    )      [                  s        +                  1          4                            (        s        +                  1          4                          )          2                +        (                  47          15                )              −                  5        4                    (        s        +                  1          4                          )          2                +        (                  47          15                )              ]  We know:      L        [          e              −        a        t              c    o    s    (    b    t    )    ]    =            s      +      a              (      s      +      a              )        2            +              b        2                  L        [          e              −        a        t              s    i    n    (    b    t    )    ]    =      b          (      s      +      a              )        2            +              b        2            therefore:  f  (  t  )  =  (      1    2    )      e                            −          t                4                  [    c    o    s    (                  47            4        t    )    −          5              47              s    i    n    (                  47            4        t    )    ]

How do I compute the following transform? (s−1)/(2s^2+s+6)

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