Try to find the inverse Laplace transform of: F ( S ) = e − π s + 2 + s s 2 + 2 s + 2

Name: Try to find the inverse Laplace transform of: F ( S ) = e − π s + 2 + s s 2 + 2 s + 2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Try to find the inverse Laplace transform of: F ( S ) = e − π s + 2 + s s 2 + 2 s + 2

Question

Try to find the inverse Laplace transform of:  F  (  S  )  =                    e                  −          π          s                    +      2      +      s                      s        2            +      2      s      +      2

Hamnetmj · Accepted Answer

f  (  t  )  =            L              −      1        {  F  (  s  )  }  =            L              −      1            (                            e                      −            π            s                          +        2        +        s                              s          2                +        2        s        +        2              )                      (1)                    =                              L                                −            1                                    (                                    e                              −                π                s                                                                    s                2                            +              2              s              +              2                                )                +                              L                                −            1                                    (                                    s              +              2                                                      s                2                            +              2              s              +              2                                )                    Now            L              −      1            (                  s        +        2                              s          2                +        2        s        +        2              )    =            L              −      1            (                  s        +        1                    (        s        +        1                  )          2                +        1              )    +            L              −      1            (          1              (        s        +        1                  )          2                +        1              )    =      e          −      t           cos  ⁡  t     +         e          −      t           sin  ⁡  tand            L              −      1            (                  e                  −          π          s                                      s          2                +        2        s        +        2              )    =  H      (    t    −    π    )        e          −              (        t        −        π        )              sin  ⁡      (    t    −    π    )      (by second shifting property )where       H  (  x  )  =      {                            1                          ,          if                                                                 x          ≥          0                                      0                          ,          if                                                                 x          &amp;lt;          0                             be the Heaviside function.Now from (1) ,  f  (  t  )  =            L              −      1        {  F  (  s  )  }  =      e          −      t           cos  ⁡  t     +         e          −      t           sin  ⁡  t     +     H      (    t    −    π    )        e          −              (        t        −        π        )              sin  ⁡      (    t    −    π    )  Remarks:      L    =  {      e          a      t        }  =      1          p      −      a          ⟹              L              −      1            {          1              p        −        a              }    =      e          a      t            L    =  {  sin  ⁡  (  a  t  )  }  =      a                  p        2            +              a        2                ⟹              L              −      1            {          a                        p          2                +                  a          2                      }    =  sin  ⁡  (  a  t  )      L    =  {  cos  ⁡  (  a  t  )  }  =      p                  p        2            +              a        2                ⟹              L              −      1            {          p                        p          2                +                  a          2                      }    =  cos  ⁡  (  a  t  )Second Shifting Property: If                 L              −      1            {    F    (    p    )    }    =  f  (  t  )    , then                 L              −      1            {    F    (    p    )               e              −        a        p              }    =  g  (  t  )     where     g  (  t  )  =      {                            f          (          t          −          a          )                    if                                           t                     &amp;gt;                     a                                      0                                                                                                                                                        if                                           t                     &amp;lt;                     a                           which can also be written as            L              −      1        {      e          −      a      p           F  (  p  )  }  =  f  (  t  −  a  )     H  (  t  −  a  )where       H  (  x  )  =      {                            1                          ,          if                                                                 x          ≥          0                                      0                          ,          if                                                                 x          &amp;lt;          0                             be the Heaviside function.

Mattie Monroe · Accepted Answer

Break the numerator into three terms and use linearity:  −      1    2    i      e          (      −      1      −      i      )      (      t      −      π      )            (    −    1    +          e              2        i        (        t        −        π        )              )    θ  (  t  −  π  )  ,  −  i      e          (      −      1      −      i      )      t            (    −    1    +          e              2        i        t              )    ,and      (          1      2        +          i      2        )        e          (      −      1      −      i      )      t            (          e              2        i        t              −    i    )    .Add them.

Try to find the inverse Laplace transform of: F(S)=(e^(-pi s)+2+s)/(s^2+2s+2)

Answered question

Answer & Explanation

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