Calculate L − 1 { 1 s 2 ( s 2 + a 2 ) } using the convolution theorem L { f ∗ g } = L { f } ⋅ L { g }My approach: f ( t ) = L − 1 { 1 s 2 } = t g ( t ) = L − 1 { 1 s 2 + a 2 } = 1 a sin ⁡ ( a t ) L − 1 { 1 s 2 ( s 2 + a 2 ) } = f ∗ g = ∫ − ∞ ∞ f ( t − τ ) g ( τ ) d τ = 1 a ∫ − ∞ ∞ ( t − τ ) sin ⁡ ( a τ ) d τbut the last integral is clearly divergent.

Name: Calculate L − 1 { 1 s 2 ( s 2 + a 2 ) } using the convolution theorem L { f ∗ g } = L { f } ⋅ L { g }My approach: f ( t ) = L − 1 { 1 s 2 } = t g ( t ) = L − 1 { 1 s 2 + a 2 } = 1 a sin ⁡ ( a t ) L − 1 { 1 s 2 ( s 2 + a 2 ) } = f ∗ g = ∫ − ∞ ∞ f ( t − τ ) g ( τ ) d τ = 1 a ∫ − ∞ ∞ ( t − τ ) sin ⁡ ( a τ ) d τbut the last integral is clearly divergent.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Calculate L − 1 { 1 s 2 ( s 2 + a 2 ) } using the convolution theorem L { f ∗ g } = L { f } ⋅ L { g }My approach: f ( t ) = L − 1 { 1 s 2 } = t g ( t ) = L − 1 { 1 s 2 + a 2 } = 1 a sin ⁡ ( a t ) L − 1 { 1 s 2 ( s 2 + a 2 ) } = f ∗ g = ∫ − ∞ ∞ f ( t − τ ) g ( τ ) d τ = 1 a ∫ − ∞ ∞ ( t − τ ) sin ⁡ ( a τ ) d τbut the last integral is clearly divergent.

Question

Calculate             L              −      1            {          1                        s          2                (                  s          2                +                  a          2                )              }   using the convolution theorem       L    {  f  ∗  g  }  =      L    {  f  }  ⋅      L    {  g  }My approach:  f  (  t  )  =            L              −      1            {          1              s        2              }    =  t  g  (  t  )  =            L              −      1            {          1                        s          2                +                  a          2                      }    =      1    a    sin  ⁡  (  a  t  )            L              −      1            {          1                        s          2                (                  s          2                +                  a          2                )              }    =  f  ∗  g  =      ∫          −      ∞              ∞        f  (  t  −  τ  )  g  (  τ  )        d    τ  =      1    a        ∫          −      ∞              ∞        (  t  −  τ  )  sin  ⁡  (  a  τ  )        d    τbut the last integral is clearly divergent.

Nancy Phillips · Accepted Answer

Note that we have  f  (  t  )  =            L              −      1            {          1              s        2              }    =  t  u  (  t  )and  g  (  t  )  =            L              −      1            {          1                        s          2                +                  a          2                      }    =            sin      ⁡      (              |            a              |            t      )                      |            a              |              u  (  t  )Then, application of the convolution theorem yields                                          L                                −            1                                    {                      1                                          s                2                            (                              s                2                            +                              a                2                            )                                }                                    =        (        f        ∗        g        )        (        t        )                                                        =                  ∫                      −            ∞                    ∞                f        (        t        −        τ        )        g        (        τ        )                d        τ                                                        =                  ∫                      −            ∞                    ∞                (        t        −        τ        )        u        (        t        −        τ        )                              sin            ⁡            (                          |                        a                          |                        τ            )                                              |                        a                          |                                      u        (        τ        )                d        τ                                                             =                  ∫          0          t                (        t        −        τ        )                              sin            ⁡            (                          |                        a                          |                        τ            )                                              |                        a                          |                                              d        τ

Calculate L^(-1){(1)/(s^2(s^2+a^2))} using the convolution theorem L{f ** g}=L{f} * L{g}

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