What is the inverse laplace transform of s s + 1 ?My try: X ( s ) = s s + 1 X ( s ) = s 1 s + 1 x ( t ) = d d t e − t = − e − t

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What is the inverse laplace transform of       s          s      +      1      ?My try:  X  (  s  )  =      s          s      +      1          X  (  s  )  =  s      1          s      +      1          x  (  t  )  =      d          d      t            e          −      t        =  −      e          −      t

skilningugdK · Accepted Answer

HINT:      s          s      +      1        =  1  −      1          s      +      1      We know,       L    (      e          a      t        )  =      1          s      −      a      and we can prove       L    (  δ  (  t  −  b  )  )  =      e          −      b      s      , put b=0

RyszardsJh · Accepted Answer

Simple way to solve this problem                                   =                  s                      s            +            a                                                            =                              (            s            +            a            )            −            a                                s            +            a                                                            =                              s            +            a                                s            +            a                          −                  a                      s            +            a                                                            =        1        −                  a                      s            +            a                                                            =                  δ                (        t        )        −                  (                a                  e                      −            a            t                          )

What is the inverse laplace transform of (s)/(s+1)?

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