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142         LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS                  [CHAP.  3



           3.21.  Find the inverse  Laplace transform of

                                            2 + 2se-"  + 4eP4'
                                                                     Re(s) > - 1
                                    X(s) =
                                               s2 + 4s + 3
                    We  see that  X(s) is a sum



                 where

                                       2                     2 s                   4
                                                 X2(s) =               XAs) =
                           XI($)  = s2  + 4s + 3         st + 4s + 3           s2 + 4s + 3
                 If
                                 xl(f )-Xds)       x2(t) #XZ(S)      ~3(f -X3(s)
                 then by  the  linearity property (3.15) and the time-shifting property (3.16) we obtain

                                          ~(t) =xl(t) +x2(t -2) +x3(t-4)                     (3.85)
                 Next, using partial-fraction expansions and from Table  3-1, we obtain























           3.22.  Using the differentiation in  s property (3.211, find the inverse  Laplace transform of





                    We have





                 and from  Eq. (3.9) we  have
                                                       1
                                         e-"u(t)  o -  Re(s) > -a
                                                     s+a
                 Thus, using the differentiation  in  s property (3.21), we obtain
                                                  X(I) = te-"'u(t)
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