Page 128 - Circuit Analysis II with MATLAB Applications
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Chapter 4  The Laplace Transformation


                                    ­  d   ½                    f  d     – st
                                 L ----- ft   ¾  =  sF s   –  f0       =  ³  ----- ft  e  dt
                                    ®
                                    ¯  dt  ¿                    0  dt
         Taking the limit of both sides by letting s o  f , we get

                                                                 T  d
                                lim  >  sF s   f0  –       @  =  lim  lim  ³  ----- ft  e – st dt
                               s o  f               s o  f  T o  f  H  dt
                                                           H o  0

         Interchanging the limiting process, we get

                                                           T  d
                                lim  >  sF s   f0         @  =  lim  ³  ----- ft   lim  e – st  dt
                                          –
                                s o  f              T o  f  H  dt  s o  f
                                                    H o  0
         and since
                                                  lim  e – st  =  0
                                                 s o  f
         the above expression reduces to
                                             lim  >  sF s   f0  –       @  =  0
                                            s o  f
         or
                                               lim  sF s   =  f 0
                                              s o  f
         11. Final Value Theorem

         The final value theorem states that the final value f f     of the time function f t    can be found from
         its Laplace transform multiplied by s, then, letting s o 0 . That is,

                                       lim  ft   =  lim  sF s   =  f f                          (4.33)
                                       t o  f    s o  0
         Proof:

         From the time domain differentiation property,

                                             d
                                            ----- ft   œ sF s   –  f0
                                            dt
         or

                                    ­  d   ½                    f  d     – st
                                 L ----- ft   ¾ dt  =  sF s   –  f0       =  ³  ----- ft  e  dt
                                    ®
                                                                  dt
                                    ¯      ¿                    0
         Taking the limit of both sides by letting s o  0 , we get




        4-10                                                Circuit Analysis II with MATLAB Applications

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