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book   Mobk070    March 22, 2007  11:7








                     108  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                       Differentiate once
                                                        1

                                                φ =              φ (−2) = 1 = A 1
                                                     (s + 1) 2
                                                       −2

                                               φ =               φ (−2) = 2 = A 0
                                                     (s + 1) 3
                       To find C,multiplyby(s + 1) and take s =−1(in theoriginalequation).

                                                            C =−1.

                       Thus
                                                    2         1         2         1
                                           F(s ) =       +         +         −
                                                  (s + 2)  (s + 2) 2  (s + 2) 3  (s + 1)
                                                                      m
                       and noting the shifting theorem and the theorem on t ,
                                                                       2 −2t
                                                f (t) = 2e  −2t  + te −2t  + 2t e  + e  −t

                       6.5.3  Quadratic Factors: Complex Roots
                       If q(s ) has complex roots and all the coefficients are real this part of q(s ) can always be written
                       in the form
                                                                 2
                                                          (s − a) + b 2                             (6.62)
                       This is a shifted form of

                                                              2
                                                             s + b 2                                (6.63)
                       This factor in the denominator leads to sines or cosines.

                       Example 6.9. Quadratic factors
                            Find the inverse transform of

                                                  2(s − 1)         2s            1
                                         F(s ) =            =             −
                                                                                  2
                                                 2
                                                                    2
                                                s + 2s + 5    (s + 1) + 4   (s + 1) + 4
                       Because of the shifted s in the denominator the numerator of the first term must also be shifted
                       to be consistent. Thus we rewrite as
                                                         2(s + 1)         3
                                                F(s ) =            −
                                                                           2
                                                             2
                                                       (s + 1) + 4   (s + 1) + 4
                       The inverse transform of
                                                               2s
                                                              2
                                                             s + 4
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