Page 115 - Advanced engineering mathematics
P. 115

3.3 Shifting and the Heaviside Function  95


                                        Here is a rationale for Heaviside’s formula. The partial fractions expansion of p(s)/q(s) has the
                                        form
                                                               p(s)   A 1     A 2         A n
                                                                   =      +       + ··· +    .
                                                               q(s)  s − a 1  s − a 2   s − a n
                                        All we need are the numbers A 1 ,··· , A n to write
                                                                  −1
                                                                                         a n t
                                                                L [F](t) = A 1 e a 1 t  + ··· + A n e .
                                        We will find A 1 . The other A j ’s are found similarly. Notice that
                                                                 p(s)         s − a 1       s − a 1
                                                           (s − a 1 )  = A 1 + A 2  + ··· + A n  .
                                                                 q(s)         s − a 1       s − a n
                                        Because the a j ’s are assumed to be distinct, then
                                                                               p(s)
                                                                     lim(s − a 1 )  = A 1
                                                                     s→a 1     q(s)
                                        with all the other terms on the right having zero limit as s → a 1 . But in this limit, (s −
                                        a 1 )p(s)/q(s) is exactly the quotient of p(s) with the polynomial obtained by deleting s − a 1
                                        from q(s). This yields Heaviside’s formula.
                                           For those familiar with complex analysis, in Section 22.4, we will present a general formula
                                                                     tz
                                            −1
                                        for L [F] as a sum of residues of e F(z) at singularities of F(z). In that context, Heaviside’s
                                        formula is the special case that F(z) is a quotient of polynomials with simple poles at a 1 ,··· ,a n .

                               SECTION 3.3        PROBLEMS


                                                                               t
                            In each of Problems 1 through 15, find the Laplace trans-  12. e (1 − cosh(t))
                            form of the function.
                                                                                    t − 2for 0 ≤ t < 16
                                                                           13. f (t) =
                                                                                    −1    for t ≥ 16
                                 3
                             1. (t − 3t + 2)e −2t

                             2. e −3t (t − 2)                              14. f (t) =  1 − cos(2t) for 0 ≤ t < 3π
                                                                                    0         for t ≥ 3π

                                      1     for 0 ≤ t < 7
                             3. f (t) =                                    15. e −5t (t + 2t + t)
                                                                                      2
                                                                                  4
                                      cos(t) for t ≥ 7
                             4. e −4t (t − cos(t))                         In each of Problems 16 through 25, find the inverse Laplace
                                                                           transform.
                                      t     for 0 ≤ t < 3
                             5. f (t) =
                                      1 − 3t  for t ≥ 3                           1
                                                                           16.
                                                                               2
                                                                              s + 4s + 12
                                      2t − sin(t) for 0 ≤ t <π
                             6. f (t) =                                           1
                                      0        for t ≥ π                   17.
                                                                              s − 4s + 5
                                                                               2
                                −t
                                      2
                             7. e (1 − t + sin(t))                         18. e −5s /s  3
                                                                                −2s
                                      t  2     for 0 ≤ t < 2                   e
                             8. f (t) =                                    19.  2
                                      1 − t − 3t  2  for t ≥ 2                s + 9
                                                                               3
                                                                                   −4s
                                      cos(t)  for 0 ≤ t < 2π               20.    e
                             9. f (t) =                                       s + 2
                                      2 − sin(t) for t ≥ 2π                       1
                                                                           21.
                                                                               2
                                    ⎧                                         s + 6s + 7
                                    ⎪−4   for 0 ≤ t < 1
                                    ⎨                                            s − 4
                            10. f (t) = 0  for 1 ≤ t < 3                   22.
                                                                               2
                                    ⎪                                         s − 8s + 10
                                    ⎩ −t
                                      e   for t ≥ 3
                                                                                s + 2
                                 −t
                            11. te cos(3t)                                 23.
                                                                              s + 6s + 1
                                                                               2
                      Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
                      Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
                                   October 14, 2010  14:14   THM/NEIL   Page-95         27410_03_ch03_p77-120
   110   111   112   113   114   115   116   117   118   119   120