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Chapter 6  Fourier, Taylor, and Maclaurin Series



                                                          a =   4A
                                                                -------
                                                           n
                                                                nπ
                                   ,
                             ,,
                and for n =  3 7 11 15 , and so on, it becomes
                                                               – 4A
                                                          a =  ----------
                                                           n
                                                                nπ
                Then, the trigonometric Fourier series for the square waveform with even symmetry is

                                                                                  (  n1 )
                                                                                    –
                               4A ⎛       1          1           ⎞   4A           ----------------
                                                                                    2 1
                                           -
                                                     -
                         ft() =  ------- cos ωt --– cos 3ωt +  -- cos 5ωt –  … =  -------  ∑  – (  1 )  -- cos nωt  (6.32)
                                                                                       -
                                π ⎝       3          5           ⎠    π                n
                                                                        n =  odd
                Alternate Solution:
                Since the waveform of Example 6.3 is the same as of Example 6.1, but shifted to the right by π 2⁄
                radians, we can use the result of Example 6.1, i.e.,


                                             4A          1         1
                                                 ⎛
                                                         -
                                                                   -
                                      ft() =  ------- sin ωt +  -- sin 3ωt +  -- sin 5ωt +  … ⎞        (6.33)
                                              π ⎝        3         5           ⎠
                                                                           ⁄
                                             ⁄
                and substitute ωt  with ωt +  π 2 , that is, we let ωt =  ωτ +  π 2 . With this substitution, relation
                (6.33) becomes
                                   4A     ⎛     π⎞   1    ⎛     π⎞   1    ⎛     π⎞
                                                                     -
                                                     -
                            f τ() =  ------- sin  ωτ +  --- +  -- sin 3 ωτ +  --- +  --sin 5 ωτ +  --- +  …
                                    π     ⎝     2 ⎠  3    ⎝     2 ⎠  5    ⎝     2 ⎠
                                                                                                       (6.34)
                                   4A     ⎛     π⎞   1   ⎛      3π⎞   1   ⎛       5π⎞
                                                     -
                                                                       -
                                 =  ------- sin  ωτ +  --- +  -- sin  3ωτ +  ------ +  -- sin  5ωτ +  ------ +  …
                                    π     ⎝     2 ⎠  3   ⎝       2  ⎠  5  ⎝       2  ⎠
                                                    )
                                                  ⁄
                                                                           )
                                                                         ⁄
                and using the identities  sin (  x +  π 2 =  cos x ,  sin (  x +  3π 2 =  – cos x , and so on, we rewrite
                (6.34) as
                                             4A          1          1
                                                         -
                                                                    -
                                      f τ() =  ------- cos ωτ –  -- cos 3ωτ +  -- cos 5ωτ …            (6.35)
                                                                            –
                                             π           3          5
                and this is the same as (6.27).
                Therefore, if we compute the trigonometric Fourier series with reference to one ordinate, and
                afterwards we want to recompute the series with reference to a different ordinate, we can use the
                above procedure to save time.







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