Page 240 - Numerical Analysis Using MATLAB and Excel
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Waveforms in Trigonometric Form of Fourier Series


                By inspection, the average is a non−zero value. We choose the period of the input sinusoid so that
                the output will be expressed in terms of the fundamental frequency. We also choose the limits of
                integration as  π  and +π , we observe that the waveform has even symmetry.
                              –
                Therefore, we expect only cosine terms to be present. The a n  coefficients are found from


                                                         1
                                                                       d
                                                   a =  ---  ∫  2π ft()cos nt t
                                                    n
                                                        π
                where for this example,                    0
                                           1   π               2A   π
                                                                              d
                                      a =  --- ∫ – π Asin tcos nt t =  ------- ∫ 0  sin tcos nt t      (6.55)
                                                           d
                                                                π
                                       n
                                           π
                and from tables of integrals,
                                                                              )
                                                                        (
                                                                 )
                                                                     cos
                                                                         m +
                                                           ( cos
                                                              –
                                                           mn x
                                 ∫ (  sin mx cos nx x =  ------------------------------ – -------------------------------    m ≠(  n n x  2  n )  2
                                                 )
                                          (
                                         )
                                                  d
                                                                        (
                                                                              )
                                                          (
                                                                       2m +
                                                                )
                                                              m
                                                         2n –
                Since
                                                             )
                                                )
                                             –
                                                                         y
                                           ( cos  xy =  ( cos  y –  x =  cos xcos +  sin xsiny
                we express (6.55) as
                                       2A 1⎧      ( cos  n –  1 t  ( cos  n +  1 t  π ⎫
                                                        )
                                                                    )
                                  a =  ------- -- - ⎩  --------------------------- – ----------------------------  0 ⎬  ⎭
                                          ⋅ ⎨
                                   n
                                        π
                                           2
                                                              n +
                                                                  1
                                                     1
                                                  n –
                                       A⎧     ( cos  n1 π  –  )  ( cos  n +  1 π  )  1  1  ⎫
                                     =  ---- ⎨  ----------------------------- –  ----------------------------- –  ------------ –  ------------ ⎬  (6.56)
                                       π ⎩    n1           n + 1       n –  1  n +  1  ⎭
                                                –
                                                                     )
                                       A 1 –  cos  n + )  π(  π  cos (  nππ –  1
                                                                  –
                                     =  ---- --------------------------------------- +  --------------------------------------
                                       π       n +  1          n – 1
                To simplify the last expression in (6.56), we make use of the trigonometric identities
                                                              π
                                       cos  n + )  π(  π  =  n cos –  sin nπsinπ =  – cos nπ
                                                          π cos
                and
                                                              π
                                        cos  n – )  π(  π  =  n cos +  sin nπsinπ =  – cos nπ
                                                          π cos
                Then, (6.56) simplifies to
                                                                       )
                                                                                      )
                                                                     –
                                                               2
                              A 1 +  cos nπ  1 +  cos nπ  A – +   (  n1 cos nπ –  (  n +  1 cos nπ
                         a =  ---- ------------------------- –  ------------------------- =  ---- -------------------------------------------------------------------------------------
                          n
                              π
                                                          π
                                       1
                                   n +
                                                                           2
                                               n –
                                                  1
                                                                          n –
                                                                              1
                                  (
                              – 2A cos nπ + 1 )                                                        (6.57)
                            =  -----------------------------------------   n ≠  1
                                  π n –(  2  1 )
                Now, we can evaluate all the a n  coefficients, except  , from (6.57). First, we will evaluate   to
                                                                   a
                                                                                                        a
                                                                    1
                                                                                                         0
               Numerical Analysis Using MATLAB® and Excel®, Third Edition                             6−23
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