Fourier Series    (continued)         Click here to skip to another example.         

  

 

Example:     Find the Fourier Series representation of  f(t)     where

 

                               0          t ≤  0               Click here for a figure of this function

             f(t)  =                                                  and for help with this problem.

                               t2       0    t ≤  π

 

 

   The Fourier series expansion of f(t)  is:

                                                 

                      f(t)  =     ao /2  + ∑   an cos nt   +  bn sin nt   

                                              n = 1

 

 

Now calculate the Fourier coefficients        ao,     an,    bn

 

Note:  The period of f(t) is 2π  and  f(t) is zero from   ˗ π      t    0 .   So one needs to

only evaluate the integrals from 0    t ≤  π.

 

 

                       π                          π                                 π

   ao  = (1/ π)  ∫ f(t) dt   =  (1/ π)  ∫ t2 dt  =  (1/ π)   t3 /3 |    =   π2 /3

                      0                           0                                 0

                                                                                                           

 

                        π                                     π                                                                     

     an  = (1/ π)  ∫ f(t) cos nt dt  = (1/ π)  ∫ t2 cos nt dt     now integrate by parts:

0                                             0

 

                           u  =  (1/ π) t2               dv  =  cos nt  dt

 

                       du  =  (2t/π)  dt            v =  (1/n) sin nt

                                                    π                     π                                       π

     an  = (1/ π)  [ ( t2  /n ) sin nt ]     -  (2/ )  ) ∫ t sin nt dt   =  (-2/ )  ) ∫ t sin nt dt  

                                                    0                    0                                       0

 

                               u  =  2t/π               dv  =  -sin nt  dt

 

                            du  =  2dt/ π              v =  (1/n) cos nt

 

                                        π                   π                                                 π

      an   =   [ 2t/n2 cos nt ]    -   (2/ n2π) ∫ cos nt dt   =   [ (2t/n2π) cos nt ]

                                       0                   0                                                  0

 

        an   =  2 cos / n2

 

 

  Click here to continue with this problem and find   bn  

 

 

 




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