Fourier equation for given part of cos function between -pi and pi

Discussion in 'Math' started by Michael Zerafa, Sep 6, 2017.

  1. Michael Zerafa

    Thread Starter New Member

    Mar 22, 2007
    I was following a post by SM in another section related to Networking but had no feedback. I am trying my luck in this section as the title is more related to Maths than electronics at the stage.

    Question and a0


    an, bn, f(x) and plot
    an.jpg bn.jpg f(x).jpg plots.JPG

    Following the general equation as quoted from various reputable sources, such as John Bird, John Bird Elec Cct theory and tech 2nd edition.JPG although the cos(x)/2 term makes perfect sense, I cannot understand its origin. I have proved it graphically to be the missing term, but mathematically I cannot.

    Please follow my reasoning as I am confident in my working but I am just as convinced I am missing something fundamental.

    Thanks for your help so far

    ps. the graph x axis is plotted as a function of pi, i.e. 1 is actually pi, 0.5 is pi/2
  2. MrAl

    Distinguished Member

    Jun 17, 2014
    Hi Mike,

    I had answered this question here:

    and the simple answer is that when we calculate the coefficient for a1 we have to use the concept of limits as we can not calculate it directly because we get zero in the denominator.

    If you are also asking or instead asking about a0, then we can do the same thing using 'an' and then we might again have to resort to limits in order to calculate a0. So a0 can come from the usual expression for a0 or from the 'an' by either setting n=0 if that works, or taking the limit as n goes to zero. The calculation is similar to doing a1 and is shown in that Homework section where you had asked about a1 previously.
    Saviour Muscat likes this.
  3. Michael Zerafa

    Thread Starter New Member

    Mar 22, 2007

    I have worked it out with this method and using the tabular method derived from the trapezium rule. Both answers co-incided quite well, as did they do graphically.

    So bottom line, thanks a million as I have learnt something new from this problem and will defiantly remember it when in need.