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# Derivation of Fourier Coefficients Equation

Module by: Michael Haag. E-mail the author

Summary: The module goes through the steps of deriving the Fourier coefficient equation from the general Fourier series equation.

## Introduction

You should already be familiar with the existence of the general Fourier Series equation, which is written as:

ft= n = c n ei ω 0 nt f t n c n ω 0 n t
(1)
What we are interested in here is how to determine the Fourier coefficients, c n c n , given a function ft f t . Below we will walk through the steps of deriving the general equation for the Fourier coefficients of a given function.

## Derivation

To solve Equation 1 for c n c n , we have to do a little algebraic manipulation. First of all we will multiply both sides of Equation 1 by e(i ω 0 kt) ω 0 k t , where kZ k .

fte(i ω 0 kt)= n = c n ei ω 0 nte(i ω 0 kt) f t ω 0 k t n c n ω 0 n t ω 0 k t
(2)
Now integrate both sides over a given period, TT:
0Tfte(i ω 0 kt)d t =0T n = c n ei ω 0 nte(i ω 0 kt)d t t T 0 f t ω 0 k t t T 0 n c n ω 0 n t ω 0 k t
(3)
On the right-hand side we can switch the summation and integral along with pulling out the constant out of the integral.
0Tfte(i ω 0 kt)d t = n = c n 0Tei ω 0 (nk)td t t T 0 f t ω 0 k t n c n t T 0 ω 0 n k t
(4)
Now that we have made this seemingly more complicated, let us focus on just the integral, 0Tei ω 0 (nk)td t t T 0 ω 0 n k t , on the right-hand side of the above equation. For this integral we will need to consider two cases: n=k n k and nk n k . For n=k n k we will have:
n ,n=k:0Tei ω 0 (nk)td t =T n n k t T 0 ω 0 n k t T
(5)
For nk n k , we will have:
n ,nk:0Tei ω 0 (nk)td t =0Tcos ω 0 (nk)td t +i0Tsin ω 0 (nk)td t n n k t T 0 ω 0 n k t t T 0 ω 0 n k t t T 0 ω 0 n k t
(6)
But cos ω 0 (nk)t ω 0 n k t has an integer number of periods, nk n k , between 00 and TT. Imagine a graph of the cosine; because it has an integer number of periods, there are equal areas above and below the x-axis of the graph. This statement holds true for sin ω 0 (nk)t ω 0 n k t as well. What this means is
0Tcos ω 0 (nk)td t =0 t T 0 ω 0 n k t 0
(7)
as well as the integral involving the sine function. Therefore, we conclude the following about our integral of interest:
0Tei ω 0 (nk)td t ={T  if  n=k0  otherwise   t T 0 ω 0 n k t T n k 0
(8)
Now let us return our attention to our complicated equation, Equation 4, to see if we can finish finding an equation for our Fourier coefficients. Using the facts that we have just proven above, we can see that the only time Equation 4 will have a nonzero result is when kk and nn are equal:
n ,n=k:0Tfte(i ω 0 nt)d t =T c n n n k t T 0 f t ω 0 n t T c n
(9)
Finally, we have our general equation for the Fourier coefficients:
c n =1T0Tfte(i ω 0 nt)d t c n 1 T t T 0 f t ω 0 n t
(10)

### Finding Fourier Coefficients Steps

To find the Fourier coefficients of periodic ft f t :

1. For a given kk, multiply ft f t by e(i ω 0 kt) ω 0 k t , and take the area under the curve (dividing by TT).
2. Repeat step (1) for all kZ k .

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