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.
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
k∈Z
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
(n−k)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
(n−k)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
n≠k
n
k
. For
n=k
n
k
we will have:
∀
n
,n=k:∫0Tei
ω
0
(n−k)td
t
=T
n
n
k
t
T
0
ω
0
n
k
t
T
(5)
For
n≠k
n
k
, we will have:
∀
n
,n≠k:∫0Tei
ω
0
(n−k)td
t
=∫0Tcos
ω
0
(n−k)td
t
+i∫0Tsin
ω
0
(n−k)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
(n−k)t
ω
0
n
k
t
has an integer number of periods,
n−k
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
(n−k)t
ω
0
n
k
t
as well. What this means is
∫0Tcos
ω
0
(n−k)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
(n−k)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
=1T∫0Tfte−(i
ω
0
nt)d
t
c
n
1
T
t
T
0
f
t
ω
0
n
t
(10)
To find the Fourier coefficients of periodic
ft
f
t
:
-
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).
-
Repeat step (1) for all
k∈Z
k
.
