Finding the coefficients of the Fourier series expansion involves some algebraic manipulation of the synthesis formula.
First of all we will multiply both sides of the equation by
e−(j
ω
0
kt)
j
ω
0
k
t
, where
k∈Z
k
.
fte−(j
ω
0
kt)=∑
n
=−∞∞
c
n
ej
ω
0
nte−(j
ω
0
kt)
f
t
j
ω
0
k
t
n
c
n
j
ω
0
n
t
j
ω
0
k
t
(3)
Now integrate both sides over a given period,
TT:
∫0Tfte−(j
ω
0
kt)d
t
=∫0T∑
n
=−∞∞
c
n
ej
ω
0
nte−(j
ω
0
kt)d
t
t
T
0
f
t
j
ω
0
k
t
t
T
0
n
c
n
j
ω
0
n
t
j
ω
0
k
t
(4)
On the right-hand side we can switch the summation and
integral and factor the constant out of the
integral.
∫0Tfte−(j
ω
0
kt)d
t
=∑
n
=−∞∞
c
n
∫0Tej
ω
0
(n−k)td
t
t
T
0
f
t
j
ω
0
k
t
n
c
n
t
T
0
j
ω
0
n
k
t
(5)
Now that we have made this seemingly more complicated, let us
focus on just the integral,
∫0Tej
ω
0
(n−k)td
t
t
T
0
j
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:
∫0Tej
ω
0
(n−k)td
t
=T ,
n=k
n
n
k
t
T
0
j
ω
0
n
k
t
T
(6)
For
n≠k
n
k
, we will have:
∫0Tej
ω
0
(n−k)td
t
=∫0Tcos
ω
0
(n−k)td
t
+j∫0Tsin
ω
0
(n−k)td
t
,
n≠k
n
n
k
t
T
0
j
ω
0
n
k
t
t
T
0
ω
0
n
k
t
j
t
T
0
ω
0
n
k
t
(7)
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
(8)
which also holds for the integral involving the sine function.
Therefore, we conclude the following about our integral of
interest:
∫0Tej
ω
0
(n−k)td
t
={T if n=k0 otherwise
t
T
0
j
ω
0
n
k
t
T
n
k
0
(9)
Now let us return our attention to our complicated equation,
Equation 5, 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 5 will have a nonzero
result is when
kk and
nn are equal:
∫0Tfte−(j
ω
0
nt)d
t
=T
c
n
,
n=k
n
n
k
t
T
0
f
t
j
ω
0
n
t
T
c
n
(10)
Finally, we have our general equation for the Fourier
coefficients:
c
n
=1T∫0Tfte−(j
ω
0
nt)d
t
c
n
1
T
t
T
0
f
t
j
ω
0
n
t
(11)
Consider the square wave function given by
x
(
t
)
=
1
/
2
t
≤
1
/
2
-
1
/
2
t
>
1
/
2
x
(
t
)
=
1
/
2
t
≤
1
/
2
-
1
/
2
t
>
1
/
2
(12)on the unit interval t∈Z[0,1)t∈Z[0,1).
c
k
=
∫
0
1
x
(
t
)
e
-
j
2
π
k
t
d
t
=
∫
0
1
/
2
1
2
e
-
j
2
π
k
t
d
t
-
∫
1
/
2
1
1
2
e
-
j
2
π
k
t
d
t
=
j
-
1
+
e
j
π
k
2
π
k
c
k
=
∫
0
1
x
(
t
)
e
-
j
2
π
k
t
d
t
=
∫
0
1
/
2
1
2
e
-
j
2
π
k
t
d
t
-
∫
1
/
2
1
1
2
e
-
j
2
π
k
t
d
t
=
j
-
1
+
e
j
π
k
2
π
k
(13)Thus, the Fourier coefficients of this function found using the Fourier series analysis formula are
c
k
=
-
j
/
π
k
k
odd
0
k
even
.
c
k
=
-
j
/
π
k
k
odd
0
k
even
.
(14)
"My introduction to signal processing course at Rice University."