We will begin by plugging our above function,
ft
f
t
, into
Equation 3.
Our interval of integration will now change to match the
interval specified by the function.
c
n
=1T∫-
T
1
T
1
1
ⅇ-ⅈ
ω
0
ntdt
c
n
1
T
t
T
1
T
1
1
ω
0
n
t
Notice that we must consider two cases:
n=0
n
0
and
n≠0
n
0
. For
n=0
n
0
we can tell by inspection that we will get
∀n,n=0:
c
n
=2
T
1
T
n
n
0
c
n
2
T
1
T
For
n≠0
n
0
, we will need to take a few more steps to
solve. We can begin by looking at the basic integral of
the exponential we have. Remembering our calculus, we
are ready to integrate:
c
n
=1T1ⅈ
ω
0
nⅇ-ⅈ
ω
0
nt|t=-
T
1
T
1
c
n
1
T
1
ω
0
n
t
T
1
T
1
ω
0
n
t
Let us now evaluate the exponential functions for the
given limits and expand our equation to:
c
n
=1T1-ⅈ
ω
0
nⅇ-ⅈ
ω
0
n
T
1
-ⅇⅈ
ω
0
n
T
1
c
n
1
T
1
ω
0
n
ω
0
n
T
1
ω
0
n
T
1
Now if we multiple the right side of our equation by
2ⅈ2ⅈ
2
2
and distribute our negative sign into the
parenthesis, we can utilize
Euler's Relation to
greatly simplify our expression into:
c
n
=1T2ⅈⅈ
ω
0
nsin
ω
0
n
T
1
c
n
1
T
2
ω
0
n
ω
0
n
T
1
Now, recall earlier that we defined
ω
0
=2πT
ω
0
2
T
. We can solve this equation for
T
T and substitute in.
c
n
=2ⅈ
ω
0
ⅈ
ω
0
n2πsin
ω
0
n
T
1
c
n
2
ω
0
ω
0
n
2
ω
0
n
T
1
And finally, if we make a few simple cancellations we
will arrive at our final answer for the Fourier
coefficients of
ft
f
t
:
∀n,n≠0:
c
n
=sin
ω
0
n
T
1
nπ
n
n
0
c
n
ω
0
n
T
1
n
"My introduction to signal processing course at Rice University."