The Fourier Transform
As we have seen, any (sufciently smooth) Function
f
(
t
) that is periodic can be built out oF sin’s and
cos’s. We have also seen that complex exponentials may be used in place oF sin’s and cos’s. We shall now
use complex exponentials because they lead to less writing and simpler computations, but yet can easily be
converted into sin’s and cos’s. IF
f
(
t
) has period 2
ℓ
, its (complex) ±ourier series expansion is
f
(
t
) =
∞
s
k
=
−∞
c
k
e
ik
π
ℓ
t
with
c
k
=
1
2
ℓ
i
ℓ
−
ℓ
f
(
t
)
e
−
ik
π
ℓ
t
dt
(1)
Not surprisingly, each term
c
k
e
ik
π
ℓ
t
in this expansion also has period 2
ℓ
, because
c
k
e
ik
π
ℓ
(
t
+2
ℓ
)
=
c
k
e
ik
π
ℓ
t
e
i
2
kπ
=
c
k
e
ik
π
ℓ
t
. We now develop an expansion For nonperiodic Functions, by allowing complex
exponentials (or equivalently sin’s and cos’s) oF all possible periods, not just 2
ℓ
, For some ²xed
ℓ
. So, From
now on, do not assume that
f
(
t
) is periodic.
±or simplicity we’ll only develop the expansions For Functions that are zero For all sufciently large

t

.
With a little more work, one can show that our conclusions apply to a much broader class oF Functions. Let
L >
0 be sufciently large that
f
(
t
) = 0 For all

t
 ≥
L
. We can get a ±ourier series expansion For the part
oF
f
(
t
) with
−
L < t < L
by using the periodic extension trick. De²ne
F
L
(
t
) to be the unique Function
determined by the requirements that
i
)
F
L
(
t
) =
f
(
t
) For
−
L < t
≤
L
ii
)
F
L
(
t
) is periodic oF period 2
L
Then, For
−
L < t < L
,
f
(
t
) =
F
L
(
t
) =
∞
s
k
=
−∞
c
k
(
L
)
e
ik
π
L
t
where
c
k
(
L
) =
1
2
L
i
L
−
L
f
(
t
)
e
−
ik
π
L
t
dt
(2)
IF we can somehow take the limit
L
→ ∞
, we will get a representation oF
f
that is is valid For all
t
’s , not
just those in some ²nite interval
−
L < t < L
. This is exactly what we shall do, by the simple expedient
oF interpreting the sum in (2) as a Riemann sum approximation to a certain integral. ±or each integer
k
,
de²ne the
k
th
Frequency to be
ω
k
=
k
π
L
and denote by Δ
ω
=
π
L
the spacing,
ω
k
+1
−
ω
k
, between any two
successive Frequencies. Also de²ne
ˆ
f
(
ω
) =
I
∞
−∞
f
(
t
)
e
−
iωt
dt
. Since
f
(
t
) = 0 For all

t
 ≥
L
c
k
(
L
) =
1
2
L
i
L
−
L
f
(
t
)
e
−
ik
π
L
t
dt
=
1
2
L
i
∞
−∞
f
(
t
)
e
−
i
(
k
π
L
)
t
dt
=
1
2
L
i
∞
−∞
f
(
t
)
e
−
iω
k
t
dt
=
1
2
L
ˆ
f
(
ω
k
) =
1
2
π
ˆ
f
(
ω
k
) Δ
ω
In this notation,
f
(
t
) =
F
L
(
t
) =
1
2
π
∞
s
k
=
−∞
ˆ
f
(
ω
k
)
e
iω
k
t
Δ
ω
(3)
For any
−
L < t < L
. As we let
L
→ ∞
, the restriction
−
L < t < L
disappears, and the right hand
side converges exactly to the integral
1
2
π
I
∞
−∞
ˆ
f
(
ω
)
e
iωt
dω
. To see this, cut the domain oF integration into
ω
y
y
=
1
2
π
ˆ
f
(
ω
)
e
iωt
0
ω
1
ω
2
ω
3
Δ
ω
February 25, 2007
The Fourier Transform
1