ft=∑n∈Z
c
n
ei
ω
0
nt
f
t
n
n
c
n
ω
0
n
t
(1)
c
n
=1T∫0Tfte−(i
ω
0
nt)d
t
=〈ft,ei
ω
0
nt〉
c
n
1
T
t
T
0
f
t
ω
0
n
t
f
t
ω
0
n
t
(2) is equal to "the strength of
ft
f
t
along the
ei
ω
0
nt
ω
0
n
t
direction" as well as "the similarity between
ft
f
t
and
ei
ω
0
nt
ω
0
n
t
."
Multiply and integrate (Reference)
and (Reference) (real and imaginary parts) to get
c
2
c
2
, which is a complex number.
Signal
ft
f
t
: Figure 2.
Also known as
c
n
c
n
. In Figure 3,
c
0
=1T∫0Tftd
t
c
0
1
T
t
0
T
f
t
and is the average value of
ft
f
t
and
c
3
=1T∫0Tfte−(i
ω
0
3t)d
t
c
3
1
T
t
0
T
f
t
ω
0
3
t
, where
e−(i
ω
0
3t)
ω
0
3
t
is the real part (Figure 4).
We can plot
ℜ
c
n
c
n
and
ℑ
c
n
c
n
as well.
Figure 5.
c
n
=1T∫0Tfte−(i
ω
0
nt)d
t
=1T∫0T4e−(i
ω
0
nt)d
t
+1T∫3TTTe−(i
ω
0
nt)d
t
=1−(i
ω
0
Tn)(e−(i
ω
0
T4n)−1)+1−(i
ω
0
Tn)(e−(i
ω
0
Tn)−e−(i
ω
0
T34n))=1−(i2πn)(e−(i2π4n)−1)+1−(i2πn)(e−(i2πn)−e−(i2π34n))=1−(i2πn)(e−(iπ2n)−1)+1−(i2πn)(1−e−(i3π2n))=1i2πn(e−(i3π2n)−e−(iπ2n))=1i2πn(eiπ2n−e−(iπ2n))=1πneiπ2n−e−(iπ2n)2i=1πnsinπ2n=12sinπ2nπ2n=sincπ2n2
c
n
1
T
t
0
T
f
t
ω
0
n
t
1
T
t
0
T
4
ω
0
n
t
1
T
t
3
T
T
T
ω
0
n
t
1
ω
0
T
n
ω
0
T
4
n
1
1
ω
0
T
n
ω
0
T
n
ω
0
T
3
4
n
1
2
n
2
4
n
1
1
2
n
2
n
2
3
4
n
1
2
n
2
n
1
1
2
n
1
3
2
n
1
2
n
3
2
n
2
n
1
2
n
2
n
2
n
1
n
2
n
2
n
2
1
n
2
n
1
2
2
n
2
n
sinc
2
n
2
(3)
where
e−(i3π2n)=eiπ2n
3
2
n
2
n
by the periodicity of complex exponentials. To
check
Equation 3, we see that
c
0
=12
c
0
1
2
. This is correct since it is the average value of
ft
f
t
.
Figure 6 is a plot of
c
n
c
n
.
- Only odd harmonics (average value
of
c
0
c
0
doesn't count as even).
-
sinc
sinc envelope on harmonics.
sinπ2n={0 if
n
even
1 if n∈159…-1 if n∈3711…
2
n
0
n
even
1
n
1
5
9
…
-1
n
3
7
11
…
(4)
Therefore,
c
n
={12 if n=01πn if n∈±1±5±9…−1πn if n∈±3±7±11…0 if
n
even
c
n
1
2
n
0
1
n
n
±
1
±
5
±
9
…
1
n
n
±
3
±
7
±
11
…
0
n
even
(5)
