Real signals have a conjugate symmetric Fourier series.
If
ft
f
t
is real it implies that
ft=ft¯
f
t
f
t
(
ft¯
f
t
is the complex conjugate of
ft
f
t
),
then
c
n
=
c
-
n
¯
c
n
c
-
n
which implies that
ℜ
c
n
=ℜ
c
-
n
c
n
c
-
n
,
i.e. the real part of
c
n
c
n
is even, and
ℑ
c
n
=-ℑ
c
-
n
c
n
c
-
n
,
i.e. the imaginary part of
c
n
c
n
is odd. See Figure 1. It also
implies that
|
c
n
|=|
c
-
n
|
c
n
c
-
n
,
i.e. that magnitude is even, and that
∠
c
n
=∠-
c
-
n
∠
c
n
∠
c
-
n
,
i.e. the phase is odd.
c
-
n
=1T∫0Tftⅇⅈ
ω
0
ntdt=∀t,ft=ft¯:1T∫0Tft¯ⅇ-ⅈ
ω
0
ntdt¯=1T∫0Tftⅇ-ⅈ
ω
0
ntdt¯=
c
n
¯
c
-
n
1
T
t
0
T
f
t
ω
0
n
t
t
f
t
f
t
1
T
t
0
T
f
t
ω
0
n
t
1
T
t
0
T
f
t
ω
0
n
t
c
n
(1)
Real and even signals have real and even Fourier series.
If
ft=ft¯
f
t
f
t
and
ft=f-t
f
t
f
t
,
i.e. the signal is real and even,
then
c
n
=
c
-
n
c
n
c
-
n
and
c
n
=
c
n
¯
c
n
c
n
.
c
n
=1T∫-T2T2ftⅇ-ⅈ
ω
0
ntdt=1T∫-T20ftⅇ-ⅈ
ω
0
ntdt+1T∫0T2ftⅇ-ⅈ
ω
0
ntdt=1T∫0T2f-tⅇⅈ
ω
0
ntdt+1T∫0T2ftⅇ-ⅈ
ω
0
ntdt=2T∫0T2ftcos
ω
0
ntdt
c
n
1
T
t
T
2
T
2
f
t
ω
0
n
t
1
T
t
T
2
0
f
t
ω
0
n
t
1
T
t
0
T
2
f
t
ω
0
n
t
1
T
t
0
T
2
f
t
ω
0
n
t
1
T
t
0
T
2
f
t
ω
0
n
t
2
T
t
0
T
2
f
t
ω
0
n
t
(2)
ft
f
t
and
cos
ω
0
nt
ω
0
n
t
are both real which implies that
c
n
c
n
is real. Also
cos
ω
0
nt=cos-
ω
0
nt
ω
0
n
t
ω
0
n
t
so
c
n
=
c
-
n
c
n
c
-
n
.
It is also easy to show that
ft=2∑n=0∞
c
n
cos
ω
0
nt
f
t
2
n
0
c
n
ω
0
n
t
since
ft
f
t
,
c
n
c
n
,
and
cos
ω
0
nt
ω
0
n
t
are all real and even.
Real and odd signals have Fourier Series that are odd and
purely imaginary.
If
ft=-f-t
f
t
f
t
and
ft=ft¯
f
t
f
t
,
i.e. the signal is real and odd,
then
c
n
=-
c
-
n
c
n
c
-
n
and
c
n
=-
c
n
¯
c
n
c
n
,
i.e.
c
n
c
n
is odd and purely imaginary.
If
ft
f
t
is odd, then we can expand it in terms of
sin
ω
0
nt
ω
0
n
t
:
ft=∑n=1∞2
c
n
sin
ω
0
nt
f
t
n
1
2
c
n
ω
0
n
t
In summary, we can find
f
e
t
f
e
t
,
an even function, and
f
o
t
f
o
t
,
an odd function, such that
ft=
f
e
t+
f
o
t
f
t
f
e
t
f
o
t
(3)
which implies that, for any
ft
f
t
,
we can find
a
n
a
n
and
b
n
b
n
such that
ft=∑n=0∞
a
n
cos
ω
0
nt+∑n=1∞
b
n
sin
ω
0
nt
f
t
n
0
a
n
ω
0
n
t
n
1
b
n
ω
0
n
t
(4)
ft
f
t
is real and odd.
c
n
=4Aⅈπ2n2ifn=…-11-7-3159…-4Aⅈπ2n2ifn=…-9-5-13711…0ifn=…-4-2024…
c
n
4
A
2
n
2
n
…
-11
-7
-3
1
5
9
…
4
A
2
n
2
n
…
-9
-5
-1
3
7
11
…
0
n
…
-4
-2
0
2
4
…
Does
c
n
=-
c
-
n
c
n
c
-
n
?
We can often gather information about the
smoothness of a signal by examining its
Fourier coefficients.
Take a look at the above examples. The pulse and sawtooth
waves are not continuous and there Fourier series' fall off like
1n
1
n
.
The triangle wave is continuous, but not differentiable and its
Fourier series falls off like
1n2
1
n
2
.
The next 3 properties will give a better feel for this.
"My introduction to signal processing course at Rice University."