New Basis Functions
Recall that the Fourier series builds a representation
composed of a weighted sum of the following basis
functions.
Section 8
1 (i.e. a constant term)
cos(t) cos(2t) cos(3t)
sin(t) sin(2t) sin(3t)
Complex Fourier Series
cos(4t)
sin(4t)
...
...
Computing the weights an , bn and c often involves
some nasty integration.
The complex Fourier series is presented first with period 2π, then with general period.
The connection with the real-valued Fourier series is
explained and formulae are given for converting between the two types of representation.
Examples are given of computing the complex Fourier
series and converting between complex and real serieses.
We now present an alternative representation based
on a different set of basis functions:
1 (i.e. a constant term)
eit
e2it
e3it
e4it
e−it e−2it e−3it e−4it
...
...
These can all be represented by the term
eint
with n taking integer values from −∞ to +∞. Note
that the constant term is provided by the case when
n = 0.
111
112
Complex Fourier Series
Series of Complex Exponentials
A representation based on this family of functions is
called the “complex Fourier series”.
f (t) =
∞
X
f (t) = d +
∞
X
[an cos(nt) + bn sin(nt)]
n=1
∞ "
X
eint + e−int
an
=d+
2
n=1
cneint
n=−∞
The coefficients, cn, are normally complex numbers.
It is often easier to calculate than the sin/cos Fourier
series because integrals with exponentials in are usually easy to evaluate.
We will now derive the complex Fourier series equations, as shown above, from the sin/cos Fourier series
using the expressions for sin() and cos() in terms of
complex exponentials.
113
!
eint − e−int
+ bn
2i
!#
∞
∞
X
X
(an + ibn) −int
(an − ibn) int
e +
e
=d+
2
2
n=1
n=1
=
∞
X
cneint
n=−∞
where
,n=0
cn = (an − ibn) /2
, n = 1, 2, 3, . . .


(a−n + ib−n) /2 , n = −1, −2, −3, . . .

 d
Note that a−n and b−n are only defined when n is
negative.
114
1 π cos(nt)f (t) dt
an = π
−π
Rπ
1
bn = π −π sin(nt)f (t) dt
R
1 R π f (t) dt
d = 2π
−π
thus for n positive
1
(an − ibn)
2 Z
1 π
=
[cos(nt) − i sin(nt)] f (t) dt
2π Z−π
1 π −int
=
e
f (t) dt
2π −π
cn =
for n negative
1
cn =
(a−n + ib−n)
2 Z
1 π
=
[cos(−nt) + i sin(−nt)] f (t) dt
2π Z−π
1 π −int
e
f (t) dt
=
2π −π
Complex Fourier Series Summary
1 π −int
e
f (t) dt
cn =
2π −π
Z
f (t) =
∞
X
cneint
n=−∞
and for n = 0
c0 = d
Z
1 π −0
e f (t) dt
=
2π −π
115
116
Complex Series Example 1
Find the complex Fourier series to model f (t) = sin(t).
1 π −int
e
f (t) dt
cn =
2π −π
Z
1 π −int
e
sin(t) dt
=
2π −π
Z
1 einπ − e−inπ
=
2π
n2 − 1
"
1 einπ − e−inπ
cn =
2π
n2 − 1
"
#
1
2i
−1
c−1 =
2i
Which means the complex Fourier series for f (t) =
c1 =
f (t) =
=
∞
X
#
Set n = 1 + ǫ and let ǫ tend to zero.
Which is zero when n does not equal 1 or −1. For
these two special cases we have to set n = 1 + ǫ
and calculate the limit of cn as ǫ tends to zero. This
gives us
sin(t) is
Finding the limit as n tends to 1
cneint
n=−∞
eit − e−it


1  eiπ(1+ǫ) − e−iπ(1+ǫ) 
c1 =
2π
(1 + ǫ)2 − 1
1 −eiπǫ + e−iπǫ
=
2π (1 + ǫ)2 − 1
"
#
1 −1 − iπǫ + 1 − iπǫ
≈
2π
1 + 2ǫ − 1
1 −2iπǫ
≈
2π
2ǫ
≈
−i
2
≈
1
2i
2i
117
118
Complex Series Example 2
Find the complex Fourier
f(x)
series to model f (x) that 1
has a period of 2π and is 1
when 0 < x < T and zero
when T < x < 2π.
T
Converting c to a, b and d
From our example on the previous page.
2π
1 π −int
cn =
e
f (t) dt
2π −π
i
i h −inT
e
− 1 , when n 6= 0
=
2πn
T
1
area =
, when n = 0
=
2π
2π
Z
x
cn =

h
i
i
−inT


− 1 , when n 6= 0
 2πn e


 1 area = T
2π
2π
We wish to calculate the coefficients for the equivalent
Fourier series in terms of sin() and cos().
T . For n > 0
Clearly d = c0 = 2π
cn = (an − ibn)/2
⇒ an = 2 Re{cn}
So the Fourier series is
f (t) =
∞
X
and bn = −2 Im{cn}
cneint
converting our expression for cn into sin() and cos():
n=−∞

1 
=
T+
2π 
i
[cos(nT ) − i sin(nT ) − 1]
πn
1
=
[sin(nT ) + i(cos(nT ) − 1)]
πn
1 − cos(nT )
sin(nT )
and bn =
.
so an =
nπ
nπ
2cn =
−1
X
i
i h −inT
e
− 1 eint
n=−∞ n

∞

h
i
X
i −inT
e
− 1 eint
+

n=1 n
, when n = 0
119
120
Converting from Real to Complex
f(t)
Convert the real Fourier se1
ries of the square wave
−1
f (t) to a complex series.
−2π −π
π
Complex Fourier Series

1 
T+
f (t) =
2π 
−1
X
i
i h −inT
e
− 1 eint
n=−∞ n
+
∞
X
i h −inT
n=1 n
e
t
2π
For the real series, we know that d = an = 0 and
− 1 eint
i



Real Fourier Series
∞
X
T
sin(nT )
f (t) =
+
cos(nt)
2π
nπ
n=1
∞
X
1 − cos(nT )
+
sin(nt)
nπ
n=1
1 π
4
bn =
sin(nt)f (t) dt =
, n odd
π −π
nπ
Z
i
h
sin(5t)
sin(3t)
4
giving f (t) = π sin(t) + 3 + 5 + . . .
To convert
 to a complex series, use

,n=0
 d
cn = (an − ibn) /2
, n = 1, 2, 3, . . .


(a−n + ib−n) /2 , n = −1, −2, −3, . . .
so we have
c0 = 0
cn = −2i/(nπ) , n positive and odd
cn = 2i/(−nπ) , n negative and |n| odd
e−5it
e−3it
e−it
−2i
... +
+
+
⇒ f (t) =
π
−5
−3
−1
#
eit
e3it
e5it
+ +
+
+ ...
1
3
5
"
Both serieses converge as 1/n.
121
122
Example 1
General Complex Series
A even function f (t) is periodic with period L = 2,
and f (t) = cosh(t − 1) for 0 ≤ t ≤ 1. Find a
complex Fourier series representation for f (t).
For period of 2π
1 2π −int
e
f (t) dt
cn =
2π 0
Z
∞
X
f (t) =
f(t)
cneint
n=−∞
1
Similarly, for period L
t
1 L −inx 2π
L f (x) dx
cn =
e
L 0
−2
Z
f (x) =
∞
X
−1
0
1
2
3
4
2π
cneinx L
1 L −int 2π
L f (t) dt
e
cn =
L 0
Z
n=−∞
The fraction 2π
L is often written as ω0 and called the
fundamental angular frequency.
123
1 2 −intπ
=
e
cosh(t − 1) dt
2 0
Z
=
sinh(1)
1 + n2 π 2
124
Example 2
Hence the complex Fourier series is
∞
X
f (t) =
Find the complex Fourier
series of the the square
wave f (x).
2π
cneint L
n=−∞
∞
X
sinh(1)eintπ
=
2 2
n=−∞ 1 + n π
1
−1
−L
L
x
Note that the mean of the function is zero, so c0 = 0.
We can check this answer by computing the equivalent real Fourier series which we calculated at the
start of section 7.
an = 2 Re{cn}
, n = 1, 2, 3, . . .
bn = −2 Im{cn} , n = 1, 2, 3, . . .
d = c0
1 L −inx 2π
L f (x) dx
cn =
e
L 0
Z
1
=
L
=
"Z
L/2
dx −
e
0
h
1 − e−inπ
∞
X
n = −∞
n 6= 0
2 sinh(1)
, n = 1, 2, 3, . . .
1 + n2 π 2
−inx 2π
L
Z L
L/2
−inx 2π
L
e
dx
inπ
i
2π

2π
einx L
2π
2π
2 
e−5ix L
e−3ix L
e−ix L
f (x) =
... +
+
+
iπ
−5
−3
−1
bn = 0
d = sinh(1)
+
125
#
i
1 h −2inπ
e
+ 1 − 2e−inπ
2inπ
f (x) =
In this case, as cn is entirely real,
an = 2cn =
f(x)
ix 2π
L
e
1
+
3ix 2π
L
e
3
+
5ix 2π
L
e
5

+ . . .
126
Converting to a Real Series
Section 8: Summary
We wish to convert the complex general range square
wave series into a series with real coefficients.
cn =
(
For period L
1 L −inx 2π
L f (x) dx
e
cn =
L 0
Z
2/(inπ) , |n| odd
0
, |n| even
f (x) =
Clearly d = c0 = 0. For a and b use:
⇒ an = 2 Re{cn} = 0
Relationship with the cos/sin Fourier series.
4
and bn = −2 Im{cn} =
, n odd
nπ
Which gives us the real series:
sin 3x 2π
2π
4
L
+
sin x
f (t) =
π
L
3
+


 d
,n=0
cn =
, n = 1, 2, 3, . . .
(an − ibn) /2


(a−n + ib−n) /2 , n = −1, −2, −3, . . .
sin 5x 2π
L
2π
cneinx L
n=−∞
cn = (an − ibn)/2

∞
X
5

+ . . .
127
an = 2 Re{cn}
, n = 1, 2, 3, . . .
bn = −2 Im{cn} , n = 1, 2, 3, . . .
d = c0
128