EM203/MM283
Tony Holohan
S323, x5107, holohana@eeng.dcu.ie
www.eeng.dcu.ie/~holohana
Week 7 Fourier Series
Week 8 Fourier Series Applications
Week 9 Fourier Transform
Week 10 Fourier Transform Applications
Week 11 Z-transform
Week 12 Z-transform Applications
1
Fourier Series
A function f (t) is called periodic if
f (t + T ) = f (t)
for all T
The period is the smallest such number T
Such functions can generally be written as a sum . . .
f (t) = a0 +
∞
X
an cos(nωt) + bn sin(nωt)
n=1
where
ω=
2π
T
The sum of the two terms for n = 1, namely, a1 cos(ωt)+
b1 sin(ωt) is called the fundamental harmonic
and the sum of the two terms for n = 2, namely, a2 cos(2ωt)+
b2 sin(2ωt) is called the second harmonic, and so on
2
Optimality
How best approximate f (t), with f (t + 2L) = f (t), by a
constant?
Z
+L
J=
(f (t) − a0 )2 dt
−L
View f (t) as given. Seek the best possible value for a0
Z +L
dJ
2 (f (t) − a0 ) (−1) dt
=0=
da0
−L
Z +L
Z +L
dt
f (t) dt = a0
⇒
−L
−L
=
a0 t|+L
−L
= 2La0
Z +L
1
⇒ a0 =
f (t) dt
2L −L
3
Optimality
How best approximate f (t), with f (t + 2L) = f (t), by
an cos(nωt) ?
Z
+L
J=
(f (t) − an cos(nωt))2 dt
−L
View f (t) as given. Seek the best possible value for an
Z +L
dJ
2 (f (t) − an cos(nωt)) (− cos(nωt)) dt
=0=
dan
−L
Z +L
Z +L
cos2 (nωt) dt
f (t) cos(nωt) dt = an
⇒
−L
−L
Using cos2 A =
Z
+L
⇒
−L
1 + cos 2A
2
an
f (t) cos(nωt) dt =
2
Z
+L
(1 + cos(2nωt)) dt
−L
+L
an
sin(2nωt) =
+
L
4nω −L
an
=
+ 0, (2nωL = n2π)
L
Hence,
+L
1
⇒ an =
L
Z
1
⇒ bn =
L
Z
f (t) cos(nωt) dt
−L
Similarly,
4
+L
f (t) sin(nωt) dt
−L
Orthogonality
It can be shown that, when n and p are integers,
Z +L
cos(nωt) cos(pωt) dt = 0 when n 6= p
−L
Z
+L
sin(nωt) sin(pωt) dt = 0 when n 6= p
−L
Z
+L
sin(nωt) cos(pωt) dt = 0
−L
Z
+L
sin(nωt) dt = 0
−L
Z +L
cos(nωt) dt = 0
−L
In other words, when any two different functions from
1, cos(nωt), sin(nωt), n = 1, 2, . . .
are multiplied together and integrated over [−L, +L], the
result is zero.
5
Parceval’s Theorem
Consider
f (t) = a0 +
∞
X
an cos(nωt) + bn sin(nωt)
n=1
It’s convenient to let b0 = 0. Then
f (t) =
∞
X
an cos(nωt) + bn sin(nωt)
n=0
Multiply both sides by f (t) and integrate over [−L, +L].
+L
Z
Z
2
f (t) dt =
∞
+L X
−L
−L
f (t) (an cos(nωt) + bn sin(nωt)) dt
n=1
Interchange summation and integration
Z
∞ Z
X
+L
2
f (t) dt =
−L
=
f (t) (an cos(nωt) + bn sin(nωt))
−L
n=1
∞ X
+L
+L
Z
Z
f (t) cos(nωt) dt + bn
an
f (t) sin(nωt)
−L
−L
n=1
+L
giving
1
2L
6
Z
∞
+L
2
f (t) dt =
−L
a20
1X 2
+
an + b2n
2 n=1
Fourier Series - Summary
f (t) = a0 +
∞
X
an cos(nωt) + bn sin(nωt)
n=1
where
Z +L
1
f (t) dt
a0 =
2L −L
Z
1 +L
an =
f (t) cos(nωt) dt
L −L
Z
1 +L
f (t) sin(nωt) dt
bn =
L −L
Optimality
Orthogonality
Parceval’s Theorem
1
2L
Z
∞
+L
2
f (t) dt =
−L
Even and odd functions
7
a20
1X 2
+
an + b2n
2 n=1
Complex Form
Recall that
ejθ = cos θ + j sin θ
and this easily leads to
ejθ + e−jθ
cos θ =
2
ejθ − e−jθ
and sin θ =
2j
Each harmonic can be written in complex form,
an cos(nωt) + bn sin(nωt)
jnωt
ejnωt + e−jnωt
e
− e−jnωt
= an
+ bn
2
2j
a
−
jb
a
+
jb
n
n
n
n
+ e−jnωt
= ejnωt
2
2
Let
c 0 = a0 ,
cn =
an − jbn
2
c−n =
an + jbn
2
Then
an cos(nωt) + bn sin(nωt) = cn ejnωt
So a Fourier series can also be written in the form
+∞
X
f (t) =
cn ejnωt
−∞
where
1
cn =
2L
Z
+L
f (t)e−jnωt dt
−L
Then
1
2L
8
Z
+L
2
f (t) dt =
−L
+∞
X
n=−∞
|cn |2
Example 1 - The Square Wave
Compute the Fourier series of
−1 when
f (t) =
+1 when
− 21 < t < 0
0 < t < 21
f (t + 1) = f (t)
Z +0.5
1
f (t)e−jnωt dt
cn =
2L −0.5
Z 0.5
Z 0
e−jnωt dt
e−jnωt dt +
=−
−0.5
0
0
0.5
e−jnωt e−jnωt =−
+
−jnω −0.5
−jnω 0
e0
ejnω0.5
e−jnω0.5
e0
=−
+
+
−
−jnω
−jnω
−jnω
−jnω
1
=
(−2 + 2 cos(nω0.5))
−jnω
2π
Since ω =
,
1
1
cn =
(−2 + 2 cos(nπ))
−jnω
2j
(−1 + (−1)n )
nω
When n is even, cn = 0. When n is odd,
=
cn =
9
−2j
−4j
=
nω
nπ
Example 2 - The Sawtooth Wave
Compute the Fourier series of
f (t) = t for
−1≤t≤1
where f (t + 2) = f (t) ∀t
1
cn =
2L
Z
+L
f (t)e−jnωt dt
−L
1 +1 −jnωt
=
te
dt
2 −1
Use integration by parts . . .
Z
Z
u dv = uv − v du
Z
Let u = t and then du = dt
−jnωt
Let dv = e
e−jnωt
and then v =
−jnω
+1 Z +1 −jnωt
e
te−jnωt −
cn =
dt
−jnω −1
−1 −jnω
+1
te+jnω
e−jnωt e−jnω
=
+
−
−jnω −jnω
(−jnω)2 −1
+1
e−jnω
e+jnω
e−jnωt =
+
+
−jnω −jnω
(nω)2 −1
e−jnω
ejnω
e+jnω
e−jnω
=
−
+
+
−jnω −jnω (nω)2 (nω)2
10
=
2 cos(nω) 2j sin(nω)
+
−jnω
(nω)2
=
2 cos(nπ) 2j sin(nπ)
+
−jnπ
(nπ)2
But ω = π,
2(−1)n
=
−jnπ
Also, c0 = 0
11
Example 3 - Periodic Train of Pulses
The signal f (t) consists of a rectangular pulse of width
2 and of unit area, and is periodic with period T = 1
sec. Compute its Fourier series.
f (t) =
1
2
for
− ≤ t ≤ +
and f (t + 1) = f (t) ∀t
1
cn =
2L
Z
+L
Z
f (t)e−jnωt dt
−L
+0.5
f (t)e−jnωt dt
=
−0.5
1
=
2
Z
+
e−jnωt dt
−
+
1 e−jnωt =
2 −jnω −
1
=
2
e−jnω − ejnω
−jnω
=
12
sin(nω)
nω
Since ω = 2π,
cn =
sin(2nπ)
2nπ
Also, c0 = 1
In the limit as → 0, since
lim
θ→0
sin θ
θ
then
cn → 1
This is a train of impulses, with a periodicity of 1 second.
The figures below show various Fourier series finite sums
for the square wave and the impulse train.
See also the fabulous Fourier Series Approximation utility at
www.jhu.edu/~signals
13
Fourier Series - Summary
f (t) =
+∞
X
cn ejnωt
n=−∞
where
1
cn =
2L
Z
+L
f (t)e−jnωt dt
−L
Optimality
Orthogonality
Parceval’s Theorem
1
2L
Z
+L
f (t) dt =
−L
Even and odd functions
14
2
+∞
X
n=−∞
c2n
n=1:100
error
1
error
f(t)~
1
0
−1
−0.5
0
−1
0
0.5
−0.5
n=1
0
−1
0
−1
0
0.5
−0.5
0
n=3
f(t)~
f(t)
1
0
−1
15
0.5
n=1:3
1
−0.5
0.5
n=1:1
1
error
f(t)~
0
error
1
−0.5
n=1:100
0
−1
0
t
0.5
−0.5
0
t
0.5
error
n=1:3
n=5
1
f(t)~
error
1
0
−1
−0.5
0
−1
0
0.5
−0.5
n=1:5
0
−1
16
0.5
n=1:5
1
error
f(t)~
1
−0.5
0
error
0
−1
0
t
0.5
−0.5
0
t
0.5
n=0:2
6
2
4
f(t)~
f(t)~
n=0:1
3
1
0
−1
−0.5
0
0
t
−2
−0.5
0.5
n=0:3
2
5
0
t
n=0:10
0.5
0
t
n=1:30
f(t)~
10
f(t)~
4
0
−2
−0.5
0
0
t
−5
−0.5
0.5
n=1:30
30
f(t)~
50
10
0
0
−10
−0.5
17
0.5
100
20
f(t)~
2
0
t
0.5
−50
−0.5
0
t
0.5