ELG3175 Introduction to
Communication Systems
Frequency-Domain
Representation:
Fourier Series and
Fourier Transform
Generalized Fourier Series
• We are given a set of functions {φn(t)}n=1,2,…,N which
are mutually orthogonal on the interval to ≤ t ≤ to + T.
t o +T
∫
to
i≠ j
 0,
φi (t )φ (t )dt = 
cn , i = j = n
*
j
• If cn = 1 for all n, we call this an orthonormal set of
functions.
Lecture 2
Generalized Fourier Series (2)
• We can find an approximation for any function x(t), on
the interval (to, to + T) by xa(t) which is a weighted sum
of the functions in the orthogonal set of functions :
N
x a (t ) =
∑X
n φ n (t )
n =1
• We wish to find the approximation that minimises the
mean square error between x(t) and xa(t):
t o +T
εN =
∫
2
x(t ) − x a (t ) dt
to
Lecture 2
Generalized Fourier Series (3)
• We can show that
1
Xn =
cn
t o +T
∫ x(t )φ
*
n
(t )dt
(1)
to
• And the best approximation is
N
x a (t ) =
∑X
n φ n (t )
n =1
• Where Xn is given by (1).
Lecture 2
The complex exponential function
e j 2πnf ot = cos(2πnf o t ) + j sin( 2πnf o t )
n is an integer
This is a periodic fiunction with period Tp.
e
j 2πnf o t
=e
j 2πnf o (t +T p )
=e
j 2πnf o t
e
j 2πnf oT p
The second exponential equals 1 when nfoTp is an integer.
Lecture 2
Orthogonality and cn
On the interval to ≤ t ≤ to+T, for fo = 1/T.
t o +T
t o +T
∫
∫
φ n (t )φ m* (t )dt =
t o +T
t o +T
∫e
to
t o +T
=
∫ dt = T
to
∫
e j 2π ( n − m ) f ot dt
to
For m≠n
For m=n
cn =
e j 2πnf ot e − j 2πnf ot dt =
to
to
j 2π ( n − m ) f o t
t o +T
∫e
dt
to
m=n
j 2π ( n − m ) f o t
e j 2π ( n − m ) f o (to +T ) − e j 2π ( n − m ) f oto
dt =
j 2π (n − m) f o
e j 2π ( n − m ) f oto e j 2π ( n − m ) f oT − e j 2π ( n − m ) f oto
=
j 2π (n − m) f o
=0
Lecture 2
Complex exponential Fourier Series
On the interval to ≤ t ≤ to + T the complex exponential Fourier
Series of x(t) is given by :
∞
x(t ) =
j 2πnf o t
X
e
=
∑ n
n = −∞
where
1
Xn =
T
∞
∑ X ne
j
2πnt
T
n = −∞
t o +T
∫
x(t )e − j 2πnf ot dt
to
Lecture 2
Complex Exponential Fourier
Series for periodic signals
∞
• Consider the signal
t ≤ ∞.
∑
X n e j 2πnf ot
on the interval -∞ ≤
n = −∞
• We know that complex exponential functions are
periodic.
• The fundamental period of a complex exponential is
T/|n|.
• The period of a sum of periodic signals is the lowest
common multiple (LCM) of the individual periods
• In this case, the fundamental period is T.
Lecture 2
Complex Exponential Fourier
Series for periodic signals (2)
∞
•
∑
X n e j 2πnf ot
is periodic with period T = 1/fo.
n = −∞
• The fundamental frequency fo is the inverse of the period.
• Therefore if x(t) is periodic with period T, then
∞
∑X
ne
j 2πnf o t
=x(t) pour -∞ < t < ∞
n = −∞
• Therefore if x(t) is periodic with period T then its Fourier
series is an exact representation . In other words,
∞
x(t) = ∑ X n e j 2πnf ot
n = −∞
Lecture 2
Complex Exponential Fourier
Series for periodic signals (3)
• We can use any period to calculate the Fourier
coefficients.
– Therefore we write
1
Xn =
T
∫
x(t )e − j 2πnf o t dt
T
Lecture 2
Example
x(t)
…
A
…
0.25
0.5
0.75
t
-A
Trouvez la série de Fourier exponentielle complexe du
signal périodique x(t)
Find the complex exponential Fourier Series of the
periodic signal x(t).
Lecture 2
Solution
• We need to find
– The period of x(t) as well as fo.
– The coefficients Xn
– Then the Fourier series itself
∞
∑ X n e j 2πnf t
o
n = −∞
Lecture 2
Solution (2)
• In our example, the period is 0.5, therefore fo = 2.
• The set of functions is therefore ej4πnt.
• Therefore 0.5
∫
X n = 2 x(t )e − j 4πnt dt
0
 0.25
= 2


∫ Ae
0
0.5
− j 4πnt
dt −
∫ Ae
0.25
0.25
− j 4πnt
− j 4πnt

dt 


0.5 

1
1

= 2 A −
e
+
e − j 4πnt
 j 4πn

j 4πn
0
0
.
25




1
1
1
1
e − jπn +
e − j 2πn −
e − jπnt 
= 2 A −
+
j 4πn j 4πn
j 4πn
 j 4πn

 1

1
A
= 2 A
−
e − jπn  =
1 − (− 1)n
 j 2πn j 2πn
 jπn
[
]
Lecture 2
Solution continued
• Forn = 0, we have X0 = 0/0.
0.5
∫
X 0 = 2 x(t )dt = 0
0
∞
x(t ) =
∑
n = −∞
n is odd
∞
2 A j 4πnt
2A
e
=∑
e j 4π ( 2i −1)t
jπn
i = −∞ jπ ( 2i − 1)
Lecture 2
Hermetian symmetry for real
valued signals
• Let us assume that x(t) is a real-valued signal.
• In other words Im{x(t)} = 0.
• Then Xn* is given by :
*
1

− j 2πnf o t
*
Xn = 
x(t )e
dt 
 T T

1
− j 2πnf o t *
=
x(t )e
dt
T
T
1
=
x * (t )e j 2πnf ot dt
T
T
1
=
x(t )e − j 2π ( − n) f o t dt = X − n
T
∫
∫(
)
∫
∫
T
Lecture 2
Trigonometric Fourier Series
• Assuming x(t) is real, the real part of Xn is :
 1

− j 2πnf o t 
Re{ X n } = Re 
x(t )e
dt 
 T T

 1

= Re 
x(t )(cos 2πnf o t − j sin 2πnf o t )dt 
 T T

 1

1
= Re 
x(t ) cos 2πnf o tdt − j
x(t ) sin 2πnf o tdt 
T
 T T

T
1
=
x(t ) cos 2πnf o tdt
T
∫
∫
∫
∫
∫
T
Lecture 2
Trigonometric Fourier Series (2)
• Therefore the imaginary part of Xn is:
Im{X n } = −
1
T
∫ x(t ) sin 2πnf tdt
o
T
• The complex exponential Fourier series can be written
as :
∞
x(t ) =
∑
X n e j 2πnf o t
n = −∞
= X0 +
∞
∑ (X
− j 2πnf o t
j 2πnf o t
e
+
X
e
n
−n
)
n =1
Lecture 2
Trigonometric Fourier Series (3)
• Since x(t) is real, X-n = Xn*,
∞
x(t ) = X 0 +
∑ {(Re{ X
n =1
j Im{ X n })(cos(2πnf o t ) + j sin(2πnf o t ) )
+ (Re{ X n } − j Im{ X n })(cos(2πnf o t ) − j sin(2πnf o t ) )}
∞
= X0 +
n}+
∑ (2 Re{X
n } cos( 2πnf o t ) − 2 Im{ X n } sin( 2πnf o t )
n =1
∞
= a0 +
∑a
n =1
)
∞
n
cos(2πnf o t ) +
∑b
n =1
n
sin(2πnf o t )
a0 = X 0 =
1
T
∫ x(t )dt
T
2
x(t ) cos 2πnf o tdt
T
T
2
bn = −2 Im{ X n } =
x(t ) sin 2πnf o tdt
T
a n = 2 Re{ X n } =
∫
∫
T
Lecture 2
Example
x(t)
A
…
…
0.25
0.5
0.75
t
-A
∞
x(t ) =
∑
n = −∞
n impaire
∞
2 A j 4πnt
2 A  j 4πnt

e
= ∑ − j
e
jπn
πn 

n = −∞
n impaire
X0 = 0, Re{Xn} = 0 and Im{Xn} = -2A/πn for odd values of n.
Therefore bn = 4A/πn for odd values of n.
Lecture 2
Example continued
∞
x(t ) =
∑
n =1
n impaire
N
Let x N (t ) =
∑
i =1
4A
sin 4πnt =
πn
∞
∑
i =1
4A
sin 4π (2i − 1)t
π (2i − 1)
4A
sin 4π (2i − 1)t
π (2i − 1)
represent the first N
terms of x(t).
Lecture 2
Lecture 2
Fourier Transform
• The Fourier transform is a frequency-dependent
function that is an extension of the Fourier series to non
periodic functions.
• It describes the spectral content of a signal
– In other words it is the frequency domain
representation of a signal.
Lecture 2
Fourier Transform (2)
• La fonction X(f) est la transformée de Fourier de x(t).
• X(f) décrit le contenu spectral de x(t).
∞
• X(f) = F{x(t)} =
∫
x(t )e − j 2πft dt
−∞
∞
• x(t) = F-1{X(f)} =
∫
X ( f )e j 2πft df
−∞
Lecture 2
Example
• Find the Fourier Transform of x(t) = Π(t).
• Solution
• The Fourier Transform of x(t) is :
∞
1/ 2
−∞
−1 / 2
X ( f ) = ∫ Π (t )e − j 2πft dt =
1 − j 2πft
=−
e
j 2πf
− j 2πft
e
dt
∫
1/ 2
−1 / 2
1
=−
(e − jπf − e jπf )
j 2πf
e jπf − e − jπf sin πf
=
=
= sinc( f )
j 2πf
πf
Lecture 2
Example 2
• Trouvez la transformée de Fourier de x(t) = δ(t).
• Solution
∞
X(f ) =
∫
δ (t )e − j 2πft dt = e − j 2πft
t =0
=1
−∞
Lecture 2
Fourier Transform Properties
1. Linearity
• The Fourier Transform is a linear function. In other
words, if X1(f) =F{x1(t)} et X2(f) = F{x2(t)}, then for x3(t)
= ax1(t) + bx2(t), X3(f) = F{x3(t)}=aX1(f) + bX2(f).
2. Time Delay
• If the Fourier Transform of x1(t) is X1(f) then the Fourier
transform of x2(t) = x1(t-to) is X ( f ) = X ( f )e − j 2πfto
2
1
3. Time Scaling
• If F{x(t)} = X(f), then F{x(at)} = (1/|a|)X(f/a).
4. Duality
• If F{x(t)} = X(f), then F{X(t)} = x(-f).
Lecture 2
Properties (2)
5. Frequency shift
• If X(f) = F-{x(t)}, then X(f-fo) = F{x(t) e j 2πf o t }
6. Convolution
• If z(t) = x(t)*y(t), then Z(f) = X(f)Y(f).
7. Multiplication
• If z(t) = x(t)y(t), then Z(f) = X(f)*Y(f).
8. Derivative
• F{ dx (t ) } = j2πfX(f)
dt
9. Integration
•
 t

F  x(λ )dλ  =
−∞

∫
1
j 2πf
X ( f ) + 12 X (0)δ ( f )
Lecture 2
Transform Properties (3)
10.Complex conjugate
• F{x*(t)} = X*(-f)
Lecture 2