The Fourier Series
Representing a Signal
• The convolution method for finding the
response of a system to an excitation takes
advantage of the linearity and timeinvariance of the system and represents the
excitation as a linear combination of impulses
and the response as a linear combination of
impulse responses
• The Fourier series represents a signal as a
linear combination of complex sinusoids
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2
Linearity and Superposition
If an excitation can be expressed as a sum of complex sinusoids
the response can be expressed as the sum of responses to
complex sinusoids.
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3
Real and Complex Sinusoids
e jx + e − jx
cos( x ) =
2
e jx − e − jx
sin( x ) =
j2
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4
Jean Baptiste Joseph Fourier
3/21/1768 - 5/16/1830
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ContinuousTime
Fourier
Series
Concept
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ContinuousTime
Fourier
Series
Concept
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ContinuousTime
Fourier
Series
Concept
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8
CT Fourier Series Definition
The Fourier series representation, x F (t ), of a signal, x(t),
over a time, t 0 < t < t 0 + TF , is
x F (t ) =
∞
∑
k =−∞
X[ k ]e j 2π ( kf F ) t
where X[k] is the harmonic function, k is the harmonic
number and fF = 1 / TF (pp. 212-215). The harmonic function
can be found from the signal as
1
X[ k ] =
TF
t 0 + TF
∫
x(t )e − j 2π ( kf F ) t dt
t0
The signal and its harmonic function form a Fourier series
FS
pair indicated by the notation, x(t ) ←
→ X[ k ] .
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CTFS of a Real Function
It can be shown (pp. 216-217) that the continuous-time
Fourier series (CTFS) harmonic function of any real-valued
function, x(t), has the property that
X[ k ] = X* [− k ]
One implication of this fact is that, for real-valued functions,
the magnitude of the harmonic function is an even function
and the phase is an odd function.
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The Trigonometric CTFS
The fact that, for a real-valued function, x(t),
X[ k ] = X* [− k ]
also leads to the definition of an alternate form of the CTFS,
the so-called trigonometric form.
∞
{
}
x F ( t ) = X c [ 0 ] + ∑ X c [ k ] cos ( 2π ( kfF ) t ) + X s [ k ] sin ( 2π ( kfF ) t )
k =1
where
2
Xc [ k ] =
TF
2
Xs [k ] =
TF
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t 0 + TF
∫ x(t ) cos(2π ( kf )t )dt
F
t0
t 0 + TF
∫ x(t )sin(2π ( kf )t )dt
F
t0
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11
The Trigonometric CTFS
Since both the complex and trigonometric forms of the
CTFS represent a signal, there must be relationships
between the harmonic functions. Those relationships are
Xc [0 ] = X[0 ]




X
=
0
0
[
]
s


 , k = 1, 2, 3,K

*
X
X
X
k
=
k
+
k
[ ]
[ ] 
 c[ ]
 X s [ k ] = j ( X[ k ] − X* [ k ])




X[0 ] = Xc [0 ]




Xc [ k ] − j X s [ k ]
X[ k ] =
 , k = 1, 2, 3,K

2


X c [ k ] + j X s [ k ]

*

 X[− k ] = X [ k ] =
2
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12
Periodicity of the CTFS
It can be shown (pg. 218) that the CTFS representation, x F (t )
of a function, x(t), is periodic with fundamental period, TF .
Therefore, if x(t) is also periodic with fundamental period,
T0 and if TF is an integer multiple of T0 then the two functions
are equal for all t, not just in the interval, t 0 < t < t 0 + TF .
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13
CTFS Example #1
Let a signal be defined by x(t ) = 2 cos( 400πt ) and let
TF = 5 ms which is the same as T0
t 0 + TF
t 0 + TF
2
2
X
k
=
x ( t ) sin ( 2π ( kfF ) t ) dt
2
Xc [ k ] =
x
t
cos
π
kf
t
dt
( ) ( ( F) )
s[ ]
∫
∫
TF t 0
TF t0
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CTFS Example #1
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CTFS Example #2
Let a signal be defined by x(t ) = 2 cos( 400πt ) and let
TF = 10 ms which is 2T0
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CTFS Example #2
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CTFS Example #3
Let x(t ) =
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1 3
1
− cos( 20πt ) + sin( 30πt ) and let TF = 200 ms
2 4
2
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CTFS Example #3
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CTFS Example #3
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CTFS Example #3
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Linearity of the CTFS
These relations hold only if the harmonic functions, X, of all
the component functions, x, are based on the same
representation time.
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CTFS Example #4
Let the signal be a 50% duty-cycle square wave with an
amplitude of one and a fundamental period , T0 = 1
x(t ) = rect( 2t ) ∗ comb(t )
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CTFS Example #4
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CTFS Example #4
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CTFS Example #4
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CTFS Example #4
A graph of the magnitude and phase of the harmonic function
as a function of harmonic number is a good way of illustrating it.
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CTFS Example #5
Let x(t ) = 2 cos( 400πt ) and let TF = 7.5 ms which is
1.5 periods of this signal.
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CTFS Example #5
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CTFS Example #5
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CTFS Example #5
The CTFS representation of this cosine is the signal
below, which is an odd function, and the discontinuities
make the representation have significant higher harmonic
content. This is a very inelegant representation.
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31
CTFS of Even and Odd Functions
• For an even function
– The complex CTFS harmonic function, X[ k ] , is
purely real
– The sine harmonic function, X s [ k ], is zero
• For an odd function
– The complex CTFS harmonic function, X[ k ] , is
purely imaginary
– The cosine harmonic function, Xc [ k ] , is zero
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CTFS Example #6
This signal has no known functional description but it
can still be represented by a CTFS.
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CTFS Example #6
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CTFS Example #6
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CTFS Example #6
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36
CTFS Properties
Let a signal, x(t), have a fundamental period, T0 x and let a
signal, y(t), have a fundamental period, T0 y . Let the CTFS
harmonic functions, each using the fundamental period as
the representation time, TF , be X[k] and Y[k]. In the
properties which follow the two fundamental periods are the
same unless otherwise stated.
Linearity
FS
α x(t ) + β y(t ) ←
→ α X[ k ] + β Y[ k ]
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CTFS Properties
Time Shifting
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FS
x(t − t 0 ) ←
→ e − j 2π ( kf 0 ) t 0 X[ k ]
FS
x(t − t 0 ) ←
→ e − j ( kω 0 ) t 0 X[ k ]
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38
CTFS Properties
Frequency Shifting
(Harmonic Number
Shifting)
FS
e j 2π ( k 0 f 0 ) t x(t ) ←
→ X[ k − k0 ]
FS
e j ( k 0ω 0 ) t x(t ) ←
→ X[ k − k0 ]
A shift in frequency (harmonic number) corresponds to
multiplication of the time function by a complex exponential.
Time Reversal
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FS
x( − t ) ←
→ X[− k ]
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39
CTFS Properties
Time Scaling
Let z(t ) = x( at ) , a > 0
T0 x
= T0 z for z(t)
a
Z[ k ] = X[ k ]
Case 1. TF =
Case 2. TF = T0 x for z(t)
If a is an integer,
 k  k
 X   , an integer
Z[ k ] =   a  a
0
, otherwise
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CTFS Properties
Time Scaling (continued)
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CTFS Properties
Change of Representation Time
FS
→ X[ k ]
With TF = T0 x , x(t ) ←
FS
With TF = mT0 x , x(t ) ←
→ Xm [k ]
k
 k
an integer
X  ,
Xm [k ] =   m  m
0
, otherwise
(m is any positive integer)
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CTFS Properties
Change of Representation Time
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CTFS Properties
FS
←
→
Time Differentiation
d
FS
→ j 2π ( kf0 ) X[ k ]
(x(t )) ←
dt
d
FS
→ j ( kω 0 ) X[ k ]
(x(t )) ←
dt
FS
←
→
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CTFS Properties
Time Integration
Case 1
Case 1. X[0 ] = 0
Case 2
X[ k ]
∫−∞ x(λ )dλ ←→ j 2π ( kf0 )
t
FS
X[ k ]
∫−∞ x(λ )dλ ←→ j( kω 0 )
t
FS
Case 2. X[0 ] ≠ 0
t
∫ x(λ )dλ is not periodic
−∞
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CTFS Properties
Multiplication-Convolution Duality
FS
x(t ) y(t ) ←
→ X[ k ] ∗ Y[ k ]
(The harmonic functions, X[k] and Y[k], must be based
on the same representation period, TF .)
x(t )
FS
y(t ) ←
→ T0 X[ k ] Y[ k ]
The symbol, , indicates periodic convolution.
Periodic convolution is defined mathematically by
x(t )
x(t )
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y(t ) = ∫ x(τ ) y(t − τ )dτ
T0
y(t ) = x ap (t ) ∗ y(t )
where x ap (t ) is any single period of x(t )
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46
CTFS Properties
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CTFS Properties
Conjugation
FS
x* (t ) ←
→ X* [ − k ]
Parseval’s Theorem
∞
1
2
2
x(t ) dt = ∑ X[ k ]
∫
T
T0 0
k =−∞
The average power of a periodic signal is the sum of the
average powers in its harmonic components.
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Convergence of the CTFS
Partial CTFS Sums
x N (t ) =
For continuous signals,
convergence is exact at
every point.
N
∑
k =− N
X [ k ] e j 2 π ( kf0 )t
A Continuous Signal
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Convergence of the CTFS
Partial CTFS Sums
For discontinuous signals,
convergence is exact at
every point of continuity.
Discontinuous Signal
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Convergence of the CTFS
The Gibbs Phenomenon
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DiscreteTime
Fourier
Series
Concept
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The Discrete-Time Fourier Series
The formal derivation of the discrete-time Fourier series (DTFS)
is on pages 259-262. The results are
x F [n ] =
∑
k= NF
X[ k ]e j 2π ( kFF ) n
1
X[ k ] =
NF
n0 + N F −1
∑
n = n0
x[n ]e − j 2π ( kFF ) n
1
where N F is the representation time, FF =
, and the notation,
NF
∑
k= NF
means a summation over any range of consecutive k’s exactly
N F in length.
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The Discrete-Time Fourier Series
Notice that in
x F [n ] =
∑
k= NF
X[ k ]e j 2π ( kFF ) n
the summation is over exactly one period, a finite summation.
This is because of the periodicity of the complex sinusoid,
e − j 2π ( kFF ) n
in harmonic number, k. That is, if k is increased by any integer
multiple of N F the complex sinusoid does not change.
e − j 2π ( kFF ) n = e − j 2π ( ( k + mN F ) FF ) n
(m an integer)
This occurs because discrete time, n, is always an integer.
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The Discrete-Time Fourier Series
In the very common case in which the representation time is
taken as the fundamental period, N 0 , the DTFS is
x[n ] =
∑ X[k ]e
j 2π ( kF0 ) n
k= N0
1
− j 2π ( kF0 ) n
←→ X[ k ] =
x
n
e
[ ]
∑
N0 n= N 0
FS
or in terms of radian frequency
x[n ] =
∑ X[k ]e
k= N0
j ( kΩ 0 ) n
1
− j ( kΩ 0 ) n
←→ X[ k ] =
x
n
e
[
]
∑
N0 n= N 0
FS
2π
where Ω0 = 2πF0 =
N0
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DTFS Example
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DTFS Properties
Let a signal, x[n], have a fundamental period, N 0 x, and let a
signal, y[n], have a fundamental period, N 0 y . Let the DTFS
harmonic functions, each using the fundamental period as
the representation time, N F , be X[k] and Y[k]. In the
properties to follow the two fundamental periods are the same
unless otherwise stated.
Linearity
FS
α x(t ) + β y(t ) ←
→ α X[ k ] + β Y[ k ]
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DTFS Properties
Time Shifting
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FS
x[ n − n0 ] ←
→ e − j 2π ( kF0 ) n0 X[ k ]
FS
x[ n − n0 ] ←
→ e − j ( kΩ 0 ) n0 X[ k ]
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DTFS Properties
Frequency Shifting
(Harmonic Number
Shifting)
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FS
e j 2π ( k 0 F0 ) n x[n ]←
→ X[ k − k0 ]
FS
e j ( k 0Ω 0 ) n x[n ]←
→ X[ k − k0 ]
Conjugation
FS
x* [n ]←
→ X* [ − k ]
Time Reversal
FS
x[− n ]←
→ X[− k ]
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DTFS Properties
Time Scaling
Let z[n ] = x[an ] , a > 0
If a is not an integer, some values of z[n] are undefined
and no DTFS can be found. If a is an integer (other than
1) then z[n] is a decimated version of x[n] with some
values missing and there cannot be a unique relationship
between their harmonic functions. However, if
n
 n
an integer
x   ,
z[n ] =   m  m
0
, otherwise
then
1
Z [ k ] = X [ k ] , N F = mN 0
m
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DTFS Properties
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DTFS Properties
Change of Representation Time
FS
→ X[ k ]
With N F = N x 0, x[n ]←
FS
With N F = qN x 0, x[n ]←
→ Xq [ k ]
 k  k
 X   , an integer
Xq [ k ] =   q  q
0
, otherwise

(q is any positive integer)
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DTFS Properties
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DTFS Properties
X[ k ]
∑ x[m]←→ 1 − e− j 2π ( kF0 ) , k ≠ 0
m =−∞
n
X[ k ]
FS
x[ m ]←→
, k≠0
∑
− j ( kΩ 0 )
1− e
m =−∞
n
Accumulation
Parseval’s
Theorem
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FS
1
2
2
n
k
x
=
X
[ ] ∑ [ ]
∑
N0 n= N 0
k= N0
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DTFS Properties
MultiplicationConvolution
Duality
First Backward
Difference
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FS
x[n ] y[n ]←
→ Y[ k ]
x[n ]
X[ k ] =
∑ Y[q] X[k − q]
q= N 0
FS
y[n ]←
→ N 0 Y[ k ] X[ k ]
(
x[n ] − x[n − 1]←→(1 − e
)
FS
x[n ] − x[n − 1]←
→ 1 − e − j 2π ( kF0 ) X[ k ]
FS
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− j ( kΩ 0 )
) X[k ]
65
DTFS Properties
FS
←
→
FS
←
→
FS
←
→
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Convergence of the DTFS
• The DTFS converges exactly with a finite
number of terms. It does not have a “Gibbs
phenomenon” in the same sense that the
CTFS does
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LTI Systems with Periodic
Excitation
The differential equation describing an RC lowpass filter is
RC v′out (t ) + vout (t ) = vin (t )
If the excitation, vin (t ) , is periodic it can be expressed as a
CTFS,
∞
vin (t ) = ∑ Vin [ k ]e j 2π ( kf 0 ) t
k =−∞
The equation for the kth harmonic alone is
RC v′out ,k (t ) + vout ,k (t ) = vin,k (t ) = Vin [ k ]e j 2π ( kf 0 ) t
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LTI Systems with Periodic
Excitation
If the excitation is periodic, the response is also, with the
same fundamental period. Therefore the response can be
expressed as a CTFS also.
vout ,k (t ) = Vout [ k ]e j 2π ( kf 0 ) t
Then the equation for the kth harmonic becomes
j 2 kπf0 RC Vout [ k ]e j 2π ( kf 0 ) t + Vout [ k ]e j 2π ( kf 0 ) t = Vin [ k ]e j 2π ( kf 0 ) t
Notice that what was once a differential equation is now
an algebraic equation.
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LTI Systems with Periodic
Excitation
Solving the kth-harmonic equation,
Vin [ k ]
Vout [ k ] =
j 2 kπf0 RC + 1
Then the response can be written as
vout (t ) =
∞
∑ V [k ]e
out
k =−∞
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j 2π ( kf 0 ) t
=
∞
∑
k =−∞
Vin [ k ]
e j 2π ( kf 0 ) t
j 2 kπf0 RC + 1
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LTI Systems with Periodic Excitation
Vout [ k ]
The ratio,
, is the frequency response of the system.
Vin [ k ]
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