Circuit Network Analysis - [Chapter3] Fourier Analysis

Network Analysis
Chapter 3
Fourier Series and Fourier Transform
Chien-Jung Li
Department of Electronic Engineering
National Taipei University of Technology

Department of Electronic Engineering, NTUT
In This Chapter
• Periodic signal analysis – Fourier Series
• Non-periodic signal analysis – Fourier Transform
• We will start with some interesting voice
examples, and see the importance of spectral
analysis.
• Very useful techniques based on symmetric
conditions make it easy for you to know the
spectral components of the periodic waveforms.
2/61

Fourier Series
• Fourier series represents a periodic signal as the
sum of harmonically related sinusoidal functions.
• It means that any periodic signal can be
decomposed into sinusoids.
• Example: Periodic function
Fundamental frequency
Harmonics
( )x t
x(t)
T 2T 3T
t
-T-2T
=1
1
f
T
3/61

Joseph Fourier (1768-1830)
FourierFourierFourierFourier was born at Auxerre (now in the Yonne
département of France), the son of a tailor. He was
orphaned at age eight. Fourier was recommended
to the Bishop of Auxerre, and through this
introduction, he was educated by the Benvenistes
of the Convent of St. Mark. Fourier went with
Napoleon Bonaparte on his Egyptian expedition in
1798, and was made governor of Lower Egypt and
secretary of the Institut d'Égypte. He also
contributed several mathematical papers to the
Egyptian Institute (also called the Cairo Institute)
which Napoleon founded at Cairo, with a view of
weakening English influence in the East. After the
British victories and the capitulation of the French
under General Menou in 1801, Fourier returned to
France, and was made prefect of Isère, and it was
while there that he made his experiments on the
propagation of heat. (from WIKIPEDIA)
4/61

Why Spectral Analysis
• Spectral analysis provides you another perspective
on a signal.
• Once we know the spectral components of a
signal, it becomes easier for us to process the
signal. For example, you can use the filtering
techniques to filter-out any frequency-component
you don’t want.
• Spectral analysis helps you to identify the
frequency components. (It is difficult to identify the
frequency components from looking at a time-domain waveform)
5/61

Voices from Man and Woman
陳海茵主播
謝向榮主播
Time-domain waveform Frequency-domain Spectrum
女生: 陳海茵主播的聲音
男生: 謝向榮主播的聲音
With Fourier analysis, one can easily
know the spectral components of a
signal.
6/61

Lowpass Filtering
500Hz
陳海茵主播
謝向榮主播
低通濾波器
f
濾波前
濾波後
500Hz
低通濾波器
f
濾波前
濾波後
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Highpass Filtering
濾波前
濾波後
濾波前
濾波後
1 kHz
陳海茵主播
謝向榮主播
高通濾波器
f
高通濾波器
f
1 kHz
8/61

Music Time : Crowd in the Palace
200 Hz
高通濾波器
f
200Hz
低通濾波器
f
1 kHz
高通濾波器
f
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Music Time : When I’m Sixty-four
200Hz
低通濾波器
f
600 Hz
高通濾波器
f
10/61

Listen to the Tones
100 Hz Tone 200 Hz Tone 500 Hz Tone
700 Hz Tone 1 kHz Tone 5 kHz Tone
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Sound of 1-tone and 2-tones
periodically repeat
periodically repeat
100 Hz
100 Hz 200 Hz
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Sound of 3-tones and 4-tones
periodically repeat
periodically repeat
100 Hz 200 Hz 500 Hz
100 Hz 200 Hz 500 Hz
700 Hz
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Sound of 5-Tones and 6-Tones
periodically repeat
periodically repeat
100 Hz 200 Hz 500 Hz
700 Hz
1 kHz
100 Hz 200 Hz 500 Hz
700 Hz
1 kHz
5 kHz
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Feel that
• We’ve observed that the combination of
harmonically related sinusoids is periodically
repeating. On the other hand, we can also say
that any periodic waveform must be the
combination of harmonically related sinusoids.
• When you see a periodic signal, you can know
that it is a combination of harmonically related
sinusoids and it has many spectral component
discretely appearing in the spectrum.
• In this chapter, we firstly discuss the periodic
signal and use the Fourier series to analyze it.
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Periodic Square Wave
t
x(t)
t
t
t
( )X jω
ω
1f 13f 15f
decomposition
.etc
T1
=1
1
1
f
T
is the fundamental frequency
=1
1
n
nf
T
is the harmonic frequency
(n is integer)
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Fourier Series Representations
• There are three forms to represent the Fourier
Series:
Sine-cosine form
Amplitude-phase form
Complex exponential form
17/61

Sine-Cosine Form (I)
( ) ( )0 1 1
1
cos sinn n
n
x t A A n t B n tω ω
∞
=
= + +∑
1 1
2
2 f
T
π
ω π= =
1 1
2
2
nt
n t nf t
T
π
ω π= =
• A periodic signal is presented as a sum of sines and
cosines in the form:
( )x t
where
is the fundamental angular frequency in rads/s
is the nth harmonic frequencyω π=1 12n nf
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(A complete cycle can also be noted
from )
Sine-Cosine Form (II)
( )= =∫0 0
area under curve in one cycle
period T
1 T
A x t dt
T
( ) ω= ≥ =∫ 10
2
cos , for 1 but not for 0
T
nA x t n tdt n n
T
( ) ω= ≥∫ 10
2
sin , for 1
T
nB x t n tdt n
T
is the DC term
(average value over one cycle)
• Other than DC, there are two components appearing at a given
harmonic frequency in the most general case: a cosine term with an
amplitude and a sine term with an amplitudenA nB
( )
( ) ( )ω ω
ω ω
 −
= − ⋅ = + = 
 
∫ ∫
1 1
1 10 0
cos 2 1 cos2 2
cos 1 cos 0
2 2
T T
n
n t t
A n t n tdt dt
T T
( )ω
ω ω
 
= ⋅ = − = 
 
∫ ∫ 1
1 10 0
sin 02 2 sin2
sin cos 0
2 2
T T
n
n t
A n t n tdt dt
T T
( )ω
ω ω
 
= ⋅ = + = 
 
∫ ∫ 1
1 10 0
cos 02 2 cos2
cos cos 1
2 2
T T
n
n t
A n t n tdt dt
T T
− ~
2 2
T T
19/61

Amplitude-Phase Form
( ) ( )ω φ
∞
=
= + +∑0 1
1
cosn n
n
x t C C n t
( ) ( )ω θ
∞
=
= + +∑0 1
1
sinn n
n
x t C C n t
2 2
n n nC A B= +
• Sine-cosine form is presented with two separate components (sine
term and cosine term) at a given frequency, each of which has two
separate amplitude.
• The sum of two or more sinusoids of a given frequency is equivalent to
a single sinusoid at the same frequency.
• The amplitude-phase form of the Fourier series can be expressed as
either
or
=0 0C A is the DC value
is the net amplitude of a given component at frequency
nf1, since sine and cosine phasor forms are always
perpendicular to each other.
where
20/61

Complex Exponential Form (I)
1
1 1cos sinjn t
e n t j n tω
ω ω= +
1
1 1cos sinjn t
e n t j n tω
ω ω−
= −
1 1
1cos
2
jn t jn t
e e
n t
ω ω
ω
−
+
=
ω ω
ω
−
−
=
1 1
1sin
2
jn t jn t
e e
n t
j
cos sinjx
e x j x= +
cos sinjx
e x j x−
= −
cos
2
jx jx
e e
x
−
+
=
−
−
=sin
2
jx jx
e e
x
j
Recall that we’ve learned
in Chapter 2.
• Euler’s formula
ω1
n is called the positive frequency, and ω− 1
n the negative frequency
From Euler’s formula, we know that both positive-frequency and negative-
frequency terms are required to completely describe the sine or cosine
function with complex exponential form.
Here
ω1jn t
e
ω− 1jn t
e
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Complex Exponential Form (II)
ω ω−
−+1 1jk t jk t
k kX e X e
( )− = kkX X
( ) ω
∞
=−∞
= ∑ 1jn t
n
n
x t X e
( ) ω−
= ∫ 1
0
1 T
jn t
nX x t e dt
T
• The general form of the complex exponential form of the Fourier series
can be expressed as
where Xn is a complex value
• At a given real frequency kf1, (k>0), that spectral representation
consists of
The first term is thought of as the “positive frequency” contribution, whereas the second is
the corresponding “negative frequency” contribution. Although either one of the two terms
is a complex quantity, they add together in such a manner as to create a real function, and
this is why both terms are required to make the mathematical form complete.
where the negative frequency coefficient X-k is the complex conjugate of the
corresponding positive frequency coefficient Xk.
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Complex Exponential Form (III)
−
= ≠, for n 0
2
n n
n
A jB
X
( )= = =∫0 0 00
1 T
X x t dt A C
T
• The coefficient Xn can be calculated from
( ) ω−
= ∫ 1
0
1 T
jn t
nX x t e dt
T
it turns out that Xn can be also calculated directly from An and Bn of the
sine-cosine form. The relationship reads
Even though An and Bn are interpreted only for positive n in the sine-
cosine form, their functional forms may be extended for both positive
and negative n in applying the above equation. Use to determined
the corresponding coefficients for negative n.
( )− = nnX X
( )φ
φ= = ∠nj
n n n nX X e X
• The DC component X0 is simply
which is the same in all the Fourier forms.
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Example – Conversion of the Forms
• A certain periodic bandlimited signal has only three frequencies in its
Fourier series representation: dc, 1kHz, and 2kHz. The signal can be
expressed in sine-cosine form as
( ) ( ) ( ) ( ) ( )π π π π= + − − +18 40cos 2000 30sin 2000 24cos 4000 10sin 4000x t t t t t
Express the signal in (a) amplitude-phase form (b) complex exponential form
( ) ( ) ( )π φ π φ= + + + +1 1 2 218 cos 2000 cos 4000x t C t C t
=0 18C
= + = ∠1 40 30 50 36.87C j
= − − = ∠ −2 24 10 26 157.38C j
( ) ( ) ( )π π= + + + −18 50cos 2000 36.87 26cos 4000 157.38x t t t
( ) ( ) ( )π π= + + + −18 50sin 2000 126.87 26sin 4000 67.38x t t t
( ) ( ) ( ) ( ) ( )π π π π+ − + − − −
= + + + +
2000 36.87 2000 36.87 4000 157.38 4000 157.3850 50 26 26
18
2 2 2 2
j t j t j t j t
x t e e e e
24/61

Example – Periodical Rectangular Wave (I)
2
T
T−
2
T
−T
( )x t
t
= = =0
area under curve in one cycle 2
2
AT A
A
T T
( )
ω
ω ω
ω ω
 
= = = − 
 
∫
2
2 1
1 10
1 10
2 2 2
cos sin sin 0
2
T
T
n
n TA A
A A n t dt n t
T n T n T
( )
ω
ω ω
ω ω
− −  
= = = − 
 
∫
2
2 1
1 10
1 10
2 2 2
sin cos cos 1
2
T
T
n
n TA A
B A n t dt n t
T n T n T
0
Determine the Fourier series representation for the following waveform.
A
25/61

Example – Periodical Rectangular Wave (II)
ω π=1 2n T n
π
π
= = ≠
2
sin 0, for 0
2
n
A
A n n
n
( )π π
π


= − = 

for odd
1 cos
for even
0
n
A
nA
B n n
n n
π
 −
=  
+ 
1 for odd
cos
1 for even
n
n
n
( ) ω ω ω ω
π π π π
= + + + + +⋯1 1 1 1
2 2 2 2
sin sin3 sin5 sin7
2 3 5 7
A A A A A
x t t t t t
ω
π
∞
=
= + ∑ 1
1
odd
2
sin
2 n
n
A A
n t
n
• Let
It is noted that the periodical rectangular wave only contains odd-numbered
spectral components.
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Example – Periodical Rectangular Wave (III)
= =0 0
2
A
X A
ω ω
ω
− −−
= =∫ 1 1
2
2
0
1 0
1
T
T
jn t jn
n
A
X Ae dt e t
T jn T
( ) ( )ω π
π π
− −−
= − = −1 2
1 1
2 2
jn T jnA A
e e
j n j n
( )π π
π
= − +1 cos sin
2
A
n j n
j n
π π
−
= = for oddn
A jA
X n
jn n
( ) ω ω ω ω
π π π π
− −
= − − − + +⋯ ⋯1 1 1 13 3
2 3 3
j t j t j t j tA A A A A
x t j e j e j e j e
• Exponential form
DC Positive frequency
contribution
Negative frequency
contribution
27/61

Frequency Spectrum Plots
+   
= + = =   
   
2 2 2 2
2 2 2 2
n nn n n
n
A BA B C
X for
0 0X C=
One-sided amplitude frequency spectrum Two-sided amplitude frequency spectrum
≠ 0n
f f
0 1f 12f 13f 14f 0 1f 12f 13f 14f− 1f− 12f− 13f− 14f
nXnC
28/61

Example – One-sided and Two-sided Spectra
18
50
26
18
25
13
25
13
0 Hz 1 kHz 2 kHz
0 Hz 1 kHz 2 kHz−1 kHz−2 kHz
One-sided amplitude frequency spectrum
Two-sided amplitude frequency spectrum
29/61

Spectra of Periodical Rectangular Wave
2
A π
2A
π
2
3
A
π
2
5
A
π
2
7
A
π
2
9
A
2
A
π
A
π3
A
π5
A
π7
A
π9
A
π9
A
π7
A
π5
A
π3
A
π
A
One-sided amplitude frequency spectrum
Two-sided amplitude frequency spectrum
30/61

Fourier Series Symmetry Conditions
• Even Function
• Odd Function
• Half-wave Symmetric
• Full-wave Symmetric
31/61

Even and Odd Functions (I)
Even function ( ) ( )− =x t x t
Odd function ( ) ( )− = −x t x t
One-sided forms have only cosine terms.
One-sided forms have only sine terms.
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Even and Odd Functions (II)
Even function ( ) ( )− =x t x t
Odd function ( ) ( )− = −x t x t
33/61

Half-wave Symmetry Condition (I)
Half-wave Symmetry ( ) 
+ = − 
 2
T
x t x t
Shifts T/2
T
T/2 3T/2
2T
One-sided forms have both cosine and sine
terms, and only odd-numbered harmonics
appear.
Define that f=1/T is the fundamental frequency
of this waveform.
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Half-wave Symmetry Condition (II)
T
T/2
+ = − 
 2
T
x t x t
Cosine waveform is half-wave symmetric.
Shifts T/2
T
T/2
Sine waveform is half-wave symmetric.
Shifts T/2
cosine
sine
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Half-wave Symmetry Condition (III)
TT/2
2nd harmonic (T2nd=T/2) is not half-wave
symmetric. (same for even-harmonics)
Shifts T/2
cosine
T
T/2
Shifts T/2
cosine
+ = − 
 2
T
x t x t
3rd harmonic (T3rd=T/3)is half-wave
symmetric. (same for odd-harmonics)
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Full-wave Symmetry Condition
Full-wave Symmetry ( ) 
+ = 
 2
T
x t x t
Shifts T/2
TT/2 3T/2 2T
One-sided forms have both cosine and sine
terms, and only even-numbered harmonics
appear.
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Homework
• Explain why the one-sided form of a full-wave
symmetric signal has both cosine and sine
terms and only even-numbered harmonics
appear. (Please also carefully read the topic of full-wave
symmetry on page-591 in the textbook)
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ConditionConditionConditionCondition CommentsCommentsCommentsComments
One-sided forms have
only cosine terms. Xn
terms are real.
only sine terms. Xn
terms are imaginary.
Odd-numbered
harmonics only
Even-numbered
harmonics only
Summary of Symmetry Conditions
( ) ( )0 1 1
1
cos sinn n
n
x t A A n t B n tω ω
∞
=
= + +∑ 1 1
2
2 f
T
π
ω π= =
( ) ( ) ( )ω φ ω θ
∞ ∞
= =
= + + = + +∑ ∑0 1 0 1
1 1
cos sinn n n n
n n
x t C C n t C C n t 2 2
n n nC A B= +
( ) ω
∞
=−∞
= ∑ 1jn t
n
n
x t X e
−
= ≠,for n 0
2
n n
n
A jB
X = =0 0 0X A C
( ) ω∫ 10
2
cos
T
x t n tdt
T
( ) ω∫ 10
2
sin
T
x t n tdt
T
( ) ω−
∫ 1
0
1 T
jn t
x t e dt
T
General
Even function
( ) ( )− =x t x t ( ) ω∫
2
10
4
cos
T
x t n tdt
T
0 ( ) ω∫
2
10
2
cos
T
x t n tdt
T
Odd function
( ) ( )− = −x t x t
0 ( ) ω∫
2
10
4
sin
T
x t n tdt
T
( ) ω
−
∫
2
10
2
sin
Tj
x t n tdt
T
Half-wave symm.
( ) 
+ = − 
 2
T
x t x t
( ) ω∫
2
10
4
cos
T
x t n tdt
T
( ) ω∫
2
10
4
sin
T
x t n tdt
T
( ) ω−
∫ 1
2
0
2 T
jn t
x t e dt
T
Full-wave symm.
( ) 
+ = − 
 2
T
x t x t ( ) ω∫
2
10
4
cos
T
x t n tdt
T
( ) ω∫
2
10
4
sin
T
x t n tdt
T
( ) ω−
∫ 1
2
0
2 T
jn t
x t e dt
T
( )=except 0nA n nB nX
Sine-cosine form:
Amplitude-phase form:
Complex exponential form:
39/61

Some Common Periodic Signals (I)
ω ω ω ω
π
 
− + − + 
 
⋯1 1 1 1
4 1 1 1
cos cos3 cos5 cos7
3 5 7
A
t t t t
ω ω ω
π
 
+ + + 
 
⋯1 1 12
8 1 1
cos cos3 cos5
9 25
A
t t t
ω ω ω ω
π
 
− + − + 
 
⋯1 1 1 1
2 1 1 1
sin sin2 sin3 sin4
2 3 4
A
t t t t
Square wave
Triangular wave
Sawtooth wave
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Some Common Periodic Signals (II)
π
ω ω ω ω
π
 
+ + − + − 
 
⋯1 1 1 1
2 2 2
1 cos cos2 cos4 cos6
2 3 15 35
A
t t t t
ω ω ω
π
 
+ − + − 
 
⋯1 1 1
2 2 2 2
1 cos2 cos4 cos6
3 15 35
A
t t t
π π π
ω ω ω
π π π
  
+ + + +  
  
⋯1 1 1
sin sin2 sin3
1 2 cos cos2 cos3
2 3
d d d
Ad t t t
d d d
τ
=d
T
Half-wave rectified cosine
Full-wave rectified cosine
Pulse wave
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Period Becomes Infinite
T 2T 3T 4T 5T
( )x t
f
nX
T 2T
T
T
f
nX
f
nX
f
nX
Single pulse → ∞T
42/61

Fourier Transform
( ) ( ) =  X f F x tF
( ) ( )−
 =  
1
x t F X fF
( ) ( ) ω
∞
−
−∞
= ∫
j t
X f x t e dt
( ) ( ) ω
∞
−∞
= ∫
j t
x t X f e df
• The process of Fourier transformation of a time function is
designated symbolically as:
• The inverse operation is designated symbolically as
• The actual mathematical processes involved in these
operations are as follows:
ω π= 2 f
43/61

Frequency Spectrum
( ) ( ) ( )
( ) ( )φ
φ= = ∠
j f
X f X f e X f f
• The Fourier transform X(f) is, in general, a complex function
and has both a magnitude and an angle. Thus, X(f) can be
expressed as
where represents the amplitude spectrum and is
the phase spectrum.
( )X f ( )φ f
( )X f
f
• A typical amplitude spectrum
For the nonperiodic signal, its
spectrum is continuous, and, in
general, it consists of
components at all frequencies
in the range over which the
spectrum is present.
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Fourier Transform Symmetry Conditions
ConditionConditionConditionCondition CommentsCommentsCommentsComments
only cosine terms. Xn
terms are real.
only sine terms. Xn
terms are imaginary.
( ) ω
∞
−
−∞∫
j t
x t e dtGeneral
Even function
( ) ( )− =x t x t ( ) ω
∞
∫0
2 cosx t tdt
Odd function
( ) ( )− = −x t x t
( ) ω
∞
− ∫0
2 sinj x t tdt
nX
• The results indicate that for either an even or an odd
function, one need integrate only over half the total interval
and double the result.
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Example – Rectangular Pulse
• Derive the Fourier transform of the rectangular pulse
function shown.
τ
−
2
τ
2
A
( )x t
( )
τ τ− < <
= 

for 2 2
0 elsewhere
A t
x t
( )
π τ
τ
π τ
=
sin f
X f A
f
τA
( )X f
τ
1
τ
2
τ
3
f
( )
τ
τ
ωτ
ω ω
ω ω
= = =∫
2
2
0
0
2 2
2 cos sin sin
2
A A
X f A tdt t
ω π= 2 f
t
Fourier
Transform
46/61

Fourier
Transform
Example – Exponential Function
• Derive the Fourier transform of the exponential function
given by
( )
α−
 >
= 
<
for 0
0 for 0
t
Ae t
x t
t
α > 0where
t
A
( )x t
( )
( )
( )
α ω
α ω
α ω α ω
∞
− +
∞
− −
= = = +
− + +∫0
0
0
j t
t j t Ae A
X f Ae e dt
j j
( )
( )α ω α π
= =
+ +
2 2 22
2
A A
X f
f
( )
ω π
φ
α α
− −
= − = −1 1 2
tan tan
f
f
f
αA
( )X f
( )φ f
f
47/61

Example – Impulse Function
• One property of the impulse function not considered earlier is
( ) ( ) ( )δ
∞
−∞
=∫ 0g t t dt g
where g(t) is any continuous function. Derive the Fourier
transform of the impulse function
( ) ( ) ω
δ δ
∞
−
−∞
  =  ∫
j t
F t t e dtF
( )δ  =  1F tF
t
1
( )δ t
f
1
( )δ  =  1F tFFourier
Transform
48/61

Common Nonperiodic Waveforms
τ
−
2
τ
2
A
( )x t
( )
( )
( )
sin
sinc
f
X f A A f
f
π τ
τ τ π τ
π τ
= = ⋅
τ− τ
A
( )x t
( )
( )
2
sin f
X f A
f
π τ
τ
π τ
 
=  
 
τ
A
( )x t
( ) π τπ τ
π π τ
− 
= −  
sin
1
2
j fjA f
X f e
f f
A
( )x t
τ
−
2
τ
2
( )
τ π τ
π τ
=
− 2 2
2 cos
1 4
A f
X f
f
Rectangular pulse Sawtooth pulse
Triangular pulse Cosine pulse
49/61

• The notation indicates that x(t) and X(f) are corresponding
transform pair
Fourier Transform Operation Pairs (I)
( ) ( )↔x t X f
Operation 1: Superposition principle ( ) ( ) ( ) ( )+ ↔ +1 2 1 2ax t bx t aX f bX f
Operation 2: Differentiation
( )
( )π↔ 2
dx t
j fX f
dt
( ) ( )F x t X f  = F
f f
( )
( )π
 
= 
 
2
dx t
F fX f
dtF
50/61

Fourier Transform Operation Pairs (II)
Operation 3: Integration ( )
( )
π−∞
↔∫ 2
t X f
x t dt
j f
( ) ( )  = F x t X fF
f f
( )
( )
π−∞
  =
  ∫ 2
t X f
F x t dt
fF
Operation 4: Time delay ( ) ( )π τ
τ −
− ↔ 2j f
x t e X f
( )x t
t t
τ
( )τ−x t
51/61

Fourier Transform Operation Pairs (III)
Operation 5: Modulation ( ) ( )π
↔ −02
0
j f t
e x t X f f
( ) ( )F x t X f  = F
f f
( ) ( )π
  = − 
02
0
j f t
F x t e X f fF
1f−1f +0 1f f−0 1f f 0f
52/61

( )  
↔  
 
1 f
x at X
a a
Operation 6: Time scaling
( )x t
t f
( ) ( )F x t X f  = F
( )x t
t
( )x t
t
<1a
>1a
( ) ( )F x t X f  = F
f
( ) ( )F x t X f  = F
Fourier Transform Operation Pairs (IV)
f
53/61

Spectrum Roll-off Rate (I)
• Spectral roll-off rate is an important factor that
can be used qualitatively in estimating the
relative bandwidths of different signals.
• The basic way to specify the rolloff rate is a 1/fk
variation for a Fourier transform or a 1/nk
variation for a Fourier series, where k is an
integer. As k increases, the spectrum diminishes
rapidly. (a signal with a 1/f3 rolloff rate would normally have
narrower bandwidth than a signal with a 1/f2 rate)
54/61

Spectrum Roll-off Rate (II)
• Time functions that are relatively smooth (no
discontinuities) tend to have higher rolloff rates
and corresponding narrower bandwidths.
• Time functions with discontinuities in the signal
tend to have lower rolloff rates and
corresponding wider bandwidths.
• An example of a smooth signal is the sinusoidal
whose bandwidth is so narrow that it is only one
components. Conversely, a square wave has
finite discontinuities in each cycle, and its
spectrum is very wide.
55/61

Spectrum Roll-off Rate (III)
ConditionConditionConditionCondition RollRollRollRoll----off Rateoff Rateoff Rateoff Rate
Fourier Transform Fourier Series
x(t) has impulses
1
f
No spectral roll-off No spectral roll-off
x(t) has finite discontinuities or -6dB/octave
1
n
or -6dB/octave
2
1
f
x(t) is continuous,
x’(t) has finite discontinuities or -12dB/octave 2
1
n
or -12dB/octave
3
1
f
x(t) and x’(t) are continuous,
x’(t) has finite discontinuities
or -18dB/octave 3
1
n
or -18dB/octave
56/61

Fourier
Transform
given by
( )
α−
 >
= 
<
for 0
0 for 0
t
Ae t
x t
t
α > 0where
t
A
( )x t
( )
( )
( )
α ω
α ω
α ω α ω
∞
− +
∞
− −
= = = +
− + +∫0
0
0
j t
t j t Ae A
X f Ae e dt
j j
( )
( )α ω α π
= =
+ +
2 2 22
2
A A
X f
f
f
αA
( )X f
x(t) has a finite discontinuity at t = 0 Rolloff rate = -6dB/octave
57/61

Fourier
Transform
given by
( )
α
α
−
 >
= 
<
for 0
for 0
t
t
Ae t
x t
Ae t
α > 0where
t
A
( )x t
( ) ( ) ω α ω α ω
∞ ∞
− − − −
−∞ −∞
= = +∫ ∫ ∫
0
0
j t t j t t j t
X f x t e dt Ae e dt Ae e dt
f
α2A
( )X f
x’(t) has a finite discontinuity at t = 0 Rolloff rate = -12dB/octave
( )
( )
( )
( )
α ω α ω
α
α ω α ω α ω α ω α ω
∞
− − +
−∞
= + = + =
− − + − + +
0
2 2
0
2
j t j t
Ae Ae A A A
j j j j
58/61

Example – Impulse Function
• One property of the impulse function not considered earlier is
( ) ( ) ( )δ
∞
−∞
=∫ 0g t t dt g
where g(t) is any continuous function. Derive the Fourier
transform of the impulse function
( ) ( ) ω
δ δ
∞
−
−∞
  =  ∫
j t
F t t e dtF
( )δ  =  1F tF
t
1
( )δ t
f
1
( )δ  =  1F tF
Fourier
Transform
No Roll-off
59/61

Example – Periodical Rectangular Wave (I)
2
T
T−
2
T
−T
( )x t
t
= = =0
area under curve in one cycle 2
2
AT A
A
T T
( )
ω
ω ω
ω ω
 
= = = − 
 
∫
2
2 1
1 10
1 10
2 2 2
cos sin sin 0
2
T
T
n
n TA A
A A n t dt n t
T n T n T
( )
ω
ω ω
ω ω
− −  
= = = − 
 
∫
2
2 1
1 10
1 10
2 2 2
sin cos cos 1
2
T
T
n
n TA A
B A n t dt n t
T n T n T
0
Determine the Fourier series representation for the following waveform.
A
Two-discontinuities in
one cycle –6dB/ Octave
60/61

Example – Periodical Rectangular Wave (II)
ω π=1 2n T n
π
π
= = ≠
2
sin 0, for 0
2
n
A
A n n
n
( )π π
π


= − = 

for odd
1 cos
for even
0
n
A
nA
B n n
n n
π
 −
=  
+ 
1 for odd
cos
1 for even
n
n
n
( ) ω ω ω ω
π π π π
= + + + + +⋯1 1 1 1
2 2 2 2
sin sin3 sin5 sin7
2 3 5 7
A A A A A
x t t t t t
ω
π
∞
=
= + ∑ 1
1
odd
2
sin
2 n
n
A A
n t
n
• Let
x(t) has a finite discontinuity Rolloff rate = -6dB/octave
61/61

Circuit Network Analysis - [Chapter3] Fourier Analysis

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Circuit Network Analysis - [Chapter3] Fourier Analysis