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 voiceexamples, and see the importance of spectralanalysis.

• Very useful techniques based on symmetricconditions make it easy for you to know thespectral components of the periodic waveforms.

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Fourier Series

• Fourier series represents a periodic signal as the sum of harmonically related sinusoidal functions.

• It means that any periodic signal can bedecomposed into sinusoids.

• Example: Periodic function

� Fundamental frequency

� Harmonics

( )x t

x(t)

T 2T 3Tt

-T-2T

=1

1f

T

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Joseph Fourier (1768-1830)

FourierFourierFourierFourier was born at Auxerre (now in the Yonnedépartement of France), the son of a tailor. He wasorphaned at age eight. Fourier was recommendedto the Bishop of Auxerre, and through thisintroduction, he was educated by the Benvenistesof the Convent of St. Mark. Fourier went withNapoleon Bonaparte on his Egyptian expedition in1798, and was made governor of Lower Egypt andsecretary of the Institut d'Égypte. He alsocontributed several mathematical papers to theEgyptian Institute (also called the Cairo Institute)which Napoleon founded at Cairo, with a view ofweakening English influence in the East. After theBritish victories and the capitulation of the Frenchunder General Menou in 1801, Fourier returned toFrance, and was made prefect of Isère, and it waswhile there that he made his experiments on thepropagation of heat. (from WIKIPEDIA)

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Why Spectral Analysis

• Spectral analysis provides you another perspective on a signal.

• Once we know the spectral components of asignal, it becomes easier for us to process thesignal. For example, you can use the filteringtechniques to filter-out any frequency-componentyou don’t want.

• Spectral analysis helps you to identify thefrequency components. (It is difficult to identify the

frequency components from looking at a time-domain waveform)

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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.

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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

f1 kHz

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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

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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

Time-domain waveform Frequency-domain Spectrum100 Hz

100 Hz 200 Hz

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Sound of 3-tones and 4-tones

periodically repeat

periodically repeat

Time-domain waveform Frequency-domain Spectrum100 Hz 200 Hz 500 Hz

100 Hz 200 Hz 500 Hz700 Hz

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Sound of 5-Tones and 6-Tones

periodically repeat

periodically repeat

Time-domain waveform Frequency-domain Spectrum

100 Hz 200 Hz 500 Hz700 Hz

1 kHz

100 Hz 200 Hz 500 Hz700 Hz

1 kHz5 kHz

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Feel that

• We’ve observed that the combination ofharmonically related sinusoids is periodicallyrepeating. On the other hand, we can also saythat any periodic waveform must be thecombination of harmonically related sinusoids.

• When you see a periodic signal, you can knowthat it is a combination of harmonically relatedsinusoids and it has many spectral componentdiscretely appearing in the spectrum.

• In this chapter, we firstly discuss the periodicsignal and use the Fourier series to analyze it.

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Periodic Square Wave

t

x(t)

t

t

t

( )X jω

ω1f 13f 15f

decomposition

.etc

T1

=11

1f

Tis the fundamental frequency

=11

nnf

Tis 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

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Sine-Cosine Form (I)

( ) ( )0 1 11

cos sinn nn

x t A A n t B n tω ω∞

=

= + +∑

1 1

22 f

Tπω π= =

1 1

22

ntn t nf t

Tπω π= =

• A periodic signal is presented as a sum of sines andcosines 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 TA x t dt

T

( ) ω= ≥ =∫ 10

2cos , for 1 but not for 0

T

nA x t n tdt n nT

( ) ω= ≥∫ 10

2sin , for 1

T

nB x t n tdt nT

is the DC term(average value over one cycle)

• Other than DC, there are two components appearing at a givenharmonic frequency in the most general case: a cosine term with anamplitude and a sine term with an amplitudenA nB

( ) ( ) ( )ω ωω ω

−= − ⋅ = + =

∫ ∫

1 11 10 0

cos 2 1 cos2 2cos 1 cos 0

2 2

T T

n

n t tA n t n tdt dt

T T

( )ωω ω

= ⋅ = − =

∫ ∫ 11 10 0

sin 02 2 sin2sin cos 0

2 2

T T

n

n tA n t n tdt dt

T T

( )ωω ω

= ⋅ = + =

∫ ∫ 11 10 0

cos 02 2 cos2cos cos 1

2 2

T T

n

n tA n t n tdt dt

T T

− ~2 2T T

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Amplitude-Phase Form

( ) ( )ω φ∞

=

= + +∑0 11

cosn nn

x t C C n t

( ) ( )ω θ∞

=

= + +∑0 11

sinn nn

x t C C n t

2 2n n nC A B= +

• Sine-cosine form is presented with two separate components (sineterm and cosine term) at a given frequency, each of which has twoseparate amplitude.

• The sum of two or more sinusoids of a given frequency is equivalent toa single sinusoid at the same frequency.

• The amplitude-phase form of the Fourier series can be expressed aseither

or

=0 0C A is the DC value

is the net amplitude of a given component at frequencynf1, since sine and cosine phasor forms are alwaysperpendicular to each other.

where

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Complex Exponential Form (I)

11 1cos sinjn te n t j n tω ω ω= +

11 1cos sinjn te n t j n tω ω ω− = −

1 1

1cos2

jn t jn te en t

ω ω

ω−+=

ω ω

ω−−=

1 1

1sin2

jn t jn te en t

j

cos sinjxe x j x= +

cos sinjxe x j x− = −

cos2

jx jxe ex

−+=

−−=sin2

jx jxe ex

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 teω− 1jn te

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Complex Exponential Form (II)

ω ω−−+1 1jk t jk t

k kX e X e

�( )− = kkX X

( ) ω∞

=−∞

= ∑ 1jn tn

n

x t X e

( ) ω−= ∫ 1

0

1 T jn tnX x t e dt

T

• The general form of the complex exponential form of the Fourier seriescan be expressed as

where Xn is a complex value

• At a given real frequency kf1, (k>0), that spectral representationconsists of

The first term is thought of as the “positive frequency” contribution, whereas the second isthe corresponding “negative frequency” contribution. Although either one of the two termsis a complex quantity, they add together in such a manner as to create a real function, andthis is why both terms are required to make the mathematical form complete.

where the negative frequency coefficient X-k is the complex conjugate of thecorresponding positive frequency coefficient Xk.

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Complex Exponential Form (III)

−= ≠, for n 02

n nn

A jBX

( )= = =∫0 0 00

1 TX x t dt A C

T

• The coefficient Xn can be calculated from

( ) ω−= ∫ 1

0

1 T jn tnX x t e dt

Tit turns out that Xn can be also calculated directly from An and Bn of thesine-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 positiveand negative n in applying the above equation. Use to determinedthe corresponding coefficients for negative n.

�( )− = nnX X

( )φ φ= = ∠njn n n nX X e X

• The DC component X0 is simply

which is the same in all the Fourier forms.23/61


Example – Conversion of the Forms

• A certain periodic bandlimited signal has only three frequencies in itsFourier series representation: dc, 1kHz, and 2kHz. The signal can beexpressed 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

182 2 2 2

j t j t j t j tx t e e e e

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Example – Periodical Rectangular Wave (I)

2T

T−2T

−T

( )x t

t

= = =0

area under curve in one cycle 22

AT AA

T T

( ) ωω ωω ω

= = = −

∫2

2 11 10

1 10

2 2 2cos sin sin 0

2

TT

n

n TA AA A n t dt n t

T n T n T

( ) ωω ωω ω

− − = = = −

∫2

2 11 10

1 10

2 2 2sin cos cos 1

2

TT

n

n TA AB A n t dt n t

T n T n T

0

Determine the Fourier series representation for the following waveform.

A

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Example – Periodical Rectangular Wave (II)

ω π=1 2n T n

ππ

= = ≠2sin 0, for 0

2n

AA n n

n

( )π ππ

= − =

for odd 1 cos

for even0

n

AnA

B n nn n

π −= +

1 for oddcos

1 for even

nn

n

( ) ω ω ω ωπ π π π

= + + + + +⋯1 1 1 1

2 2 2 2sin sin3 sin5 sin7

2 3 5 7A A A A A

x t t t t t

ωπ

∞

=

= + ∑ 11

odd

2sin

2 nn

A An 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 2A

X A

ω ω

ω− −−= =∫ 1 1

22

01 0

1T

Tjn t jn

n

AX Ae dt e t

T jn T( ) ( )ω π

π π− −−= − = −1 2 1 1

2 2jn T jnA A

e ej n j n

( )π ππ

= − +1 cos sin2A

n j nj n

π π−= = for oddn

A jAX n

jn n

( ) ω ω ω ω

π π π π− −= − − − + +⋯ ⋯1 1 1 13 3

2 3 3j 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 frequencycontribution

Negative frequencycontribution

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Frequency Spectrum Plots

+ = + = =

2 2 2 2

2 2 2 2n nn n n

n

A BA B CX for

0 0X C=

One-sided amplitude frequency spectrum Two-sided amplitude frequency spectrum

≠ 0n

f f0 1f 12f 13f 14f 0 1f 12f 13f 14f− 1f− 12f− 13f− 14f

nXnC

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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

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Spectra of Periodical Rectangular Wave

2A π

2A

π23

A

π25

A

π27

A

π29

A

2A

πA

π3A

π5A

π7A

π9A

π9A

π7A

π5A

π3A

πA

One-sided amplitude frequency spectrum

Two-sided amplitude frequency spectrum

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Fourier Series Symmetry Conditions

• Even Function

• Odd Function

• Half-wave Symmetric

• Full-wave Symmetric

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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

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Half-wave Symmetry Condition (I)

Half-wave Symmetry ( ) + = − 2

Tx 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


Tx 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


Tx 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

Tx 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.

37/61


Homework

• Explain why the one-sided form of a full-wavesymmetric signal has both cosine and sineterms and only even-numbered harmonicsappear. (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 haveonly cosine terms. Xn

terms are real.

One-sided forms haveonly sine terms. Xn

terms are imaginary.

Odd-numbered harmonics only

Even-numbered harmonics only

Summary of Symmetry Conditions

( ) ( )0 1 11

cos sinn nn

x t A A n t B n tω ω∞

== + +∑ 1 1

22 f

Tπω π= =

( ) ( ) ( )ω φ ω θ∞ ∞

= =

= + + = + +∑ ∑0 1 0 11 1

cos sinn n n nn n

x t C C n t C C n t 2 2n n nC A B= +

( ) ω∞

=−∞

= ∑ 1jn tn

n

x t X e−= ≠,for n 02

n nn

A jBX = =0 0 0X A C

( ) ω∫ 10

2cos

Tx t n tdt

T( ) ω∫ 10

2sin

Tx t n tdt

T( ) ω−

∫ 1

0

1 T jn tx t e dtT

General

Even function

( ) ( )− =x t x t ( ) ω∫2

10

4cos

Tx t n tdt

T0 ( ) ω∫

2

10

2cos

Tx t n tdt

T

Odd function

( ) ( )− = −x t x t 0 ( ) ω∫2

10

4sin

Tx t n tdt

T( ) ω−

∫2

10

2sin

Tjx t n tdt

T

Half-wave symm.

( ) + = − 2

Tx t x t

( ) ω∫2

10

4cos

Tx t n tdt

T( ) ω∫

2

10

4sin

Tx t n tdt

T( ) ω−

∫ 12

0

2 T jn tx t e dtT

Full-wave symm.

( ) + = − 2

Tx t x t ( ) ω∫

2

10

4cos

Tx t n tdt

T( ) ω∫

2

10

4sin

Tx t n tdt

T( ) ω−

∫ 12

0

2 T jn tx t e dtT

( )= except 0nA n nB nX

Sine-cosine form:

Amplitude-phase form:

Complex exponential form:

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Some Common Periodic Signals (I)

ω ω ω ωπ − + − +

⋯1 1 1 1

4 1 1 1cos cos3 cos5 cos7

3 5 7A

t t t t

ω ω ωπ

+ + +

⋯1 1 12

8 1 1cos cos3 cos5

9 25A

t t t

ω ω ω ωπ − + − +

⋯1 1 1 1


2 3 4A

t t t t

� Square wave

� Triangular wave

� Sawtooth wave

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Some Common Periodic Signals (II)

π ω ω ω ωπ + + − + −

⋯1 1 1 1

2 2 21 cos cos2 cos4 cos6

2 3 15 35A

t t t t

ω ω ωπ + − + −

⋯1 1 1

2 2 2 21 cos2 cos4 cos6

3 15 35A

t t t

π π πω ω ωπ π π

+ + + +

⋯1 1 1

sin sin2 sin31 2 cos cos2 cos3

2 3d d d

Ad t t td d d

τ=dT

� 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

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Fourier Transform

( ) ( ) = X f F x tF

( ) ( )− = 1 x t F X fF

( ) ( ) ω∞ −

−∞= ∫

j tX f x t e dt

( ) ( ) ω∞

−∞= ∫

j tx t X f e df

• The process of Fourier transformation of a time function isdesignated symbolically as:

• The inverse operation is designated symbolically as

• The actual mathematical processes involved in theseoperations are as follows:

ω π= 2 f

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Frequency Spectrum

( ) ( ) ( ) ( ) ( )φ φ= = ∠j fX f X f e X f f

• The Fourier transform X(f) is, in general, a complex functionand has both a magnitude and an angle. Thus, X(f) can beexpressed as

where represents the amplitude spectrum and isthe phase spectrum.

( )X f ( )φ f

( )X f

f

• A typical amplitude spectrum

For the nonperiodic signal, itsspectrum is continuous, and, ingeneral, it consists ofcomponents at all frequencies

in the range over which thespectrum is present.

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Fourier Transform Symmetry Conditions

ConditionConditionConditionCondition CommentsCommentsCommentsComments

One-sided forms haveonly cosine terms. Xn

terms are real.

One-sided forms haveonly sine terms. Xn

terms are imaginary.

( ) ω∞ −

−∞∫j tx t e dtGeneral

Even function

( ) ( )− =x t x t ( ) ω∞

∫02 cosx t tdt

Odd function

( ) ( )− = −x t x t ( ) ω∞

− ∫02 sinj x t tdt

nX

• The results indicate that for either an even or an oddfunction, one need integrate only over half the total intervaland double the result.

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Example – Rectangular Pulse

• Derive the Fourier transform of the rectangular pulsefunction shown.

τ−2

τ2

A

( )x t

( ) τ τ− < <=

for 2 2

0 elsewhere

A tx t

( ) π ττπ τ

= sin fX f A

f

τA

( )X f

τ1

τ2

τ3

f

( )τ

τ ωτω ωω ω

= = =∫2

2

00

2 22 cos sin sin

2A A

X f A tdt t

ω π= 2 f

t

FourierTransform

46/61


FourierTransform

Example – Exponential Function

• Derive the Fourier transform of the exponential functiongiven by

( )α− >

= <

for 0

0 for 0

tAe tx t

t

α > 0where

t

A

( )x t

( )( )

( )α ω

α ω

α ω α ω

∞− +∞ − −= = = +− + +∫0

0

0j t

t j t Ae AX f Ae e dt

j j

( )( )α ω α π

= =+ +2 2 22 2

A AX f

f

( ) ω πφα α

− −= − = −1 1 2tan tan

ff

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 Fouriertransform of the impulse function

( ) ( ) ωδ δ∞ −

−∞ = ∫

j tF t t e dtF

( )δ = 1F tF

t

1

( )δ t

f

1

( )δ = 1F tFFourierTransform

48/61


Common NonperiodicWaveforms

τ−2

τ2

A

( )x t

( ) ( ) ( )sinsinc

fX f A A f

f

π ττ τ π τ

π τ= = ⋅

τ− τ

A

( )x t

( ) ( ) 2sin f

X f Af

π ττ

π τ

=

τ

A

( )x t

( ) π τπ τπ π τ

− = −

sin1

2j fjA f

X f ef f

A

( )x t

τ−2

τ2

( ) τ π τπ τ

=− 2 2

2 cos1 4

A fX 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 tj fX f

dt

( ) ( ) F x t X f = F

f f

( ) ( )π

=

2dx t

F fX fdtF

50/61


Fourier Transform Operation Pairs (II)

� Operation 3: Integration ( ) ( )π−∞

↔∫ 2

t X fx t dt

j f

( ) ( ) = F x t X fF

f f

( ) ( )π−∞

= ∫ 2

t X fF x t dt

fF

� Operation 4: Time delay ( ) ( )π ττ −− ↔ 2j fx t e X f

( )x t

t tτ

( )τ−x t

51/61


Fourier Transform Operation Pairs (III)

� Operation 5: Modulation ( ) ( )π ↔ −020

j f te x t X f f

( ) ( ) F x t X f = F

f f

( ) ( )π = − 02

0j f tF x t e X f fF

1f− 1f +0 1f f−0 1f f 0f

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( ) ↔

1 fx 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 thatcan be used qualitatively in estimating therelative 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 aninteger. As k increases, the spectrum diminishesrapidly. (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 (nodiscontinuities) tend to have higher rolloff ratesand corresponding narrower bandwidths.

• Time functions with discontinuities in the signaltend to have lower rolloff rates andcorresponding wider bandwidths.

• An example of a smooth signal is the sinusoidalwhose bandwidth is so narrow that it is only onecomponents. Conversely, a square wave hasfinite discontinuities in each cycle, and itsspectrum 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

1f

No spectral roll-off No spectral roll-off

x(t) has finite discontinuities or -6dB/octave1n

or -6dB/octave

2

1f

x(t) is continuous,x’(t) has finite discontinuities or -12dB/octave

2

1n

or -12dB/octave

3

1f

x(t) and x’(t) are continuous,x’(t) has finite discontinuities

or -18dB/octave3

1n

or -18dB/octave

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FourierTransform



( )α− >

= <

for 0

0 for 0

tAe tx t

t

α > 0where

t

A

( )x t

( )( )

( )α ω

α ω

α ω α ω

∞− +∞ − −= = = +− + +∫0

0

0j t

t j t Ae AX f Ae e dt

j j

( )( )α ω α π

= =+ +2 2 22 2

A AX f

f

f

αA

( )X f

x(t) has a finite discontinuity at t = 0 Rolloff rate = -6dB/octave

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FourierTransform



( )α

α

− >=

<

for 0

for 0

t

t

Ae tx t

Ae t

α > 0where

t

A( )x t

( ) ( ) ω α ω α ω∞ ∞− − − −

−∞ −∞= = +∫ ∫ ∫

0

0

j t t j t t j tX 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

2j t j tAe Ae A A Aj j j j

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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 Fouriertransform of the impulse function

( ) ( ) ωδ δ∞ −

−∞ = ∫

j tF t t e dtF

( )δ = 1F tF

t

1

( )δ t

f

1

( )δ = 1F tFFourier

Transform

No Roll-off

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Example – Periodical Rectangular Wave (I)

2T

T−2T

−T

( )x t

t

= = =0

area under curve in one cycle 22

AT AA

T T

( ) ωω ωω ω

= = = −

∫2

2 11 10

1 10

2 2 2cos sin sin 0

2

TT

n

n TA AA A n t dt n t

T n T n T

( ) ωω ωω ω

− − = = = −

∫2

2 11 10

1 10

2 2 2sin cos cos 1

2

TT

n

n TA AB 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

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Example – Periodical Rectangular Wave (II)

ω π=1 2n T n

ππ

= = ≠2sin 0, for 0

2n

AA n n

n

( )π ππ

= − =

for odd 1 cos

for even0

n

AnA

B n nn n

π −= +

1 for oddcos

1 for even

nn

n

( ) ω ω ω ωπ π π π

= + + + + +⋯1 1 1 1


2 3 5 7A A A A A

x t t t t t

ωπ

∞

=

= + ∑ 11

odd

2sin

2 nn

A An t

n

• Let

x(t) has a finite discontinuity Rolloff rate = -6dB/octave

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Circuit Network Analysis - [Chapter3] Fourier Analysis

Engineering

Transcript of Circuit Network Analysis - [Chapter3] Fourier Analysis