Polytechnic University - Department of Electrical & Computer

EE3054 Signals and Systems

Fourier Series and Spectrum

Yao WangPolytechnic University

Most of the slides included are extracted from lecture presentations prepared by McClellan and Schafer

3/25/2008 © 2003, JH McClellan & RW Schafer 2

License Info for SPFirst Slides

� This work released under a Creative Commons Licensewith the following terms:

� Attribution� The licensor permits others to copy, distribute, display, and perform

the work. In return, licensees must give the original authors credit.

� Non-Commercial� The licensor permits others to copy, distribute, display, and perform

the work. In return, licensees may not use the work for commercial purposes—unless they get the licensor's permission.

� Share Alike� The licensor permits others to distribute derivative works only under

a license identical to the one that governs the licensor's work.� Full Text of the License� This (hidden) page should be kept with the presentation

What is Fourier Series?

� Any real, periodic signal with fundamental freq. f0=1/T0 can be represented as the sum of complex exponential signals with freq= k f0

�� SPECTRUM: SPECTRUM: plot of ak, Complex Amplitude for k-th Harmonic�� ANALYSIS:ANALYSIS: Determine coefficients ak from x(t)

�� SYNTHESIS:SYNTHESIS: Generating x(t) from a_k

∫−=

0

0

0

0

)/2(1 )(T

dtetxa tTkjTk

π

{ }∑=

−∗++=N

k

tfjk

tfjk

kk eaeaatx1

220)( ππ


Example:

� Write as sum of sin(3 pi t) and sin( 9pi t)� Then expand using complex exponential

)3(sin)( 3 ttx π=

tjtjtjtj ejejejejtx ππππ 9339

883

83

8)( −−

−+

+

−+

=


Example )3(sin)( 3 ttx π=


883

83

8)( −−

−+

+

−+

=


Example

In this case, analysisjust requires picking off the coefficients.

)3(sin)( 3 ttx π=


883

83

8)( −−

−+

+

−+

=

3=k1=k1−=k

3−=k

ka


Analysis: x(t) � ak

� Step 1: determine the fundamental period of the signal, T0� Shortest interval where signal repeats or satisfy

x(t+T0)=x(t)� Step 2: using the following formula to compute

a_k:

∫−=

0

0

0

0)(1

T

dtetxa tkjTk

ω


Ex. 1: SQUARE WAVE

0–.02 .02 0.04

1

t

x(t)

.01

sec. 04.0for 0

01)(

0

0021

021

=

<≤

<≤=

TTtT

Tttx


FS for a SQUARE WAVE {ak}

)0()(1 0

0

0

)/2(

0≠= ∫

− kdtetxT

aT

ktTjk

π

02.

0)04./2(

)04./2(04.1

02.

0

)04./2(104.1 ktj

kjktj

k edtea ππ

π −−

− == ∫

kje

kj

kkj

πππ

2)1(1)1(

)2(1 )( −−=−

−= −


DC Coefficient: a0

)0()(1 0

0

0

)/2(

0== ∫

− kdtetxT

aT

ktTjk

π

)Area(1)(1

0000

0

Tdttx

Ta

T

== ∫

21

02.

00 )002(.

04.11

04.1 =−== ∫ dta


Fourier Coefficients ak

� ak is a function of k� Complex Amplitude for k-th Harmonic� This one doesn’t depend on the period, T0

=

±±=

±±=

=−−=

0

,4,20

,3,11

2)1(1

21 k

k

kkj

kja

k

k l

l

π

π


Spectrum from Fourier Series

=

±±=

±±=−

=

0

,4,20

,3,1

21 k

k

kkj

ak l

l

π)25(2)04.0/(20 ππω ==


Ex. 2: Rectified Sine Wave {ak}

)1()(1 0

0

0

)/2(

0±≠= ∫

− kdtetxT

aT

ktTjk

π

2/

0))1)(/2((2

)1)(/2(2/

0))1)(/2((2

)1)(/2(

2/

0

)1)(/2(21

2/

0

)1)(/2(21

2/

0

)/2()/2()/2(

1

2/

0

)/2(21

0

00

00

00

0

0

0

0

0

0

0

0

000

0

0

0

00

2

)sin(

T

kTjTj

tkTjT

kTjTj

tkTj

TtkTj

Tj

TtkTj

Tj

TktTj

tTjtTj

T

TktTj

TTk

ee

dtedte

dtejee

dteta

+−

+−

−−

−−

+−−−

−−

−

−=

−=

−=

=

∫∫

∫

∫

π

π

π

π

ππ

πππ

ππ

Half-Wave Rectified Sine


( ) ( )( ) ( )

( )( )

±==−−−=

−−−=

−−−=

−=

−−

−−−+

+−+

−−−

+−+

−−−

+−

+−

−−

−−

even 1?

odd 01)1(

11

11

)1(1

)1(4)1(1

)1()1(4

1)1()1(4

1

2/)1)(/2()1(4

12/)1)(/2()1(4

1

2/

0))1)(/2((2

)1)(/2(2/

0))1)(/2((2

)1)(/2(

2

2

0000

0

00

00

00

0

kkk

ee

ee

eea

k

kk

kk

kjk

kjk

TkTjk

TkTjk

T

kTjTj

tkTjT

kTjTj

tkTj

k

π

π

ππ

ππ

ππ

ππ

π

π

π

π

FS: Rectified Sine Wave {ak}

41j±


Spectrum

� Show plot


Fourier Series Synthesis� HOW do you APPROXIMATE x(t) ?

� Use FINITE number of coefficients

∫−=

0

0

00

)/2(1 )(T

tkTjTk dtetxa π

real is )( when* txaa kk =−tfkj

N

Nkkeatx 02)( π∑

−==


Fourier Series Synthesis


Synthesis: 1st & 3rd Harmonics))75(2cos(

32))25(2cos(2

21)( 22

ππ ππ

ππ

−+−+= ttty


Synthesis: up to 7th Harmonic)350sin(

72)250sin(

52)150sin(

32)50cos(2

21)( 2 ttttty π

ππ

ππ

ππ

ππ +++−+=


Fourier Synthesisl+++= )3sin(

32)sin(2

21)( 00 tttxN ω

πω

π


Gibbs’ Phenomenon

� Convergence at DISCONTINUITY of x(t)� There is always an overshoot� 9% for the Square Wave case


Fourier Series Demos

� Fourier Series Java Applet� Greg Slabaugh

� Interactive

� http://users.ece.gatech.edu/mcclella/2025/Fsdemo_Slabaugh/fourier.html

� MATLAB GUI: fseriesdemo

� http://users.ece.gatech.edu/mcclella/matlabGUIs/index.html


fseriesdemo GUI


Fourier Series Java Applet


Alternate Forms of FS

� Generally, we have the FS representation

� If x(t) is real, a_k=a^*_{-k} (Conjugate symmetry)

� Proof

∑∞

−∞=

+=k

tfjk

keaatx π20)(

∑∞

=

−++=1

2*20)(

k

tfjk

tfjk

kk eaeaatx ππ


Alternate Forms of FS

∑∞

=

−++=1

2*20)(

k

tfjk

tfjk

kk eaeaatx ππ

kkkk

jkk

kkkk

aaA

eAa

tfAatx

k

∠==

=

++= ∑∞

=

θ

θπ

θ

;2

;21

)2cos()(1

0


HISTORY

� Jean Baptiste Joseph Fourier� 1807 thesis (memoir)

� On the Propagation of Heat in Solid Bodies� Heat !� Napoleonic era

� http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Fourier.html


READING ASSIGNMENTS

� This Lecture:� Fourier Series in Ch 3, Sects 3-4, 3-5 & 3-6

� Review: entire Chap 3

Documents

Polytechnic University - Department of Electrical & Computer