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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
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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
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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)( −−
−+
+
−+
=
3/25/2008 © 2003, JH McClellan & RW Schafer 5
Example )3(sin)( 3 ttx π=
tjtjtjtj ejejejejtx ππππ 9339
883
83
8)( −−
−+
+
−+
=
3/25/2008 © 2003, JH McClellan & RW Schafer 6
Example
In this case, analysisjust requires picking off the coefficients.
)3(sin)( 3 ttx π=
tjtjtjtj ejejejejtx ππππ 9339
883
83
8)( −−
−+
+
−+
=
3=k1=k1−=k
3−=k
ka
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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
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Ex. 1: SQUARE WAVE
0–.02 .02 0.04
1
t
x(t)
.01
sec. 04.0for 0
01)(
0
0021
021
=
<≤
<≤=
TTtT
Tttx
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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 )( −−=−
−= −
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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
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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
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Spectrum from Fourier Series
=
±±=
±±=−
=
0
,4,20
,3,1
21 k
k
kkj
ak l
l
π)25(2)04.0/(20 ππω ==
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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
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( ) ( )( ) ( )
( )( )
±==−−−=
−−−=
−−−=
−=
−−
−−−+
+−+
−−−
+−+
−−−
+−
+−
−−
−−
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±
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Spectrum
� Show plot
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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∑
−==
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Fourier Series Synthesis
3/25/2008 © 2003, JH McClellan & RW Schafer 18
Synthesis: 1st & 3rd Harmonics))75(2cos(
32))25(2cos(2
21)( 22
ππ ππ
ππ
−+−+= ttty
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Synthesis: up to 7th Harmonic)350sin(
72)250sin(
52)150sin(
32)50cos(2
21)( 2 ttttty π
ππ
ππ
ππ
ππ +++−+=
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Fourier Synthesisl+++= )3sin(
32)sin(2
21)( 00 tttxN ω
πω
π
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Gibbs’ Phenomenon
� Convergence at DISCONTINUITY of x(t)� There is always an overshoot� 9% for the Square Wave case
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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
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fseriesdemo GUI
3/25/2008 © 2003, JH McClellan & RW Schafer 24
Fourier Series Java Applet
3/25/2008 © 2003, JH McClellan & RW Schafer 25
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
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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
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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
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READING ASSIGNMENTS
� This Lecture:� Fourier Series in Ch 3, Sects 3-4, 3-5 & 3-6
� Review: entire Chap 3