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The Fourier Series
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Joseph Fourier 1768 - 1830
Joseph Fourier
Fourier wasobsessed with thephysics of heat anddeveloped theFourier series and
transform to modelheat-flow problems.
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Anharmonic waves are sums of sinusoids.
Consider the sum of two sine waves (i.e., harmonic waves)of different frequencies:
The resulting wave is periodic, but not harmonic.Essentially all waves are anharmonic.
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Fourierdecomposing
functions
Here, we write asquare wave asa sum of sinewaves.
sin(wt)
sin(3wt)
sin(5wt)
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A Fourier series is an expansion of aperiodicfunctionf(t) in terms of an infinite sum
of cosines and sines
Introduction
1
0 )sincos(2
)(
n
nn tnbtnaa
tf ww
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In other words, any periodicfunction can beresolved as a summation ofconstant value and cosine and sine functions:
1
0
)sincos(2)( nnn tnbtna
a
tf ww
)sincos( 11 tbta ww 2
0a
)2sin2cos( 22 tbta ww
)3sin3cos( 33 tbta ww
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The computation and study of Fourierseries is known as harmonic analysisandis extremely useful as a way to break upan arbitrary periodic function into a set ofsimple terms that can be plugged in,solved individually, and then recombinedto obtain the solution to the original
problem or an approximation to it towhatever accuracy is desired or practical.
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=
+ +
+ + +
Periodic Function
2
0a
ta wcos1
ta w2cos2
tb wsin1
tb w2sin2
f(t)
t
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1
0
)sincos(2)( nnn tnbtna
a
tf ww
where
T
dttfT
a0
0 )(2
frequencylFundementa2
T
w
T
n tdtntfTa0
cos)(2
w
T
n tdtntfTb0
sin)(2
w
*we can also use the integrals limit .
T
dttfT
a0
0 )(2
2/
2/
T
T
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Example 1
Determine the Fourier series representation of thefollowing waveform.
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Solution
First, determine the period & describe the one periodof the function:
T= 2
21,0
10,1)(
t
ttf )()2( tftf
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Then, obtain the coefficients a0, an and bn:
10101)(2
2)(
22
1
1
0
2
00
0 dtdtdttfdttfTa
T
Or, sincey =f(t) over the interval [a,b], hence
b
a
dttf )( is the total area below graph
1)11(22
],0[overgraphbelowArea2)(2
0
0
TT
dttfT
a
T
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w
n
n
n
tndttdtn
tdtntfT
an
sinsin0cos1
cos)(2
1
0
2
1
1
0
2
0
Notice that n is integer which leads ,since
0sin n03sin2sinsin
Therefore, .0na
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w
n
n
n
tndttdtn
tdtntfTbn
cos1cos0sin1
sin)(
2
1
0
2
1
1
0
2
0
15cos3coscos 16cos4cos2cos
Notice that
Therefore,
even,0
odd,/2)1(1
n
nn
nb
n
n
orn
n )1(cos
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ttt
tnn
tnbtnaa
tf
n
n
n
nn
ww
5sin5
23sin
3
2sin
2
2
1
sin)1(1
2
1
)sincos(2
)(
1
1
0
Finally,
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Some helpful identities
For n integers,n
n )1(cos 0sin n
02sin
n 12cos
n
xx sin)sin( xx cos)cos(
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[Supplementary]
The sum of the Fourier series terms canevolve (progress) into the original
waveform
From Example 1, we obtain
ttttf
5sin5
2
3sin3
2
sin
2
2
1
)(
It can be demonstrated that the sum willlead to the square wave:
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tttt
7sin7
25sin
5
23sin
3
2sin
2ttt
5sin5
23sin
3
2sin
2
tt
3sin3
2sin
2t
sin2
(a) (b)
(c) (d)
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ttttt
9sin9
27sin
7
25sin
5
23sin
3
2sin
2
ttt
23sin23
23sin
3
2sin
2
2
1
(e)
(f)
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Example 2
Given ,)( ttf 11 t
)()2( tftf
Sketch the graph off(t) such that .33 t
Then compute the Fourier series expansion off(t).
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Solution
The function is described by the following graph:
T= 2
w
T
2We find that
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Then we compute the coefficients:
02
11
22
2
)(2
1
1
21
1
1
1
0
ttdt
dttfT
a
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0coscos
)cos(cos0
cos)]sin([sin
sinsin
coscos)(2
22
22
1
1
22
1
1
1
1
1
1
1
1
w
n
nnn
nn
n
tn
n
nn
dtn
tn
n
tnt
tdtnttdtntf
T
an
since xx cos)cos(
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w
nnn
n
n
nn
n
n
n
tn
n
nn
dtn
tn
n
tnt
tdtnttdtntfT
b
nn
n
1
22
1
1
22
1
1
1
1
1
1
1
1
)1(2)1(2cos2
)sin(sincos2
sin)]cos([cos
coscos
sinsin)(2
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ttt
tnn
tnbtnaa
tf
n
n
n
nn
ww
3sin
3
22sin
2
2sin
2
sin)1(2
)sincos(2
)(
1
1
1
0
Finally,
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Example 3
Given
42,0
20,2)(
t
tttv
)()4( tvtv
Sketch the graph of v (t) such that .120 t
Then compute the Fourier series expansion of v (t).
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Solution
The function is described by the following graph:
T= 4
2
2 w
T
We find that
0 2 4 6 8 10 12t
v (t)
2
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Then we compute the coefficients:
1
2
2
2
1)2(
2
1
0)2(42
)(2
2
0
22
0
4
2
2
0
4
0
0
ttdtt
dtdtt
dttvT
a
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222222
2
0
22
2
0
2
0
4
2
2
0
4
0
])1(1[2)cos1(2
2
2cos1
cos
2
10
sin
2
1sin)2(
2
1
0cos)2(21cos)(2
w
w
w
w
w
w
w
w
ww
nn
n
n
n
n
tn
dt
n
tn
n
tnt
tdtnttdtntvT
a
n
n
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ww
w
w
w
w
w
w
w
w
w
ww
nnnn
n
n
tn
n
dtn
tn
n
tnt
tdtnttdtntvTbn
212
2sin1
sin
2
11
cos
2
1cos)2(
2
1
0sin)2(2
1sin)(
2
22
2
0
22
2
0
2
0
4
2
2
0
4
0
since 0sin2sin w nn
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122
1
0
2sin
2
2cos
])1(1[2
2
1
)sincos(2
)(
n
n
n
nn
tn
n
tn
n
tnbtnaa
tv
ww
Finally,
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Symmetry Considerations
Symmetry functions:
(i) even symmetry
(ii) odd symmetry
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Even symmetry
Any functionf(t) is even if its plot issymmetrical about the vertical axis, i.e.
)()( tftf
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Even symmetry (cont.)
The examples of even functions are:
2)( ttf
t t
t
||)( ttf
ttf cos)(
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Even symmetry (cont.)
The integral of an even function from A to+A is twice the integral from 0 to +A
t
AA
Adttfdttf
0
eveneven )(2)(A +A
)(even tf
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Odd symmetry
Any functionf(t) is odd if its plot isantisymmetrical about the vertical axis, i.e.
)()( tftf
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Odd symmetry (cont.)
The examples of odd functions are:
3)( ttf
t t
t
ttf )(
ttf sin)(
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Odd symmetry (cont.)
The integral of an odd function from A to+A is zero
t 0)(odd
A
A
dttfA
+A
)(odd tf
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Even and odd functions
(even)(even) = (even)
(odd)(odd) = (even)
(even)
(odd) = (odd) (odd)(even) = (odd)
The product properties of even and oddfunctions are:
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Symmetry consideration
From the properties of even and oddfunctions, we can show that:
for even periodic function;
2/
0
cos)(4
T
n tdtntfT
a w 0nb
for odd periodic function;
2/
0
sin)(4
T
n tdtntfT
b w00 naa
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How?? [Even function]
2
T
2
T
2/
0
2/
2/
cos)(4
cos)(2
TT
T
n tdtntfT
tdtntfT
a ww
(even) (even)
| |
(even)
0sin)(2
2/
2/
T
T
n tdtntfT
b w
(even) (odd)
| |
(odd)
)(tf
t
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How?? [Odd function]
2
T
2
T
2/
0
2/
2/
sin)(4
sin)(2
TT
T
n tdtntfT
tdtntfT
b ww
(odd)(odd)
| |
(even)
0cos)(22/
2/
T
T
n tdtntfT
a w
(odd) (even)
| |(odd)
)(tf
t
0)(2
2/
2/
0
T
T
dttfT
a
(odd)
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L 0 0
-L 0 0
0
0 0
0 0
- 0
If f(x) is an even function
f(x)dx ( ) ( ) ( ) ( ) ( )
( ) ( ) 2 ( )
If f(x) is an odd function, then
( ) ( ) ( ) ( ) ( ) ( )
L L
L L
L L
L
L L
L L L
f x dx f x dx f x d x f x dx
f x dx f x dx f x dx
f x dx f x dx f x dx f x d x f x dx
00
0( ) ( ) 0
If f(x) is even and g(x) is odd, then
h(x)=f(x)g(x) is an odd function
h(x)=f(x)g(x)=f(-x)[-g(-x)]=-[f(-x)g(-x)]=-h(-x)
L
L
Lf x dx f x dx
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Example 4
Given
21,1
11,
12,1
)(
t
tt
t
tf
)()4( tftf
Sketch the graph off(t) such that .66 t
Then compute the Fourier series expansion off(t).
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Solution
The function is described by the following graph:
T= 4
2
2 w
T
We find that
046 2 4 6t
f(t)
2
1
1
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Then we compute the coefficients. Sincef(t) is
an odd function, then
0)(2
2
2
0
dttfT
a
0cos)(2
2
2
tdtntf
T
an w
and
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w
w
w
w
w
ww
w
w
w
w
w
w
w
w
w
w
ww
ww
n
n
n
n
n
nn
nn
n
tn
n
n
ntndt
ntn
ntnt
tdtntdtnt
tdtntfT
tdtntfT
bn
cos2sin2cos
cos2cossincos
coscoscos
sin1sin4
4
sin)(4
sin)(2
22
1
0
22
2
1
1
0
1
0
2
1
1
0
2
0
2
2
since 0sin2sin w
nn
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1
1
1
1
0
2sin)1(2
2sin
cos2
)sincos(
2
)(
n
n
n
n
nn
tnn
tn
n
n
tnbtnaa
tf
ww
Finally,
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Example 5
Compute the Fourier series expansion off(t).
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Solution
The function is described by
T= 3
3
22 w
Tand
32,1
21,2
10,1
)(
t
t
t
tf
)()3( tftf
T= 3
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Then we compute the coefficients.
3
81
2
32)01(
3
421
3
4)(
4)(
22/3
1
1
0
2/3
0
3
0
0
dtdtdttfT
dttfT
a
3
8)23()12(2)01(
3
2121
3
2)(
2 3
2
2
1
1
0
3
0
0
dtdtdtdttfTa
Or, sincef(t) is an even function, then
Or, simply
3
84
3
2
periodain
graphbelowareaTotal2)(
23
0
0
TdttfTa
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3
2sin
2
3
2sinsin2
2
sin2
3sin2
3
4
sin2
3sin2sin
3
4
sin2
3
4sin
3
4
cos2cos13
4
cos)(4
cos)(2
2/3
1
1
0
2/3
1
1
0
2/3
0
3
0
w
w
w
w
w
w
w
w
w
w
w
ww
ww
n
n
nn
n
nn
n
nn
nn
n
tn
n
tn
tdtntdtn
tdtntfT
tdtntfT
an
;3
2w
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1
1
1
0
3
2cos
3
2sin
12
3
4
3
2cos
3
2sin
2
3
4
)sincos(2
)(
n
n
n
nn
tnn
n
tnn
n
tnbtnaa
tf
ww
Finally,
and 0nb sincef(t) is an even function.
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Function defines over a finite interval
Fourier series only support periodic functions
In real application, many functions are non-
periodic The non-periodic functions are often can be
defined over finite intervals, e.g.
y = 1 y = 1
y = 2
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Therefore, any non-periodic function must be
extended to a periodic function first, beforecomputing its Fourier series representation
Normally, we prefer symmetry (even or odd)periodic extension instead of normal periodic
extension, since symmetry function will providezero coefficient of either an or bn
This can provide a simpler Fourier seriesexpansion
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)(ty
t
)(tf
t
)(even tf
)(odd tf
t
t
lttytf 0,)()(
)()( tfltf lT
0,)(
0,)()(tlty
lttytf
)()2( tfltf
lT 2
0,)(
0,)()(
tlty
lttytf
)()2( tfltf
lT 2
l0
l0 l2ll2
l0 l2ll2
l3l3
l3l3
l0 l2ll2 l3l3
T
T
T
Periodic extension
Even periodic extension
Odd periodic extension
Non-periodicfunction
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Half-range Fourier series expansion
The Fourier series of the even or oddperiodic extension of a non-periodic
function is called as the half-range Fourierseries
This is due to the non-periodic function is
considered as the half-range before it isextended as an even or an odd function
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If the function is extended as an even
function, then the coefficient bn= 0, hence
1
0 cos2
)(n
n tnaa
tf w
which only contains the cosine harmonics. Therefore, this approach is called as the
half-range Fourier cosine series
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If the function is extended as an odd
function, then the coefficient an= 0, hence
1
sin)(n
n tnbtf w
which only contains the sine harmonics. Therefore, this approach is called as the
half-range Fourier sine series
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Example 6
ttf 0,1)(
Compute the half-range Fourier sine series expansionoff(t), where
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Solution
Since we want to seek the half-range sine series,the function to is extended to be an odd function:
T= 2
12
T
w
0 t
f(t)
1
1
220 t
f(t)
1
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Hence, the coefficients are
00 naa
w
0
2/
0sin12
4
sin)(
4
ntdttdtntfTb
T
n
and
even,0
odd,/4)cos1(
2cos2
0 n
nnn
nn
nt
Therefore,
ntn
ntnn
tf
nnn
sin4
sin)cos1(2
)(
odd11
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Example 7
10,12)(
tttf
Determine the half-range cosine series expansionof the function
Sketch the graphs of bothf(t) and the periodicfunction represented by the series expansion for3 < t< 3.
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Solution
Since we want to seek the half-range cosine series,the function to is extended to be an even function:
T= 2
w T
2
t
f(t)
t
f(t)
1
2233 1
1
11
11
2
1
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Hence, the coefficients are
02)12(24
)(
4 1
0
2
1
0
2/
00
ttdttdttfTa
T
1
0
22
1
0
1
0
1
0
2/
0
cos4sin2
2sin
2sin)12(
2
cos)12(2
4cos)(
4
w
ntn
nn
dtn
tn
n
tnt
tdtnttdtntfT
a
T
n
even,0
odd,/8)1(cos422
22
n
nn
n
n
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0nb
Therefore,
odd1
22
odd1
22
1
0
cos18
cos8
0
cos)(
nn
nn
n
n
tnn
tnn
tnaatf
w
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Exponential Fourier series
Recall that, from the Eulers identity,
xjxejx sincos
yields
2cos
jxjx eex
2
sin
j
eex
jxjx and
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Then the Fourier series representation becomes
11
0
1
0
1
0
1
0
1
0
222
222
222
222
)sincos(2
)(
n
tjnnn
n
tjnnn
n
tjnnntjnnn
n
tjntjn
n
tjntjn
n
n
tjntjn
n
tjntjn
n
n
nn
ejba
ejbaa
ejbaejbaa
eejb
eea
a
j
eeb
eea
a
tnbtnaa
tf
ww
ww
wwww
wwww
ww
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Here, let we name
11
0
222)(
n
tjnnn
n
tjnnn ejba
ejbaa
tf ww
2
nnn
jbac
,
2
nnn
jbac
Hence,
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
n
tjn
n
ececcec
ececc
ececc
www
ww
ww
1
0
1
11
0
11
0
and .20
0
ac
c0 cncn
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Then, the coefficient cn can be derived from
T
tjn
T
TT
TT
nnn
dtetfT
dttnjtntfT
tdtntfjtdtntfT
tdtntfT
jtdtntf
T
jbac
0
0
00
00
)(1
]sin)[cos(1
sin)(cos)(1
sin)(2
2cos)(
2
2
1
2
w
ww
ww
ww
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In fact, in many cases, the complexFourier series is easier to obtain rather
than the trigonometrical Fourier series
In summary, the relationship between thecomplex and trigonometrical Fourier series
are:
2
nnn
jbac
2
nnn
jbac
T
dttfT
ac
0
00 )(
1
2
T
tjn
n dtetfT
c0
)(1
w
nn cc or
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Example 8
Obtain the complex Fourier series of the followingfunction
2 44 2 0
2e
1
)(tf
t
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Since , . Hence
Solution
2
1
2
1
2
1
)(1
22
0
2
0
0
0
ee
dte
dttf
T
c
t
t
T
1w2T
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)1(21
)1(21
)1(21
12
1
2
1
2
1
)(1
222)1(2
2
0
)1(
2
0
)1(
2
0
0
jn
e
jn
ee
jn
e
jn
e
dtedtee
dtetf
T
c
njjn
tjn
tjnjntt
T
tjn
n
w
since1012sin2cos
2 njne
nj
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jnt
nn
tjn
n ejn
eectf
)1(2
1)(
2
w
Therefore, the complex Fourier series off(t) is
0
2
0
2
0 2
1
)1(2
1
c
e
jn
e
cn
nn
*Notes: Even though c0 can be found by substitutingcn with n = 0, sometimes it doesnt works (as shownin the next example). Therefore, it is always better to
calculate c0 alone.
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2
2
121n
ecn
cn is a complex term, and it depends on n.Therefore, we may plot a graph of |cn| vs n.
In other words, we have transformed the functionf(t)in the time domain (t), to the function cn in thefrequency domain (n).
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Example 9
Obtain the complex Fourier series of the function inExample 1.
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Solution
2
11
2
1)(
11
00
0 dtdttfTc
T
w
)1(22
1
012
1)(
1
1
0
2
1
1
00
w
jntjn
tjn
T
tjn
n
en
j
jn
e
dtedtetfT
c
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)1(2
jn
n en
jc
But njn nnjne )1(cossincos
Thus,
even,0
odd,/]1)1[(2 n
nnjnj n
Therefore,
odd0
2
1)(
nn
n
tjn
n
tjn
n en
jectf
w
*Here notice that .00
ccn
n
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even,0
odd,1
n
nncn
The plot of |cn| vs n is shown below
2
10 c
0.5
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Exercise 1
Obtain the complex Fourier series of thefunction bellow;
f(x) =1, - x < 0
0, 0 x <
S l ti
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dxedxedxec inxinxinxn
0
0
0
)1(2
1)0(
2
1)1(
2
1
0
02
1)1(
2
1
dxc
)cos1(2
12
nnie
nic
in
n
Solution
Kita tinjau untuk n = 0 dan n 0
a. Untuk n = 0
b. Untuk n 0
= n
i
n ganjil
0, n genap
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...5
1
3
1
3
1
5
1
...2
1
)(5335 ixixixixixix
eeeeee
i
xf
...
5
5sin
3
3sin
1
sin2
2
1)(
xxxxf
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Parsevals Theorem
Parservals theorem states that the
average power in a periodic signal is equal
to the sum of the average power in its DCcomponent and the average powers in itsharmonics
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=
+ +
+ + +
2
0a
ta wcos1
ta w2cos2
tb wsin1
tb w2sin2
f(t)
t
PavgPdc
Pa1 Pb1
Pa2 Pb2
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For sinusoidal (cosine or sine) signal,
R
V
R
V
R
VP
2
peak
2peak
2
rm s
2
12
For simplicity, we often assumeR= 1,
which yields2
peak21 VP
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For sinusoidal (cosine or sine) signal,
2
2
2
2
2
1
2
1
2
0
dcavg
2
1
2
1
2
1
2
1
2
2211
babaa
PPPPPP baba
1
222
0avg )(2
1
4
1
n
nn baaP
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)()2/(2
1)(
1 22
1
20
2
n
n
n
La
a
baaxfL
inpx
necxf
)(
22
)(1
n
n
La
a
cxfL
Parsevals Identity
For exponential form
The Parsevals identity
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Example 10
Use Parseval theorem to determine sum of number series in
the exercise 1
ii
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C0= )cos1(2
12
nn
ie
n
ic
in
n
22
)(1
n
n
La
a
cxfL
2
10
2
11
2
1)(
2
12
0
202
dxdxdxxf
Parseval identity
2
2
2
)cos1(2
12/1
2
1
nn
cn
n
.. .
7
1
5
1
3
11
2
2
12222
2
12
S