fourier series and transform
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1
Fourier Series and Fourier Transform
2
What is a Fourier series ?
0
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) can be represented by atrigonometric series:
( ) ( cosn
f x f xf x
f x a a nx
π
= + +1
0
0
sin )
where 1 ( ) and
21 ( ) cos , 1, 2,3,...and
1 ( )sin , 1, 2,3,...
and the , ' and ' are the so-called Four
nn
n
n
n n
b nx
a f x dx
a f x nxdx n
b f x nxdx n
a a s b s
π
π
π
π
π
π
π
π
π
∞
=
−
−
−
=
= =
= =
∑
∫
∫
∫ier coefficients.
3
Derivation of the Fourier Coefficients
0
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) can be represented by atrigonometric series:
( ) ( cosn
f x f xf x
f x a a nx
π
= + +1
0- - - -1
0-
sin ).................(*)
Note that if we integrate both sides of (*) from to , we have:
( ) ( cos sin )
sin ( ) 2 (
nn
n nn
n
b nx
x x
f x dx a dx a nxdx b nxdx
nxf x dx a a
π π π π
π π π π
π
π
π π
π
∞
=
∞
=
= − =
= + +
⇒ = +
∑
∑∫ ∫ ∫ ∫
∫ 01
0 -
cos ) 2
1 ( )2
nn
nxb an n
a f x dx
π π
π π
π
π
π
π
∞
= − −
−⎤ ⎤+ =⎥ ⎥⎦ ⎦
⇒ =
∑
∫
4
Derivation of the Fourier Coefficients (cont’d)0
1
0- -
Now that ( ) ( cos sin ).................(*)
If we multiply both sides of (*) with cos and then integrate both sides from to , we have:
( ) cos cos ( cos c
n nn
n
f x a a nx b nx
mxx x
f x mxdx a mxdx a nxπ π
π π
π π
∞
=
= + +
= − =
= +
∑
∫ ∫ - -1
-
- - -
- -
os sin cos )....(**)
But: cos = 0 and
0 for 1 1cos cos cos( ) cos( ) and for 2 2
1 1sin cos sin( ) sin(2 2
nn
mxdx b nx mxdx
mxdx
n mnx mxdx n m xdx n m xdx
n m
nx mxdx n m xdx n
π π
π π
π
π
π π π
π π π
π π
π π
π
∞
=
+
≠⎧= + + − = ⎨ =⎩
= + +
∑ ∫ ∫
∫
∫ ∫ ∫
∫ ∫ -
0-
-
) 0
Thus, (**)
( ) cos 0 0
1 = ( ) cos , 1, 2,...
m
m
m xdx
f x mxdx a a
a f x mxdx m
π
π
π
π
π
π
π
π
− =
⇒ = ⋅ + +
⇒ =
∫
∫
∫
5
Derivation of the Fourier Coefficients (cont’d)0
1
0- -
Now that ( ) ( cos sin ).................(*)
If we multiply both sides of (*) with sin and then integrate both sides from to , we have:
( )sin sin ( cos s
n nn
n
f x a a nx b nx
mxx x
f x mxdx a mxdx a nxπ π
π π
π π
∞
=
= + +
= − =
= +
∑
∫ ∫ - -1
-
- - -
- - -
in sin sin )....(**)
But: sin = 0 and
1 1cos sin sin( ) sin( ) 0 and2 2
0 for 1 1sin sin cos( ) cos( )2 2
nn
mxdx b nx mxdx
mxdx
nx mxdx m n xdx m n xdx
n mnx mxdx n m xdx n m xdx
π π
π π
π
π
π π π
π π π
π π π
π π π π
∞
=
+
= + + − =
≠= − − + =
∑ ∫ ∫
∫
∫ ∫ ∫
∫ ∫ ∫
0-
-
and for
Thus, (**)
( ) sin 0 0
1 = ( )sin , 1, 2,...
m
m
n m
f x mxdx a b
b f x mxdx m
π
π
π
π
π
π
⎧⎨ =⎩
⇒ = ⋅ + +
⇒ =
∫
∫
6
Example of Fourier Series
01
0 -
if 0Consider ( ) and ( 2 ) ( ).
if 0Consider the Fourier series representation of
( ) ( cos sin ).................(*)
1 1( ) [2 2
n nn
k xf x f x f x
k x
f x a a nx b nx
a f x dx kdπ
π
ππ
π
π π
∞
=
− − < <⎧= + =⎨ < <⎩
= + +
= = −
∑
∫0
- 0
0
- - 0
0
- - 0
0
0
] ( ) 02
1 1( )cos [ cos cos ] 0
1 1( )sin [ sin sin ]
cos cos 2 ( ) [1 ( 1) ( 1) 1)] [1 ( 1) ]
Theref
n
n
n n n
kx kdx
a f x nxdx k nxdx k nxdx
b f x nxdx k nxdx k nxdx
k nx nx k kn n n n
π
π
π π
π π
π π
π π
π
π
π ππ
π π
π π
π π π−
+ = − =
= = − + =
= = − +
⎤ ⎤= − = − − − − + = − −⎥ ⎥⎦ ⎦
∫ ∫
∫ ∫ ∫
∫ ∫ ∫
0 for 2, 4,6,...ore, 4 for 1,3,5,...
4 1 1Thus, ( ) (sin sin 3 sin 5 ...)3 5
n
nb k n
nkf x x x x
π
π
=⎧⎪= ⎨
=⎪⎩
= + + +
7
Fourier series for Odd functions
01
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) can be represented by a
Fourier series: ( ) ( cos sin )
If ( ) is
n nn
f x f xf x
f x a a nx b nx
f x
π
∞
=
= + +∑
0 * 0
0 0 * 0
* *
* 0 0 * 0 0
an odd function, i.e. ( ) ( ) for all , then:1 1 1( ) [ ( ) ( ) ] [ ( *) ( *) ( ) ]
2 2 21 1[ ( *) * ( ) ] [ ( *) * ( ) ] 0
2 2A
x
x
x x
x x
f x f x x
a f x dx 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
π π π
π π π
π π π π
π π π
π π
=
− − =
= =
= =
− = −
= = + = − − +
= − + = − + =
∫ ∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
0
0
* 0
* 0
* 0
* 0
lso, 1 1( ) cos [ ( ) cos ( ) cos ]
1 [ ( *)cos( *) ( *) ( ) cos ]
1 [ ( *)cos( *) * ( ) cos ] 0 for 1, 2,3,...
Thus, if ( ) is an od
n
x
x
x
x
a f x nxdx f x nxdx f x nxdx
f x nx d x f x nxdx
f x nx dx f x nxdx n
f x
π π
π π
π
π
π
π
π π
π
π
− −
=
=
=
=
= = +
= − − − +
= − − + = =
∫ ∫ ∫
∫ ∫
∫ ∫
1
d periodic function with period of 2 , then
( ) sinnn
f x b nx
π∞
=
=∑
8
Fourier series for Even functions
01
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) can be represented by a
Fourier series: ( ) ( cos sin )
If ( ) is
n nn
f x f xf x
f x a a nx b nx
f x
π
∞
=
= + +∑
0
0
* 0
* 0
* 0
an even function, i.e. ( ) ( ) for all , then:1 1( )sin [ ( )sin ( )sin ]
1 [ ( *)sin( *) ( *) ( )sin ]
1 [ ( *)( 1)sin( *) * ( )cos ]
n
x
x
x
f x f x x
b f x nxdx f x nxdx f x nxdx
f x nx d x f x nxdx
f x nx dx f x nxdx
π π
π π
π
π
π
π
π π
π
π
− −
=
=
=
− =
= = +
= − − − +
= − − +
∫ ∫ ∫
∫ ∫
∫* 0
01
0 for 1, 2,3,...
Thus, if ( ) is an even periodic function with period of 2 , then
( ) cos
x
nn
n
f x
f x a a nx
π
=
∞
=
= =
= +
∫
∑
9
Representation by a Fourier series
Theorem:If a periodic function ( ) with period 2 is piecewise continuousin the interval , and has a left-hand derivative and right-hand derivative at each point of that interval, then the Four
f xx
ππ π− ≤ ≤
0
0
ier series (*) of ( ) is convergent and its sum is ( ), except at a point at which ( ) is discontinuous and thesum of the series is the average of the left- and right-hand limitsof ( ) at .
f x f xx f x
f x x
10
Functions of any period p=2L
0
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) has a Fourier series given by:
( ) ( cos sin )..n n
f x L f xf x
n nf x a a x b xL Lπ π
= + +1
0
.................(***)
where 1 ( ) and
21 ( ) cos , 1, 2,3,...and
1 ( )sin , 1, 2,3,...
Proof:
Substitute
n
L
L
L
n L
L
n L
a f x dxL
n xa f x dx nL L
n xb f x dx nL L
x Lvv xL
π
π
ππ
∞
=
−
−
−
=
= =
= =
= ⇔ =
∑
∫
∫
∫
, then corresponds to
thus, let ( ) = ( ). Then: ( 2 )( 2 ) ( )= ( +2L)= ( 2 ) ( ) ( )
( ) has a period of 2 .We can therefore expand ( ) using the previous results of (*).
x L v
g v f xL v Lvg v f f f x L f x g v
g vg v
π
πππ π
π
= ± = ±
++ = + = =
⇒
11
Functions of any period p=2L (Example)
0
Let ( ) be a PERIODIC function of period 2 , and ( ) satisfies some"commonly practical" conditions, then, ( ) has a Fourier series given by:
( ) ( cos sin )..n n
f x L f xf x
n nf x a a x b xL Lπ π
= + +1
0
.................(***)
1 1 1where ( ) , ( ) cos , ( )sin2
0 if 2 1Find the Fourier series for ( ) if 1 1 2 4 2
0 if 1 2Obs
n
L L L
n nL L L
n x n xa f x dx a f x dx b f x dxL L L L L
xf x k x period L L
x
π π
∞
=
− − −= = =
− < < −⎧⎪= − < < = = ⇒ =⎨⎪ < <⎩
∑
∫ ∫ ∫
2 1
0 2 1
1
1
erve that ( ) is an even function ' 01 1 1( ) ( )
2 4 4 2
0 if is even1 1 2 2( )cos cos sin if 1,5,9,...
2 2 22 if 3,7,11,...
Th
n
L
L
L
n L
f x b ska f x dx f x dx kdx
L
nn x n x k n ka f x dx k dx n
L L n nk n
n
π π ππ π
π
− − −
− −
⇒ =
= = = =
⎧⎪⎪⎪= = = = =⎨⎪−⎪ =⎪⎩
∫ ∫ ∫
∫ ∫
2 1 3 1 5 1 7us, ( ) (cos cos cos cos ...)2 2 3 2 5 2 7 2k kf x x x x xπ π π π
π= + − + − +
12
Half-range Expansion
1 1 1
1
Consider a function ( ) which is only defined for the interval0 , then we can construct
( ) for 0 ( ) and ( 2 ) ( )
( ) for 0Then, ( ) is a periodic, ODD, function of per
f xx L
f x x Lf x f x L f x
f x L xf x
≤ ≤
≤ ≤⎧= + =⎨− − ≤ ≤⎩
1
1 01 1
1
iod 2 , and thus, we can obtainthe Fourier series expansion for ( ), i.e.
( ) ( cos sin ) sin ..............(**)
1where ( )sin ,
n n nn n
n
Lf x
n n nf x a a x b x b xL L L
n xb f x dxL L
π π π
π
∞ ∞
= =
= + + =
=
∑ ∑
0
1 10
* 0
1* 0
1 0
1, 2,3,...
1Also that [ ( )sin ( )sin ]
1 ( *) [ ( *)sin ( *) ( )sin ]
1 ( *) [ ( *)sin ( *) ( )sin ]
L
L
L
n L
x L
x L
L
n
n x n xb f x dx f x dxL L L
n x n xf x d x f x dxL L L
n x n xf x d x f x dxL L L
π π
π π
π π
−
−
=
=
=
= +
−= − − +
= − +
∫
∫ ∫
∫ ∫* 0
*
*
1* 0 0
0
1
1 ( *) = [ ( *)sin ( *) ( )sin ]
2 = ( )sin
Once ' are determined, (**) can be used to represent ( ) for 0because ( ) (
x
x L
x L L
x
L
n
n x n xf x d x f x dxL L L
n xf x dxL L
b s f x x Lf x f
π π
π
=
=
=
=+
≤ ≤=
∫ ∫
∫ ∫
∫
) for 0 .x x L= ≤ ≤
13
Half-range Expansion (cont’d)
2 2 2
2
Alternatively, for the function ( ) which is only defined for the interval0 , we can construct
( ) for 0 ( ) and ( 2 ) ( )
( ) for 0Then, ( ) is a periodic, EVEN, functio
f xx L
f x x Lf x f x L f x
f x L xf x
≤ ≤
≤ ≤⎧= + =⎨ − − ≤ ≤⎩
2
2 0 01 1
n of period 2 , and thus, we can obtainthe Fourier series expansion for ( ), i.e.
( ) ( cos sin ) cos ..............(**)
where
n n nn n
Lf x
n n nf x a a x b x a a xL L Lπ π π∞ ∞
= =
= + + = +∑ ∑
0 2 0
2 0
1 1 ( ) ( )21 2 ( ) cos ( ) cos 1, 2,3,...
Once the ' are determined, (**) can be used to represent ( ) for 0because ( )
L L
L
L L
n L
n
a f x dx f x dxL L
n x n xa f x dx f x dx nL L L L
a s f x x Lf x f
π π−
−
= =
= = =
≤ ≤=
∫ ∫
∫ ∫
2 ( ) for 0 .x x L= ≤ ≤
14
Complex Fourier series
01
0
Consider the Fourier series representation for a PERIODIC function ( ) of period 2 :
( ) ( cos sin )....................(*)
1 1where ( ) , ( ) c2
n nn
n
f x
f x a a nx b nx
a f x dx a f xπ
π
π
π π
∞
=
−
= + +
= =
∑
∫1os , ( )sin
It can be shown that ( ) can also be represented in the following alternative form:
1 ( ) where ( )2
n
inx inxn n
n
nxdx b f x nxdx
f x
f x c e c f x e dx
π π
π π
π
π
π
π
− −
∞−
−=−∞
=
= =
∫ ∫
∑ ∫
15
Complex Fourier series (cont’d)0
1
0
Proof: Consider ( ) ( cos sin )....................(*)
1 1 1where ( ) , ( ) cos , ( )sin2
Recall that cos sin ; cos sin1cos2
n nn
n n
inx inx
f x a a nx b nx
a f x dx a f x nxdx b f x nxdx
e nx i nx e nx i nx
nx
π π π
π π ππ π π
∞
=
− − −
−
= + +
= = =
= + = −
⇒ =
∑
∫ ∫ ∫
( ) ( ) ( )
( ) ( )
0 0
1 and sin2 2
1 1( cos sin ) = ( ) ( )2 2 2 2
1 1Define = ; and = ( ) and = ( ); (*) becomes:2 2
inx inx inx inx inx inx
inx inx inx inx inx inxn nn n n n n n
n n n n n n
ie e nx e e e ei
a ba nx b nx e e i e e a ib e a ib e
c a c a ib k a ib
− − −
− − −
−+ = − = −
⇒ + = + − − − + +
− +
01
( ) ( )...................(**)
1 1 1Also that = ( ) ( ( ) cos ( )sin ) ( )2 2 2
1 1 1and = ( ) ( ( ) cos ( )sin )2 2 2
inx inxn n
n
inxn n n
n n n
f x c c e k e
c a ib f x nx i f x nx dx f x e dx
k a ib f x nx i f x nx dx
π π
π π
π
π
π π
π π
∞−
=
−
− −
−
= + +
− = − ⋅ =
+ = + ⋅ =
∑
∫ ∫
∫ ( )
Finally, by defining and substitute by in (**) to get:1 ( ) where ( ) , 0, 1, 2,...
2
inx
n n n n
inx inxn n
n
f x e dx
c k k c
f x c e c f x e dx n
π
π
π
ππ
−
− −
∞−
−=−∞
=
= = = ± ±
∫
∑ ∫
16
Complex Fourier series (cont’d)
For a periodic function ( ) of period =2 ,1 ( ) where ( ) , 0, 1, 2,...
2is the so-called complex form of Fourier series or just complex Fourier ser
inx inxn n
n
f x
f x c e c f x e dx nπ
π
π
π
∞−
−=−∞
= = = ± ±∑ ∫ies
For a periodic function ( ) with period = 2 , the corresponding complexFourier series is given by:
( )
1where ( ) , 0,2
ni xL
nn
ni xLL
n L
f x L
f x c e
c f x e dx nL
π
π
∞
=−∞
−
−
=
= =
∑
∫ 1, 2,...± ±
17
Complex Fourier series: An Example
(1
Consider a periodic function ( ) of period =2 , where ( ) for 1Let ( ) where ( ) , 0, 1, 2,...
2then
1 1 1( )2 2 2
x
inx inxn n
n
inx x inxn
f x f x e x
f x c e c f x e dx n
ec f x e dx e e dx
π
π
π π
π π
π π π
π
π π π
∞−
−=−∞
−− −
− −
= − < <
= = = ± ±
= = =
∑ ∫
∫ ∫)
2
2 2
1 ( )1 2 1
And since cos sin cos ( 1)1 1 1and , and sinh
1 (1 )(1 ) 1 2( 1) sinh (1 ) sinh ( 1) (1 )( ) ,
(1 ) (1 )
in x inx
in n
n nx inx
nn
e e ein in
e n n nin in e e
in in in nin inc f x e e
n n
π π π
π
π
π π
π
π π π
π
π ππ π
− −
−
±
−
∞
=−∞
⎤ −=⎥− −⎦
= ± = = −
+ + −= = =
− − + +
− + − +⇒ = ⇒ = =
+ +∑
Furthermore, (1 ) (1 )(cos sin ) (cos sin ) ( cos sin )and (1 ) (1 )(cos sin ) (cos sin ) ( cos sin )
(1 ) (1 ) 2(cos sin )
Thus,
inx
inx
inx inx
x
in e in nx i nx nx n nx i n nx nxin e in nx i nx nx n nx i n nx nx
in e in e nx n nx
f
π π
−
−
− < <
+ = + + = − + +
− = − − = − − +
⇒ + + − = −
21
sinh ( 1) 2(cos sin )( ) [1 ], (1 )
nx
n
nx n nxx e xn
π π ππ
∞
=
− −= = + − < <
+∑
18
Representation for Aperiodic function: Fourier TransformConsider an APERIODIC function ( ). Define ( ) be the Fourier Transform of ( ) as follows:
( ) ( )
It can be shown that: 1 ( ) (
2
i x
f xF i f x
F i f x e dx
f x F i
ω
ω
ω
π
∞ −
−∞=
=
∫
{ }
{ }{ }1
1
)
Let's define as the Fourier transform operator, then
( ) ( ) = ( )
also define as the Inverse Fourier transform operator, then
( )
i x
i x
e d
F i f x f x e dx
f x
ω
ω
ω ω
ω
∞
−∞
∞ −
−∞
−
−
=
=
∫
∫
F
F
F
F { } 1( ) ( ) 2
i xF i F i e dωω ω ωπ
∞
−∞= ∫
19
Representation for Aperiodic function: Fourier TransformConsider an APERIODIC function ( ) defined for the interval ,
and ( ) 0 otherwise. Define a periodic function ( ) of period =2 , i.e. ( ) ( 2 )
and ( ) ( ) for
Then, the Com
f x L x L
f x f x L f x f x L
f x f x L x L
− ≤ ≤
= = +
= − < <
plex Fourier series for ( ) is given by:
1 1( ) .........(1) where ( ) ( ) ......(2)2 2
Define ( ) ( ) ......................(3)
Sub.
n n ni x i x i xLL L L
n n Ln
i x
f x
f x c e c f x e dx f x e dxL L
F i f x e dx
π π π
ωω
∞ − −∞
− −∞=−∞
∞ −
−∞
= = =
=
∑ ∫ ∫
∫
0 0
0 0
0 00
(3) into (2) 2 ( ) ( ) where = ,
( ) ( ) = ( ) .......(4)2 2
As , ( ) approaches ( ) and consequently (4) becomes :
n
in x in x
n n
nL c F i F inL L
F inf x e F in eL
L f x f x
f
ω ω
π πω ω
ω ω ωπ
∞ ∞
=−∞ =−∞
⇒ ⋅ = =
⇒ =
→∞
∑ ∑
000
0 0 0
( ) ( ) ...............(5)2
Also, since = , 0 as , let , and
thus, the summation in (5) becomes an integral:1 ( ) ( ) ......
2
in x
n
n
i x
x F in e
L nL
f x F i e d
ω
ω
ω ωπ
πω ω ω ω ω ω
ω ωπ
∞
=−∞
∞
−∞
=
→ →∞ = ∆ =
⇒ =
∑
∫ .........(6)
From (3), ( ) is the so-called Fourier transform of a function ( ).F i f xω
20
Fourier transform: An example
{ }0
, for 0Find the Fourier transform of ( )
0 otherwise
( ) ( ) ( ) = ( 1) ( 1)
Note that ( ) is in general a complex-value function.
ai x i x i a i a
k x af x
k i kF i f x f x e dx ke dx e ei
F i
ω ω ω ωωω ω
ω
∞ − − − −
−∞
< <⎧= ⎨⎩
⋅= = = − = −
−∫ ∫F
21
Some useful Properties of Fourier transform
{ } { } { }
{ } { }
{ }
{ }
0
0
0
0
1. Linearity: ( ) ( ) [ ( ) ( )] ( ) ( )
2. Time-shifting: ( ) ( )
3. Frequency-shifting: ( ) ( ( ))
14. Time and Frequency Scaling: ( ) ( )
i x
i x
i x
af x bg x af x bg x e dx a f x b g x
f x x e f x
e f x F i
if ax Fa a
ω
ω
ω ω ω
ω
∞−
−∞
−
+ = + = +
− =
= −
=
∫F F F
F F
F
F
22
Some useful Properties of Fourier transform (cont’d)
{ } { }
{ } { }2
5. Differentiation in Time: If ( ) 0 as , then
'( ) '( ) ( ) ( ) ( ) ( )
and ''( ) ( )
16. Integration: ( *) * ( ) (0) (
i x i x i x
x
f x x
f x f x e dx f x e i f x e dx i f x
f x f x
f x dx F i Fi
ω ω ωω ω
ω
ω π δω
∞ ∞∞− − −
−∞−∞ −∞
−∞
→ →∞
⎤= = − − =⎦
= −
⎧ ⎫= +⎨ ⎬
⎩ ⎭
∫ ∫
∫
F F
F F
F
{ }
)
( )7. Differentiation in Frequency: ( )
8. Convolution: Let the convolution * of functions and be defined by:
( ) ( * )( ) ( ) ( ) ( ) ( )
then:
dF ixf x id
f g f g
h x f g x f p g x p dp f x p g p dp
ω
ωω
∞ ∞
−∞ −∞
=
= = − = −∫ ∫
F
{ } { } { } ( * )( ) ( ) ( )f g x f x g x=F F F
23
Fourier Transform of Periodic functions
{ }
0
0
0
10
The concept of Fourier transform also applies to periodic functions as well. 1Consider ( ) 2 ( ) ( ) ( )
2
Thus, for ( ) 2 ( ), ( ) ( )
o
i xi x
in xn n
n n
F i f x F i e d e
F i c n f x F i c e
ωω
ω
ω πδ ω ω ω ωπ
ω πδ ω ω ω
∞
−∞
∞ ∞−
=−∞ =−∞
= − ⇒ = =
= ⋅ − = =
∫
∑ ∑F
00r 2 ( )................(*)
For any period function ( ), say of period = 2 , since we can express ( ) as
a Complex Fourier series: ( )
in xn n
n n
ni xL
n nn n
c e c n
f x L f x
f x c e c
ω
π
πδ ω ω∞ ∞
=−∞ =−∞
∞
=−∞ =−∞
⎧ ⎫ = ⋅ −⎨ ⎬⎩ ⎭
= =
∑ ∑
∑
F
0
0
0
0
where
From (*), the Fourier transform of this periodic function ( ) is given by:
( ) 2 ( )
where 's are the coefficients of
in x
in xn n
n n
n
eL
f x
F i c e c n
c
ω
ω
πω
ω πδ ω ω
∞
∞ ∞
=−∞ =−∞
=
⎧ ⎫= = ⋅ −⎨ ⎬⎩ ⎭
∑
∑ ∑F
the Complex Fourier series of ( ).f x
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