Fourier Series and Transform
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Transcript of Fourier Series and Transform
1Fourier Series and Fourier Transform
2What 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
= + +1
0
0
sin )
where 1 ( ) and
21 ( ) cos , 1, 2,3,...and
1 ( )sin , 1, 2,3,...
and the , ' and ' are the socalled 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
=
=
= =
= =
ier coefficients.
3Derivation 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
= + +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
=
=
= == + +
= +
0
1
0 
cos ) 2
1 ( )2
nn
nxb an n
a f x dx
=
+ =
=
4Derivation of the Fourier Coefficients (contd)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 nx
== + +
= == +
 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
=+
= + + = == + +

0

) 0
Thus, (**)
( ) cos 0 0
1 = ( ) cos , 1, 2,...
m
m
m xdx
f x mxdx a a
a f x mxdx m
=
= + + =
5Derivation of the Fourier Coefficients (contd)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 nx
== + +
= == +
 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
=+
= + + == + =
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
=
= + + =
6Example 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 kd
=
<
7Fourier 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
== + +
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
= == == =
= = = + = +
= + = + =
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
==
==
= = +
= +
= + = =
1
d periodic function with period of 2 , then
( ) sinnn
f x b nx
==
8Fourier 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
== + +
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
==
=
== = +
= +
= +
* 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
=
=
= =
= +
9Representation by a Fourier series
Theorem:If a periodic function ( ) with period 2 is piecewise continuousin the interval , and has a lefthand derivative and righthand derivative at each point of that interval, then the Four
f xx
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 righthand 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 L = + +
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
=
=
= =
= =
= =
, 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
= =
++ = + = =
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 L = + +
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
=
= = = < < = < < = = = <
12
Halfrange 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 xn n nf x a a x b x b xL L L
n xb f x dxL L
= == + + =
=
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
==
=
= += +
= +
* 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
==== +
=
) for 0 .x x L=
13
Halfrange Expansion (contd)
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