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MATHS

### 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 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

=

=

= =

= =

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 left-hand derivative and right-hand 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 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 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

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 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

Half-range 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