Fourier Series and Transform

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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