Fourier .Fourier Transforms The complex Fourier series has an important ... The Fourier transform

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Transcript of Fourier .Fourier Transforms The complex Fourier series has an important ... The Fourier transform

  • Fourier Transforms

    The complex Fourier series has an important limiting form when

    the period approaches infinity, i.e., T0 or L . Suppose that in this limit

    (1) k =nLremains large (ranging from to ) and

    (2) cn 0 since it is proportional to 1/L, but

    g(k) = limLcn0

    Lcn =

    12

    f (x)e ikx

    dx = finite

    then we have

    f (x) = cneikx

    n=

    = limLcn0

    12

    Lg(k)eikx

    n=

    where k =nL. The sum over n is in steps of n = 1. Thus, we can

    write using

    k =

    Ln

    which becomes infinitesimally small when L becomes large, as a sum over k, which becomes an integral in the limit

    f (x) = limLcn0

    12

    nL

    g(k)eikx =n=

    limk0

    12

    kg(k)eikxk=

    = 12

    g(k)eikxdk

    We call g(k) the Fourier Transform of f(x)

    g(k) = 1

    2f (x)e ikx

    dx = F( f )and the last equation is the so-called Fourier inversion formula.

    We can now obtain an integral representation of the delta-function. This corresponds to the orthogonality condition for the complex exponential Fourier series.

    We substitute the definition of g(k) into the inversion formula to get

    Page 1

  • f (x) = 12

    g(k)eikxdk

    = dkeikx1

    2dx '

    e ikx ' f (x ')

    = dx ' f (x ')

    1

    2dk

    eik (x x ')

    = dx ' f (x ')

    (x x ')

    where

    (x x ') = 1

    2dk

    eik (x x ')

    Properties

    The evaluation of the integrals involved in many Fourier transforms involves complex integration, which we shall learn later. We will just state some properties

    Examples:

    The Fourier transform of the box function

    f (x) =

    1 x a0 x a

    is

    F( f ) = 1

    2dx '

    e ikx ' f (x ') =12

    dx '

    e ikx ' =12

    e ikx '

    ik

    =12

    2sin kk

    The Fourier transform of the derivative of a function is

    Fdfdx

    =

    12

    dx '

    e ikx 'df (x ')dx '

    =12

    f (x ')e ikx '

    (ik) dx '

    e ikx ' f (x ')

    = ik2

    dx '

    e ikx ' f (x ') = ikF( f )

    where we have assumed that f (x) 0 as x . This generalizes to

    F

    dn fdxn

    = (ik)n F( f )

    Other useful properties of the Fourier transform are:

    F( f (x)) = g(k) , F( f (x a)) = e ikag(k) , F( f (x)eax ) = g(k + ia)

    A short table of Fourier Transforms is shown below:

    Page 2

  • f (x) g(k)

    (x) 12

    0 x>0ex x

  • InvFT F( )G( )[ ] = 1

    2f (t) g(t) f (t) g(t)[ ] = 2F( )G( )

    An example will help us visualize the real complexity of the convolution operation. We consider the convolution of two functions

    h(t) = f (t) g(t) = d f ( )g(t )

    Where we choose f(t) = square pulse and g(t) = triangular pulse as shown below.

    Now h(t) = area under product integrand f ( )g(t ) as a function of t.

    Procedure(must be done carefully):

    (1) plot f ( ) and g(t ) versus ; h(t) is given by this process for all possible t values.

    (2) t 0, say t=-2; In this case it is clear that there is never any overlap of the functions and therefore their product is zero and thus h(t) = 0 for t 0 .

    Page 4

  • Also from the picture below it is clear that h(t) = 0 for t 1 (say 0.5)

    and also for 7 < t (say t = 8) as shown below;

    Page 5

  • (4) 1 < t < 2 , say t = 1.5; Here the functions only overlap in the range 1 < t < 2 as shown below.

    The red curve is f(tau)*g(tau-t) for t = 1.5. The area under the red curve is h(t).

    In this overlap region,

    Page 6

  • h(t) = d f ( )g(t )1

    t

    f ( ) = 2

    g(t ) =25

    (t ) 0 < (t ) < 5

    0 otherwise

    The range 0 < (t ) < 5 is equivalent to (t 5) < < t . Over this range we have

    g(t ) = 2

    5(t )

    and

    h(t) = f (t) g(t) = d 45

    (t )1

    t

    1 < t < 2

    = 25t 2

    45t +

    25

    1 < t < 2

    (5) 2 < t < 6 , say t = 4 Here the square pulse is entirely inside the triangular pulse as shown below.

    Page 7

  • We have

    h(t) = f (t) g(t) = d 45

    (t )1

    2

    2 < t < 6

    = 45t

    65

    2 < t < 6

    (6) Final region 6 < t < 7 , say t = 6.5; Triangular pulse has moved to the right of the square pulse as shown below.

    We have

    h(t) = f (t) g(t) = d 45

    (t )t5

    2

    6 < t < 7

    = 25t 2 +

    85t +

    425

    6 < t < 7

    We finally get

    h(t) =

    0 t < 125t 2

    45t +

    25

    1 < t < 2

    45t

    65

    2 < t < 6

    25t 2 +

    85t +

    425

    6 < t < 7

    0 7 < t

    Page 8

  • which looks like

    A very tricky and tedious process to comprehend !!

    Correlation

    The correlation process gives a measure of the similarity of two signals. One of its most important applications is to pick a known signal out of a sea of noise. The cross-correlation between f(t) and g(t) is defined as

    fg (t) = d f ( )g( t)

    The cross-correlation of a function with itself

    ff (t) = d f ( ) f ( t)

    is called an autocorrelation.

    This operation is similar to the convolution operation except that the second function is not inverted. It is just as tricky as the convolution operation.

    We can write

    FT d f ( )g( t)

    = 2F( )G( )

    Page 9

  • Examples:

    (1) The Square Pulse - Consider the function

    f (t) =

    1 T / 2 < t < T / 2 0 otherwise

    f(t) is absolutely integrable so it has a valid Fourier transform. It is given by

    F( ) = 12

    dte i tT /2

    T /2

    =2

    sinT2

    which looks like (for T=1)

    In the limit T we have (T = 50, in fact, here)

    Page 10

  • We get a sharp spike, but the area remains constant. This implies that as T

    F( ) ( )Formally, we have

    F( ) = lim

    T

    12

    dte i t =T /2

    T /2

    12

    dte i t =

    2 ( )

    (2) Transform of a Delta-Function - Consider the function

    f (t) = (t). The transform is

    F( ) = 1

    2dte i t (t) =

    12

    The inverse transform is

    12

    dte i t12

    =

    (t)Now

    FT

    dfdt

    = iFT ( f ) = iF( )

    Therefore for the square pulse we have

    dfdt

    = (t + T / 2) (t T / 2)

    dfdt

    = (t + T / 2) (t T / 2)( )

    But

    FT ( f (t t0 )) = e i t0FT ( f (t))

    Thus,

    FTdfdt

    = e i (T /2) e i (T /2)( )FT (t)( ) = i 2

    sinT

    2= iF( )

    F( ) = 2

    sinT2

    for the square pulse ( as before)

    Remember this only makes sense inside an integral.

    (3) Transform of a Gaussian - Consider the function

    f (t) =

    e

    2 t2 = normalized Gaussian pulse

    Page 11

  • We choose = 1. The peak is at / . The 1/2 maximum points are separated by t = 1 / . The area under the curve is = 1.

    The Fourier transform is

    F( ) = 1

    2dt

    e2 t2 e i t =

    2

    dt

    e(2 t2 + i t )

    We complete the square to evaluate the integral. We have

    2t2 + it = 2t2 + it + = t + ( )2

    2t = it = i2

    = 2 = 2

    4 2

    We thus have

    F( ) =

    2e 2

    4 2 dt

    e( t+ i

    2)2

    Let

    x = t + i

    2 dx = dt

    F( ) = 1

    2e 2

    4 2 dx

    e x2

    =12

    e 2

    4 2

    which is a different Gaussian.

    An important feature is

    f (t) t 1

    F( ) 2t 2

    In general for any f(t) we have t c = constant . In the wave theory of quantum mechanics, this corresponds to the Heisenberg Uncertainty Principle.

    The Laplace Transform

    Another important integral transform is the Laplace transform.

    For a function f(t), we define the Laplace transform by

    F(s) = dte st

    0

    f (t) = L( f (t))

    Page 12

  • The Laplace transform is a linear operator so that

    L(af (t) + bg(t)) = aL( f (t)) + bL(g(t))The Laplace transform has a first shifting property expressed as

    if L( f (t)) = F(s) , then L(eat f (t)) = F(s a)

    The Laplace transform has a second shifting property expressed as

    if L( f (t)) = F(s) , and g(t) =

    f (t a) t>a0 t

  • Ha (t) =

    1 t>a0 t

  • L(J0 (t)) =

    (1)k

    22k (k!)2(2k)!s2k+1k=0

    = 1s 1+(1)k

    22k k!1 3 5 ....... (2k 1)

    s2kk=1

    Now using the binomial theorem

    (1+ x)n = 1+ n1!x +

    n(n 1)2!

    x2 + ....+ n(n 1)......(n k +1)k!

    xk