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Math Help - Fourier Transform Help Needed

  1. #1
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    Fourier Transform Help Needed

    DEAR GROUP MEMBERS,

    I am trying to find the Fourier transform of f(x) defined as,
    1, for |x| < a
    f(x) =
    0, for |x| > a.

    SOLUTION :

    In the final steps it resolves to :

    2/pi INT [(1/s.sinas.cosxs) ds ]from 0 to INFINITY = f(x) ---- (1)

    INT [{1/s.sinas.cosxs}ds ]from 0 to INFINITY = pi/2, for |x| < a
    = 0, for |x| > a
    ---- (2)

    I would like to know, how it resolves from Equation 1 to 2 . I have thought a lot about it but still unable to solve this.

    And how is the integration of sin s / s from 0 to inifinity = pi / 2.

    Million thanks for solving this query in advance.
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  2. #2
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    Quote Originally Posted by rufi View Post
    DEAR GROUP MEMBERS,

    I am trying to find the Fourier transform of f(x) defined as,
    1, for |x| < a
    f(x) =
    0, for |x| > a.

    SOLUTION :

    In the final steps it resolves to :

    2/pi INT [(1/s.sinas.cosxs) ds ]from 0 to INFINITY = f(x) ---- (1)

    INT [{1/s.sinas.cosxs}ds ]from 0 to INFINITY = pi/2, for |x| < a
    = 0, for |x| > a
    ---- (2)

    I would like to know, how it resolves from Equation 1 to 2 . I have thought a lot about it but still unable to solve this.

    And how is the integration of sin s / s from 0 to inifinity = pi / 2.

    Million thanks for solving this query in advance.
    I'm not clear about what you have done, but depending on how your
    course defines the FT the FT of f(x) is:

    F(w) = integral_{x=-infty to infty} f(x) exp(i w x) dx

    where your definition may differe from this by a scattering of 2's and pi's
    in random places and a -'s sign here or there depending which is takes as
    the forward or backward trasform. But working with the above we have:

    F(w) = integral_{x=-infty to infty} f(x) exp(i w x) dx = integral_{x=-a to a} exp(i w x) dx

    Now (for w !=0, w=0 has to be considered seperatly but this works for the limit as w->0):

    integral_{x=-a to a} exp(i.w.x) dx = [1/(i.w)] [exp(i.w.a)-exp(-i.w.a)]

    ............... = [1/(i.w)] [2.i sin(w.a)]

    ............... = 2 sin(w.a)/w.

    So: F(w) = 2 sin(w.a)/w

    Now your solution will differe in detail due to the difference in how your FT
    is defined, but you should be able to obtain the answer you need easily
    following the approach above.

    (Though on rereading what you have written it looks more like you are trying
    to find the inverse FT of F(x) = sin(a.x)/x)

    RonL
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  3. #3
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    Thanks a lot CAPTAIN for your reply. Let me write complete question :

    f(x) = 1 , for |x| < a
    = 0, for |x| > a

    and hence find the value of INT [ { sinx/x } dx ] from 0 to infinity.


    Yes, you are very right that after taking the Fourier transform of the problem, which is the same as your's, i was trying to find the inverse Fourier transform and having the difficulty in resolving this.

    Could you please tell me step by step how to find inverse Fourier transform for this. The approach which i was following was :

    -1
    F { 2/s . sinas } = f(x)

    1/2pi INT [ {2/s . sinas. e^ixs} ds] from - INF to + INF = f(x)

    1/pi INT [ {1/s.sinas (cosxs +i sinxs)} ds] from -INF to + INF= f(x).

    2/pi INT [ { 1/s.sinas.cosxs } ds ] from 0 to INF = f(x).

    Now after this , i exactly don't know how to solve this... if we are supposed to use integration by parts, how it can be done. In a book, it solves like this after the above step :

    INT [ { 1/s.sinas.cosxs} ds ] from 0 to infinity = pi/2, for |x|<a
    = 0, for |x|<a


    Or Forget everything and give me your way of solving this.
    Thanks again CAPTAIN.
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  4. #4
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    Quote Originally Posted by rufi View Post
    Thanks a lot CAPTAIN for your reply. Let me write complete question :

    f(x) = 1 , for |x| < a
    = 0, for |x| > a

    and hence find the value of INT [ { sinx/x } dx ] from 0 to infinity.


    Yes, you are very right that after taking the Fourier transform of the problem, which is the same as your's, i was trying to find the inverse Fourier transform and having the difficulty in resolving this.

    Could you please tell me step by step how to find inverse Fourier transform for this. The approach which i was following was :

    -1
    F { 2/s . sinas } = f(x)

    1/2pi INT [ {2/s . sinas. e^ixs} ds] from - INF to + INF = f(x)

    1/pi INT [ {1/s.sinas (cosxs +i sinxs)} ds] from -INF to + INF= f(x).

    2/pi INT [ { 1/s.sinas.cosxs } ds ] from 0 to INF = f(x).

    Now after this , i exactly don't know how to solve this... if we are supposed to use integration by parts, how it can be done. In a book, it solves like this after the above step :

    INT [ { 1/s.sinas.cosxs} ds ] from 0 to infinity = pi/2, for |x|<a
    = 0, for |x|<a


    Or Forget everything and give me your way of solving this.
    Thanks again CAPTAIN.
    To do this properly I would need to know the exact form of the Fourier
    Transform that you are working with (one of the curses of the FT is that
    every one defines it differently). But lets use this definition:

    FT(f(x)) = F(w) = integral_{x=-infty to +infty} f(x) exp(-i.x.w) dx

    Then the inverse Fourier Transform is:

    FT^-1(F(w)) = 1/(2.pi) integral__{w=-infty to +infty} F(x) exp(i.x.w) dx.

    With:

    f(x) = 1 for |x|<a

    and

    f(x) = 0 otherwise,

    we have:

    F(w) = 2 sin(a.w)/w

    (you can proove this using what is in my earlier post)

    so:

    f(x) = 1/(2.pi) integral__{w=-infty to +infty} 2 sin(a.w)/w exp(i.x.w) dx

    and so (as f is continuous at x=0):

    f(0) = 1/(2.pi) integral__{w=-infty to +infty} 2 sin(a.w)/w dx = 1

    Hence:

    integral__{w=-infty to +infty} sin(w)/w dx = pi

    and as sin(w)/w is symetric

    integral__{w=0 to +infty} sin(w)/w dx =(1/2) integral__{w=-infty to +infty} sin(w)/w dx = pi= pi/2

    ie:


    integral__{w=0 to +infty} sin(w)/w dx = pi/2

    RonL
    Last edited by CaptainBlack; April 28th 2007 at 06:18 AM.
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  5. #5
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    Thanks a million again Captain for your reply.

    You are much more knowledgeable then atleast me so i request you to please help me understand this method of solving.

    Well, calculating FT is relatively easy and the FT of the function is 2/s.sinas

    Now, to prove the second part of the question (i.e to find the value of INT [ {sinx/x} dx ] from 0 to infinity ), i m taking the Inverse Fourier Transform . I m doing it like this :

    Inverse Fourier [ 2/s.sinas ] = f(x)
    = 1/2pi INTEGRAL [ { 2/s.sinas.e^ixs }ds from - inf to + inf ] = f(x)

    =1/pi[{1/s.sinas(cosxs+i sinxs)ds from -inf to +inf]=f(x)
    Here [ e^ixs = cosxs + sinxs ]

    = 2/pi [ { 1/s.sinas.cosxs } ds from 0 to INF] = f(x)
    [ b/c 1/s/sinas.sinxs is odd ]

    =[ {1/s.sinas.cosxs] ds from 0 to inf ]= pi/2 for |x| < a
    = 0 for |x|> a.

    Now, here if you can explain how the last step is reached from the previous step. This is my only difficulty which i m trying to get a solution to.

    Thanks a lot.
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  6. #6
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    Quote Originally Posted by rufi View Post
    Thanks a million again Captain for your reply.

    You are much more knowledgeable then atleast me so i request you to please help me understand this method of solving.

    Well, calculating FT is relatively easy and the FT of the function is 2/s.sinas

    Now, to prove the second part of the question (i.e to find the value of INT [ {sinx/x} dx ] from 0 to infinity ), i m taking the Inverse Fourier Transform . I m doing it like this :

    Inverse Fourier [ 2/s.sinas ] = f(x)
    = 1/2pi INTEGRAL [ { 2/s.sinas.e^ixs }ds from - inf to + inf ] = f(x)

    =1/pi[{1/s.sinas(cosxs+i sinxs)ds from -inf to +inf]=f(x)
    Here [ e^ixs = cosxs + sinxs ]

    = 2/pi [ { 1/s.sinas.cosxs } ds from 0 to INF] = f(x)
    [ b/c 1/s/sinas.sinxs is odd ]

    =[ {1/s.sinas.cosxs] ds from 0 to inf ]= pi/2 for |x| < a
    = 0 for |x|> a.

    Now, here if you can explain how the last step is reached from the previous step. This is my only difficulty which i m trying to get a solution to.

    Thanks a lot.
    You don't need to ever take a full inverse FT as you need it only at one point, and you know its value there.

    You use your knowlege of the forward transform of f(x) to get
    F(w) = 2 sin(a.w)/w, then you need never compute the inverse transform
    (at least at the points of continuity of f(x)) as you know it from the
    properties of the Fourier Transform Inverse Fourier Transform pair.

    RonL
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