Results 1 to 4 of 4

Math Help - short time Fourier transform

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    3

    short time Fourier transform

    Hey, I have a question about localization a signal in time domain by windowing.
    would anyone explain to me these phrases:
    1) The basis functions sin(ωt) and cos(ωt) are not localized in time!(support region in frequency = 0)
    2) The δ(t) is not localized in frequency! (support region in time = 0)
    cheers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Re: short time Fourier transform

    1) The basis functions are nonzero for arbitrarily large times. For example, \sin \left( {(4n+1)\pi\over 2}\right) = 1 for arbitrarily large n\in\mathbb N.
    2) The delta distribution is nonzero for artitrarily large frequencies. Note that \delta(t) = \int d\omega e^{i\omega t}, and hence in the freqency domain, \delta(t) is the constant function 1 which does not have compact support (take nonzero values over an infinite range).
    Last edited by vincisonfire; December 16th 2011 at 12:17 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2011
    Posts
    3

    Re: short time Fourier transform

    Quote Originally Posted by vincisonfire View Post
    1) The basis functions are nonzero for arbitrarily large times. For example, \sin \left( {(4n+1)\pi\over 2}\right) = 1 for arbitrarily large n\in\mathbb N.
    2) The delta distribution is nonzero for artitrarily large frequencies. Note that \delta(t) = \int d\omega e^{i\omega t}, and hence in the freqency domain, \delta(t) is the constant function 1 which does not have compact support (take nonzero values over an infinite range).
    Thanks for reply, but I don't get the point of view
    what it means: support region in frequency = 0 for sin & cos?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Re: short time Fourier transform

    The support is the set of times t such that the function takes nonzero values \{t|f(t)\neq 0\}. For a \delta-function, the support is \{0\}.
    So what the statement above is saying is: note that even though a \delta-function's support is a single number, its Fourier transform is completely unbounded.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laplace transform and Fourier transform what is the different?
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: December 29th 2010, 10:51 PM
  2. [SOLVED] Continuous-Time Fourier Transform of a Response (matlab)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 1st 2010, 11:04 AM
  3. Fourier Transform of rect and time shifting.
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: June 5th 2010, 11:12 AM
  4. Replies: 3
    Last Post: April 22nd 2010, 05:39 AM
  5. Replies: 0
    Last Post: April 23rd 2009, 05:44 AM

Search Tags


/mathhelpforum @mathhelpforum