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Is a fourier transform a rotoation?

  1. May 15, 2010 #1
    From my undergraduate textbook: Circuits, Signals, and Systems by Siebert, p 453


    ====================================================
    Consider the two principal waveform representations schemes ...

    [tex]
    x(t) = \int x(\tau)\delta(t - \tau)d\tau
    [/tex]

    [tex]
    x(t) = \int X(f)e^{j2\pi f t}df
    [/tex]

    If we consider the set of delayed impulses as determining one set of orthogonal vectors and the set of complex exponentials as determining another set, then [itex] x(\tau) d\tau [/itex] and [itex] X(f) df [/itex] are the components of [itex] x(t) [/itex] along the corresponding coordinates. The frequency-domain representations of [itex] x(t) [/itex] thus amounts to picking a coordinate system that is rotated from the time-domain coordinate system. And Parseval's Theorem

    [tex]

    \int x^2 (t) dt = \int |X(f)|^2 df

    [/tex]
    is just a statement of the fact that the length of a vector is independent of the coordinate system in which it is described
    ====================================================

    Is this true? If so, then it should be possible to the find the axis of rotation, right? How does one go about that? Does the question make sense?

    Thanks
    Roy
     
  2. jcsd
  3. May 15, 2010 #2

    Hurkyl

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    An "axis of rotation" only makes sense in a three-dimensional vector space. e.g. to make sense of it you could do the following:

    Form the rotation matrix. Its eigenvalues will be one of the following:
    • Three 1's
    • One 1, two -1's
    • One 1, one complex-conjugate pair of norm 1

    The eigenspace associated to -1, or the space generated by the eigenspaces for the complex-conjugate pair describe a plane that is rotated by the rotation.

    The eigenspace associated to 1 is a line that is fixed by the rotation.


    The fact that the line uniquely determines a plane perpendicular to it is a unique feature of three-dimensional Euclidean space. That the list of possible sets of eigenvalues is so short is another unique feature of three-dimensional Euclidean space. (there simply aren't enough dimensions for the behavior of a rotation to be complex*)


    In finite dimensions, the space breaks up into a fixed space (which can be zero-dimensional in even dimension!), and planes that are rotated. (And some extra cases where the eigenvalues are repeated) I don't know if we can even say that in the infinite dimensional case you're considering.


    *: used as an English word rather than a technical term
     
    Last edited: May 15, 2010
  4. May 15, 2010 #3
    Hurkyl,

    Thanks for responding.

    Perhaps my question was too literal. If this relation is true, is there some was to understand the transform as a rotation, besides the Parseval theorem being analogous to preserving a "length" in some infinite dimensional space? What if we restricted ourselves to a discrete time / discrete Fourier transform, to keep the dimensions finite?

    - Roy
     
  5. May 15, 2010 #4

    Hurkyl

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    FWIW, wikipedia's article lists the eigenvalues and one eigenbasis for the Fourier transform.
     
  6. May 17, 2010 #5
    Hurkyl,

    Thanks for the link. I really should look more closely at wikipedia. According to

    http://en.wikipedia.org/wiki/Fracti...pretation_of_the_Fractional_Fourier_Transform

    and

    http://en.wikipedia.org/wiki/Linear_canonical_transformation,

    the Fourier transform is a rotation by 90 deg in the time-frequency domain. This may be related to the concept that position and momentum of fourier transforms of one another according to quantum physics theory.

    - Roy
     
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