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Math Help - Orthonormal Conjugate Quadrature Filter, Fourier Transform

  1. #1
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    Orthonormal Conjugate Quadrature Filter, Fourier Transform

    This question is strange enough so I'm beginning to suspect a typo/misinterpretation.

    Let w be defined by
    \hat{w}=\{ \begin{array}{c}<br />
1 \text{ if } \xi \in (0.5,1] \\<br />
-1 \text{ if } \xi \in [-1,-0.5) \\<br />
0 \text{ otherwise }\end{array} \}

    Show that w is admissible and compute its normalization constant.

    First of all it is very strange to think of the fourier transform to be a function on the reals as opposed to a sequence, right? (I suspect this might be a typo)

    Anyway, I interpret admissible to mean: satisfies the Orthonormal Conjugate Quadrature Filter Conditions.

    So first I actually find w by taking the inverse fourier transform:

    w=\check{\hat{w}}=\sum_{\xi=-\infty}^{\infty}\hat{w(\xi)}e^{2\pi i \xi x}
    =e^{2\pi i x}-e^{-2\pi i x}=-2isin(2\pi x)

    and this is zero at all integer values, which makes it a very strange trivial case (which is not orthonormal).

    Any thoughts?
    Thanks.
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  2. #2
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    Quote Originally Posted by verdi View Post
    First of all it is very strange to think of the fourier transform to be a function on the reals as opposed to a sequence, right? (I suspect this might be a typo)
    You may be confusing Fourier transform and Fourier series. The Fourier transform of a function is a function on the reals.
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  3. #3
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    Good point.

    That gives me

    Edit: I spotted a miscalculation...I'll get back to this

    w(\xi)=\check{\hat{w(\xi)}}=\int_{-\infty}^{\infty}\hat{w(x)}e^{2\pi i \xi x} dx=\int_{\frac{1}{2}}^{1}e^{2\pi i \xi x} dx-\int_{-1}^{-\frac{1}{2}}e^{2\pi i \xi x} dx

    =(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))

    =(\frac{1}{\pi \xi})(sin(2\pi \xi)+ i cos(\pi \xi))

    which takes on \frac{-i}{\pi \xi} at odd \xi, \frac{i}{\pi \xi} at even \xi

    and at 0, we get 2+\infty.

    Is that really a good result. It's technically non-zero at infinitely many integers and I don't like what's going on at 0.

    Does this all make sense?
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  4. #4
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    <br />
=(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))<br />
    <br />
=(\frac{1}{\pi i \xi})(cos(2\pi \xi)- cos(\pi \xi))<br />
    <br />
\xi \text{ even } \rightarrow 0
    \xi \text{ odd } \rightarrow \frac{2}{\pi i \xi}
    \xi \text{ 0 (special case of even)}\rightarrow 0

    I don't believe that this counts as finite support. Correct me if I am wrong.
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  5. #5
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    Quote Originally Posted by verdi View Post
    <br />
=(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))<br />
    <br />
=(\frac{1}{\pi i \xi})(cos(2\pi \xi)- cos(\pi \xi))<br />
    <br />
\xi \text{ even } \rightarrow 0
    \xi \text{ odd } \rightarrow \frac{2}{\pi i \xi}
    \xi \text{ 0 (special case of even)}\rightarrow 0

    I don't believe that this counts as finite support. Correct me if I am wrong.
    Remember \xi is real, not just integer. The support certainly isn't finite. I don't know however what "admissible" and "normalizing constant" mean in this context.
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  6. #6
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    Well my impression based on the context is that we're trying to construct something analogous to the Haar filter:
    h(k)=\frac{1}{\sqrt{2}} \text{ if k=0 or k=1}
    =0 if k\neq{0} and k\neq{1}

    Which we at least only care about on integer values, even if it could be extended to a larger domain.
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  7. #7
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    I think I might have cleared some things up for myself.
    Continuous wavelet transform - Wikipedia, the free encyclopedia
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