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Math Help - Walsh Functions

  1. #1
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    Walsh Functions

    Hi,

    I have a little question. Why Walsh functions are orthonormal basis in L2[0,1] ?
    Last edited by BlackWarrior; January 16th 2009 at 04:22 AM.
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  2. #2
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    Apart from the first Walsh function, which is identically 1, all the others are equal to +1 on half of the interval [0,1] and 1 on the remaining half. Also, the product of any two Walsh functions is again a Walsh function (the first 2^n functions form a multiplicative group. It follows that the integral over [0,1] of the product of two Walsh functions is zero. Hence they form an orthogonal set.
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  3. #3
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    Hi,

    First thank you for your response..

    I get that w1(x)=1 , w2(x)=w3(x) ... so how walsh functions can be orthogonal ?
    Last edited by BlackWarrior; January 16th 2009 at 04:24 AM.
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  4. #4
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    You will find schematic graphs of the first eight Walsh functions here. In fact, W_2(x) is +1 on [0,1/2] and 1 on [1/2,1], whereas W_3(x) is +1 on [0,1/4] and [3/4,1], and 1 on [1/4,3/4].

    See also this link.
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  5. #5
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    Hi,

    Oplag , thank you !

    Can you write a definition of these functions ???

    Last edited by BlackWarrior; January 14th 2009 at 04:41 AM.
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  6. #6
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    Quote Originally Posted by BlackWarrior View Post
    Can you write a clear definition of these functions ??? ( i didn't understand the definition in wiki)
    The mathematical definition is complicated, using something called the Gray code. But there is a reasonably understandable inductive way to define these functions, taking them in blocks of 2^n at a time.

    Start with W_1, which is the constant function 1. To get W_2, you split the unit interval in two, and define W_2(x) = W_1(x) on the first half of the interval and W_2(x) = W_1(x) on the second half. (I'm ignoring what happens at the point x=1/2, because the value at a single point is irrelevant when it comes to integration. If you want to be fussy, replace closed intervals by half-open intervals in what follows.)

    To get W_3 and W_4, you apply the same process to W_2 and W_1 respectively. (Note the reversal of the ordering: W_3 comes from W_2, and W_4 from W_1.) In more detail, here's how you construct W_3 from W_2. Take each of the two intervals on which W_2 is constant, split it in half, and define W_3 to be the same as W_2 on the first of each pair of half-intervals, and W_2 on the second one.

    The inductive step works like this. Suppose that W_k has been defined for 1≤k≤2^n, and that it is constant on each interval \left[\tfrac{j-1}{2^n},\tfrac{j}{2^n}\right]\ (1\leqslant j\leqslant2^n). For 1≤k≤2^n, define W_{2^n+k} to be equal to W_{2^n+1-k} on each interval \left[\tfrac{2j-2}{2^{n+1}},\tfrac{2j-1}{2^{n+1}}\right], and equal to -W_{2^n+1-k} on each interval \left[\tfrac{2j-1}{2^{n+1}},\tfrac{2j}{2^{n+1}}\right], for 1\leqslant j\leqslant2^n. In that way, W_k is defined for all k up to and including k=2^{n+1}.

    Notice again the reversal of the ordering: W_{2^n+1} comes from W_{2^n}, and so on until W_{2^{n+1}} comes from W_1.
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  7. #7
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    Thank you man
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