# Walsh Functions

• Jan 12th 2009, 12:25 PM
BlackWarrior
Walsh Functions
Hi,

I have a little question. Why Walsh functions are orthonormal basis in L2[0,1] ?
• Jan 13th 2009, 12:07 AM
Opalg
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.
• Jan 13th 2009, 02:45 AM
BlackWarrior
Hi,

First thank you for your response..

I get that w1(x)=1 , w2(x)=w3(x) ... so how walsh functions can be orthogonal ?
• Jan 13th 2009, 03:11 AM
Opalg
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].

• Jan 13th 2009, 03:33 AM
BlackWarrior
Hi,

Oplag , thank you ! (Clapping)

Can you write a definition of these functions ???

(Wink)
• Jan 13th 2009, 08:19 AM
Opalg
Quote:

Originally Posted by BlackWarrior
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 $W_1$.
• Jan 13th 2009, 08:40 AM
BlackWarrior
Thank you man (Clapping) (Happy)