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I'm trying to solve the following problem where I need to express $x(t)$ in terms of the given $φ_i(t)$ functions. (It is proven that the $φ_i(t)$ functions are orthonormal).:


enter image description here


Here's what I tried:

In order to find the coefficients, the following definition of the geometric representation of a signal was used. (Simon Hagkin - Communication Systems)

enter image description here

Using equation 5.5 above

$$ x(t) = \sum_{j=1}^{3} x_{j} φ_j(t) $$

Referring to the equation 5.6 above, I have tried to find coefficients ($x_{j}$) of each basis function. However, all three coefficients became zero.

This is how it was solved for $x_{1} $:

$$ x_{1} =\int_{0}^{4} {x(t) φ_1(t)} dt $$ $$ x_{1} =\int_{0}^{1} {-1 . (1/2)} dt +\int_{1}^{2} {1 . (1/2)} dt + \int_{2}^{3} {1 . (-1/2)} dt + \int_{3}^{4} {-1 . (-1/2)} dt$$ $$ x_{1} = 0 $$

for $x_{2} $

$$ x_{2} =\int_{0}^{4} {x(t) φ_2(t)} dt $$ $$ x_{2} =\int_{0}^{1} {-1 . (1/2)} dt +\int_{1}^{3} {1 . (1/2)} dt + \int_{3}^{4} {-1 . (1/2)} dt $$ $$ x_{2} = 0 $$

for $x_{3} $

$$ x_{3} =\int_{0}^{4} {x(t) φ_3(t)} dt $$ $$ x_{3} =\int_{0}^{1} {-1 . (1/2)} dt +\int_{1}^{2} {1 . (-1/2)} dt + \int_{2}^{3} {1 . (1/2)} dt + \int_{3}^{4} {-1 . (-1/2)} dt$$ $$ x_{3} = 0 $$

What am I doing wrong?

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  • $\begingroup$ Yes the coefficients turn out to be zero... (unless there's a typo somewhere) you are not doing anything wrong. $\endgroup$ – Fat32 Apr 25 at 13:11
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Here is a graphical explanation. Sorry, I have depicted $-x(t) $ in red, and the $\psi_k$ in black. In gray, the area of the product on sub-intervals. Positive when $-x(t)$ and $\psi_k$ have the same sign, negative otherwise. As you can see, the areas sum to zero. So $-x(t)$ and $x(t)$ therefore are orthogonal to the other three functions. Now, you have a set of four orthogonal functions, known as Walsh functions or Hadamard bases.

graphical explanation for orthogonality

Ypu can find an illustration at Wolfram: Walsh Functions or orthogonal trains of square pulses.

Walsh functions consist of trains of square pulses

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  • $\begingroup$ Thank you very much for the detailed explanation. Any reason to use $ -x(t) $ instead of $ x(t) $ for the demonstration? $\endgroup$ – big sad Apr 25 at 17:30
  • $\begingroup$ Not, I did the drawing in my backyard without a computer without looking at your post, by heart. My drawing started at 1 like traditionallyfor this Walsh function. Then I came back to post it, realizing my mistake, and was a bit lazy. As the result is zero, I hope you would not mind ;) $\endgroup$ – Laurent Duval Apr 25 at 17:36
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Feels like a bit of a trick question. Your answer is indeed correct. $\Psi_1$ ...$\Psi_3$ are an orthonormal basis but it's an incomplete basis. Any signal you can construct with this basis has 4 degrees of freedom, but you only have three basis functions. In order to have a complete basis, you need 4 functions. Turns $x(t)$ (scaled properly) would make the 4th basis function, since it's orthogonal to the other three.

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  • $\begingroup$ Nice point about the dofs. I am wondering whether the mention of incomplete basus is clear: a basis ought to be complete, here indeed there are several spaces. $\endgroup$ – Laurent Duval Apr 26 at 4:09

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