# Geometric representation of a signal using basis functions

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).:

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)

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?

• Yes the coefficients turn out to be zero... (unless there's a typo somewhere) you are not doing anything wrong. Apr 25, 2021 at 13:11

## 2 Answers

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.

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

• Thank you very much for the detailed explanation. Any reason to use $-x(t)$ instead of $x(t)$ for the demonstration? Apr 25, 2021 at 17:30
• 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 ;) Apr 25, 2021 at 17:36

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.

• 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. Apr 26, 2021 at 4:09