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?
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.