Given: I am given four different signals
\begin{align} s_0(t) = \begin{cases} 2, \ \ \ \ 0 < t \leq 1 \\ -2, \ 1 < t \leq 2 \\ 2, \ \ \ \ 2 < t \leq 3 \\ \end{cases} , \ \ \ s_2(t) = \begin{cases} 1, \ \ \ \ 0 < t \leq 1 \\ -2, \ 1 < t \leq 2 \\ \end{cases} \end{align} \begin{align} s_1(t) = \begin{cases} -1, \ 0 < t \leq 1 \\ 3, \ \ \ \ 1 < t \leq 2 \\ 1, \ \ \ \ 2 < t \leq 3 \end{cases} ,\ \ \ s_3(t) = \begin{cases} -1, \ 0 < t \leq 2 \\ -3, \ 2 < t \leq 3 \end{cases} \end{align}
Question: I want to find the orthonormal basis functions using Gram-Schmidt procedure.
Solution: I pretty much know how the Gram-Schmidt algorithm works, but I am note sure how to do the calculations. If I let $s_1$ and $s_2$ form a basis I can calculate the norm as
\begin{align} e_1(t) = \frac{s_1}{\langle s_1,s_1 \rangle} = \frac{s_1}{\int_0^{\infty}s_1^2dt} \end{align} But what i wonder here is what is the integral of $s_1^2$ ?
For $e_2(t)$ we have
\begin{align} e_2(t) = \frac{s_2}{\langle s_2,s_2 \rangle} = \frac{s_2}{\int_0^{\infty}s_2^2dt} \end{align}
Also here I don't know how to calculate the integral of $s_2^2$? So, basically I know the procedure but lack knowledge how to calculate integrals of functions that are not continuous. Is there somebody who can give me a hint? The rest should be straightforward.
It is easier to do it expressing the signals as vectors
\begin{align} \vec{s_0} = \begin{bmatrix} 2 \\ -2 \\ 2 \\ \end{bmatrix} , \ \ \ \vec{s_2} = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} \end{align} \begin{align} \vec{s_1} = \begin{bmatrix} -1 \\ 3 \\ 1 \end{bmatrix} ,\ \ \ \vec{s_3} = \begin{bmatrix} -1 \\ -1 \\ -3 \end{bmatrix} \end{align}
Now, using Gram-Schmidt procedure we have
\begin{align} \vec{u_0} = \vec{s_0} = \begin{bmatrix} 2 \\ -2 \\ 2 \end{bmatrix} \end{align}
and
\begin{align} \vec{u_1} = \vec{s_1} - \frac{\vec{u_0}\cdot\vec{s_1}}{\vec{u_0}\cdot\vec{u_0}}\vec{u_0} = \begin{bmatrix} -1 \\ 3 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix} \end{align}
next, we have
\begin{align} \vec{u_2} = \vec{s_2} - \frac{\vec{u_0}\cdot\vec{s_2}}{\vec{u_0}\cdot\vec{u_0}}\vec{u_0} - \frac{\vec{u_1}\cdot\vec{s_2}}{\vec{u_1}\cdot\vec{u_1}}\vec{u_1} = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} - \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{align} Then {$v_1,v_2$} is a orthogonal basis for W. Normalizing gives \begin{align} \vec{v1} = \frac{1}{} \end{align}