I have the following three signals:
$$s_1(t)=\begin{cases}Acos\left(\frac{2πt}{T}\right) & 0<t<\frac{T}{2} \\ 0 & \text{otherwise}\end{cases}$$ $$s_2(t)=\begin{cases}Acos\left(\frac{2πt}{T}\right) & 0<t<T \\ 0 & \text{otherwise}\end{cases}$$ $$s_3(t)=\begin{cases}Acos\left(\frac{2πt}{T}\right) & \frac{T}{2}<t<T \\ 0 & \text{ otherwise}\end{cases}$$
"Build an orthonormal basis and plot the corresponding signal constellation."
-The professor says.
It is apparent that it is the same signal, only applied at various subsets of time. I tried using the Gram-Schmidt method to break down the signal as: $s_i(t)=\sum\limits_{i=1}^3a_iφ_i(t)$ so I can also plot the signal constellation.
..only to realize that there is clearly a more suitable method I haven't figured out yet.
Thank you for your help.
Orthonormal basis $\{\phi_k(t)\}^n$ means $$\int\phi_i(t)\phi_j(t)dt=\begin{cases}0,&i\neq j\\ 1, &i=j \end{cases}$$ It is not difficult to realize that $s_1(t)$ and $s_3(t)$ are orthogonal, and $s_2(t)=s_1(t)+s_3(t)$. So we build the basis using $s_1(t)$ and $s_3(t)$:
$$\int_{0}^{\frac{T}{2}}\left(B\cos\left(\frac{2πt}{T}\right)\right)^2dt=\frac{B^2T}{4}=1\Rightarrow B=\frac{2}{\sqrt{T}}$$ Hence, $$\phi_1(t)=\begin{cases}\frac{2}{\sqrt{T}}\cos\left(\frac{2πt}{T}\right)& 0<t<\frac{T}{2} \\ 0 & \text{otherwise}\end{cases}$$ and similarly, $$\phi_2(t)=\begin{cases}\frac{2}{\sqrt{T}}\cos\left(\frac{2πt}{T}\right)& \frac{T}{2}<t<T \\ 0 & \text{otherwise}\end{cases}$$ So we have $$s_1(t)=\frac{A\sqrt{T}}{2}\phi_1(t)+0\phi_2(t)\Rightarrow \mathbf{s}_1=[\frac{A\sqrt{T}}{2},0]$$ $$s_2(t)=\frac{A\sqrt{T}}{2}\phi_1(t)+\frac{A\sqrt{T}}{2}\phi_2(t)\Rightarrow \mathbf{s}_2=[\frac{A\sqrt{T}}{2},\frac{A\sqrt{T}}{2}]$$ $$s_3(t)=0\phi_1(t)+\frac{A\sqrt{T}}{2}\phi_2(t)\Rightarrow \mathbf{s}_3=[0,\frac{A\sqrt{T}}{2}]$$