I have a quick question,
This is an exam question for the course System Identification:
Consider the following periodic input signal $u(t)$:
$u(t) = 0, \quad t = 0,3$
$u(t) = 1, \quad t = 1,4$
$u(t) = -1, \quad t = 2,5$
The question is "Determine the maximum order of persistence of excitation of the input signal $u$".
The minimal order is normally determined by checking when the toeplitz matrix of correlations $R$ becomes singular.
In order to construct the matrix $R$, I've detrmined $R_u(0)=\frac{2}{3}$ , $R_u(1)=-\frac{1}{3}$, $R_u(2)=\frac{1}{3}$, $R_u(3)=\frac{2}{3}$, $R_u(4)=-\frac{1}{3}$ and $R_u(5)=-\frac{1}{3}$
$R_2=\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix}, \quad$ $R_3=\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\end{bmatrix}, \quad$ $R_4 =\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\end{bmatrix}$ and $R_5 =\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{bmatrix}$
However, none of these matrices are singular and $R_6$ becomes a bit to cumbersome to do by hand during a test.
Normally the matrix becomes singular at $R_3$ or at most $R_4$ from which you can then conclude that the order of persistent excitation is $2$ and $3$ respectively.
So I don't know how the maximum order is determined. Any help will be greatly appreciated.
$R_u(2)=-1/3$ resulting in a signular $\mathbf{R}_3$.