I have the following LTI system
$\dot{x} = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix} x + \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} $
$y = \begin{pmatrix} 1 & 1& 1\end{pmatrix}x(t)$
I am wondering if its possible to choose an initial state at t=0 so that the output of the system has the form $y(t) = te^{-t}$ when $t>0$.
I believe the solution is impossible because out of any system you must know 2 of these to solve for the other: input, output, current state. Is this correct?
If you mean for any $u(t)$ then the answer is no. I think you are correct about "if you know initial condition and input you can find output, if you know input and output you can estimate initial condition and if you know initial condition and output you can find input".
But if you mean for zero input whether we can get just $te^{-t}$ term at the output we need to check:
$x(t) = e^{At}x_0$
and as A is in Jordan form
$y(t) = Cx(t) = Ce^{At}x_0 = \begin{bmatrix} 1 && 1 && 1 \end{bmatrix} \begin{bmatrix} e^{-t} && te^{-t} && 0 \\ 0 && e^{-t} && 0 \\ 0 && 0 && e^{3t} \end{bmatrix}x_0 = \begin{bmatrix} e^{-t} && e^{-t} + te^{-t} && e^{3t}\end{bmatrix}x_0$
Now, as you can see, if we choose $x_0$ as $\begin{bmatrix} -1 && 1 && 0 \end{bmatrix}^T$, then $y(t) = te^{-t}$ for $t>0$.