Suppose I have a discrete LTI state space system: $$x(t+1)=Ax(t)+Bu(t),$$ $$y(t)=Cx(t)+Du(t),$$
What i am courious is that under what conditions the input $u(t)$ satisfies would make $lim_{t\rightarrow+\infty}\|x(t)\|=+\infty$,but $lim_{t\rightarrow+\infty}\|y(t)\|<\delta,$
Or we just put it succinctly, the system output remains bounded, while the system states are unbounded.
I think intuitively to make the states unbounded the input must be unbouded,but I don't know the next step for analysis.
So could you please recommend some relative papers or just give me some results to help me solve this question?
There are multiple ways to address your question. The first one is through observability/detectability. Some people have already commented on that.
The second one is through the zeros of the system. The definition of the zeros is the dynamics of the system when placed conditions where the output is identically zero. That is, we look at the internal dynamics for a specific choice for the initial condition $x_0$ and the input $u(t)$ for which $y\equiv 0$. The dynamics of the resulting system will be the zero dynamics. Therefore, if one of the zeros is unstable, then the state will blow up even though the output is zero.
For instance, consider the system $$ x(t+1)=\begin{bmatrix}0.3 & -0.02\\1 &0\end{bmatrix}x(t)+\begin{bmatrix}1\\0\end{bmatrix}u(t),\quad y(t)=\begin{bmatrix}1&-2\end{bmatrix}x(t) $$
If we pick the initial condition $x_0=(2,1)$ and the input $u(t)=\begin{bmatrix}1.7&0.02\end{bmatrix}x(t)$, we get that $y\equiv0$ and
$$ x(t+1)=\begin{bmatrix}2 & 0\\1 &0\end{bmatrix}x(t) $$ which means that $$ x(t)=\begin{bmatrix} 2^{t}\\2^{t-1} \end{bmatrix} $$ which obviously blows up with time. As a final though, please note that the transfer function associated with the system is given by
$$ H(z)=\dfrac{z-2}{z^2-2.1z+0.2} $$ where we retrieve the unstable zero at 2.