Consider the following equation:

$\ddot{x} = -a x - b \dot{x}$

which we interpret as saying that we are trying to control $x$ by setting $\ddot{x}$.

We can rewrite this with $X = \begin{bmatrix} x \\ \dot{x} \end{bmatrix}$ and $K= \begin{bmatrix} 0 & -1 \\ a & b \end{bmatrix}$ as:

$\dot{X}=-K X$.

We have a choice of $a$ and $b$. For any value $a>0$ and $b>0$, the system is stable in the sense that both eigenvalues of K have positive real parts.

Now let's assume that the control cannot be applied instantly, i.e. $-ax-b\dot{x}$ is our target $y$ for $\ddot{x}$. The system becomes:

$\dddot{x} = -k(\ddot{x}-y) = -k \ddot{x} - a k x - b k \dot{x}$

I am interested in the values of $a$ and $b$ such that this system, ie I am interested in the values of $a$ and $b$ such that the eigenvalues of

$K = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ a k & b k & k \end{bmatrix}$ all have positive real parts.

Looking at numerical examples, it seems that the solution is just that $a < b k$, but I cannot prove it simply. This would be somewhat intuitive since I can rewrite it as the sum of the inverse of the eigenvalues of the 2x2 matrix $K$ is greater than $1\over{k}$, ie that the sum of the 2 characteristic times has to be greater than the characteristic time of the lag.