Stability of linear controller in the presence of a lag 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.
 A: If $k>0$, it becomes elementary algebra. As Arthur pointed out, the equation is $P(z)=z^3-kz^2+(bk)z-ak=0$.
On the one hand, assume that all roots have positive real part. Then we either have 3 positive roots, or one positive root and two non-zero complex conjugate ones. In every case, the product of roots is positive, so we must have $ak>0$, i.e., $a>0$. Now, since the sum of the roots is $k$ (Vieta), the largest positive root should be strictly less than $k$, so $P(z)$ must preserve sign on $[k,+\infty)$ and, in particular, we must have $P(k)>0$ whence $bk>a$. Thus, $0<a<bk$ is a necessary condition.
On the other hand, assume $0<a<bk$ holds. Then, clearly, the equation has no roots on $(-\infty,0]$ ($P(z)<0$ there). Thus we either have three positive roots, which is fine for us, or one positive root and two complex conjugate ones. Again, we have $P(z)=(z-k)(z^2+bk)+(bk-a)k>0$ on $[k,+\infty)$, so the positive root $z_+$ is $<k$, whence (by Vieta again), the common real part of the two complex conjugate roots is $\frac 12(k-z_+)>0$.
Thus, indeed, the condition $0<a<bk$ is both necessary and sufficient.
A: The eigenvalues are the solutions of the cubic equation
$$-a k + b k \lambda - k \lambda^2 + \lambda^3 = 0$$.
Given that the solutions depend continuously on the parameters $a, b, k$, on the boundary of the region of valid parameters, at least one of the root must have real part 0.
If that root is exactly 0, then either $a$ or $k$ is 0. In any case the other two roots are:
$$\frac{1}{2}(k\pm\sqrt{k^2-4 b k})$$
if $k^2 < 4 b k$ the root is imaginary and since $k \ge 0$ the real part is positive. If $k^2 > 4 b k$ we still have $\sqrt(k^2 - 4 b k)$ < k since $b > 0$ so this works too.
If a = 0 or k = 0 there are no roots with negative real part.
If 0 is not a root, then there is a pure imaginary root, but the conjugate must also be a root. Expanding:
$$(\lambda^2 + C^2)(\lambda - \lambda_0) = 
\lambda^3 - \lambda^2 \lambda_0^2 + \lambda C^2 - C^2 \lambda_0$$
Identifying the roots: $\lambda_0 = k$ $C^2= a$ and finally $a = b k$.
