# 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.

• You should be a bit careful about how you state it because for $a=bk$, the eigenvalue equation factors as $(\lambda-k)(\lambda^2+bk)=0$, which is a borderline case as one would expect only if $k> 0, b>0$. Do you always assume that $a,b,k$ are all positive and are interested in the stability condition under that assumption? – fedja Oct 14 '20 at 21:11
• $k>0$ since 1/k is the characteristic time of the lag. – Bernard Oct 14 '20 at 22:46

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 is a necessary condition.

On the other hand, assume $$0 holds. Then, clearly, the equation has no roots on $$(-\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_+$$ is $$, 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 is both necessary and sufficient.

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 $$a = b k$$.