Why is high-gain controller undesirable

Question

In control theory and for example the scalar plant:

$\dot{x}=ax+u+d$

where $x$ is the state and $u$ is input and $d$ is disturbance

If the following control law is chosen:

$u=-kx$

where

$k\geq |a|$ and $|d|\geq d_0$

Then the steady-state value for $x$ is bounded as following:

$|x|\leq\dfrac{d_0}{k-a}$

by increasing the value of $k$ we can make the steady-state value of $x$ as small as we like. But this leads to high-gain controller which is undesirable.

My question is Why the high-gain controller is undesirable in this case and in general?

Answer 1

Because of different practical reasons:

• In reality, you can't make $u$ infinitly large, so $k \rightarrow \infty$ which would make $|x| \rightarrow 0$ is not feasible.
• In reality, you also have measurement noise which gets also amplified by large $k$.
• Also, in reality, large $u$ come with a cost (like energy consumption) so large $k$ can be too expensive.

But there are more problems with high gain. In many cases you dont really have a continous controller like in your system, but a digital. For the continous scalar system:

$$ \dot{x} = a x + u \tag{1} $$

and $u = -k x$ with $k > |a|$ you have $a - k < 0$ so the system is stable, no matter how large you take $k$ as long as its larger than $|a|$.

But if you have a digital controller, you end up with a discrete system. For example $(1)$ has the transfer function

$$ G(s) = \frac{1}{s - a} $$

which is discretized with sample-and-hold element and sample time $T$:

$$ G(z) = \frac{e^{a T} - 1}{a(z - e^{a T})} $$

Closed loop with static state feedback and gain $k$:

$$ G_c(z) = \frac{1 - e^{a T}}{k + a e^{a T} - a z - k e^{a T}} $$

This transfer function has the pole

$$ \frac{k(1 - e^{a T})}{a} + e^{a T} $$

that is outside the unit circle for large $k$. So high gain feedback can make your system unstable even if the continous system is stable, because of discretization.