Stability of LTI systems with control

[JoshLagoon]

Hello,

I'm interessed by the following LTI system with control u :
x' = Ax +Bu and x(k+1) = Ax(k) + Bu(k).

In this paper (section 3): https://proceedings.neurips.cc/paper/2020/file/9cd78264cf2cd821ba651485c111a29a-Paper.pdf

They seems to say that only A needs to be stable to get a stable system.

Why is B not considered in the stability of these systems ?

Thank you.

 

[collinsmark]

Hello @JoshLagoon, welcome to PF!

Hello,

I'm interessed by the following LTI system with control u :
x' = Ax +Bu and x(k+1) = Ax(k) + Bu(k).

In this paper (section 3): https://proceedings.neurips.cc/paper/2020/file/9cd78264cf2cd821ba651485c111a29a-Paper.pdf

They seems to say that only A needs to be stable to get a stable system.

Why is B not considered in the stability of these systems ?

Thank you.

I haven't studied the paper besides glancing through it, but for what it's worth, here's my quick take.

First let me define the terms I'm using. In the discrete time LTI system, we have the following difference equation:

$$y_t≡x_{t+1} = Ax_t+Bu_t$$
where,

• $y_t$ is the current output vector.
• $x_{t+1}$ is the next state vector.
• $x_t$ is the current state vector. This is the vector that defines the current state of the system. So if you were to ask yourself, "what state is the system at this particular time?" The answer is $x_t$. That's your answer. That and only that.
• $A$ is called the "state matrix," but don't let its name fool you. It doesn't define the state of the system (recall the state of the system is $x_t$). Instead, $A$ is a matrix that operates on $x_t$, the current state of the system. In other words, $A$ is an operator that works on the current state of the system to produce something that gets used later. But it has no effect on the current state of the system. $A$ doesn't change over time, btw. Think of it as a constant matrix once the designer of the system finishes designing the system. (I believe the whole point of the paper is how to create an optimal $A$. But once you have a good $A$ there's no need to change it. Thus it becomes a constant matrix in normal use.)
• $u_t$ is the vector containing all the inputs. Think of these as a set of dials a "user" can change while the system is running, assuming the system even has inputs at all. For systems that have inputs, $u_t$ can be anything. And these inputs can change to anything else from one time step to the next. The system designer doesn't have any control over the inputs themselves.
• $B$ is a matrix that operates on the inputs. Like $A$, once the system designer finishes designing the system, $B$ is a constant matrix.

So why is $A$ so important to stability? Because $A$ is the only thing that defines the relationship between the current state of the system and the next state of the system. Nothing else effects this very narrow and specific relationship. (And also, knowing the current state of the system gives you some [although not necessarily complete] information about the previous state of the system.)

One might argue, "but the next state of the system is also influenced by the inputs, $u_t$ and the control matrix $B$. Yes, that's true. But neither of these affect the current state of the system, nor do they affect future or past inputs.

The previous state of the system, the current state of the system, and the future state of the system are all related to each other by the difference equation, and specifically, by the state matrix $A$. It is this relationship that gives rise to stability concerns.

In contrast, we can't say anything like that regarding the inputs, $u_t$ and the control matrix $B$. If you know the current inputs, $u_t$, what does that tell you about the inputs in the next time step? Nothing. the difference equation doesn't offer any information in this regard. They are independent. Likewise, if you know the current inputs, $u_t$, does that give you any information about the previous inputs (even in the slightest)? No. Nothing. There is no defined relationship between previous, present, and future inputs. So, even though $B$ has an effect on future outputs, it has no effect on future, present, or past inputs, on which $B$ operates. Thus $B$ does not affect stability.