# Does every control system contain an integrator?

Hi,

Let's call the box in green a plant. Does every plant consist of an integrator? It has to because when $\overset{\cdot }{x}(t)$ passes through the green box, it becomes x(t). It's an operation of integrator; integrator converts the derivative back to the variable. Could you please guide me?

BvU
Homework Helper
You designate an integrator 'the plant' and then you are surprised that a 'plant' integrates?

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PainterGuy
DaveE
Gold Member
No, not every control system has an integrator. But yours does.

However, most control systems contain an integrator. That's how you make it achieve zero DC error.

PainterGuy and AlexCaledin
tech99
Gold Member
It seems to me that a control system without an integrator will feed back the actual signal, so becomes a negative feedback system, which has different properties.

DaveE
anorlunda
Staff Emeritus
It seems to me that a control system without an integrator will feed back the actual signal, so becomes a negative feedback system, which has different properties.
If you want something nearly universal about all control systems, negative feedback is the closest you will get. Seldom zero feedback and never net positive feedback.

But many control systems have negative feedback and proportional output. No integrator. For example, James Watts' flyball governor.

PainterGuy
DaveE
Gold Member
Hi,

Let's call the box in green a plant. Does every plant consist of an integrator? It has to because when $\overset{\cdot }{x}(t)$ passes through the green box, it becomes x(t). It's an operation of integrator; integrator converts the derivative back to the variable. Could you please guide me?

View attachment 282402
Sorry, I miss read your question. No, not every plant acts as an integrator, in fact that's common in my experience. Most plants are proportional. The integrator is normally applied as part of the control system.

However, this partly depends on how you define the inputs and outputs. So, for example, a motor may have speed proportional to applied voltage, but position is the integral of applied voltage. So, same motor, but different behavior depending on what you want to control.

PainterGuy, eq1 and Merlin3189
Thank you, everyone!

I don't think my question was phrased properly. Let me try again.

You designate an integrator 'the plant' and then you are surprised that a 'plant' integrates?

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It's not me who designated an integrator as the plant rather this is how the state representation is shown. Please see the figure below. The representation suggests that an integrator is always there present and when $\overset{\cdot }{x}(t)$ passes through it, it becomes x. In other words, an integrator is a natural part of state space representation. Where am I going wrong?

https://en.wikipedia.org/wiki/State-space_representation#Linear_systems

DaveE
Gold Member
Thank you, everyone!

I don't think my question was phrased properly. Let me try again.

It's not me who designated an integrator as the plant rather this is how the state representation is shown. Please see the figure below. The representation suggests that an integrator is always there present and when $\overset{\cdot }{x}(t)$ passes through it, it becomes x. In other words, an integrator is a natural part of state space representation. Where am I going wrong?

View attachment 282469
https://en.wikipedia.org/wiki/State-space_representation#Linear_systems
I suspect your question is bordering on semantics. A system with no integrators (or derivatives) can be described in the state space format, such as:

\begin{align} &\dot x = Ax + bu \ &y = Cx + du \end{align}

But without integrators (or derivatives), there is no state of the system. It's not dynamical and has no memory. So, A=0, b=0, C=0. It is the zero order trivial case that reduces to a set of linear equations. So in practice, no one talks about state space unless there is a state, i.e. unless there is at least one derivative.

Also, of course, the canonical form shown differentiates the equations to eliminate integrals in favor of derivatives.

I think there may be some confusion in the responses between what is the plant, what is the compensation, and what is the system. My take on your block diagram is that it is the canonical form of a generic linear dynamic system in state space form. It really doesn't matter at the system level which is the plant and which is the compensation when you are solving a previously designed system. Of course it makes a tremendous difference when you are designing that system.

PainterGuy, eq1 and BvU
anorlunda
Staff Emeritus
Also that Wiki page talks about the representation of a linear system. Not all systems are control systems.

tech99 and PainterGuy
Also that Wiki page talks about the representation of a linear system. Not all systems are control systems.
They're not even all continuous,
https://en.wikipedia.org/wiki/Bang–bang_control
Some don't even sense the output,
https://en.wikipedia.org/wiki/Feed_forward_(control)

But I totally agree with DaveE, the question is basically a semantics one. Once the definitions are agreed upon it will be pretty easy to check the question against the definition. The trick is agreeing on a definition. :)

DaveE and PainterGuy