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For example, I use Routh-Hurwitz Cirterion to examine the following system with P controller gain \$K\$:

$$\frac{{K\left( {2 - s} \right)}}{{{s^3} + 3{s^2} + 5s + 3}}$$

I closed loop it with \$H=1\$ and found that \$K\$'s BIBO stability range is

$$ - \frac{3}{2} < K < \frac{{12}}{5}$$

I think when \$K=1\$ then controller will be just ignored.

But what happened when \$K=0\$? Does that cause system unlink together?

What does it mean when \$K=0\$?

Why is \$K=0\$ still stable on the Routh-Hurwitz criterion?

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This is a different way to draw the proportionally-compensated closed-loop block diagram:

enter image description here

The assumption behind the closed-loop block diagrams you know (like the one shown in PetPaulsen's answer above) is that K1=K2=Kp If that's true you can simply replace K1 and K2 with a gain of Kp after the summer and you have the same mathematical relationship of inputs to ouptuts but just in the form you're used to.

So what happens when Kp=1? Since Kp=K1=K2 we already know that K1 and K2 are 1. The system is still compensated because the feedback gain (K2) is not zero. Thus, in this form it's apparent that when Kp=1 the system is still compensated and the controller will not be ignored. The above diagram shows what the canonical form can't show as clearly. In this example, it's easy to see that the system only collapses to an uncompensated system if K2=0 and K1=1, but not if they're both 1.

Another of your questions was basically 'Why is Kp=0 considered 'stable'? Shouldn't that be 'disconnected' instead?'

It's something of a controls convention that zero gain represents an uncompensated system. When you redraw the block diagram in the form I did you can begin to see the outlines of their thinking: if Kp=0, then K2=0 and feedback is broken and we're left with the uncompensated system. But isn't K1 supposed to equal K2, which is 0? Here's where a bit of fudging takes place (in my opinion). They like mathematical continua and zero almost fits perfectly: if Kp is the smallest positive number, or the smallest negative number then you have a proportional controller. In fact if it's anything other than 0 it works - just not quite zero. Unless you adjust your thinking: K1 only has to equal K2 when if you want a proportional feedback controller. If you break the feedback you're under no obligation to have K1=K2, so let's just make it 1 and call it a day.

That still doesn't make it practical though. Trust me, if the 'correct' answer to a problem is a gain of Kp=0 you're having your leg pulled.

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There will be no input to the plant, if \$K = K_p = 0\$ (and \$K_i = 0, K_d = 0)\$. The controller is ignored in this case (not if \$K = 1\$, because you still feedback the output).

Block diagram

(Original picture taken from Wikipedia and modified)

With \$K = 0\$ the system is stable, because the equilibriums of your plant are stable.


Expanding based on your comment: The block diagram and the transfer funtions are just models for the system you want to describe. So for \$K=0\$ this could mean, that the controller is connected, but no control signal is applied to the plant (because you multiply by 0). Setting \$K=0\$ could also mean that the controller is not connected at all. For example you want to simulate what happens, when the P-Controller fails. You could do this by setting \$K=0\$.

For \$K=1\$ the controller is not ignored. As long as you got the feedback you respond to changes in the output. Lets say \$v(t)\$ is the input to the plant. $$v(t) = K_p \cdot e(t) = K_p \cdot (u(t) - y(t))$$ For \$K=1\$ you are still feeding back \$y(t)\$. $$v(t) = e(t) = u(t) - y(t)$$ It would be wrong to say, that the controller wouldn't do anything. It just doesn't amplify the signal.

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  • \$\begingroup\$ Oh, I maybe understand. Now I think K=0 means no input to plant but still connected with plant,right? K=1 means we still need plant to be acted and plant input = e(t). But does plant input=e(t) mean we just ignore controller and just connect wire from e(t) to the plant? Because in the case of controller=1,I think that controller doesn't do anything,right? \$\endgroup\$ – sam May 5 '12 at 23:03
  • \$\begingroup\$ @sam - I tried to clarify your questions, hope this helps. If not, ask again. \$\endgroup\$ – PetPaulsen May 6 '12 at 7:25
  • \$\begingroup\$ Thank you for very detail explanation. Now I know K=0. For K=1, what you mean is controller is e(t), so controller couldn't be ignored? In case of controller = e(t), what I mean is its input = y(t) and u(t) ,and controller's operation is u(t)-y(t),so the controller's output=e(t). \$\endgroup\$ – sam May 6 '12 at 15:36
  • \$\begingroup\$ If controller is not the above comment (e(t)). Is controller = 1, and its input=e(t) and controller just wired (means not do anything change to the input), and its output = e(t)? Thank you~ \$\endgroup\$ – sam May 6 '12 at 15:39
  • \$\begingroup\$ @sam - It's hard for me to understand your last two comments. You want to know whether the summation block (u(t) - y(t)) is part of the controller, right? -- Yes it is. y(t) comes from your sensors and u(t) what you want the output to be. This is all fed into the controller. \$\endgroup\$ – PetPaulsen May 6 '12 at 20:51

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