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I wanted to plot the root locus for a system, whose plant transfer function is

$$ P(s)= \frac{k}{Ts+1}e^{-ts}. $$

A PI controller was suggested for this system, and the task is to find $K_p$ and $K_i$ values, such that it satisfies the following performance specs:

  • Maximum peak overshoot < 5%
  • Settling time (2% band) < 0.8 seconds

This is how I modified the open loop transfer function to suit the format of root locus finding.

$$ G(s)H(s) = C(s)P(s)= K_p\frac{k(s+A)(2-ts)}{s(Ts+1)(2+ts)}, $$

where $A = K_i/K_p$ and Pade approximation was used. For starters, $A$ was assumed to be 1.

But, when I plot the root locus of the above, this is what I get. enter image description here

Here, the white region is the feasible region which satisfies the performance specs.

As we can see, the root locus of the pole introduced by the integrator (at Im axis) ends up being completely inside the unfeasible region. This is the case for any $A$ value I take.

What should I do? Is my approach correct? And more importantly, how will this pole affect my performance?

I'm new to this field, and I'm a bit confused by this. Please help me out here. Thanks in advance.

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  • $\begingroup$ You do not give a value for $t$, but the zero defined by $(2 - ts)$ is unstable -- yet I don't see a zero to the right of the imaginary axis. $\endgroup$ – TimWescott Sep 13 at 0:45
  • $\begingroup$ @TimWescott zeros in the RHP aren't unstable---just poles ---right? (but agreed we should see another zero over there, looks like it is the zero at -A above consistent with his use of A =1 ) $\endgroup$ – Dan Boschen Sep 13 at 2:35
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    $\begingroup$ @DanBoschen: Even control systems engineers have slang. "Unstable zero" just means a zero in the RHP -- it doesn't mean the system is unstable (but it does mean it'll be at least a bit weird to control). $\endgroup$ – TimWescott Sep 13 at 3:53
  • $\begingroup$ @TimWescott ah that makes more sense. $\endgroup$ – Dan Boschen Sep 13 at 3:57
  • $\begingroup$ A problem is that the nominal first-order model from Pade is trying to wrap around an integrator. Personally I would drop back, especially with a system with time delays, and use Nichols plots. It is as accurate as you want and features/control of response is obvious: ocw.mit.edu/courses/aeronautics-and-astronautics/… But if a Pade approximation is insisted on; this looks to do some analysis: researchgate.net/publication/… $\endgroup$ – rrogers Sep 15 at 21:07
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The root locus will start at the open loop poles and move toward the open loop zeros as the gain is varied. Your pole cannot get past the zero that is located in the yellow region. This pole that is close to the imaginary axis will be dominant, meaning it will dominate your response times, slowing down the entire system. One approach is to cancel that zero with an additional open loop pole at it's location, unless that zero was introduced by your controller in which you have the ability to move it to be out in the white area, or eliminate it entirely.

It appears that the zero that terminates the closed loop pole as shown is due to $(s+A)$ in the numerator, in which case choosing a value for A that is within the white region should allow the pole to extend further to that point as the gain is increased. Also, the pole and zero due to the Pade approximation does not appear to be visible in the root locus.

To see how poles close to the imaginary axis are dominant, consider the simple case of $x(t) = e^{-at}$ which has a Laplace Transform $X(s)=\frac{1}{s+a}$ (which is a pole at $s= -a$. The time domain equation is a decaying exponential with time constant $a$, if a is very small this will take a very long time to decay. If $a$ is very large, the decay rate will be very fast.

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  • $\begingroup$ Couldn't he cancel that pole directly with the PI controler? $\endgroup$ – Ben Sep 12 at 15:22
  • $\begingroup$ @Ben - Cancel that zero you mean? That pole is the integrator from the PI controller I believe which starts at the origin and the open loop zero from the PI adds the second zero that is on the horizontal axis. Yes you could cancel the zero with the controller (which is what I suggested) but the pole in a true PI controller is at the origin - the integrator. He could certainly add an additional pole in addition to the P and I in PI, which would be part of the modified controller, but it would be an open loop pole that would be needed to cancel the open loop zero, right? $\endgroup$ – Dan Boschen Sep 12 at 15:47
  • $\begingroup$ yep sorry, cancel that zero. $\endgroup$ – Ben Sep 12 at 16:00
  • $\begingroup$ (@Ben An additional pole in addition to the I in PI I meant, the P is not a pole!) did what I say make sense to you? $\endgroup$ – Dan Boschen Sep 12 at 16:11

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