I'm using the control system toolbox provided by matlab to estimate the gains of my controller: using root locus design I get a graph like this one enter image description here.

My question is: what is the x on the x-axis? maybe a pole position at a previous iteration of the optimization procedure that I have run to find a gain value that satisfies my requirements? It shouldn't be the open loop pole position, because my system is formed by two integrators multiplied by a constant (1/inertia). Thanks

Edit: I add the requested details: I start from the following simulink diagram: enter image description here my trasfer function is $$G_\Theta(s) = \frac{Y(s)}{U(s)} = \frac{\Theta(s)}{\tau_\Theta} = \frac{1}{I_y s^2}$$ with Iy = 0.0054 (another little question, the point in which I'm taking out the torque is correct?) and then I select analysis , control design, compensator design. I select Kp and Kd as the gains to be tuned, and I use the root locus for specifying the constraints. Then I click in SISO Design Task, automated tuning, optimize compensator, which automatically tries to find gain values to satisfy my constraints. The white are is the area that satisfies the constraints, and I think that the pink squares are my poles position after having been completed the optimization procedure. This is correct? But in this case, what is the x(pole) shown? Thanks


1 Answer 1


I can add more detail to this answer later if you can provide your transfer function or block diagram you used to make that plot.

Regarding your question, an X is a pole, so that is what the X on your X-axis is.

Recall that, if you have a plant $G(s)$ ($ G(s)=1/s^2$ in your case), and you have a feedback loop, your transfer function becomes $G(s)/(1 +/- G(s))$, so you do in fact wind up with poles.

If you believe you are viewing outdated information on your plot, then be sure to call 'clf;' between plot commands, with a 'figure(N)' command prior to focus plot N if you have multiple plots open. 'clf' clears a figure window.

Finally, as a reminder, double check that what you are plotting is in fact the open loop system if that is that you would like to evaluate. If you are still having trouble, again, post the commands you are using to plot and I'll update this answer.

Hope this helps!

  • $\begingroup$ I have add the details, I hope to having been described all the things well. Thanks $\endgroup$
    – Daniel
    Commented Jun 7, 2015 at 9:11
  • $\begingroup$ can you give a look to the edited question? $\endgroup$
    – Daniel
    Commented Jun 14, 2015 at 13:25
  • $\begingroup$ I had looked at it earlier. What I noticed is that the root locus diagram starts with the open loop poles and zeros and plots how they change as gains vary. That you have an X implies you have an open loop pole. I don't have the optimization package so I can't duplicate your steps, which is why I didn't comment earlier. All I can say is that it looks like you are not plotting the open loop system you think you are. Also, side note, $\tau = I \alpha$, so I don't know why you're multiplying your acceleration by $1/I$. $\endgroup$
    – Chuck
    Commented Jun 14, 2015 at 14:41

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