root locos for the transfer function with matlab

I have a transfer function and i want to calculate the speed of each pole and i do not know how to do that

  • $\begingroup$ Welcome to Robotics, A_S. This sounds like a homework question. What work have you done so far to attempt to solve the problem? What part of the process has you stuck? $\endgroup$ – Chuck May 9 '17 at 10:07
  • $\begingroup$ i didn't really found what speed of pole means is searched several references and did not find any of them use the name speed of pole on of the references i searched was modern control systems by richard c. dorf and robert h. bishop $\endgroup$ – A_S May 10 '17 at 19:30
  • $\begingroup$ @Chuck ??? Do you know the answer or a good reference or site to search it $\endgroup$ – A_S May 11 '17 at 12:31

What are you doing that you're being asked to find the speed of a pole? If this is a course (online or actual), then I would expect the material would have been covered already. Any time I've heard of "speed" mention with regards to a pole, it's how "fast" or "slow" the pole is, which means how far or close the pole is to the imaginary axis (respectively). A pole that is very close to the imaginary axis will take a long time to settle; one that is very far from the imaginary axis will settle very quickly.

This is how you get into the "dominant pole(s)" or "dominant response" - the "faster" dynamics (poles) settle quickly and the "slower" dynamics (poles) are left, and so at the medium- to long-term, the only observable dynamics are those of the "slower" poles.

I remember learning to compare pole speed by looking at their natural frequencies, but in researching for your question I've seen several posts stating it's the distance from the imaginary axis (-$\xi \omega_n$), which I believe makes more sense than just the natural frequency ($\omega_n$).

From this site

In this system, as the poles begin to move from the open loop poles, the "slow" pole (It doesn't move slower, but because it is closer to the imaginary axis the response due to that pole is slower.) at s = -1 initially moves to the left, indicating a faster response in that component of the response. The "fast" pole starts from s = -2, and move to the right, indicating that that portion of the response slows down. Still, the overall effect will be a speedup in the system, at least until the poles become complex. You can see that in this clip that shows the unit step response of the system with the root locus above.

Here is a PDF on control engineering that makes numerous references to how fast the system responds with respect to damping ratios and natural frequencies.

So again, I would imagine you didn't ask yourself to do this problem, so talk to the person that gave you the problem and ask them what they mean when they say speed. If I had to guess, I would say to find $-\xi \omega_n$ for each pole.

| improve this answer | |
  • $\begingroup$ Chuck, do you mean to find $\Re{\omega_n}$? $\endgroup$ – SteveO May 11 '17 at 22:35
  • $\begingroup$ @SteveO - The magnitude of the real value of the pole, yes. $\endgroup$ – Chuck May 12 '17 at 4:43
  • $\begingroup$ Thanks. That kind of makes sense for the "speed of the poles." $\endgroup$ – SteveO May 12 '17 at 5:29
  • $\begingroup$ @SteveO - Yeah, kind of a mis-applied terminology; really it's the speed of the system response, but that response happens because of the poles and their locations. Kind of like a "fast" driver haha - the car did all the work. The system response is what matters, but the poles drive that response. $\endgroup$ – Chuck May 12 '17 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.