1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coursework : Root Locus

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to design a controller for "Cruise Control System" using Root Locus method.


    2. Relevant equations
    We havnt studied that method yet and nor do I think it is in the course outline. Actually that is why I selected root locus so that I can learn something beyond the course for Control System.


    3. The attempt at a solution
    Luckily I found the project online. http://www.engin.umich.edu/class/ctms/examples/cruise/ccrl.htm"

    The problem is in understanding the method itself. I have referred to many books but I am still not able to catch the real deal. I will start with the basic problems I have.

    1. How many poles/zeros do we move to make the system stable?

    2. How do we know that we have to move poles only or zeros only or both ?

    3. Referring to the above link. Doesnt the two dotted lines (under "Proportional controller") and the dotted semi circle bound all the points which will give the desired rise time and overshoot ? Why do they don't when the criteria for those lines and semicircle is set?

    4. Referring to the above link. I do not understand the part under "Proportional controller" where after plotting the root locus (I understand the plotting part quite fine) it is written to chose "-0.4" as the pole. Now the confusion is that there are million points between the dotted lines and the dotted semi-circle, so how do I know that which one will give the desired rise time and overshoot.

    Will be waiting anxiously for reply....

    Thanks in advance
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 26, 2009 #2

    CEL

    User Avatar

    Answers to your questions:
    1. Your system is stable for every gain (no part of the RL is in the RHP).
    2. Adding zeros tend to shift the RL to the left, while adding poles tend to shift it to the right. If you want more stability, you should add zeros to your system, but since you cannot add a simple zero, you should add a compensator with a pole and a zero, the zero beeing nearer the origin than the pole. This is called a lead compensator.
    3. I don't understand your question.
    4. Since the RL is all in the real axis, you must pick a real value for the location of your closed loop pole. They picked -0.4 because it is a rounder number than -0.3873, for example.
     
    Last edited by a moderator: Apr 24, 2017
  4. May 27, 2009 #3
    3. My question was that what does the region bound by the semicircle and the two dotted lines represnt ?

    4. RL all in real axis ? I dun get it...Wasnt the semi-circle and the dotted lines the locus ? And if you pick up -0.4 then it doesnt meet the criteria actually. The rise time comes out to be more that 5 sec...You have to selecte a value much farther than -0.4...So how d o I know which one to select ?
     
  5. May 27, 2009 #4

    CEL

    User Avatar

    The semicircle represents the region of the plane where [tex]\omega_n = 0.36[/tex]. Points inside the circle have [tex]\omega_n < 0.36[/tex], those outside have [tex]\omega_n > 0.36[/tex].
    The two dotted lines are the locus of the points with [tex]\zeta=0.6[/tex]. Between the lines we have [tex]\zeta>0.6[/tex].
    So, the point -0.4 fulfills both specifications: [tex]\omega_n > 0.36[/tex] and [tex]\zeta>0.6[/tex].
    The RL is the line from the pole -0.05 to minus infinity (in blue in the graphic).
     
  6. May 27, 2009 #5
    But if the point -0.4 fulfills both specifications and the two specifications of eta and Frequency were calculated from the initial criteria of rise time and overshoot then why does the rise time doesnt still come out to be corect when plotted in MATLAB ?
     
  7. May 27, 2009 #6

    CEL

    User Avatar

    Remember that the rise time is the time the system takes to go from 10% to 90% of the final value. The paper you linked to says this criterion has been met (and the plot seems to confirm it).
     
  8. May 27, 2009 #7
    Can you please try it out in MATLAB ? MATLAB doesnt agree..
     
  9. May 27, 2009 #8
    Even the overshoot is more than 10% where as it was catered for in the calculations as well. Why is that so ?
     
  10. May 27, 2009 #9

    CEL

    User Avatar

    How do you get an overshoot in a first order system?
     
  11. May 27, 2009 #10

    CEL

    User Avatar

    I have tried it in Matlab. You are right. The rise time is a little less than 5.5 sec.
     
  12. May 27, 2009 #11
    I am not sure my self but they have calculated in that tutorial. The final velocity should have been around 10 anyway and its even less than 9 so its not good.
     
  13. May 27, 2009 #12
    Any idea where does the problem lie ? Root locus has been plotted correctly so what can be wrong ?
     
  14. May 27, 2009 #13

    CEL

    User Avatar

    There is no overshoot. Look at the plot.
    There is a steady state error in the velocity, because all zero type systems present a ss error to a step input. That is why in the following item they use a lag compensator to reduce the error.
     
  15. May 27, 2009 #14

    CEL

    User Avatar

    The formula they used to relate rise time to [tex]\omega_n[/tex] is valid for a second order system. Since the system is first order, the formula gives only an approximate relationship.
     
  16. May 28, 2009 #15
    Where do I get the relations for Zeta and Frequency for 1st order systems ?
     
  17. May 28, 2009 #16
    Is the one for Zeta applicable on 1st order systems ?
     
  18. May 28, 2009 #17

    CEL

    User Avatar

    In first order systems zeta = 1. The frequency is the absolute value of the pole.
     
  19. May 29, 2009 #18
    Why can't a first order system overshoot ? What restrains it ? Any formula to calculate it ?
     
    Last edited: May 29, 2009
  20. May 29, 2009 #19
    And what about steady state error for 1st order systems ? Where do I get the formula ?

    And does a type 0 system with one pole classify as 1st order ?
     
    Last edited: May 29, 2009
  21. May 29, 2009 #20

    CEL

    User Avatar

    Overshoot is how much the response of the system goes over the steady state value. In a first order system the response is always below ss.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Coursework : Root Locus
  1. Root locus varying H (Replies: 3)

  2. Root Locus Sketching (Replies: 2)

Loading...