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Open loop vs Closed loop Transfer Functions

  1. Oct 19, 2014 #1
    Okay, I have a simple plant which is 1/s*(s+1)

    Two poles on 0 and -1 and no zeros. This is the case for open loop. But when I close the loop with unity feedback and add a K gain, I end up with following transfer function;

    K/s^2 + s + K

    So, clearly the poles are now at somewhere else. However, for example in this video, poles are still considered where they were with open loop;



    I am very confused with this. For a closed loop system, do I have to change the poles according to second transfer function. Likewise, adding a lead compensator, something like (s+a)/(s+b) would change the transfer function with a unity feedback and the denominator of the lead compensator, which is (s+b) would appear in the denominator of the new transfer function after closing the loop with unity feedback.

    But in the same video he just puts the poles and zeros coming from the compensator.

    Appreciate if someone can clarify this.
     
  2. jcsd
  3. Oct 19, 2014 #2
    He's drawing the root locus of the closed-loop system by using his knowledge of how those roots relate to the location of the open-loop poles.

    Recall that the root locus is the path that's traced out by the closed-loop poles as you vary the open-loop gain. Each point on it corresponds to a specific value of ##k##. He's showing you both the open-loop poles and the closed-loop poles of the system (that's the root locus), and, as you can tell, they're not at the same location in the complex plane.
     
  4. Oct 20, 2014 #3
    Okay, I got it.

    O more question;

    I obtained the desired characteristic polynomial for the given design specs like rise time and overshoot. Found the roots and then used lead compensator to move the poles along this path. Found the exact K value for those roots. Rise time seems okay however maximum overshoot is more than I designed. It had to be < 20% but I got 32% overshoot. If I designed everything correctly, that should not occur, right ?
     
  5. Oct 20, 2014 #4

    You'll have to take into account the effect of zeros on the transient response as well - it's not only the pole locations that determines it (I assume you've made sure they're dominant).

    I could be of more help if you showed your work.
     
  6. Oct 20, 2014 #5
    I designed a lead compensator for 1/s*(s+1) plant to meet the desired criteria;

    Rise time < 0.5 and Max. Overshoot < 20%

    I found the Natural frequency by using w_n≥1.8/t_r formula and damping ratio by using M_p=100e-ζπ/√1-ζ2

    So, Wn = 3.6 and Damping ratio = 0.4559

    With those values my characteristic equation would be;

    s2+ 3.28s+12.96

    (s2+ 2ζw_n s+ w_n2)

    So the poles are located at x= -1.64-3.20i and x= -1.64+3.20i

    The lead compensator I ended up with is (s+2.64)/(s+6.35) and the K = 19.9 to move the poles to desired location.

    Rise time is under 0.5 when I inspect the step response. However, overshoot is about 32%. It should be %20.

    I believe there is something wrong with Wn or Damping ratio so that my characteristic equation is somehow wrong. What do you think ? I am struck there for hours
     
  7. Oct 20, 2014 #6
    You've placed a pair of complex-conjugate poles just where you wanted, but you have to realize that, due to the addition of the lead compensator, you have an additional closed-loop pole and zero that will affect the closed-loop transient response.

    It will take some iteration to get the response you're after. Try overcompensating a bit in terms of damping required and recalculate. Moving the zero further towards the left-most open-loop pole will probably help.
     
  8. Oct 20, 2014 #7
    So, you suggest considering less max. overshoot at the very beginning and recalculate the damping accordingly, right ?

    I can understand that there is additional pole and zero but they just cancels each other and also changes the path of the plant poles. So, I don't understand why it is effecting the transient response.

    Thank you.
     
  9. Oct 20, 2014 #8
    Yes.

    They don't cancel - if they did, the compensator wouldn't have any influence.

    The lead compensator introduces a closed-loop pole (this isn't the same as the pole location of the lead compensator) and a closed-loop zero (this is the same as the zero location of the lead compensator).

    Do you have MATLAB with the Control System Toolbox? If you do, 'sisotool' is a great utility for experimenting with open-loop zero/pole locations and see how they affect the closed-loop poles.
     
  10. Oct 20, 2014 #9
    Just to elaborate a bit:

    Say you have a plant ##P(s) = \frac{n(s)}{d(s)}## where ##n(s)## and ##d(s)## are polynomials in ##s##.

    The closed-loop transfer function ##T(s)## is then given by:
    $$
    T(s) = \frac{\frac{n(s)}{d(s)}}{1 + \frac{n(s)}{d(s)}} = \frac{n(s)}{d(s) + n(s)}
    $$
    So you see, any zero of ##P(s)## is also a zero of ##T(s)##, but both the zeros and poles of ##P(s)## affect the pole locations of ##T(s)##.
     
  11. Oct 20, 2014 #10
    But when we draw the root locus we only consider the n(s)/d(s) term on the denominator near the 1. This was also the very first point confused me.
     
  12. Oct 20, 2014 #11
    Can you elaborate a bit? I'm not sure what you mean.

    The root locus is the path that's traced out in the complex plane by solutions to ##1 + \frac{n(s)}{d(s)} = 0 \Rightarrow d(s) + n(s) = 0## as you vary a single parameter, which is usually chosen to be the open-loop gain.
     
  13. Oct 20, 2014 #12
    I mean;

    Consider my above mentioned case;

    Plant is 1/s(s+1) and lead comp. is (s+a)/(s+b)

    Let's call plant A and comp. B

    So the T(s) = KAB/1+KAB

    and we only draw the zeros and poles of the KAB part on the denominator which is (s+a)/(s+b)*s(s+1)

    However, if we further simplify the T(s) what we got is = K(s+a)/K(s+a) + (s+b)*(s*(s+1))

    So clearly this time characteristic equation is s^3 + (b+1)s^2 + (b+K)s + Ka
    So there are three poles but they are different.

    Wait a second, are those poles actually the ones we already found which are -b, 0 and -1 but this time they just moved along the root locus path. So, actually they are the same poles but at different position this time due to K ?
     
  14. Oct 20, 2014 #13
    That's an expression. The characteristic equation for your closed-loop system is:
    $$s^3 + (b+1)s^2 + (b+K)s + Ka = 0$$
    Solutions to this equation are the poles of the closed-loop system.
    You're talking about two completely different systems. ##\{-b, -1, 0\}## are the poles of the open-loop system. They have a certain relationship to the poles of the closed-loop system (which you exploit when sketching its root locus), but they're not "the same".

    In your design, you've placed two (a conjugate pair) of the poles of the closed-loop system. There's another one, which you haven't dealt with at all (I assume).
     
  15. Oct 21, 2014 #14
    Okay, I kinda grasp the idea.

    One more question I have if you don't mind;

    I am Checking the phase and gain margin of the system with bode plot. I know how to interpret the bode plots but one point is not clear to me.

    My plant is P(s), compensator is C(s) and C(s) also multiplies with a K gain. When I am finding the open loop TF as L(s), do I have to multiply with K as well ? So, the question is;

    L(s) = C(s)*P(s) or L(s) = K*C(s)*P(s)

    Which one I have to consider for phase and gain margin evaluation ?
     
  16. Oct 21, 2014 #15
    When you form your closed-loop transfer function:
    $$
    T(s) = \frac{G(s)}{1 + G(s)H(s)}
    $$
    where ##G(s)## and ##H(s)## is everything in the forward and feedback path, respectively.

    The open-loop function is defined as that ##G(s)H(s)## term in the denominator, and that's the function you have to use when determining how the closed-loop system will behave in terms of stability and performance.

    In your case, the open-loop function is ##KC(s)P(s)##, assuming it's a unity-feedback configuration.
     
  17. Oct 21, 2014 #16
    Okay, thanks so much for your help.
     
  18. Oct 21, 2014 #17
    You're very welcome.
     
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