HELP- 3rd order system, find the value of K

Discussion in 'Homework Help' started by ryushin, Jul 17, 2016.

  1. ryushin

    Thread Starter New Member

    Jul 17, 2016
    I have a OPEN loop transfer function, which is 1/(s+1)(s+5)(s+20) with a unity feedback.

    Our lecturer request us to find the value of gain,K that have (for close loop)
    -Settling Time less than 8s
    -Overshoot less than 40%

    I had already figure out the value of zeta(damping ratio) by using the 2nd order system of Maximum Overshoot equation which is = 2.8, and how do i figure out the value of Wn( frequency) because when i use the 2% criterion to calculate it, it provide me the value of 1.7864(Why it is different with the freq i find in MATLAB) and how do i use this 2nd order approximation to recover back to find the value of gain K in 3rd order.

    Characteristic equation of 3rd order closed loop:s^3+26s^2+125s+(100+K)

    ps. I had use MATLAB to figure out the gain(using 3rd order cloose loop transfer function), value should be about K=860, and with damp ratio 2.8 and freq of 6.54rad/s

    sorry for my broken english and thx for helping
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014

    To start, are you sure you got the right closed loop transfer function?
    Since the closed loop transfer functions is the forward function with unity negative feedback, it looks like the denominator should have a "-K" in it rather that a "+K" in it.
    So that would make the last term 100-K not 100+K.

    Perhaps you can show a little more of your original work with this and we can go over it.
    Also, to determine the 2nd order main characteristics we often look at the dominate roots, but if you want to show a little more of your work that might help too.
    Please, please, when you write make it highly legible and clearly written. I say this because if it is too sloppy it's too hard to follow :)
  3. ryushin

    Thread Starter New Member

    Jul 17, 2016
  4. ryushin

    Thread Starter New Member

    Jul 17, 2016
    Above are some calculation about what i had right now, then i confusing how to determine the value of K, since i using 2nd order approximation
  5. MrAl

    AAC Fanatic!

    Jun 17, 2014
    Hello again,

    Ok, very good, and you are right that K is positive for this problem.

    Now you said that you are using the 2nd order approximation, so now i have to ask, where is that 2nd order approximation that you used to set the gain?

    In the picture shown, we can see the original time solution with K=860, then using just the dominant roots with a scaling factor to get both infinite time solutions the same, then with the original plus the dominate roots solution overlapped. The overlapped picture tells us that the dominant roots solution is a very close approximation to the original time function and so should be able to be used to set the original gain.

    But if you've done it another way, that's fine, as long as you got what you were after. So the remaining question is, looking at the first graph in the picture is that the time solution you were after?
    If so then you are done, but if not, then you have to find out why it is not correct.

    I dont want to say too much too soon here because this is the homework section. I will say that if you graph your function you can usually tell what is going on. Graphing helps eliminate errors.
    That's not to say this problem is right or wrong so far, just that graphing helps to show if it is right or not.

    So if you can show what 2nd order approximation you used that would be good too, even if you like the solution so far.

    There is one more small point to be made here. That is the one of optimization or pseudo optimization. That's where you are given a set of specs that read "x<N" and if x is much less than N you satisfy the specs but you end up with a much slower response than if x was just a little less than N, or some other aspect is not as good as it can be (perhaps too fast). By making x a little larger the spec might still be met while improving some other aspect of the system.
    This might not be as relevant for this problem though.
    Last edited: Jul 18, 2016
  6. ryushin

    Thread Starter New Member

    Jul 17, 2016
    Thanks for the reply.... umm I think my problem is how is about how do I find out the dominant pole when the gain value is K(the place where I hang)

    I have a 3rd order CE which is s^3+26s^2+125s+(100+K), although I have this and the value of zeta, but I not able to sub the value of zeta inside it, so I was thinking a way to find out during 2nd order approximation, how do I find out the 2ND ORDER APPROXIMATION CE without providing the gain value of K

    And sorry maybe my language problem cause u to confuse, umm currently I had no 2nd order approximation Transfer Function, the 860 value is get by using the MATLAB(since it able to handle higher order system.....)

    I think my big trouble should be just on how should i get the dominant pole of the Transfer Function , and figure out the 2nd order approximation T(s), and thanks for spending your time and explain to me :)
    Fewa likes this.
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
    Hello again,

    Well first you would get the value for zeta which you say you obtained but that doesnt look correct. If we look at a graph of several second order systems with damping ratios from 0.1 to say 1, we see a forty percent overshoot comes in with a damping ratio of about 0.30 or a little less, therefore 0.28 would be more reasonable.
    If you dont have this graph you can easily make one by setting up a second order system in the time domain and the varying zeta to see the different curves or i could post a graph.
    Alternately, using the second order system and the calculation you provided:
    you can calculate the required z from that, which i suppose you did but maybe moved the decimal point over.

    You also might want to think about what happens to the settling time when we choose a certain value for K. Since we only have one parameter to change, we can only set one thing and the other thing either comes into range or it doesnt, and if it doesnt then we have to go back and try another value for K which will satisfy criterion 1 and also that second criterion.

    With all this in mind, it is not that difficult to simply try a few values for K and see what pair of complex roots will result. We know we need a pair of complex roots because the response will be partly oscillatory.
    So the idea then is to try some values for K, solve for the complex roots, then compare the damping factor to the factor calculated (carefully) from the above. This isnt that hard to do for a 3rd order system because there are known solutions to 3rd degree equations.
    What else makes it a little simpler is we can just lump the 100+K into just k, then later subtract 100 from k to get K.
    Doing that, if we try k=1 we get three real roots, which cant be right.
    If we try k=200, we get a complex pair and one real root, which is what we want.
    If we calculate the damping factor however we do not get anything hear 0.30 or so.
    Going higher, we try 500, then maybe 1000, and we find 1000 gives us the wrong root in the other direction, so we go back down to say 750, then try that.

    The above is a sort of hands on approach. In reality we would probably do a root locus and that would show us the right value because of the placement of the two complex poles.

    So give that a shot and see how it goes, but be careful when calculating the damping factor. A damping factor of 2.8 would be a slowly evolving system that is real exponential in nature which has no oscillatory terms.

    If you like we can try other examples too, and once you see how this works i can present the solution(s) i found also.

    Just a small note, and that is that you should be able to tell from the graphs what you have, at least to some extent if not perfectly. If the graphs of the 3rd order system looks reasonable, then you are probably close to the solution or close enough in this problem.
    You should be able to judge that at some point if you havent already done so.

    Also note that there are other ways to get to the second order system but they would be much more involved. You may be put off by the length of the procedure.

    If you find that the above trial and error is not your style, then you can set up a differential:
    by calculating two values of z for two CE one with K=K1 and one with K=K2, then calculate the differential dz/dK then divide the required value of dz=z1-z by the differential and that should give you the increment in K (so Knew=Kold+dK) required to get closer to the actual solution. Doing this a second time will get you much closer, a third time even closer, etc.
    If using a starting value of k=800 for example (that's actually k=800=K+100) we'll get a value of about dk=150 so that will take us up to 950 already, which is very close to the right solution (and K=k-100).

    Using a root locus we can look at the relationship of the angle of the poles to the damping factor to get the damping factor, so if we did a root locus we'd see it pretty fast.

    If you want to do this the long way, you can vary K and repeatedly solve for the time domain solution and then compare the overshoot max to what you need. Although this takes a lot of calculations to do this is the most intuitive because you see the time domain solution right at every try. This also takes into account any deviation that would be caused by the real pole, if any, although that isnt a problem unless the real pole is comparable to the complex poles.

    Here's one exact solution but it isnt very simple:

    and solving for k may be impossible so we still have to solve it numerically.
    To see what this would come out like, we have the graph shown in the picture.
    To get the approximate value of k we just look at the graph.
    Keep in mind this is k not K, so K=k-100.

    Last edited: Jul 18, 2016
  8. MrAl

    AAC Fanatic!

    Jun 17, 2014
    Hello again,

    I thought i would add a root locus plot so we can see where the roots lie as k is varied (and again k=K-100 from the original equation). This graph is approximate.
    In this plot, k is varied from 0 to 1400, and the result is plotted in green.
    The small white dots show where K=0 on the plots.
    Note the two green plots, one to the left, one to the right.
    The one to the left shows the real root migration, the one to the left shows the
    complex pair migration.
    The red line is not part of the graph, but represents the angle of the damping factor.
    That is the angle for a damping factor of about 0.28 or so.
    It crosses the green curved part at about wn=6.5 or so, which is about 6.3 on the imaginary axis.

    Last edited: Jul 20, 2016