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Control Theory Problem

  1. Jul 15, 2009 #1
    Im trying to figure out how a certain example is working.

    in this example the function

    [tex]x = g - \frac{k}{m} \frac{(z-z0)^2}{(i-i0)^2}[/tex]

    where g, k, m, z0, i0 are fixed values and z, i are variables

    The function above is written as

    [tex]x = f(i,z) ( \sqrt{g} (z-z0) - y (i-i0) )[/tex]

    where f(i,z) is a new function as function of i, z and y is a new variable independent of i, z.

    The idea is to use linearize the system around its equilibrium point (z0 = 0.072 and i0=1). But in the example they dont show how they have found f(i,z) (and f(i0,z0)) and y.

    In order to reconstruct the example i need this so i tried to find f(i,z) by writing f(i,z) as (a+b) and solve the system as

    [tex] f(i,z) = (a+b) ( \sqrt{g} (z-z0) - y (i-i0) ) = a \sqrt{g} (z-z0) - a y (i-i0) + b \sqrt{g} (z-z0) - b y (i-i0) [/tex]

    thus if we compute a, b, y we get

    [tex] a = \frac{\sqrt{g}}{z-z0} , \quad b =\sqrt{\frac{k}{m}}\frac{i-i0}{(z-z0)^2} , \quad y = \sqrt{\frac{k}{m}} [/tex]

    thus

    [tex] f(i,z) = a+b = \frac{ \sqrt{gm} (z-z0) + \sqrt{k}(i-i0) }{ \sqrt{m}(z-z0)^2 } [/tex]

    if i now fill in i=i0 and z=z0 (equilibrium point) then the answer is infinity, which cant be correct.

    Anyone knows where i make a mistake?
    Thanks in advance
    Azizz
     
    Last edited: Jul 15, 2009
  2. jcsd
  3. Jul 15, 2009 #2
    Please rewrite this a little more clearly so I can understand it to help you. Double spacing (between text and equations) would be nice so everything isn't on top of each other.
     
  4. Jul 15, 2009 #3
    Sure no problem. I didnt notice the spaces were so small. Hopefully this is more clear.
     
  5. Jul 15, 2009 #4
    If you're going to linearize this, you should do it using the Jacobian matrices method.
     
  6. Jul 16, 2009 #5
    Ok thats absolutely true. But i think i misunderstood the question. Apparently they are not linearising, but they use the nominal value (that is f(i0,z0)) and let this function vary in a certain range. Then stability of the system is analysed for the variations.

    So we search for the nominal value (z=z0, i=i0) for the function

    [tex] f(i,z) = \frac{ \sqrt{gm}(z-z0) - \sqrt{k} (i-i0) } { \sqrt{m}(z-z0)^2 } [/tex]

    which is infinity if computed directly. However, there should exists a trajectory which yields a finite solution.

    The known equations are

    [tex] \frac{d^2}{dt^2}(z-z0) = g - \frac{k}{m} \frac{(i-i0)^2}{(z-z0)^2} = f(i,z) (\sqrt{g} (z-z0) - \sqrt{\frac{k}{m}} (i-i0) ) [/tex]

    [tex] \frac{d}{dt}(i-i0) = \frac{1}{l}(v-v0) - \frac{r}{l}(i-i0) [/tex]

    [tex] z0 = \sqrt{\frac{k}{mg}} i0 [/tex]

    Also v0=2 and i0=1.

    So now the question is how do i find that trajectory. I think i have to find i as a function of z (or vv) and substitute that in the equation for f(i,z) which causes some cancellations. Any suggestions how to do this?

    Azizz
     
  7. Jul 16, 2009 #6
    Can I ask you where did you get this example? And do they cover Linear Fractional Transformations?

    Because if so, things get different. You perturb your system by these additional [itex]\Delta z, \Delta i [/itex] etc. And try to guarantee stability by let say, small gain condition or some other condition.
     
  8. Jul 16, 2009 #7
    This is an example from reader which covers the robust control theory.

    Yes it does. But those perturbations does not influence the nominal value, does it?
     
  9. Jul 16, 2009 #8
    I am sorry but I can not understand the equations. So maybe i can describe in words what you might want to do.

    You isolate the nonlinearities/uncertainties in your model such that the remaining model is an LTI system. Then you form the LFT representation of this, by simply obtaining two systems interconnected to each other.

    Can you tell me the name of the book or the URL? maybe I have it...
     
  10. Jul 16, 2009 #9
    The problem of the reader is that it is in Dutch. It is called "Robuust Regelen". But I doubt if this helps any.

    Anyway, you are right about the problem. The first step is to factorize the equation with the f(i,z) term. Then f(i,z) includes the non-linearities and the remaining is just an LTI system. Then f(i,z) is assumed to be a uncertainty, call if [tex] \delta(t) [/tex] if you like. The nonimal value of the uncertainty is [tex] \delta0(t) = f(i0,z0) [/tex]. The range of nonlinearities is defined as [tex] \delta(t) \in [-1.5 \delta0 , 1.5 \delta0 ] [/tex].

    Thats why infinity cannot be the nominal value of the uncertainty.

    And i dont think they make a lft interconnection of the system. The full system is described by three equations, or in state space form:

    [tex] \frac{d}{dt} \begin{pmatrix} z-z0 \\ \dot{z} \\ i-i0 \end{pmatrix} = \begin{pmatrix} 0 & 1& 0 \\ \sqrt{g} f(i,z) & 0 & -yf(i,z) \\ 0&0& - \frac{r}{l} \end{pmatrix} \begin{pmatrix} z-z0 \\ \dot{z} \\ i-i0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ \frac{1}{l} \end{pmatrix} (v-v0) [/tex]

    which is equal to the equations i gave a few posts ago.

    Hopefully this is more clear?
     
  11. Jul 16, 2009 #10
    Ah, I have not done Robust control yet. So, you are outside my realm :)
     
  12. Jul 17, 2009 #11
    azizz, maybe quadratic stability with a common Lyapunov function is asked if they don't use LFT representation. Can you give me some hint about the subject so that I don't drive to irrelevant directions.
     
  13. Jul 17, 2009 #12
    The idea is first to make the system robust against the nonlinearities, as we have discussed so far. Then in the next step a controller u=K(z-r) is found which stabilizes the system (the state space equation is now denoted as z=Gu). The controller is found by Hinfinity synthesis.

    If i have know how to describe the system with the [tex] \delta(t) [/tex] factor, then i think i can solve the rest.
     
    Last edited: Jul 17, 2009
  14. Jul 17, 2009 #13
    Oh but what you describe is a implicit LFT representation. ;)

    Here is the idea. It is really simple. As you see only the second state is multiplied by f(i,z) right?

    Imagine a virtual signal is connected to [itex]\Delta[/itex] block, getting multiplied and coming back to compute [tex]\frac{d}{dt}\dot z[/tex]. You get what I am saying?

    Let an artificial signal pair p,q. Then

    [itex]
    \begin{align*}
    \frac{d}{dt} \begin{pmatrix} \Delta z \\ \dot{z} \\ \Delta i \end{pmatrix} &= \begin{pmatrix} 0 & 1& 0 \\ 0 &0 &0 \\ 0&0& - \frac{r}{l} \end{pmatrix} \begin{pmatrix} \Delta z \\ \dot{z} \\ \Delta i \end{pmatrix} &+\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}q &+ \begin{pmatrix} 0 \\ 0 \\ \frac{1}{l} \end{pmatrix} \Delta v\\
    p &= \begin{pmatrix}\sqrt{g} & 0 &-y\end{pmatrix}\begin{pmatrix} \Delta z \\ \dot{z} \\ \Delta i \end{pmatrix} &+0q &+ 0 \Delta v\\
    q &= f(i,z)p
    \end{align*}
    [/itex]

    Now I have an LTI system with an additional output p and an additional input q which they are related via the uncertainty/nonlinearity that you want to isolate then the rest is small gain condition together with the size of your uncertainty/nonlinearity. Hope you get the idea
     
  15. Jul 17, 2009 #14
    what you say is true. and i certainly can use that representation to continue, but it is still unknown what the factor f(i,z) exactly is representing. in the example they clearly write [tex] f(i,z) = \delta(t) [/tex] and use [tex] f(i0,z0) = \delta0(t) [/tex] as nominal value to denote the system z=Gu. they first search a Hinfinity controller for the nominal value and then check stability for the range in which [tex] \delta(t) [/tex] is defined (50% deviation from its nominal value).

    if i want to do the Hinfinity synthesis i need to describe the system with numerical values, but i dont know what value to use for [tex] f(i0,z0) = \delta0(t) [/tex], as my calculation says it is infinity.

    i've read something about this problem, and it seems that there exists a certain trajectory in the z/i plane for which [tex] f(i0,z0) = \delta0(t) [/tex] does exist. so ive tried to rewrite the system with the given equations such that the [tex] \Delta z [/tex] term in the denumerator drops out, without success so far.
     
  16. Jul 17, 2009 #15
    But it should not be like this in the first place anyhow. From your first equation in your first post, maybe you should find the limit at the singularity to see if it is removable or not.
     
  17. Jul 17, 2009 #16
    well thanks a lot for your help. i will try to solve it with the information youve posted :)
     
  18. Jul 21, 2009 #17
    I might have found something useful: a book in which they use the same problem as an exercise.

    http://w3.ele.tue.nl/fileadmin/ele/MBS/CS/Files/Courses/DISClmi/lmichap5.pdf [Broken] (see last page)

    Perhaps this helps a bit, since I still have not been able to solve the problem, or even get any closer to the solutions.
     
    Last edited by a moderator: May 4, 2017
  19. Jul 22, 2009 #18
    Let's go piece by piece.

    First of all, did you plug in the value v_0 = 2 and obtained a positive z_0? It is like a spring mass system nominally. You plug in some voltage and it stays at some position close enough...

    Then we get a nominal operating point which in case of zero uncertainty + zero disturbance it will float in air...
     
  20. Jul 22, 2009 #19
    ok so first part a: we search for the nominal value (or equilibrium) for v0=2. in equilibrium we know that all derivates are equal to zero (no change in position, current or voltage), so what we get is

    [tex] v_0 = 2 [/tex]

    such that

    [tex] L \frac{d}{dt} i_0 + R i_0 = v_0 \Longrightarrow R i_0 = v_0 \Longrightarrow i_0 = \frac{v_0}{R} = 1 [/tex]

    such that

    [tex] M \frac{d^2}{dt^2} z_0 = Mg - k\frac{i_0^2}{z_0^2} \Longrightarrow 0=Mg - k \frac{1}{z_0^2} \Longrightarrow z_0 = \pm \sqrt{\frac{k}{Mg}} i_0 [/tex]

    (but ofcourse we only consider the positive solution for z0)

    Then we can proceed with part b. This is where my problem occurs. I can give the solution i first found, but perhaps this only works misleading?
     
  21. Aug 6, 2009 #20
    what you say is true indeed. sorry for the inconvience
     
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