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Control Theory State-Space method with derivative input

  1. Mar 29, 2015 #1
    Hi,

    I'm reading Ogata's Modern Control Engineering, and when he talks about the representation of a differential equation in state space he divides the method in two. The first one is when the input of the differential equation involves no derivative term, for example:

    x'(t)+x(t)=u(t)

    The next step is doing it with a differential equation that has inputs that have derivatives. For example:

    x'(t)+x(t)=u(t)+u'(t)

    He then mention that the state varibles will be

    x1=y-β0u
    x2=y'-β1u-β0u' and so on...

    I've tried finding a reason for this and the nearest I've come is the following PDF, which has errors:

    http://www.ece.rutgers.edu/~gajic/psfiles/canonicalforms.pdf

    Anybody got an idea how that can be derived?
     
  2. jcsd
  3. Mar 30, 2015 #2

    donpacino

    User Avatar
    Gold Member

    note: this is not mine...
    http://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SS.html

    Consider the third order differential transfer function:

    img7D.gif

    We start by multiplying by Z(s)/Z(s) and then solving for Y(s) and U(s) in terms of Z(s). We also convert back to a differential equation.

    img58.gif

    We can now choose z and its first two derivatives as our state variables

    img5C.gif

    Now we just need to form the output

    img89.gif

    Unfortunately, the third derivative of z is not a state variable or an input, so this is not a valid output equation. However, we can represent the term as a sum of state variables and outputs:

    img8A.gif

    and

    img8C1.gif

    From these results we can easily form the state space model:

    img8E.gif

    In this case, the order of the numerator of the transfer function was less than that of the denominator. If they are equal, the process is somewhat more complex.
     
    Last edited by a moderator: Apr 19, 2017
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