1. Limited time only! Sign up for a free 30min personal 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!

State Transition Matrix, Determining States

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data

    An LTI system is given in state-space form,

    [itex]\left( \begin{array}{cc}
    \dot{x_{1}} \\
    \dot{x_{2}}
    \end{array} \right)
    =
    \left( \begin{array}{cc}
    -1 & 0.5 \\
    1 & 0
    \end{array} \right)
    \left( \begin{array}{cc}
    x_{1} \\
    x_{2}
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    0.5 \\
    0
    \end{array} \right)
    u
    [/itex]

    A unit-step signal is applied to the input of the system. If,

    [itex]x_{1}(0) = 1, \quad x_{2}(0) = 0[/itex]

    determine the state of the system after t = 0.1 sec.

    2. Relevant equations



    3. The attempt at a solution

    [itex]\underline{\dot{x}} = \underline{A} \underline{x} + \underline{B}u[/itex]

    [itex]\mathcal{L} \Rightarrow s \underline{X(s)} - \underline{x(0)} = \underline{A}\underline{X(s)} + \underline{B} U(s)[/itex]

    [itex]\Rightarrow \underline{X(s)} = (s\underline{I} - \underline{A})^{-1} \underline{x(0)} + (s\underline{I} - \underline{A})^{-1}\underline{B}U(s)[/itex]

    Working through the simplification I obtain,

    [itex]\left( \begin{array}{cc}
    X_{1}(s) \\
    X_{2}(s)
    \end{array} \right)
    =
    \left( \begin{array}{cc}
    \frac{s}{s^{2}+s-0.5}\\
    \frac{1}{s^{2}+s-0.5}
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    \frac{0.5}{s^{2}+s-0.5}\\
    \frac{0.5}{s(s^{2}+s-0.5)}
    \end{array} \right)
    [/itex]
    Thus,
    [itex]
    X_{1}(s) = \frac{s}{s^{2}+s-0.5} + \frac{0.5}{s^{2}+s-0.5}
    [/itex]
    [itex]
    X_{2}(s) = \frac{1}{s^{2}+s-0.5} + \frac{0.5}{s(s^{2}+s-0.5)}
    [/itex]

    How can I put this in a form that I can easily pull out of the s-domain back into time domain using inverse Laplace transform? If weren't for the -0.5 in the denominator I think I could work something out by reworking it into one of the following two forms,
    [itex]
    \frac{\omega}{(s+a)^{2} + \omega^{2}} \quad \text{ or } \quad \frac{s+a}{(s+a)^{2} + \omega^{2}}
    [/itex]

    Any ideas? Did I make a mistake in my simplification perhaps?
     
  2. jcsd
  3. Mar 9, 2014 #2

    donpacino

    User Avatar
    Gold Member

    1.) what are the initial conditions for the system??? i am assuming x(0) equals zero. If x(0) equals zero you need to remove those second sections from your equations for X1 and X2.

    2.)
    below is one of the forms you were talking about
    e^(-at)*sin(wt) = invlap{ω /( (s+a)^2+ω^2)}

    why don't you try this one?
    e^(-at)*sinh(wt) = invlap{ω /( (s+a)^2-ω^2)}
     
  4. Mar 9, 2014 #3

    donpacino

    User Avatar
    Gold Member

    or this


    e^(at)*sin(wt) = invlap{ω /( (s-a)^2+ω^2)}
     
  5. Mar 9, 2014 #4

    donpacino

    User Avatar
    Gold Member

    or this


    e^(at)*sinh(wt) = invlap{ω /( (s-a)^2-ω^2)}
     
  6. Mar 9, 2014 #5
    They are provided in the original post.
     
  7. Mar 10, 2014 #6

    donpacino

    User Avatar
    Gold Member

    My apologies. For some reason I though that was the input vector.

    due to the fact that X2(0)=0, your expression for X2(s) is not correct. Eliminate the second part of the statement
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted