State Space Question

Discussion in 'Homework Help' started by Kayne, May 15, 2010.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Hi All,

    Just wondering if I have done this correctly. To write the system below into a state space form with a unit step.
    100(s+5)/ (s+2)(s+3)


    100s+500/s^2+5s+6

    d^2y/dt^2 + 5dy/dt + 6y = 100du/dt+500u

    Y(s) = 100s^-1+500s^-2/ 1+5s^-1+6s^2
    Y(s) = (100s^-1+500s^-2)E(s)
    E(s)= U(s)/1+5s^-1+6s^-2 therefore
    E(s) = U(s) - 5s^-1E(s)-6s^-2E(s)

    dx1/dt=x2
    dx2/dt2=x3
    dx3/dt3= -6x3-5x2+u

    y= 500x1 +100x2

    d/dt
    :x1: :0, 1, 0: :x1: :0:
    :x2:=:0, 0, 1: :x2:+:0: u(t)
    :x3: :0,-6,-5: :x3: :1:

    y=:500, 100, 0:


    I would also like to know where I can find the information to write the above in correct equations so it make more sence to the reader.


    Thanks
     
    Last edited: May 15, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I think it should be ....

    \frac{Y(s)}{U(s)}=\frac{100(s+5)}{s^2+5s+6}

    \frac{Y(s)}{U(s)}=\frac{100(s^{-1}+5s^{-2})}{1+5s^{-1}+6s^{-2}}

    E(s)=\frac{U(s)}{1+5s^{-1}+6s^{-2}

    E(s)=U(s)-E(s)(5s^{-1}+6s^{-2})

    and so on ....
     
  3. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
  4. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    I will re do the question using the link for correct formatting and re submit.
     
  5. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Thanks for the heads up on the code I have managed to write the answer so it is more understandable

     \frac {Y(s)}{U(s)}=\frac {100s+500}{s^{2}+5s+6}

    Y(s)=\frac {100s^{-1}+500s^{-2}}{1+5s^{-1}+6s^{-2}}

    Y(s)=(100s^{-1}+500s^{-2})E(s)

    E(s)=\frac {U(s)}{1+5s^{-1}+6s^{-2}}

    E(s)=U(s)-5s^{-1}E(s)-6s^{-2}E(s)

     \frac {dx_{1}}{dt}= x_{2}

     \frac {dx_{2}}{dt}= x_{3}

     \frac {dx_{3}}{dt}= -6x_{2}-5x_{3}

     y = 500x_{1}+100x_{2}



    \frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\\x_{3}\end{array})=(\begin{array}{cc}0&1&0\\0&0&1\\-6&-5&0\end{array})(\begin{array}{cc}x_{1}\\x_{2}\\x_{3}\end{array})+(\begin{array}{cc}0\\0\\1\end{array})u(t)

     y=(\begin{array}{cc}500&100&0\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\\x_{3}\end{array})

    Hopefully what I have done is correct, thanks for your help
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You've obtained what appears to be a 3rd order representation of a 2nd order system. Is this how you were taught to do it?

    The normal representation would be

    d(X(t))/dt=A.X(t)+B.u(t)
    y(t)= C.X(t)+D.u(t)

    In a 2nd order system there would be two state variables, so A would be a 2x2 matrix .....
     
  7. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    This is why i may be having trouble with the the rest of the question. I found a question that is similar and followed the example. Will go back and have another look

    Thanks for the info
     
  8. Ethan

    New Member

    Feb 7, 2009
    6
    1
    t_n_k is correct. You need to select a minimal set of variables that completely describe the system. In your case you only need 2 not 3 because it is a second order system.
     
  9. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Have had another look at the question and found another example that is similar so hopefully this may be correct.


     \frac {Y(s)}{U(s)}=\frac {100s+500}{s^{2}+5s+6}

    Y(s)=\frac {100s^{-1}+500s^{-2}}{1+5s^{-1}+6s^{-2}}

    Y(s)=(100s^{-1}+500s^{-2})E(s)

    E(s)=\frac {U(s)}{1+5s^{-1}+6s^{-2}}

    E(s)=U(s)-5s^{-1}E(s)-6s^{-2}E(s)

     \frac {dx_{1}}{dt}= x_{2}

     \frac {dx_{2}}{dt}= -6x_{1}-5x_{2}

     y = 500x_{1}+100x_{2}



    \frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\0\end{array})u(t)

     y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})

    thanks for the help
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Probably an oversight - You need a '1' in the first state space equation B matrix (the equation second from the bottom) to include u(t) in the function.
     
  11. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    your correct, Looking back in my notes i did miss that when typing that in thanks TNK
     
  12. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Thanks for pointing that out TNK. You are correct it was an oversight.


    \frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\1\end{array})u(t)

     y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})

    To continue on from this to a discrete time system in the form

     X(k+1)=\Phi X(k)=\Gamma u(k) ,
     y=CX(k)

    I have done the following

    A= (\begin{array}{cc}0&1\\-6&-5\end{array}) B=(\begin{array}{cr}0\\1\end{array})


    \Phi(t)= L^-1[(\begin{array}{cc}s&0\\0&s\end{array})- (\begin{array}{cr}0&1\\-6&-5\end{array})]^-1


    = L^-1[(\begin{array}{cc}s&-1\\6&s+5\end{array})]^-1

    = L^-1[(\begin{array}{cc}\frac{1}{s}&\frac{-1}{s^2}\\6&\frac{1}{s+5}\end{array})]

    \Phi(t)=[(\begin{array}{cc}1&-t\\6&e^-5t)\end{array})]

    I would like to know if what I have worked out for \Phi(t) is correct I am not sure about changing  \frac {1}{s+5} that is all and this is required to work out \Gamma


    Thanks for any help
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Keeping in mind that ....

    A_{d}=e^{AT}

    Where Ad is the discretized form of continuous matrix A. You called this \Phi(t) rather than A_{d} which is just a matter of convention.

    Are you confident that ...

    A_{d}=e^{AT}=L^{-1}\left\{(sI-A)^{-1}\right\}_{t=T}

    is satisfied in your derivation?

    You might show how you derived

    (sI-A)^{-1}

    which leads to Ad or \Phi(t) as you have it in the post.
     
  14. Ethan

    New Member

    Feb 7, 2009
    6
    1
    [​IMG] I believe this is not accurate. You need to do partial fraction expansion.
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I agree.

    You can do the expansion of

    e^{AT}

    using Taylor's expansion or do the inversion of the matrix [sI-A] and apply the inverse Laplace transform at t=T.
     
  16. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Have taken on board what has been said and have work though the question again.

    \frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\1\end{array})u(t)

     y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})

    To continue on from this to a discrete time system in the form

     X(k+1)=\Phi X(k)=\Gamma u(k) ,
     y=CX(k)

    I have done the following

    A= (\begin{array}{cc}0&1\\-6&-5\end{array}) B=(\begin{array}{cr}0\\1\end{array})


    \Phi(t)= L^-1[(\begin{array}{cc}s&0\\0&s\end{array})- (\begin{array}{cr}0&1\\-6&-5\end{array})]^-1


    = L^-1[(\begin{array}{cc}s&-1\\6&s+5\end{array})]^-1

    =L^-1\frac{1}{s^2+5s+6} (\begin{array}{cc}\ s+5&1\\-6&s\end{array})

    =L^-1[(\begin{array}{cc}\frac{s+5}{s^2+5s+6}&\frac{1}{s^2+5s+6}\\\frac{-6}{s^2+5s+6}&\frac{s}{s^2+5s+6}\end{array})]

    =L^-1[(\begin{array}{cc}\frac{3}{(s+2)}-\frac{2}{(s+3)}&\frac{1}{(s+2)}-\frac{1}{(s+3)}\\\frac{-6}{(s+2)}+\frac{6}{(s+3)}&\frac{-2}{(s+2)}+\frac{3}{(s+3)}\end{array})]


    \Phi=[(\begin{array}{cc}\3e^-2T-2e^-3T&e^-2T-e^-3T\\-6e^-2T+6e^-3T&-2e^-2T+3e^-3T\end{array})]

    \Gamma=\int^{T}_{0}e^Aq Bdq=\int^{T}_{0}[(\begin{array}{cc}\3e^-2q-2e^-3q&e^-2q-e^-3q\\-6e^-2q+6e^-3q&-2e^-2q+3e^-3q\end{array})][(\begin{array}{cc}\0\\1\end{array})]dq

    =[(\begin{array}{cc}\\\frac{1}{2}-e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]


    So the original equation
     X(k+1)=\Phi X(k)=\Gamma u(k) ,
     y=CX(k)

    is then worked out to be

     X(k+1)=[(\begin{array}{cc}\3e^-2T-2e^-3T&e^-2T-e^-3T\\-6e^-2T+6e^-3T&-2e^-2T+3e^-3T\end{array})][(\begin{array}{cc}\\x_{1}(k)\\x_{2}(k)\end{array})] +[(\begin{array}{cc}\\\frac{1}{2}-e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]u(k)

    y=(\begin{array}{cc}\500&100\end{array})[(\begin{array}{cc}\\x_{1}(k)\\x_{2}(k)\end{array})]


    This looks better than the first attempt so hopefully this method is correct.Thanks for the help given so far
     
  17. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Re-check this.
     
  18. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    \Gamma=\int^{T}_{0}e^Aq Bdq=\int^{T}_{0}[(\begin{array}{cc}\3e^-2q-2e^-3q&e^-2q-e^-3q\\-6e^-2q+6e^-3q&-2e^-2q+3e^-3q\end{array})][(\begin{array}{cc}\0\\1\end{array})]dq

    =[(\begin{array}{cc}\\\frac{-1}{3}e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]

    I think that I may have got the derivatives incorrect last post, hopefully this is the correct answer for \Gamma..
     
  19. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This is my solution ....
     
  20. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Thanks TNK, I thought that the 1/6 had to be subtracted from the equation not just added into the equation.
     
Loading...