Obtaining the equivalent discrete time state space model (zero order hold is used)

Discussion in 'Homework Help' started by u-will-neva-no, Dec 10, 2011.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Hello all,

    My task here is to take the state space model equations (from this thread http://forum.allaboutcircuits.com/showthread.php?t=62884) into an equivalent discrete time state space model, assuming a zero order hold is used.

    the formulas that will be used are:
    y[k] = Cx[k] + Du[k].

    From the answer from the previous forum, this give: (where t = k)

    C is:
    \left{[ \begin{array}{lml}<br />
-1 & \, & 0\\<br />
\end{array} \right\] <br />

    and D is:
    \left[ \begin{array}{lml}<br />
0<br />
\end{array} \right\] <br />

    My problem lies on the next part:
    The general formula is x[k+1] =\Phix[k]+\Gamma u[k]
    \Phi = exponential(AT)
    so: \Phi = exponential{ \left[ \begin{array}{lml}<br />
\frac{-1}{RC} & \, &\frac{-1}{2RC} \\<br />
\frac{1}{2RC} & \, & 0 \\<br />
<br />
\end{array} \right\]}T<br />

    i dont think I can simplify this anymore so this will be my value on  \Phi

    the formula for \Gamma=\int ^T_0 exponential((AT)dt)B

    This gives the result:

    \Gamma = \frac{1}{RC}\int ^T_0 {exponential{ \left[ \begin{array}{lml}<br />
\frac{-1}{RC} & \, &\frac{-1}{2RC} \\<br />
\frac{1}{2RC} & \, & 0 \\<br />
<br />
\end{array}\right\] <br />
B

    where B is \left[ \begin{array}{lml}<br />
-1<br />
0<br />
\end{array} \right\] <br />

    Im not sure how to deal with the exponential part to deal with my anlysis. Im thinking laplace but just cant spot how to simplify it. Any hints will be fab!
     
    Last edited: Dec 10, 2011
  2. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    I wanted to revive this thread. Anyone have some suggestions?
     
Loading...