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

#### u-will-neva-no

Joined Mar 22, 2011
230
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} -1 & \, & 0\\ \end{array} \right\]$$

and D is:
$$\left[ \begin{array}{lml} 0 \end{array} \right\]$$

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

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} \frac{-1}{RC} & \, &\frac{-1}{2RC} \\ \frac{1}{2RC} & \, & 0 \\ \end{array}\right\]$$B

where B is $$\left[ \begin{array}{lml} -1 0 \end{array} \right\]$$

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!

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