Zero-order hold discretization of continuous time state space system
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Zero-order hold discretization of continuous time state space system
(Detailed proof is linked here.)
Assume that the state-space model of the form
\[\dot x = Ax+Bu, \hspace 0.5cm y=Cx+Du\hspace 1cm(1)\]Our goal is find the discretization of the above equation like
\[x_{k+1} = A_dx_{k}+B_du_{k} \\ y=C_dx_{k}+Du_k\]First of all,
\[C_d = C \\ D_d = D\]Let’s find $ A_d , B_d$ Given the model in (1) the solution is
\[x(t) = e^{A(t-t_0)}x(t_0)+\int_{t_0}^{t} e^{A(t-\tau)} B u(\tau) \,d\tau, \\\]Let us $t_0=kT,\hspace 0.5cm t = kT+T$ Then,
\[x(kT+T) = e^{AT}x(kT)+\int_{kT}^{kT+T} e^{A(kT+T-\tau)} B u(\tau) \,d\tau, \\\]Let us also consider $u(t)=u(kT) \hspace 0.5cm \forall t\in[kT,kT+T] $
Using change of variable $\eta =kT+T-\tau$, we obtain
\[x(kT+T) = e^{AT}x(kT)+\int_{0}^{T} e^{A\eta} d\eta B u(kT), \\\]Since $x(kT+T) = x_{k+1}$
\[A_d =e^{AT} \\ B_d = \int_{0}^{T} e^{A\eta} d\eta B \\ C_d = C\\ D_d = D \\\]Let’s apply taylor series (ref)
Finally, we can get
\[A_d = \sum_{n=0}^{\infty}\frac{T^nA^n}{n!}=I+AT+\frac{A^2T^2}{2!} +... \\ B_d = \sum_{n=0}^{\infty}\frac{A^nT^{n+1}}{(n+1)!}B=(T+\frac{AT^2}{2!}+...)B\]Example
Let’s convert below continuous state space system to discrete state system
\[A=\left\lceil \begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{matrix} \right\rceil,\hspace{2cm} B = \left[\begin{array}{ll1} {0} \\ {0} \\ {1} \\ \end{array}\right]\]Then, the answer is
\[A_d=\left\lceil \begin{matrix} 1 & T & \frac{1}{2}T^2\\ 0 & 1 & T\\ 0 & 0 & 1 \end{matrix} \right\rceil , \hspace 2cm B_d= \left[\begin{array}{ll1} {\frac{1}{3!}T^3} \\ {\frac{1}{2}T^2} \\ {T} \\ \end{array}\right]\]