Discretization - DigitalServo

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Discretization

\begin{align} \dot{x}(t)=Ax(t)+Bu(t) \end{align}

\begin{align} x(t)={\rm exp}\left(\int^{t}_{0} Adt\right)C\ \ (C={\rm const}.) \end{align}

\begin{align} x(t)={\rm exp}\left(\int^{t}_{0} Adt\right)C(t) \end{align}

\begin{align} & \dot{x}(t)=Ax(t)+Bu(t) \\ \Leftrightarrow & A{\rm exp}\left(\int^{t}_{0} Adt\right)C(t) + {\rm exp}\left(\int^{t}_{0} Adt\right)\frac{d}{dt}C(t)=A{\rm exp}\left(\int^{t}_{0} Adt\right)C(t)+Bu(t) \\ \Leftrightarrow & {\rm exp}\left(\int^{t}_{0} Adt\right)\frac{d}{dt}C(t)=Bu(t) \\ \Leftrightarrow & \frac{d}{dt}C(t)={\rm exp}\left(-\int^{t}_{0} Adt\right)Bu(t) \\ \Leftrightarrow & C(t)=\int^{t}_{0} {\rm exp}\left(-\int^{t}_{0} Adt\right)Bu(t) dt +C_{0}\ \ (C_{0}={\rm const}.) \\ \therefore & x(t)={\rm exp}\left(\int^{t}_{0} Adt\right)\int^{t}_{0} {\rm exp}\left(-\int^{t}_{0} Adt\right)Bu(t) dt +{\rm exp}\left(\int^{t}_{0} Adt\right)C_{0} \end{align}

\begin{align} x(t)&={\rm exp}\left(\int^{t}_{0} Adt\right)\int^{t}_{0} {\rm exp}\left(-\int^{\tau}_{0} Ad\tau\right)Bu(\tau) d\tau +C_{0}{\rm exp}\left(\int^{t}_{0} Adt\right) \\ &=\int^{t}_{0} \left(\int^{t}_{\tau} Ad\tau\right)Bu(\tau){\rm exp} d\tau +{\rm exp}\left(\int^{t}_{0} Adt\right)C_{0} \\ &= {\rm exp}\left(\int^{t}_{0} Adt\right)x(0)+ \int^{t}_{0} {\rm exp}\left(\int^{t}_{\tau} Ad\tau\right)Bu(\tau) d\tau \end{align}

\begin{align} x(t)&= e^{At}x(0) + \int^{t}_{0} e^{A(t-\tau)}Bu(\tau) d\tau \end{align}

\begin{align} x(kT_{\rm s})&= e^{A(kT_{\rm s})}x(0) + \int^{kT_{\rm s}}_{0} e^{A(kT_{\rm s}-\tau)}Bu(\tau) d\tau \\ x((k+1)T_{\rm s})&= e^{A((k+1)T_{\rm s})}x(0) + \int^{(k+1)T_{\rm s}}_{0} e^{A((k+1)T_{\rm s}-\tau)}Bu(\tau) d\tau \\ &= e^{A((k+1)T_{\rm s})}x(0) + \int^{(k+1)T_{\rm s}}_{kT_{\rm s}} e^{A((k+1)T_{\rm s}-\tau)}Bu(\tau) d\tau + \int^{kT_{\rm s}}_{0} e^{A((k+1)T_{\rm s}-\tau)}Bu(\tau) d\tau \\ &= e^{AT_{\rm s}}\left( e^{A(kT_{\rm s})}x(0) + \int^{kT_{\rm s}}_{0} e^{A(kT_{\rm s}-\tau)}Bu(\tau) d\tau \right)+ \int^{(k+1)T_{\rm s}}_{kT_{\rm s}} e^{A((k+1)T_{\rm s}-\tau)}Bu(\tau) d\tau \\ &= e^{AT_{\rm s}}x(kT_{\rm s})+ \int^{(k+1)T_{\rm s}}_{kT_{\rm s}} e^{A((k+1)T_{\rm s}-\tau)} Bu(\tau) d\tau \end{align}

\begin{align} x((k+1)T_{\rm s})&= e^{AT_{\rm s}}x(kT_{\rm s})+ \left(\int^{(k+1)T_{\rm s}}_{kT_{\rm s}} e^{A((k+1)T_{\rm s}-\tau)} d\tau\right) Bu(kT_{\rm s}) \end{align}

\begin{align} \gamma&\triangleq&(k+1)T_{\rm s}-\tau\ \ (\gamma:\ T_{\rm s}\rightarrow 0,\ \ d\tau = -d\gamma) \end{align}

\begin{align} x((k+1)T_{\rm s})&= e^{AT_{\rm s}}x(kT_{\rm s})+ \left(\int^{0}_{T_{\rm s}} e^{A\gamma} (-d\gamma)\right) Bu(kT_{\rm s}) \\&= e^{AT_{\rm s}}x(kT_{\rm s})+ \left(\int^{T_{\rm s}}_{0} e^{A\gamma} d\gamma\right) Bu(kT_{\rm s}) \\ \therefore x((k+1)T_{\rm s})&=A_{\rm d}x(kT_{\rm s})+B_{\rm d}u(kT_{\rm s}) \ \ \left( A_{\rm d} \triangleq e^{AT_{\rm s}},\ \ B_{\rm d}\triangleq \int^{T_{\rm s}}_{0} e^{A\gamma} d\gamma B \right)\end{align}

CONTENTS

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