ملاحظ لوننبرگر هوتعبير عن نظام يستعمل لحساب الحالة في نظام آخر وتكون مداخل الملاحظ هي مداخل ومخارج النظام وتكون مخارج الملاحظ هي حالات النظام المراد فهم حالته. يستعمل ملاحظ لوننبرگر في التحكم عن طريق إرجاع الحالة.

ملاحظ الحالة النمطي

The state of a physical discrete-time system is assumed to satisfy

${\displaystyle \mathbf {x (k+1)=A\mathbf {x (k)+B\mathbf {u (k)$
${\displaystyle \mathbf {y (k)=C\mathbf {x (k)+D\mathbf {u (k)$

ملاحظ النمط المنزلق

${\displaystyle \mathbf {\dot {\hat {x =\left[{\frac {\partial H(\mathbf {\hat {x ) {\partial \mathbf {x \right]^{-1 M(\mathbf {\hat {x )\,\operatorname {sgn (V(t)-H(\mathbf {\hat {x ))$

where:

• The ${\displaystyle \mathrm{sgn}<!--\; \; -->(\cdot <!--\; \cdot \; -->)}$${\displaystyle n$ dimensions. That is,

${\displaystyle \operatorname {sgn (\mathbf {z )={\begin{bmatrix \operatorname {sgn (z_{1 )\\\operatorname {sgn (z_{2 )\\\vdots \\\operatorname {sgn (z_{i )\\\vdots \\\operatorname {sgn (z_{n )\end{bmatrix$

for the vector ${\displaystyle \mathbf {z \in \mathbb {R ^{n$.

• The vector ${\displaystyle H(\mathbf {x )$ has components that are the output function ${\displaystyle h(\mathbf {x )$ and its repeated Lie derivatives. In particular,

${\displaystyle H(\mathbf {x )\triangleq {\begin{bmatrix h_{1 (\mathbf {x )\\h_{2 (\mathbf {x )\\h_{3 (\mathbf {x )\\\vdots \\h_{n (\mathbf {x )\end{bmatrix \triangleq {\begin{bmatrix h(\mathbf {x )\\L_{f h(\mathbf {x )\\L_{f ^{2 h(\mathbf {x )\\\vdots \\L_{f ^{n-1 h(\mathbf {x )\end{bmatrix$

where ${\displaystyle L_f^{i}h}$${\displaystyle h}$n, ${\displaystyle H(\mathbf{x}(t))}$${\displaystyle H(\mathbf{x})}$

• The diagonal matrix ${\displaystyle M({\hat {\mathbf {x )$ of gains is such that

${\displaystyle M({\hat {\mathbf {x )\triangleq \operatorname {diag (m_{1 ({\hat {\mathbf {x ),m_{2 ({\hat {\mathbf {x ),\ldots ,m_{n ({\hat {\mathbf {x ))={\begin{bmatrix m_{1 ({\hat {\mathbf {x )&&&&&\\&m_{2 ({\hat {\mathbf {x )&&&&\\&&\ddots &&&\\&&&m_{i ({\hat {\mathbf {x )&&\\&&&&\ddots &\\&&&&&m_{n ({\hat {\mathbf {x )\end{bmatrix$

where, for each ${\displaystyle i\in \{1,2,\dots ,n\$, element ${\displaystyle m_{i ({\hat {\mathbf {x )>0$ and suitably large to ensure reachability of the sliding mode.

• The observer vector ${\displaystyle V(t)$ is such that

${\displaystyle V(t)\triangleq {\begin{bmatrix v_{1 (t)\\v_{2 (t)\\v_{3 (t)\\\vdots \\v_{i (t)\\\vdots \\v_{n (t)\end{bmatrix \triangleq {\begin{bmatrix \mathbf {y (t)\\\{m_{1 ({\hat {\mathbf {x )\operatorname {sgn (v_{1 (t)-h_{1 ({\hat {\mathbf {x (t)))\ _{\text{eq \\\{m_{2 ({\hat {\mathbf {x )\operatorname {sgn (v_{2 (t)-h_{2 ({\hat {\mathbf {x (t)))\ _{\text{eq \\\vdots \\\{m_{i-1 ({\hat {\mathbf {x )\operatorname {sgn (v_{i-1 (t)-h_{i-1 ({\hat {\mathbf {x (t)))\ _{\text{eq \\\vdots \\\{m_{n-1 ({\hat {\mathbf {x )\operatorname {sgn (v_{n-1 (t)-h_{n-1 ({\hat {\mathbf {x (t)))\ _{\text{eq \end{bmatrix$

where ${\displaystyle \mathrm{sgn}<!--\; \; -->(\cdot <!--\; \cdot \; -->)}$${\displaystyle \{\ldots \ _{\text{eq$ denotes an "equivalent value operator" of a discontinuous function in sliding mode.

The modified observation error can be written in the transformed states ${\displaystyle \mathbf {e =H(\mathbf {x )-H(\mathbf {\hat {x )$. In particular,

${\displaystyle {\begin{aligned {\dot {\mathbf {e &={\frac {\operatorname {d {\operatorname {d t H(\mathbf {x )-{\frac {\operatorname {d {\operatorname {d t H({\hat {\mathbf {x )\\&={\frac {\operatorname {d {\operatorname {d t H(\mathbf {x )-M({\hat {\mathbf {x )\,\operatorname {sgn (V(t)-H({\hat {\mathbf {x (t))),\end{aligned$

and so

${\displaystyle {\begin{aligned {\begin{bmatrix {\dot {\mathbf {e _{1 \\{\dot {\mathbf {e _{2 \\\vdots \\{\dot {\mathbf {e _{i \\\vdots \\{\dot {\mathbf {e _{n-1 \\{\dot {\mathbf {e _{n \end{bmatrix &={\mathord {\overbrace {\begin{bmatrix {\dot {h _{1 (\mathbf {x )\\{\dot {h _{2 (\mathbf {x )\\\vdots \\{\dot {h _{i (\mathbf {x )\\\vdots \\{\dot {h _{n-1 (\mathbf {x )\\{\dot {h _{n (\mathbf {x )\end{bmatrix ^{{\tfrac {\operatorname {d {\operatorname {d t H(\mathbf {x ) -{\mathord {\overbrace {M({\hat {\mathbf {x )\,\operatorname {sgn (V(t)-H({\hat {\mathbf {x (t))) ^{{\tfrac {\operatorname {d {\operatorname {d t H(\mathbf {\hat {x ) ={\begin{bmatrix h_{2 (\mathbf {x )\\h_{3 (\mathbf {x )\\\vdots \\h_{i+1 (\mathbf {x )\\\vdots \\h_{n (\mathbf {x )\\L_{f ^{n h(\mathbf {x )\end{bmatrix -{\begin{bmatrix m_{1 \operatorname {sgn (v_{1 (t)-h_{1 ({\hat {\mathbf {x (t)))\\m_{2 \operatorname {sgn (v_{2 (t)-h_{2 ({\hat {\mathbf {x (t)))\\\vdots \\m_{i \operatorname {sgn (v_{i (t)-h_{i ({\hat {\mathbf {x (t)))\\\vdots \\m_{n-1 \operatorname {sgn (v_{n-1 (t)-h_{n-1 ({\hat {\mathbf {x (t)))\\m_{n \operatorname {sgn (v_{n (t)-h_{n ({\hat {\mathbf {x (t)))\end{bmatrix \\&={\begin{bmatrix h_{2 (\mathbf {x )-m_{1 ({\hat {\mathbf {x )\operatorname {sgn ({\mathord {\overbrace {{\mathord {\overbrace {v_{1 (t) ^{v_{1 (t)=y(t)=h_{1 (\mathbf {x ) -h_{1 ({\hat {\mathbf {x (t)) ^{\mathbf {e _{1 )\\h_{3 (\mathbf {x )-m_{2 ({\hat {\mathbf {x )\operatorname {sgn (v_{2 (t)-h_{2 ({\hat {\mathbf {x (t)))\\\vdots \\h_{i+1 (\mathbf {x )-m_{i ({\hat {\mathbf {x )\operatorname {sgn (v_{i (t)-h_{i ({\hat {\mathbf {x (t)))\\\vdots \\h_{n (\mathbf {x )-m_{n-1 ({\hat {\mathbf {x )\operatorname {sgn (v_{n-1 (t)-h_{n-1 ({\hat {\mathbf {x (t)))\\L_{f ^{n h(\mathbf {x )-m_{n ({\hat {\mathbf {x )\operatorname {sgn (v_{n (t)-h_{n ({\hat {\mathbf {x (t)))\end{bmatrix .\end{aligned$

وبذلك:

1. ما دام ${\displaystyle m_{1 (\mathbf {\hat {x )\geq |h_{2 (\mathbf {x (t))|$, the first row of the error dynamics, ${\displaystyle {\dot {\mathbf {e _{1 =h_{2 ({\hat {\mathbf {x )-m_{1 ({\hat {\mathbf {x )\operatorname {sgn (\mathbf {e _{1 )$, will meet sufficient conditions to enter the ${\displaystyle e_{1 =0$ sliding mode in finite time.
2. Along the ${\displaystyle e_{1 =0$ surface, the corresponding ${\displaystyle v_{2 (t)=\{m_{1 ({\hat {\mathbf {x )\operatorname {sgn (\mathbf {e _{1 )\ _{\text{eq$ equivalent control will be equal to ${\displaystyle h_{2 (\mathbf {x )$, and so ${\displaystyle v_{2 (t)-h_{2 ({\hat {\mathbf {x )=h_{2 (\mathbf {x )-h_{2 ({\hat {\mathbf {x )=\mathbf {e _{2$. Hence, so long as ${\displaystyle m_{2 (\mathbf {\hat {x )\geq |h_{3 (\mathbf {x (t))|$, the second row of the error dynamics, ${\displaystyle {\dot {\mathbf {e _{2 =h_{3 ({\hat {\mathbf {x )-m_{2 ({\hat {\mathbf {x )\operatorname {sgn (\mathbf {e _{2 )$, will enter the ${\displaystyle e_{2 =0$ sliding mode in finite time.
3. Along the ${\displaystyle e_{i =0$ surface, the corresponding ${\displaystyle v_{i+1 (t)=\{\ldots \ _{\text{eq$ equivalent control will be equal to ${\displaystyle h_{i+1 (\mathbf {x )$. Hence, so long as ${\displaystyle m_{i+1 (\mathbf {\hat {x )\geq |h_{i+2 (\mathbf {x (t))|$, the ${\displaystyle (i+1)$^th row of the error dynamics, ${\displaystyle {\dot {\mathbf {e _{i+1 =h_{i+2 ({\hat {\mathbf {x )-m_{i+1 ({\hat {\mathbf {x )\operatorname {sgn (\mathbf {e _{i+1 )$, will enter the ${\displaystyle e_{i+1 =0$ sliding mode in finite time.

So, for sufficiently large ${\displaystyle m_i}$

In the case of the sliding mode observer for the system with the input, additional conditions are needed for the observation error to be independent of the input. For example, that

${\displaystyle {\frac {\partial H(\mathbf {x ) {\partial \mathbf {x B(\mathbf {x )$

does not depend on time. The observer is then

${\displaystyle {\dot {\mathbf {\hat {x =\left[{\frac {\partial H(\mathbf {\hat {x ) {\partial \mathbf {x \right]^{-1 M(\mathbf {\hat {x )\operatorname {sgn (V(t)-H(\mathbf {\hat {x ))+B(\mathbf {\hat {x )u.$

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