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zamofing_t
2019-01-31 16:36:05 +01:00
parent 0d0eb2fe7a
commit 5e9bc4e4c0
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python/MXTuningDoc/fastStageTune.tex
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@@ -177,21 +177,21 @@ $s \cdot iqMeas =\frac{1}{L}iqVolts - \frac{R}{L}iqMeas$\\
Input: Force $u(t)=F$\\
Output: Position $y(t)=x_1(t)$\\
total mass $m=m_1+m_2+\ldots+m_n$\\
damping: $d_1 \ldots d_n$\\
springs: $k_1 \ldots k_n$\\
mass $m=m_1+m_2+\ldots+m_n$\\
damping: $d=d_1+d_2+\ldots+d_n$\\
springs: $k=k_1+k_2+\ldots+k_n$\\
\eqref{mech1} shows the mechanical differential equations:
\begin{equation}
\begin{aligned}
m_1\ddot{x}_1 = & u(t) -k_1x_1-d_1\dot{x}_1\\
& + k_2x_2+d_2\dot{x}_2
+ k_3x_3+d_3\dot{x}_3
+ k_4x_4+d_4\dot{x}_4\\
m_2\ddot{x}_2= & -k_2x_2-d_2\dot{x}_2\\
m_3\ddot{x}_3= & -k_3x_3-d_3\dot{x}_3\\
m_4\ddot{x}_4= & -k_4x_4-d_4\dot{x}_4\\
& + k_2(x_2-x_1)+d_2(\dot{x}_2-\dot{x}_1)
+ k_3(x_3-x_1)+d_3(\dot{x}_3-\dot{x}_1)
+ k_4(x_4-x_1)+d_4(\dot{x}_4-\dot{x}_1)\\
m_2\ddot{x}_2= & k_2(x_2-x_1)+d_2(\dot{x}_2-\dot{x}_1)\\
m_3\ddot{x}_3= & k_3(x_3-x_1)+d_3(\dot{x}_3-\dot{x}_1)\\
m_4\ddot{x}_4= & k_4(x_4-x_1)+d_4(\dot{x}_4-\dot{x}_1)\\
\end{aligned}\label{mech1}
\end{equation}
@@ -230,19 +230,19 @@ x_4\\
A=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
-\frac{k_1}{m_1} & -\frac{d_1}{m_1} & -\frac{k_2}{m_1} & -\frac{d_2}{m_1} & -\frac{k_3}{m_1} & -\frac{d_3}{m_1} & -\frac{k_4}{m_1} & -\frac{d_4}{m_1}\\
-\frac{k}{m_1} & -\frac{d}{m_1} & \frac{k_2}{m_1} & \frac{d_2}{m_1} & \frac{k_3}{m_1} & \frac{d_3}{m_1} & \frac{k_4}{m_1} & \frac{d_4}{m_1}\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & -\frac{k_2}{m_1} & -\frac{d_2}{m_2} & 0 & 0 & 0 & 0 \\
-\frac{k_1}{m_2} & -\frac{d_1}{m_2} & \frac{k_2}{m_2} & \frac{d_2}{m_2} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & -\frac{k_3}{m_1} & -\frac{d_3}{m_3} & 0 & 0 \\
-\frac{k_1}{m_3} & -\frac{d_1}{m_3} & 0 & 0 & \frac{k_3}{m_1} & \frac{d_3}{m_3} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -\frac{k_4}{m_1} & -\frac{d_4}{m_4} \\
-\frac{k_1}{m_4} & -\frac{d_1}{m_4} & 0 & 0 & 0 & 0 & \frac{k_4}{m_1} & \frac{d_4}{m_4} \\
\end{bmatrix},\quad
B=\begin{bmatrix}
\frac{1}{m} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\\
\frac{1}{m_1} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\\
\end{bmatrix}
%,\quad
%C=\begin{bmatrix}
@@ -252,10 +252,51 @@ B=\begin{bmatrix}
\label{mech2}
\end{equation}
\subsection{Stage data}
This data comes from datasheets and construction information [biblio: 'Dynamics of Parker Stage', Wayne Glettig 5.12.2018].
\begin{tabular}{|r|l|}
\hline
Stage Y mass& 340g \\
Stage X mass& 950g \\
Interferometer mirrors & 51g (additional)\\
Aluminun (instead ABS) & 42g (additional)\\
\hline
\end{tabular}
\vspace{1pc}
\begin{tabular}{|r|c|l|}
\hline
Continous force && 5.51N \\
Peak force && 12N \\
Static friction && 1N\\
Viscose damping && 0.5N$\cdot$s/m\\
Motor constant &Km& 1.46N/$\sqrt{watt}$\\
Resistance &R&8.8$\Omega$\\
Inductance &L& 2.4mH\\
\hline
\end{tabular}
\vspace{1pc}
The data in the data sheet are quite confusing. But lets check the motor Konstant Km.\\
The data sheet says:\\
Stall Current Continous 0.92A, Stall force Continous 4N
\[
U=R\cdot I \qquad P=U \cdot I \quad \rightarrow \quad P=R \cdot I^2\\
\]
at a constant current of 0.92A we have $ 8.8 \cdot 0.92^2 = 7.44 $W the resulting force will be:\\
$1.46N \cdot \sqrt{7.44} = 3.98 N$
\section{Simulink/MATLAB simulations}
TODO: describe the identification process to get a model out of the bode plots. identifyFxFyStage.m\\
Simulink simulation \verb|stage_closed_loop.slx| with \verb|ServoDeltaTau_z G(z)| showed similar response
Therefore the model seems good enough