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major renaming of transfer functions

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zamofing_t
2019-02-15 10:46:29 +01:00
parent f3c5fafe5f
commit e792d0b028
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MXfastStageDoc/MXfastStage.tex
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@@ -128,6 +128,13 @@ Out of the bode plot we can read approx.:\\
Motor 1: -33dB at 130Hz\\
Motor 2: -33dB at 84Hz\\
The chirp parameters for low frequencies bode measurements (yellow) are: \verb|minFrq=1, maxFrq=15, amp=1000|.\\
In this regime of 0 to 10Hz the bode amplitude is flat, which is mainly the result of the friction:\\
Motor 1 has 9.63dB($\approx$ factor 10) at 10 Hz. This means that an amplitude of 1000um needs roughly 100 \verb|curr_bits|.
The friction as measured in figure \ref{fig:mot1_frict} is around 100 \verb|curr_bits|. At lower frequencies than 10 Hz with an amplitude of 1000um, the friction free motion consumes much less current (-40dB) than the friction.
Therefore the whole current is used to overcome the friction and the bode amplitude remains constant.
%n times higher mass $\rightarrow$ n times lower frequency for same amplitude response\\
%n times higher frequency $\rightarrow$ n times higher velocity $\rightarrow$ $n^2$ times more acceleration==current
%1um at 12Hz with 1 mA $\rightarrow$ with 2000mA $\rightarrow$ sqrt(2000)*12Hz=540Hz
@@ -385,15 +392,15 @@ springs: $k=k_1+k_2+\ldots+k_n$\\
\begin{equation}
\begin{aligned}
\dot{x} = Ax + Bu\\
y = Cx + Du
\mathbf{\dot{x} = Ax + Bu}\\
\mathbf{y = Cx + Du}
\end{aligned}\label{eq:mech2}
\end{equation}
\eqref{eq:mech2} are the general input output equations in matrix form with x and A defines as \eqref{eq:mech3}:
\begin{equation}
x=
\mathbf{x}=
\begin{bmatrix}
x_1\\
\dot{x}_1\\
@@ -404,7 +411,7 @@ x_3\\
x_4\\
\dot{x}_4\\
\end{bmatrix},\quad
\dot{x}=
\mathbf{\dot{x}}=
\begin{bmatrix}
\dot{x}_1\\
\ddot{x}_1\\
@@ -415,7 +422,7 @@ x_4\\
\dot{x}_4\\
\ddot{x}_4\\
\end{bmatrix},\quad
A=
\mathbf{A}=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
-\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}\\
@@ -440,7 +447,14 @@ B=\begin{bmatrix}
\label{eq:mech3}
\end{equation}
\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black,colbacktitle=red!50,coltitle=black,title=TODO]
$k_1=0$ The non linear friction has to be compensate outside of the control loop.\\
$d_1$ seems to be negligable (section \ref{sec:stageData}).\\
But the system is not observable. $\rightarrow$ Perhaps the integrators of the system can be removed. having only $\mathbf{x}=[ \dot{x}_1 \dot{x}_2 \dot{x}_3 \dot{x}_4]$ as the state vector
\end{tcolorbox}
\subsection{Stage data}
\label{sec:stageData}
This data comes from datasheets and construction information \cite{DynParker}.
@@ -469,16 +483,19 @@ Inductance &L& 2.4mH\\
\vspace{1pc}
The data in the data sheet are quite confusing. But lets check the motor Konstant Km.\\
The data sheet for the current are quite confusing (sine, trap, rms). But lets check the motor Konstant Km.\\
The data sheet says:\\
Stall Current Continous 0.92A, Stall force Continous 4N
Stall Current Continous 0.92A (trap, DC), 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:\\
$1.46N \cdot \sqrt{7.44} = 3.98 N$
$1.46N \cdot \sqrt{7.44} = 3.98 N$\\
Lets have a look at the viscose damping:\\
our speed is roughly 50$\mu$m/10ms. this is 5mm/s. The damping force at that speed is: $0.5\frac{Ns}{m}\cdot 5\frac{mm}{s}=2\cdot10^{-3}N$.
The used current for that force is negligable, and theerefore the viscose damopling is negligable.
\subsection{identification of stages}
@@ -500,20 +517,26 @@ The approximated transfer functions can be tweaked and edited with: e.g. \verb|c
The full transfer function is then split in individual parts that are put back into \verb|MXTuning.py|.\\
the transfer functions for each two motors are separated atomic transfer function as stated in table \ref{tab:trfFunc1}\\
tfc = transfer function current, tfp = transfer function position. Consecutive letters are simplified transfer functions. tf[1-9]=transfer function resonance.
\begin{table}[h!]
\center
\begin{tabular}{|l|l|l|l|}
\hline
key & description & motor1 (fy) & motor2 (fx) \\
key & description & motor1 (fy) & motor2 (fx) \\
\hline
tfc & curren loop tf & f=694 ? 1389 , d=0.75 & same \\
tf1 & \vtop{\hbox{\strut main mechanical tf} \hbox{\strut with -40dB/dec }}
& mag=6dB f=7.96Hz d=0.6 & mag=12dB f=3.34 d=0.4 \\
tf2 & mechanical resonance tf & f=[197,199] d=[0.02,0.02] & f=[55|61] d=[0.2|0.2] \\
tf3 & mechanical resonance tf & & f=[128|137] d=[0.05|0.05] \\
tf4 & mechanical resonance tf & & f=[410|417] d=[0.015|0.015] \\
tf5 & mechanical resonance tf & & f=[230|233] d=[0.04|0.04] \\
tfc & current loop tf & f=694 ? 1389 , d=0.75 & same \\
tfd & simplified cur.loop tf $PT_1$ & f-3dB, 45\deg at 400Hz & same \\
\hline
tfp & \vtop{\hbox{\strut main mechanical tf} \hbox{\strut with -40dB/dec }}
& mag=6dB f=7.96Hz d=0.6 & mag=12dB f=3.34 d=0.4 \\
tfq & \vtop{\hbox{\strut simplified mechanical tf} \hbox{\strut 2 integrators with -40dB/dec }}
& 0dB at 19.8Hz & 0dB at 11.84Hz\\
\hline
tf1 & mechanical resonance tf & f=[197,199] d=[0.02,0.02] & f=[55|61] d=[0.2|0.2] \\
tf2 & mechanical resonance tf & & f=[128|137] d=[0.05|0.05] \\
tf3 & mechanical resonance tf & & f=[410|417] d=[0.015|0.015] \\
tf4 & mechanical resonance tf & & f=[230|233] d=[0.04|0.04] \\
\hline
\end{tabular}
\caption{plant partial transfer functions}
@@ -521,7 +544,7 @@ tf5 & mechanical resonance tf & & f=[23
\end{table}
%\vspace{1pc}
The black line on figure \ref{fig:mot_open} shows the concatenate transfer function $tf1 \cdot tf2 \cdot tf3 \cdot tf4 \cdot tf5 \cdot tfc$ .
The black line on figure \ref{fig:mot_open} shows the concatenate transfer function $tfc \cdot tfp \cdot tf1 \cdot tf2 \cdot tf3 \cdot tf4$.\\
\vspace{1pc}
\fbox{\parbox[t]{15cm}{The current loop frequency from the MATLAB identification looks different than the used one. Matlab identifies $w_0=8727rad/sec = f0=1389Hz$ but to match the bode plot a value of half frequency need to be taken:$f_0=694$. The reason could be discretization and time delay because the servo loop is processed after the phase loop.}}
@@ -584,29 +607,17 @@ opposite to the PID cpntroller which is a ‘black-box design’, the state spa
\end{figure}
To implement a state space controller with observer the model has to be observable and controllable. The full model of the tranfer function $tf1 \cdot tf2 \cdot tf3 \cdot tf4 \cdot tf5 \cdot tfc$ is not controllable due to the various resonance frequencies. Therefore for the observer design the state space model has to be simplified.
To implement a state space controller with observer the model has to be observable and controllable. The full model of the tranfer function $tfc \cdot tfp \cdot tf1 \cdot tf2 \cdot tf3 \cdot tf4$ is not controllable due to the various resonance frequencies. Therefore for the observer design the state space model has to be simplified. (see simplified transfer function in table \ref{tab:trfFunc1}\\
In addition to the transferfunction in table \ref{tab:trfFunc1} following simplified transfer functions have been added:\\
\begin{tabular}{|l|l|l|l|}
\hline
key & description & motor1 (fy) & motor2 (fx) \\
\hline
tfd & simplified loop tf $PT_1$ & f-3dB, 45\deg at 400Hz & same \\
tf0 & \vtop{\hbox{\strut simplified mechanical tf} \hbox{\strut 2 integrators with -40dB/dec }}
& 0dB at 19.8Hz & 0dB at 11.84Hz\\
\hline
\end{tabular}
\vspace{1pc}
Figure \ref{fig:mdl_bode1} shows the bode plots of the best model to the most simplified model:\\
\begin{tabular}{ll}
plant & best model tranfer function: $tf1 \cdot tf2 \cdot tf3 \cdot tf4 \cdot tf5 \cdot tfc$\\
mdl1c & main mechanical with second order current loop: $tf1 \cdot tfc$\\
mdl1d & main mechanical with $PT_1$ current loop (first order): $tf1 \cdot tfd$\\
mdl1 & main mechanical (incl. 'viscose friction'): $tf1$\\
mdl0 & only 2 integrators: $tf0$\\
plant & best model tranfer function: $tfc \cdot tfp \cdot tf1 \cdot tf2 \cdot tf3 \cdot tf4$\\
mdlcp & main mechanical with second order current loop: $tfc \cdot tfp$\\
mdldp & main mechanical with $PT_1$ current loop (first order): $tfd \cdot tfp$\\
mdlp & main mechanical (incl. 'viscose friction'): $tfp$\\
mdlq & only 2 integrators: $tfq$\\
\end{tabular}
\vspace{1pc}
@@ -627,10 +638,10 @@ The matlab models are:\\
\begin{tabular}{ll}
\texttt{ss\_plt:} & best approach of the plant with mechanics, resonance, current loop etc.\\
\texttt{ss\_c1:} & model without resonance (only current and main mechanical)\\
\texttt{ss\_d1:} & model without resonance (simplified current and main mechanical)\\
\texttt{ss\_1:} & model without current loop, no resonance (only main mechanical)\\
\texttt{ss\_0:} & simplified mechanical, no current loop, no resonance\\
\texttt{ss\_cp:} & model without resonance (only current and main mechanical)\\
\texttt{ss\_dp:} & model without resonance (simplified current and main mechanical)\\
\texttt{ss\_p:} & model without current loop, no resonance (only main mechanical)\\
\texttt{ss\_q:} & simplified mechanical, no current loop, no resonance\\
\end{tabular}\\
\vspace{1pc}
@@ -639,13 +650,18 @@ Following code calculates parameters for a observer controller, does a simulatio
\begin{verbatim}
clear;clear global;close all;
mot=identifyFxFyStage(7);
sscType=0
for k =1:2
[ssc]=StateSpaceControlDesign(mot{k},sscType);sim('observer');
f=figure(); h=plot(desPos_actPos.Time,desPos_actPos.Data,'g');
set(h(1),'color','b'); set(h(2),'color',[0 0.5 0]);
print(f,sprintf('figures/sim_cl_observer_%d',mot{k}.id),'-depsc');
for sscType=0:1
for k=1:2
[ssc]=StateSpaceControlDesign(mot{k},sscType);sim('observer');
f=figure(); h=plot(desPos_actPos.Time,desPos_actPos.Data,'g');grid on;
set(h(1),'color','b'); set(h(2),'color',[0 0.5 0]);
print(f,sprintf('figures/sim_cl_obs_%d_%d',sscType,mot{k}.id),'-depsc');
f=bodeSamples(desPos_actPos);
print(f,sprintf('figures/sim_cl_obs_bode%d_%d',sscType,mot{k}.id),'-depsc');
end
end
\end{verbatim}
@@ -659,10 +675,10 @@ end
\begin{figure}[h!]
\center
\includegraphics[scale=.45]{../matlab/figures/sim_cl_observer_1.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_observer_bode1.eps}\\
\includegraphics[scale=.45]{../matlab/figures/sim_cl_observer_2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_observer_bode2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_0_1.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_bode0_1.eps}\\
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_0_2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_bode0_2.eps}
\caption{Observer sim: Motor 1 Motor 2}
\label{fig:mot_observer_sim}
\end{figure}
@@ -675,10 +691,10 @@ With an additional prefilter resonance and reductions can be suppressed a bit.\\
\begin{figure}[h!]
\center
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_pf_1.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_pf_bode1.eps}\\
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_pf_2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_pf_bode2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_1_1.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_bode1_1.eps}\\
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_1_2.eps}
\includegraphics[scale=.45]{../matlab/figures/sim_cl_obs_bode1_2.eps}
\caption{Observer with prefilter sim: Motor 1 Motor 2}
\label{fig:mot_observer_sim}
\end{figure}