leo/cours-m1-eea
1
0
Fork 0
Code Issues Pull requests Releases Wiki Activity

passage en tikz

Browse source
This commit is contained in:
Pierre-antoine Comby 2019-03-14 23:20:12 +01:00
parent 6d34563513
commit 438d8939f7
1 changed files with 32 additions and 12 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

44
424-Systeme_Non_Lineaires/Cours/chap4.tex
View file

@ -19,7 +19,15 @@ On prend $x=X\sin \omega t$. Dans le cas linéaire, seule la valeur de $\omega$
\begin{figure}[h!] \begin{figure}[h!]
\centering \centering
\includegraphics[scale=0.4]{2/424-1.png} \begin{tikzpicture}
\draw[-latex] (-4,0) -- (4,0)node[above]{$Re$};
\draw[-latex] (0,-4) -- (0,4)node[left]{$Im$};
\draw (0,0) to[out=110,in=0] (-2,1) to[out=180,in=80] (-4,-3) node[below]{$X_1$};
\draw (0,0) to[out=130,in=0] (-1,0.7) to[out=180,in=80] (-3,-3)node[below]{$X_2$};
\draw (0,0) to[out=150,in=0] (-0.5,0.3) to[out=180,in=80](-2,-3) node[below]{$X_3$};
\node[above] at (-2,1) {$T_{BO}$};
\end{tikzpicture}
\caption{Modification du lieu en fonction de l'amplitude}
\end{figure} \end{figure}
Puisque $H(p)$ rejette les harmoniques d'ordre supérieur à 1, on peut donc décomposer \[y(t)=P \sin \omega t + Q \cos \omega t\] Puisque $H(p)$ rejette les harmoniques d'ordre supérieur à 1, on peut donc décomposer \[y(t)=P \sin \omega t + Q \cos \omega t\]
@ -294,24 +302,36 @@ i.e. en notant $\left.\derivp[]{X}\right|_{\zero}=\left.\derivp[]{X}\right|_0$
\centering \centering
\begin{tikzpicture} \begin{tikzpicture}
\begin{axis} \begin{axis}
[axis lines= middle, [axis lines= middle,scale=0.8,
ticks=none, domain=0:10, ticks=none, domain=0:10,samples=100,
xmin=0,xmax=10,ymin=-2,ymax=2] xmin=0,xmax=10,ymin=-2,ymax=2,clip=false]
\addplot[black,smooth]{cos(2*deg(x))}; \addplot[black,smooth,dashed]{cos(2*deg(x))};
\addplot[black,smooth]{cos(2*deg(x))*(exp(x/10))}; \addplot[black,smooth]{cos(2*deg(x))*(1+0.3*exp(x/8))};
\addplot[black,smooth]{cos(2*deg(x))*(1+0.3*exp(-x/5))};
\draw[-latex] (axis cs: -0.1,1) -- (axis cs: -0.1,1.3) node[midway,left]{$\delta x >0 $};
\draw (axis cs:10,1.5) node[right]{$m>0$};
\draw (axis cs:10,0.5) node[right]{$m<0$};
\end{axis} \end{axis}
\end{tikzpicture}\qquad%
\begin{tikzpicture}
\begin{axis}
[axis lines= middle,scale=0.8,
ticks=none, domain=0:10,samples=100,
xmin=0,xmax=10,ymin=-2,ymax=2,clip=false]
\addplot[black,smooth,dashed]{cos(2*deg(x))};
\addplot[black,smooth]{cos(2*deg(x))*(1-0.3*exp(x/8))};
\addplot[black,smooth]{cos(2*deg(x))*(1-0.3*exp(-x/5))};
\draw[-latex] (axis cs: -0.1,1) -- (axis cs: -0.1,0.7) node[midway,left]{$\delta x< 0$};
\draw (axis cs:10,0) node[right]{$m>0$};
\draw (axis cs:10,0.5) node[right]{$m<0$};
\end{axis}
\end{tikzpicture} \end{tikzpicture}
\includegraphics[scale=0.4]{2/424-61.png}
\end{figure} \end{figure}
$m > 0$ et $\delta X > 0$ : CL est stable $m > 0$ et $\delta X > 0$ : CL est stable
$m < 0$ et $\delta X > 0$ : CL est instable $m < 0$ et $\delta X > 0$ : CL est instable
\begin{figure}[h!]
\centering
\includegraphics[scale=0.4]{2/424-62.png}
\end{figure}
$\delta X < 0$ et $m < 0$ : CL est stable $\delta X < 0$ et $m < 0$ : CL est stable
$\delta X < 0$ et $m > 0$ : CL est instable $\delta X < 0$ et $m > 0$ : CL est instable