☰

速度与位置图的绘图构造。相位图构造的插值？

tag-icon • tag-icon pgfplots plot graphs physics mathematics
速度与位置图的绘图构造。相位图构造的插值？

我需要绘制一些图表。首先是函数

    \begin{equation}
    x(t)= -e^{ -(0.1 \ {s}^{-1}) t} \cos \left( ( 0.995 \ {rad} / \mathrm{s})t \right)
    \end{equation}

和 $\dot{x}$ （时间导数函数）

\begin{equation}
    \dot{x}(t)= e^{-(0.1 \ {s}^{-1}) t}\left[(0.1 \ {s}^{-1}) \cos \left( ( 0.995 \ {rad} / \mathrm{s})t \right)+ ( 0.995 \ {rad} / \mathrm{s})\sin ( ( 0.995 \ {rad} / \mathrm{s})t )\right] .
\end{equation}

到目前为止，我已经通过以下方式制作了各自的情节

    \begin{figure}[ht]
    \centering
        \caption{ The plots of the position and speed versus time (underdamped oscillator).}
\begin{tikzpicture}[scale=1.9]
\begin{axis}[
    axis lines = left,
    xlabel = {$t$, $ \left[\text{s} \right]$},
    %ylabel = {$a(t)$, $ \left[\text{m/s}^2 \right]$},
    grid=major,
    ymin=-1,
    ymax=1,
]
\addplot [
    domain=0:60, 
     samples=300, 
    color=YellowGreen, 
    thick,
]
{2.71828^(-0.1*x)*cos(deg(0.995*x-3.1415))};
\addlegendentry{\tiny $ x(t)$, , $ \left[\text{cm} \right]$}
\addplot [
    domain=0:60,
     samples=300,
    color=TealBlue,
     thick,
    ]
    {-2.71828^(-0.1*x)*((0.1*cos(deg(0.995*x-3.1415))+0.995*sin(deg(0.995*x-3.1415))) };
\addlegendentry{\tiny $  \dot{x}(t)$,  $ \left[\text{cm/s} \right]$}
\end{axis}
\end{tikzpicture}
\end{figure}

得到图表

在此处输入图片描述

还剩下什么问题：问题1。我需要的第二个图是相图，即 $\dot{x}(t)$ vs $x(t)$ 图，我不确定如何构造。我在想对函数 $x(t)$ 和 $\dot{x}(t)$ 进行采样/点采集，然后使用这些点进行相图的插值构造，是否可以以某种方式实现？但是，我在 latex 论坛上找不到很多关于这类事情的信息。我的男朋友用 python 制作了他的图表，所以我知道相图必须如下所示

在此处输入图片描述

但我希望有某种方法可以仅使用乳胶来制作图表。有什么想法吗？

还剩下什么问题：问题2。我还想知道是否有任何方法可以确定在振幅低于其最大值的 $10^{-2}$ 之前系统穿过 $x=0$ 线的次数，但如果可以仅使用 latex 命令来输出这个数字。

答案1

显然，Bamboo 和我的想法非常相似。这个还计算了您在问题的第二部分中要求的交集。（涉及大量清理工作，许多更改与 Bamboo 的出色答案非常相似。）

\documentclass{article}
\usepackage{geometry}
\usepackage[fleqn]{amsmath}
\usepackage{siunitx}
\usepackage[dvipsnames]{xcolor}
\usepackage{pgfplots}
\usepgfplotslibrary{fillbetween}% loads intersections
\pgfplotsset{compat=1.17}
\begin{document}
\begin{equation}
    x(t)= -\mathrm{e}^{ -(\SI{0.1}{\per\second}) t}\, 
    \cos \left( ( \SI{0.995}{\radian\per\second})t \right)
\end{equation}
and of $\dot{x}$ (time derivative function)
\begin{equation}
    \dot{x}(t)= \mathrm{e}^{-(\SI{0.1}{\per\second}) t}
    \left[(\SI{0.1}{\per\second}) \cos \left( (\SI{0.995}{\radian\per\second})t \right)
    + ( \SI{0.995}{\radian\per\second})\sin ( ( \SI{0.995}{\radian\per\second})t )\right] .
\end{equation}

\begin{figure}[ht]
\centering
\caption{The plots of the position and speed versus time (underdamped oscillator).}
\begin{tikzpicture}[scale=1.6]
\begin{axis}[declare function={%
    pos(\x)=exp(-0.1*\x)*cos(deg(0.995*\x-pi));%
    posdot(\x)=-exp(-0.1*\x)*((0.1*cos(deg(0.995*\x-pi))+0.995*sin(deg(0.995*\x-pi)));
    },
    axis lines = left,
    xlabel = {$t$, $ \left[\text{s} \right]$},
    %ylabel = {$a(t)$, $ \left[\text{m/s}^2 \right]$},
    grid=major,
    ymin=-1,
    ymax=1,
    legend style={font=\footnotesize}
]
\addplot [
    domain=0:60, 
     samples=300, 
    color=YellowGreen, 
    thick,
]
{pos(x)};
\addlegendentry{$ x(t)~\left[\si{\centi\meter}\right]$}
\addplot [
    domain=0:60,
     samples=300,
    color=TealBlue,
     thick,
    ]
    {posdot(x)};
\addlegendentry{$\dot{x}(t)~ \left[\si{\centi\meter\per\second} \right]$}
\end{axis}
\end{tikzpicture}
\end{figure}

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=1.6]
\begin{axis}[declare function={%
    pos(\x)=exp(-0.1*\x)*cos(deg(0.995*\x-pi));%
    posdot(\x)=-exp(-0.1*\x)*((0.1*cos(deg(0.995*\x-pi))+0.995*sin(deg(0.995*\x-pi)));
    },
    axis lines = left,
    xlabel = {$x(t)~ \left[\si{\centi\meter} \right]$},
    ylabel = {$\dot x(t)~ \left[\si{\centi\meter\per\second} \right]$},
    grid=major,
    ymin=-1,
    ymax=1,
    xmax=0.75
]
\addplot [
    domain=0:60, 
     samples=601, 
    color=blue, 
    thick,smooth
]({pos(x)},{posdot(x)});
\addplot [name path=phase,
    domain=0:60, 
     samples=601, 
    draw=none]({pos(x)},{posdot(x)});
\path[name path=axis] 
 (0,1) -- (0,{abs(pos(0))/100}) 
  (0,-1) -- (0,{-abs(pos(0))/100})  
  ; 
\path[name intersections={of=phase and axis,total=\t}]
    \pgfextra{\xdef\MyNumIntersections{\t}}; 
\end{axis}
\end{tikzpicture}
\caption{Phase space diagram. The phase curve intersects 
$\MyNumIntersections$
times with the $x=0$ axis before reaching 0.01 times its maximal value.}
\end{figure}
\end{document}

在此处输入图片描述

在此处输入图片描述

笔记：

  1. 我把函数声明保留在本地，因为重新声明它们会比较困难，但并非不可能。也就是说，如果你声明pos(\x)为全局函数，你就不能轻易地声明另一个同名的函数。
  2. pipgf 知道和的值e，因此您可以使用该exp函数。
  3. 我使用不可见的、非平滑的图来计算交点，因为交点数永远不是完全可信的，并且对于平滑的图来说会变得更加不稳定。

附录：只是为了好玩：这使用了 Bamboo 的好主意，即安装一个过滤器来计算第一的图，结果更可靠。好消息是数字 14 得到了确认，因此上面的结果似乎给出了正确的数字（无论是偶然还是无意的）。分析结果是int(10*ln(100))=14，所以一切都很好。在这个版本中，我还删除了Bamboo 提出的\left和\right。无论如何，重点是计算第一个图中的交点应该非常可靠，在第二个图中我不太确定。

\documentclass{article}
\usepackage{geometry}
\usepackage[fleqn]{amsmath}
\usepackage{siunitx}
\usepackage[dvipsnames]{xcolor}
\usepackage{pgfplots}
\usepgfplotslibrary{fillbetween}% loads intersections
\pgfplotsset{compat=1.17}
\begin{document}
\begin{equation}
    x(t)= -\mathrm{e}^{ -(\SI{0.1}{\per\second}) t}\, 
    \cos \left( ( \SI{0.995}{\radian\per\second})t \right)
\end{equation}
and of $\dot{x}$ (time derivative function)
\begin{equation}
    \dot{x}(t)= \mathrm{e}^{-(\SI{0.1}{\per\second}) t}
    \left[(\SI{0.1}{\per\second}) \cos \left( (\SI{0.995}{\radian\per\second})t \right)
    + ( \SI{0.995}{\radian\per\second})\sin ( ( \SI{0.995}{\radian\per\second})t )\right] .
\end{equation}

\begin{figure}[ht]
\centering
\caption{The plots of the position and speed versus time (underdamped oscillator).}
\begin{tikzpicture}[scale=1.6]
\begin{axis}[declare function={%
    pos(\x)=exp(-0.1*\x)*cos(deg(0.995*\x-pi));%
    posdot(\x)=-exp(-0.1*\x)*((0.1*cos(deg(0.995*\x-pi))+0.995*sin(deg(0.995*\x-pi)));
    },
    axis lines = left,
    xlabel = {$t~ [\text{s} ]$},
    %ylabel = {$a(t)$, $ \left[\text{m/s}^2 \right]$},
    grid=major,
    ymin=-1,
    ymax=1,
    legend style={font=\footnotesize}
]
\addplot [
    domain=0:60, 
     samples=300, 
    color=YellowGreen, 
    thick,
]
{pos(x)};
\addlegendentry{$ x(t)~[\si{\centi\meter}]$}
\addplot [
    domain=0:60,
     samples=300,
    color=TealBlue,
     thick,
    ]
    {posdot(x)};
\addlegendentry{$\dot{x}(t)~ [\si{\centi\meter\per\second} ]$}
\addplot [name path=x,
    x filter/.expression={abs(pos(x))<abs(pos(0))/100 ? nan :x},
    domain=0:60, 
     samples=300,
    draw=none]
{pos(x)};
\path[name path=axis] (0,0) -- (60,0);
\path[name intersections={of=x and axis,total=\t}] 
foreach \X in {1,...,\t} {(intersection-\X) node[red,circle,inner sep=1.2pt,fill]{}}
(60,-1) node[above left,font=\footnotesize,
align=right,text width=6.5cm]{$x(t)$ intersects $\t$ times
with the $x=0$ axis before dropping below $1\%$ of its initial amplitude.};
\end{axis}
\end{tikzpicture}
\end{figure}

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=1.6]
\begin{axis}[declare function={%
    pos(\x)=exp(-0.1*\x)*cos(deg(0.995*\x-pi));%
    posdot(\x)=-exp(-0.1*\x)*((0.1*cos(deg(0.995*\x-pi))+0.995*sin(deg(0.995*\x-pi)));
    },
    axis lines = left,
    xlabel = {$x(t)~ [\si{\centi\meter}]$},
    ylabel = {$\dot x(t)~ [\si{\centi\meter\per\second} ]$},
    grid=major,
    ymin=-1,
    ymax=1,
    xmax=0.75
]
\addplot [
    domain=0:60, 
     samples=601, 
    color=blue, 
    thick,smooth
]({pos(x)},{posdot(x)});
\addplot [name path=phase,
    domain=0:60, 
     samples=601, 
    draw=none]({pos(x)},{posdot(x)});
\path[name path=axis] 
 (0,1) -- (0,{abs(pos(0))/100}) 
  (0,-1) -- (0,{-abs(pos(0))/100})  
  ; 
\path[name intersections={of=phase and axis,total=\t}]
    \pgfextra{\xdef\MyNumIntersections{\t}}; 
\end{axis}
\end{tikzpicture}
\caption{Phase space diagram. The phase curve intersects 
$\MyNumIntersections$
times with the $x=0$ axis before reaching 0.01 times its maximal value.}
\end{figure}
\end{document}

在此处输入图片描述

答案2

这是您的代码的更清晰的版本以及@Schrödinger's cat 提到的参数图。

请注意使用siunitx包来排版单位。此外，\left[... \right]在这种情况下确实没有必要。最后，我明确声明了您的函数，以便通过设置简化它们的使用tikz declare function。

编辑更新版本使用此信息绘制交点并在参数图中绘制节点。请注意，我使用 来x filter丢弃此图中的低振幅结果，这与薛定谔猫方法明显不同。

\documentclass[tikz,dvipsnames,border=3.14mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usepackage{siunitx}

\usetikzlibrary{intersections}

\tikzset{
    declare function={
        f(\t) = 2.71828^(-0.1*\t)*cos(deg(0.995*\t-3.1415));
        df(\t) = -2.71828^(-0.1*x)*((0.1*cos(deg(0.995*x-3.1415))+0.995*sin(deg(0.995*x-3.1415)));
    },
}


\begin{document}
\begin{tikzpicture}[scale=1.9]
\begin{axis}[
    axis lines = left,
    xlabel = {$t \quad [\si{\second}]$},
    grid=major,
    ymin=-1,
    ymax=1,
    legend cell align=left,
    legend style={font=\small},
    domain=0:60,
    samples=300,
]
\addplot [color=YellowGreen,thick] {2.71828^(-0.1*x)*cos(deg(0.995*x-3.1415))};
\addlegendentry{$x(t) \quad [\si{\centi\meter}]$}
\addplot [color=TealBlue,thick] {-2.71828^(-0.1*x)*((0.1*cos(deg(0.995*x-3.1415))+0.995*sin(deg(0.995*x-3.1415)))};
\addlegendentry{$\dot{x}(t) \quad [\si{\meter\per\second}]$}
\end{axis}
\end{tikzpicture}

\begin{tikzpicture}[scale=1.9]
    \begin{axis}[
        axis lines = left,
        xlabel = {$x(t) \quad [\si{\centi\meter}]$},
        ylabel = {$\dot{x}(t) \quad [\si{\centi\meter\per\second}]$},
        grid=major,
        ymin=-1,
        ymax=1,
        legend cell align=left,
        legend style={font=\small},
        domain=0:60,
        samples=300,
        x filter/.expression={abs(x)>1e-2 ? x : nan)},
        clip=false,
    ]
    \addplot [color=YellowGreen,thick, name path=paramplot] ({f(x)},{df(x)});
    \path[name path=yzeroline] (\pgfkeysvalueof{/pgfplots/xmin},0) -- (\pgfkeysvalueof{/pgfplots/xmax},0);
    \path[name intersections={of=paramplot and yzeroline,total=\totalintersects}] 
    foreach \nb in {1,...,\totalintersects}{
        node[circle,fill=red, inner sep=1pt] at (intersection-\nb){}
    }
    node[draw,fill=white,anchor=south west,outer sep=0pt] at (rel axis cs:0.01,0.01) {Number of intersections : \totalintersects}
    ;
    \end{axis}
    \end{tikzpicture}
\end{document}

在此处输入图片描述 在此处输入图片描述

相关内容