Latex 数学表达式对齐

Question

我在对齐一些数学表达式时遇到了问题。是否可以使某些表达式与“&”符号对齐，而某些表达式与页面中心对齐？如屏幕截图所示，我已经完成了它，但中间有一些空行。我想将第一行和倒数第二行对齐到中心，其余对齐到“&”符号，没有空格。 提前谢谢你

\begin{gather}
    |e(t)| \leq e_{lim} \Rightarrow t \in \left\langle t_{set}; \infty\right) \label{eq:settling_time}
\end{gather}
\begin{align}
    \left|\frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right]\right| &\leq e_{lim} \nonumber \\
    \frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right] &> 0; \quad t \in \mathbb{R}_0^+ \nonumber \\
    \frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right] &\leq e_{lim} \nonumber \\
    K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1 &\leq e_{lim}\cdot (K_p\cdot K_s - p_1) \nonumber \\
    K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} &\leq e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1 \nonumber \\
    e^{(p_1 - K_p\cdot K_s)\cdot t} &\leq \frac{e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{K_p\cdot K_s} \nonumber \\
    (p_1 - K_p\cdot K_s)\cdot t &\leq \ln \left[\frac{e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{K_p\cdot K_s}\right] \nonumber \\
    t &\geq \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s} \nonumber
\end{align}
\begin{gather}
    t \in \Bigg \langle \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s}; \infty \Bigg) \in \left\langle t_{set}; \infty \right) \nonumber \\
    t_{set} = \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s} \label{eq:settling_time_1st_order_P_reg}
\end{gather}

Answer 1

您可以使用aligned 在gather 中设置对齐的组件：

\documentclass{article}

\usepackage{amsmath,amssymb}

\begin{document}

\begin{gather}
  | e(t) | \leq e_{\text{lim}} \Rightarrow t \in \langle t_{\text{set}}; \infty ) \\
  \begin{aligned}
    \biggl| \frac{1}{K_p \cdot K_s - p_1} \cdot \bigl[ K_p \cdot K_s \cdot e^{(p_1 - K_p \cdot K_s) \cdot t} - p_1 \bigr] \biggr| & 
      \leq e_{\text{lim}} \\
    \frac{1}{K_p \cdot K_s - p_1} \cdot \bigl[ K_p \cdot K_s \cdot e^{(p_1 - K_p \cdot K_s) \cdot t} - p_1 \bigr] &
      > 0; \quad t \in \mathbb{R}_0^+ \\
    \frac{1}{K_p \cdot K_s - p_1} \cdot \bigl[ K_p \cdot K_s \cdot e^{(p_1 - K_p \cdot K_s) \cdot t} - p_1 \bigr] &
      \leq e_{\text{lim}} \\
    K_p \cdot K_s \cdot e^{(p_1 - K_p \cdot K_s) \cdot t} - p_1 &
      \leq e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) \\
    K_p \cdot K_s \cdot e^{(p_1 - K_p \cdot K_s) \cdot t} &
      \leq e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1 \\
    e^{(p_1 - K_p \cdot K_s) \cdot t} &
      \leq \frac{e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1}{K_p \cdot K_s} \\
    (p_1 - K_p \cdot K_s) \cdot t &
      \leq \ln \biggl[ \frac{e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1}{K_p \cdot K_s} \biggr] \\
    t &
      \geq \frac{\ln \biggl[ \dfrac{e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1}{K_p \cdot K_s} \biggr]}{p_1 - K_p \cdot K_s}
  \end{aligned} \nonumber \\
    t \in \Bigg \langle \frac{ \ln \biggl[ \dfrac{e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1}{K_p \cdot K_s} \biggr]}{p_1 - K_p \cdot K_s}; \infty \Bigg) \in \langle t_{\text{set}}; \infty ) \nonumber \\
    t_{\text{set}} = \frac{\ln \biggl[ \dfrac{e_{\text{lim}} \cdot (K_p \cdot K_s - p_1) + p_1}{K_p \cdot K_s} \biggr]}{p_1 - K_p \cdot K_s}
\end{gather}

\end{document}

Answer 2

我建议使用\IEEEeqnarray 环境。如果你使用 * 这个版本，你可以去掉所有的 \nonumber 命令（在这种情况下是 9 个）并在不需要等式 no 的地方插入 \yesnumber 命令（在这种情况下是 2 个）。

\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{IEEEtrantools}

%-------Shows page layout-------------------
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%-------------------------------------------

\begin{document}
    \begin{IEEEeqnarray*}{rcl}
        \IEEEeqnarraymulticol{3}{c}{|e(t)| \leq e_{lim} \Rightarrow t \in \left\langle t_{set}; \infty\right)} \yesnumber \label{eq:settling_time}\\
        \left|\frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right]\right| & \leq & e_{lim}  \\
        \frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right]              & >    & 0; \quad t \in \mathbb{R}_0^+  \\
        \frac{1}{K_p\cdot K_s - p_1}\cdot \left[K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1\right]              & \leq & e_{lim}  \\
        K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t} - p_1                                                             & \leq & e_{lim}\cdot (K_p\cdot K_s - p_1)  \\
        K_p\cdot K_s\cdot e^{(p_1 - K_p\cdot K_s)\cdot t}                                                                   & \leq & e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1  \\
        e^{(p_1 - K_p\cdot K_s)\cdot t}                                                                                     & \leq & \frac{e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{K_p\cdot K_s}  \\
        (p_1 - K_p\cdot K_s)\cdot t                                                                                         & \leq & \ln \left[\frac{e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{K_p\cdot K_s}\right]  \\
        t                                                                                                                   & \geq & \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s} \\
        \IEEEeqnarraymulticol{3}{c}{t \in \Bigg \langle \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s}; \infty \Bigg) \in \left\langle t_{set}; \infty \right)}  \\
        \IEEEeqnarraymulticol{3}{c}{t_{set} = \frac{\displaystyle \ln \left[\frac{\displaystyle e_{lim}\cdot (K_p\cdot K_s - p_1) + p_1}{\displaystyle K_p\cdot K_s}\right]}{\displaystyle p_1 - K_p\cdot K_s}} \yesnumber \label{eq:settling_time_1st_order_P_reg}
    \end{IEEEeqnarray*}
\end{document}  

this document 的附录 F 介绍了 IEEEeqnarray 命令的使用。

正如评论中提到的，您可能会在TEX.SE找到更快的回复