在 Beamer 块之间对齐等号

在 Beamer 块之间对齐等号

我有一个由两个包含一些方程式的块组成的 Beamer 框架:

\begin{frame}
  \frametitle{Modell}
  \begin{block}{A  $(X_{t})$}
    \begin{align*}
    X_{t} = \epsilon_{t} \quad \text{mit} \quad \epsilon_{t}
    \overset{iid}{\sim} \mathcal{N}\left(0, \sigma_{X}^{2}\right)
    \end{align*}
  \end{block}
  \begin{block}{B $(Y_{t})$}
    \begin{align*}
      Y_{t} &= \sigma_{Y,t}\: \eta_{t}, \\ \sigma_{Y,t}^{2} &= \omega + \alpha_{Y} Y_{t-1}^{2} + \beta_{Y}\sigma_{Y,t-1}^{2} \quad \text{mit} \quad  \sqrt{\frac{\nu}{\nu-2}}\cdot \eta_{t} \overset{iid}{\sim} t\left(\nu\right). 
    \end{align*}
  \end{block}

\end{frame}

我想喜欢全部等号要对齐,换句话说,我希望在两个块之间对齐。通常我会使用,intertext但由于我正在处理两个块环境,所以我认为这行不通。

谢谢!

答案1

这是一种可能性。使用幻像在块 1 中重复块 2 的最长方程。

在此处输入图片描述

代码

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{matrix}

\begin{document}

\begin{frame}
  \frametitle{Modell}
  \begin{block}{A  $(X_{t})$}
    \begin{align*}
    X_{t} &= \epsilon_{t} \quad \text{mit} \quad \epsilon_{t}
    \overset{iid}{\sim} \mathcal{N}\left(0, \sigma_{X}^{2}\right)\\
\phantom{\sigma_{Y,t}^{2}} &\phantom{= \omega + \alpha_{Y} Y_{t-1}^{2} + \beta_{Y}\sigma_{Y,t-1}^{2} \quad \text{mit} \quad  \sqrt{\frac{\nu}{\nu-2}}\cdot \eta_{t} \overset{iid}{\sim} t\left(\nu\right).} 
    \end{align*}
  \end{block}
  \begin{block}{B $(Y_{t})$}
    \begin{align*}
      Y_{t} &= \sigma_{Y,t}\: \eta_{t}, \\ \sigma_{Y,t}^{2} &= \omega + \alpha_{Y} Y_{t-1}^{2} + \beta_{Y}\sigma_{Y,t-1}^{2} \quad \text{mit} \quad  \sqrt{\frac{\nu}{\nu-2}}\cdot \eta_{t} \overset{iid}{\sim} t\left(\nu\right). 
    \end{align*}
  \end{block}

\end{frame}

\end{document}

答案2

带有动态放置且不带有\hphantom

在此处输入图片描述

\documentclass{beamer}
\usepackage{tikz,array}
\usetikzlibrary{matrix}

\newlength{\myl}
\newcolumntype{Z}{>{\rule{0pt}{1.4em}\hfill$}m{\myl}<{$}}

\newenvironment{MyAlign}{%
    \setlength{\arraycolsep}{1.5pt}%

    \smallskip

    $\begin{array}{Zcl}
    }{%
    \end{array}$

    \smallskip

    }

\begin{document}

\begin{frame}
\setlength{\myl}{1cm}

  \frametitle{Modell}
  \begin{block}{A  $(X_{t})$}
    \begin{MyAlign}
    X_{t} &=& \epsilon_{t} \quad \text{mit} \quad \epsilon_{t}
    \overset{iid}{\sim} \mathcal{N}\left(0, \sigma_{X}^{2}\right)\\
    \end{MyAlign}
  \end{block}
  \begin{block}{B $(Y_{t})$}
    \begin{MyAlign}
      Y_{t} &=& \sigma_{Y,t}\: \eta_{t}, \\
      \sigma_{Y,t}^{2} &=& \omega + \alpha_{Y} Y_{t-1}^{2} + \beta_{Y}\sigma_{Y,t-1}^{2} \quad \text{mit} \quad  \sqrt{\frac{\nu}{\nu-2}}\cdot \eta_{t} \overset{iid}{\sim} t\left(\nu\right).\\
    \end{MyAlign}
  \end{block}

\end{frame}
\end{document}

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