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如何改变乳胶中的子单元格高度？

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如何改变乳胶中的子单元格高度？

我想用 LaTeX 制作一张和图片一样的表格。

在此处输入图片描述

我已经使用过以下代码：

\begin{table}[h!]
    \begin{tabular}{|c|c|c|c|c|}\hline
        Class & Format & Constant Value & Slope Value & Slope at \\ \hline
Gain  &    K &$ 0^0 if K > 0$ &        0    &  -$\infty$ \\ \cline{3-3}& &$ 180^0 if K > 0$ & & \\ \hline
        
\makecell{Zero at \\ origin}&    $s^{n_0}$   & $90 n_0$ & -    &  -$\infty$ \\ \hline 
        
\makecell{Simple Zero \\ at  $z_i$}&     $\left(1 + \frac{s}{z_i}\right)^{n_1}$   & - & $45^o n_1$    &  $0.1 \times z_i$ \\ \cline{4-5} & & & $-45^o n_1$    &  $10 \times z_i$ \\ 
 \hline
\makecell{Quatrature \\ Zero at $z_j$}&     $\left( 1 + 2\zeta\frac{s}{z_j} + \left(\frac{s}{z_j}   \right)^2 \right)^{n_2}$   & - & $90^o n_2$    &  $0.1 \times z_j$ \\ \cline{4-5} & & & $-90^o n_2$    &  $10 \times z_j$ \\ \hline
        
\makecell{Pole at \\ origin}&    $1/s^{m_0}$   & $-90 m_0$ & -    &  -$\infty$ \\ \hline 
\makecell{Simple Pole \\ at  $p_i$}&     $\frac{1}{\left(1 + \frac{s}{p_i}\right)^{m_1}}$   & - &  $-45^o m_1$    &  $0.1 \times p_i$ \\ \cline{4-5} & & & $+45^o m_1$    &  $10 \times p_i$ \\ \hline
\makecell{Quatrature  \\ Pole at $p_j$}&     $\frac{1}{\left( 1 + 2\zeta\frac{s}{p_j} + \left(\frac{s}{p_j} \right)^2 \right)^{m_2}}$   & - &$-90^o m_2$    &  $0.1 \times p_j$ \\ \cline{4-5} & & & $+90^o m_2$    &  $10 \times p_j$  \\ \hline
    \end{tabular}
\end{table}

4-5但是，如您所见，行和行中的子单元格高度不一样7-8。我该如何匹配子单元格高度？

答案1

以下内容可以帮助您入门。请相应调整表格的其余部分：

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\usepackage{geometry}
\usepackage{makecell}
\renewcommand{\theadfont}{\normalsize}
\usepackage[column=0]{cellspace}
\setlength{\cellspacetoplimit}{\tabcolsep}
\setlength{\cellspacebottomlimit}{\tabcolsep}
\begin{document}

\begin{table}[h!]
\centering
    \begin{tabular}{|0c|>{\(}c<{\)}|>{\(}c<{\)}|@{}>{\(}c<{\)}@{}|@{}>{\(}c<{\)}@{}|}\hline
        \thead{Class} 
          & \thead{\text{Format}} 
            & \thead{\text{Constant}\\ \text{Value}} 
              & \thead{\text{Slope}\\ \text{Value}} 
                & \thead{\text{Slope at}} \\ \hline

\makecell{Zero at \\ origin}
  & s^{n_0}   
    & 90 n_0 
      & - 
        &  -\infty \\ 
\hline 
        
\makecell{Simple Zero \\ at  $z_i$}
  & \left(1 + \frac{s}{z_i}\right)^{n_1}   
    & - 
      & \begin{tabular}{>{\(}0c<{\)}} 45^o n_1 \\ \hline -45^o n_1 \end{tabular}    
        &  \begin{tabular}{>{\(}0c<{\)}} 0.1 \times z_i \\ \hline  10 \times z_i \end{tabular}\\ 
\hline
    \end{tabular}
\end{table}

\end{document}

答案2

我建议您 (a) 删除所有垂直规则，(b) 使用更少但间隔良好的水平规则（通过使用包的宏booktabs）和 (c) 使用更多\makecell指令将某些行的最后两列结果分组。或者，使用tabular*环境而不是tabular环境来确保表格适合文本块。并且，请将^0和的实例替换^o为^{\circ}。

在此处输入图片描述

\documentclass{article}
\usepackage{makecell}
\usepacakge{amsmath} % for \text macro
\usepackage{booktabs}% for \toprule, \midrule, \bottomrule, and \addlinespace macros
\begin{document}

\begin{table}[h!]
\setlength\tabcolsep{0pt} % let LaTeX figure out intercolumn whitespace amounts
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} l cccc}
\toprule
Class & Format & Constant Value & Slope Value & Slope at \\ 
\midrule
Gain  & 
  $K$ & 
  \makecell{\text{$\phantom{00}0^{\circ}$ if $K>0$}\\[0.25ex]
            \text{$180^{\circ}$ if $K>0$}} & 
  0 &  $-\infty$ \\ 
\addlinespace \addlinespace

\makecell[l]{Zero at \\ origin}&
  $s^{n_0}$ & $90 n_0$ & -- & $-\infty$ \\ 
\addlinespace
\makecell[l]{Simple Zero \\ at $z_i$} & 
  $\bigl(1+\frac{s}{z_i} \bigr)^{n_1}$ & 
  -- & 
  \makecell{$\phantom{-}45^{\circ}n_1$\\[0.25ex] $-45^{\circ}n_1$} &  
  \makecell{$0.1\cdot z_i$\\[0.25ex] $10 \cdot z_i$} \\ 
\addlinespace
\makecell[l]{Quatrature \\ Zero at $z_j$}& 
  $\bigl( 1 + 2\zeta\frac{s}{z_j} + (\frac{s}{z_j})^2 \bigr)^{n_2}$ 
  & -- & 
  \makecell{$\phantom{-}90^{\circ}n_2$\\[0.25ex] $-90^{\circ}n_2$} &
  \makecell{$0.1\cdot z_j$\\[0.25ex] $10 \cdot z_j$} \\ 
\addlinespace\addlinespace
    
\makecell[l]{Pole at \\ origin}&  $1/s^{m_0}$ & $-90 m_0$ & --  &  $-\infty$ \\ 
\addlinespace 
\makecell[l]{Simple Pole \\ at  $p_i$} &   
  $1\big/\bigl(1 + \frac{s}{p_i}\bigr)^{m_1}$ & 
  -- &  
  \makecell{$-45^{\circ}m_1$\\[0.25ex] $+45^{\circ} m_1$} &
  \makecell{$0.1 \cdot p_i$\\[0.25ex] $10 \cdot p_i$} \\ 
\addlinespace
\makecell[l]{Quatrature \\ Pole at $p_j$} &   
  $1\big/\bigl( 1 + 2\zeta\frac{s}{p_j} + (\frac{s}{z_j})^2 \bigr)^{m_2}$ & 
  -- &
  \makecell{$-90^{\circ} m_2$\\[0.25ex] $+90^{\circ} m_2$} &  
  \makecell{$0.1 \cdot p_j$ \\[0.25ex] $10 \cdot p_j$}  \\ 
\bottomrule
\end{tabular*}
\end{table}
\end{document}

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