在逐项环境中对齐方程式

Question

以下代码中的等式（=）应该对齐：

\documentclass[
    pdftex,a4paper,11pt,oneside,fleqn,
    bibliography=totoc,listof=totoc,
    headlines=2.1,headsepline,
    numbers=noenddot
]{scrreprt}

%%%----- Mathe ----------------------------------
\usepackage{amsmath,amsfonts,amssymb,bm}
\usepackage[squaren,textstyle]{SIunits}
\usepackage{icomma}
    
\usepackage{mathtools}
\usepackage[makeroom]{cancel}
                    
\begin{document}

\begin{itemize}
    \item {\textbf{P-Regler:} }
    $\begin{aligned}[t]
    \quad F_{\mathrm{R}}(s) = K_{\mathrm{P}}
    \end{aligned}$
    \item {\textbf{I-Regler: }}
    $\begin{aligned}[t]
    \quad F_{\mathrm{R}}(s) = \dfrac{K_{\mathrm{I}}}{s}
    \end{aligned}$
    \item {\textbf{PI-Regler: }}
    $\begin{aligned}[t]
    \quad F_{\mathrm{R}}(s)= K_{\mathrm{P}} + \dfrac{K_{\mathrm{I}}}{s} = K_{\mathrm{P}} \cdot \dfrac{1 + T_{\mathrm{N}} \cdot s}{T_{\mathrm{N}} \cdot s} \quad \text{mit} \quad T_{\mathrm{N}} = \dfrac{K_{\mathrm{P}}}{K_{\mathrm{I}}}
    \end{aligned}$
\end{itemize}

\end{document}

目前结果：

标志=应该对齐。

Answer 1

\documentclass[
a4paper,11pt,oneside,fleqn,
bibliography=totoc,listof=totoc,
headlines=2.1,headsepline,
numbers=noenddot
]{scrreprt}

%%%----- Mathe ----------------------------------
\usepackage{amsfonts,amssymb,bm}
\usepackage[squaren,textstyle]{SIunits}
\usepackage{icomma}

\usepackage{mathtools}
\usepackage[makeroom]{cancel}

\begin{document}
    
    \begin{itemize}
        \item \makebox[5em][l]{\textbf{P-Regler: }}
        $\begin{aligned}[t]
            \quad F_{\mathrm{R}}(s) = K_{\mathrm{P}}
        \end{aligned}$
        \item \makebox[5em][l]{\textbf{I-Regler: }}
        $\begin{aligned}[t]
            \quad F_{\mathrm{R}}(s) = \dfrac{K_{\mathrm{I}}}{s}
        \end{aligned}$
        \item \makebox[5em][l]{\textbf{PI-Regler: }}
        $\begin{aligned}[t]
            \quad F_{\mathrm{R}}(s)= K_{\mathrm{P}} + \dfrac{K_{\mathrm{I}}}{s} = K_{\mathrm{P}} \cdot \dfrac{1 + T_{\mathrm{N}} \cdot s}{T_{\mathrm{N}} \cdot s} \quad \text{mit} \quad T_{\mathrm{N}} = \dfrac{K_{\mathrm{P}}}{K_{\mathrm{I}}}
        \end{aligned}$
    \end{itemize}
    
\end{document}

顺便说一句：你不需要类选项 pdftex，而是aligned可以使用简单的数学表达式$...$

\begin{itemize}
    \item \makebox[5em][l]{\textbf{P-Regler: }}
          \quad$ F_{\mathrm{R}}(s) = K_{\mathrm{P}}$
    \item \makebox[5em][l]{\textbf{I-Regler: }}
           \quad$F_{\mathrm{R}}(s) = \dfrac{K_{\mathrm{I}}}{s}$
    \item \makebox[5em][l]{\textbf{PI-Regler: }}
           \quad$F_{\mathrm{R}}(s)= K_{\mathrm{P}} + \dfrac{K_{\mathrm{I}}}{s} = K_{\mathrm{P}} \cdot \dfrac{1 + T_{\mathrm{N}} \cdot s}{T_{\mathrm{N}} \cdot s} \quad \text{mit} \quad T_{\mathrm{N}} = \dfrac{K_{\mathrm{P}}}{K_{\mathrm{I}}}$
\end{itemize}

Answer 2

使用这个array包你将得到方程的精确对齐：

\documentclass[
    pdftex,a4paper,11pt,oneside,fleqn,
    bibliography=totoc,listof=totoc,
    headlines=2.1,headsepline,
    numbers=noenddot
                ]{scrreprt}

%%%----- Mathe ----------------------------------
\usepackage{mathtools}          % it supersede amsmath
\usepackage{amssymb,bm}
\usepackage{siunitx}    % SIunits is obsolete

\begin{document}
\[\renewcommand\arraystretch{2}
\begin{array}{>{\bullet} l >{$\bfseries}l<{$} r @{\;} l}
    & P-Regler: 
        & F_{\mathrm{R}}(s) & = K_{\mathrm{P}}              \\
    & I-Regler: 
        & F_{\mathrm{R}}(s) & = \dfrac{K_{\mathrm{I}}}{s}   \\
    & PI-Regler:
        & F_{\mathrm{R}}(s) & = K_{\mathrm{P}} + \dfrac{K_{\mathrm{I}}}{s}  
                              = \dfrac{1 + T_{\mathrm{N}}}{T_{\mathrm{N}}}
                                \quad \text{mit} \quad
                                T_{\mathrm{N}} = \dfrac{K_{\mathrm{P}}}{K_{\mathrm{I}}}
\end{array}
\]
\end{document}