![如何将一个方程分成两行或三行?](https://linux22.com/image/355435/%E5%A6%82%E4%BD%95%E5%B0%86%E4%B8%80%E4%B8%AA%E6%96%B9%E7%A8%8B%E5%88%86%E6%88%90%E4%B8%A4%E8%A1%8C%E6%88%96%E4%B8%89%E8%A1%8C%EF%BC%9F.png)
我有以下等式
\begin{equation}\label{eq:36}
\emph{$SINR^{D,UL}_{k,n}$} = \frac{p_{k,n} |h_{k,n}|^2}{\sum_{m=1}^{M} \omega_{n,m}p_{m,n}|h^{C}_{m,n}|^2 + \sum_{c^{'}=1}^{C^{'}} \sum_{m^{'}}^{M^{'}} \omega^{'}_{n,m^{'},c^{'}}p_{m^{'},n}|h^{C}_{m^{'},n,k,c^{'}}|^2 + \sum_{c^{'}=1}^{C^{'}} \sum_{k^{'}}^{K^{'}} \zeta^{'}_{n,k^{'},c^{'}}p_{k^{'},n}|h^{D}_{k^{'},n,k,c^{'}}|^2 + N_{0}B}
\end{equation}
该方程式不太适合用一行来表达。我怎样才能将其分成两行或三行?请参阅我的图表示例。
多谢。
答案1
也许使用multline
?
\documentclass{article}
\usepackage{amsmath}
\newcommand\abs[1]{\lvert#1\rvert}
\begin{document}
\begin{multline}
\label{eq:36}
\mathrm{SINR}^{D,UL}_{k,n} = p_{k,n} \abs{h_{k,n}}^2
\biggl(
\sum_{m=1}^{M} \omega_{n,m} p_{m,n} \abs{h^{C}_{m,n}}^2 \\
+ \sum_{c'=1}^{C'} \sum_{m'}^{M'} \omega'_{n,m',c'} p_{m',n} \abs{h^{C}_{m',n,k,c'}}^2 \\
+ \sum_{c'=1}^{C'} \sum_{k'}^{K'} \zeta'_{n,k',c'} p_{k',n} \abs{h^{D}_{k',n,k,c'}}^2
+ N_{0}B
\biggr)^{\!-1}
\end{multline}
\end{document}
或者引入别名:
\documentclass{article}
\usepackage{amsmath}
\newcommand\abs[1]{\lvert#1\rvert}
\begin{document}
\begin{equation}
\label{eq:36}
\mathrm{SINR}^{D,UL}_{k,n} = \frac{p_{k,n} \abs{h_{k,n}}^2}{X + Y + Z + N_{0}B}
\end{equation}
where
\begin{align*}
X &= \sum_{m=1}^{M} \omega_{n,m} p_{m,n} \abs{h^{C}_{m,n}}^2 \\
Y &= \sum_{c'=1}^{C'} \sum_{m'}^{M'} \omega'_{n,m',c'} p_{m',n} \abs{h^{C}_{m',n,k,c'}}^2 \\
Z &= \sum_{c'=1}^{C'} \sum_{k'}^{K'} \zeta'_{n,k',c'} p_{k',n} \abs{h^{D}_{k',n,k,c'}}^2
\end{align*}
\end{document}
你也可以做出问题中所展示的令人讨厌的事情,但我不建议这样做。
\documentclass{article}
\usepackage{amsmath}
\newcommand\abs[1]{\lvert#1\rvert}
\begin{document}
\begin{equation}
\label{eq:36}
\mathrm{SINR}^{D,UL}_{k,n} = \frac{p_{k,n} \abs{h_{k,n}}^2}{
\begin{pmatrix}
\sum_{m=1}^{M} \omega_{n,m} p_{m,n} \abs{h^{C}_{m,n}}^2 + {} \\
\sum_{c'=1}^{C'} \sum_{m'}^{M'} \omega'_{n,m',c'} p_{m',n} \abs{h^{C}_{m',n,k,c'}}^2 + {} \\
\sum_{c'=1}^{C'} \sum_{k'}^{K'} \zeta'_{n,k',c'} p_{k',n} \abs{h^{D}_{k',n,k,c'}}^2 + N_{0}B
\end{pmatrix}
}
\end{equation}
\end{document}