![子方程环境中的换行](https://linux22.com/image/360167/%E5%AD%90%E6%96%B9%E7%A8%8B%E7%8E%AF%E5%A2%83%E4%B8%AD%E7%9A%84%E6%8D%A2%E8%A1%8C.png)
我在子方程中使用对齐,这对于正常长度的方程很有效。但是我想将这个长方程分成两部分或三部分。
目前我使用 \\ 来拆分长公式,我的文档可以编译,但结果并不理想。我见过很多建议(多行、拆分、对齐、对齐、breqn),但我似乎被众多可能的建议淹没了。
这就是我现在所得到的,但正如你所见,等式 (1a) 看起来很糟糕。(编辑:像这样的东西会很理想https://i.stack.imgur.com/7YOU7.png)
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{align}
Z^0(e^{j\omega h^0}) &= P^0_{11}(e^{j\omega h^0})\Lambda^0(e^{j\omega h^0})+P^0_{12}(e^{j\omega h^0})\mathcal{I}_{zoh}(e^{j\omega h^0})Q(e^{j\omega h})\cdot \\ \frac{1}{F}\sum_{f=0}^{F-1}P^0_{12}(e^{j\omega h^0-\frac{f}{F}\omega^0_s})\Lambda^0(e^{j\omega h^0-\frac{f}{F}\omega^0_s}) \nonumber
\intertext{where}
Q(\omega) &= (I-C(\omega)P_{22}(\omega))^{-1}C(\omega)
\end{align}
\end{subequations}
\end{document}
答案1
我认为没有必要将两个等式中的等号对齐,因此我也提供了一个无需对齐的解决方案。
我认为最好的技巧是使用split
:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{align}
\begin{split}
Z^0(e^{j\omega h^0})
&= P^0_{11}(e^{j\omega h^0})\Lambda^0(e^{j\omega h^0}) \\
&\qquad+ P^0_{12}(e^{j\omega h^0})\mathcal{I}_{zoh}(e^{j\omega h^0})Q(e^{j\omega h})\\
&\qquad\times \frac{1}{F}\sum_{f=0}^{F-1}P^0_{12}
(e^{j\omega h^0-\frac{f}{F}\omega^0_s})\Lambda^0(e^{j\omega h^0-\frac{f}{F}\omega^0_s})
\end{split}
\intertext{where}
Q(\omega) &= (I-C(\omega)P_{22}(\omega))^{-1}C(\omega)
\end{align}
\end{subequations}
\begin{subequations}
\begin{equation}
\begin{split}
Z^0(e^{j\omega h^0})
&= P^0_{11}(e^{j\omega h^0})\Lambda^0(e^{j\omega h^0}) \\
&\qquad+ P^0_{12}(e^{j\omega h^0})\mathcal{I}_{zoh}(e^{j\omega h^0})Q(e^{j\omega h})\\
&\qquad\times \frac{1}{F}\sum_{f=0}^{F-1}P^0_{12}
(e^{j\omega h^0-\frac{f}{F}\omega^0_s})\Lambda^0(e^{j\omega h^0-\frac{f}{F}\omega^0_s})
\end{split}
\end{equation}
where
\begin{equation}
Q(\omega) = (I-C(\omega)P_{22}(\omega))^{-1}C(\omega)
\end{equation}
\end{subequations}
\end{document}
答案2
另外两种可能性:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{subequations}
\begin{align}
Z^0(e^{j\omega h^0}) &= \begin{aligned}[t]P^0_{11}(e^{j\omega h^0})\Lambda^0(e^{j\omega h^0})+P^0_{12}(e^{j\omega h^0})\mathcal{I}_{zoh}(e^{j\omega h^0})Q(e^{j\omega h}) \\
{}\times\frac{1}{F}\sum_{f=0}^{F-1}P^0_{12}(e^{j\omega h^0-\frac{f}{F}\omega^0_s})\Lambda^0(e^{j\omega h^0-\frac{f}{F}\omega^0_s}) \end{aligned}\\
\shortintertext{where}
Q(\omega) &= (I-C(\omega)P_{22}(\omega))^{-1}C(\omega)
\end{align}
\end{subequations}
\begin{subequations}
\begin{align}
Z^0(e^{j\omega h^0}) &= \begin{aligned}[t]P^0_{11}(e^{j\omega h^0})\Lambda^0(e^{j\omega h^0})+P^0_{12}(e^{j\omega h^0})\mathcal{I}_{zoh}(e^{j\omega h^0})Q(e^{j\omega h}) \\
{}\times\frac{1}{F}\sum_{f=0}^{F-1}P^0_{12}(e^{j\omega h^0-\frac{f}{F}\omega^0_s})\Lambda^0(e^{j\omega h^0-\frac{f}{F}\omega^0_s}) \end{aligned}\\
\text{where\quad}
Q(\omega)&= (I-C(\omega)P_{22}(\omega))^{-1}C(\omega)
\end{align}
\end{subequations}
\end{document}