公式使用small命令后,段后的首行出现一个小空格怎么去除???

2019-12-01 02:11发布

```tex \documentclass[10pt,twocolumn,twoside]{IEEEtran} \usepackage[dvips]{epsfig} \usepackage[dvips...

```tex \documentclass[10pt,twocolumn,twoside]{IEEEtran} \usepackage[dvips]{epsfig} \usepackage[dvips]{graphicx} \usepackage{amsmath,amsfonts,bm,amssymb,psfrag,ifthen,color,subfigure} \usepackage{algpseudocode} \usepackage{algorithmicx} \usepackage{setspace} \usepackage{cite} \usepackage{url} \usepackage{longtable} \usepackage{stfloats} % Written by Sigitas Tolusis \usepackage{graphics,booktabs,color,subfigure} \usepackage{float} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{intent} \begin{document} \begin{small} \begin{equation} \begin{split} &{f_{{\alpha_1}x_{{\rm{THO}}}^{(1)}}}(w)\otimes{f_{{\alpha_2}x_{{\rm{THO}}}^{(2)}}}(w)\!=\!\int_{-\infty}^\infty{{f_{{\alpha _1}x_{{\rm{THO}}}^{(1)}}}(l)}{f_{{\alpha_2}x_{{\rm{THO}}}^{(2)}}}(w-l)dl\\ &= \int_{-\infty }^\infty{\left[ {\frac{1}{{\sqrt {2\pi}{\alpha_1}{\sigma_1}}}\exp \left({\frac{{-{l^2}}}{{2\alpha_1^2\sigma_1^2}}}\right)u[{\mathop{\rm{sgn}}}({\alpha_1})l]+\frac{1}{2}\delta(l)}\right]}\times\\ &\left[{\frac{1}{{\sqrt{2\pi}{\alpha_2}{\sigma_2}}}\exp\left({\frac{{-{{(w-l)}^2}}}{{2\alpha_2^2\sigma_2^2}}}\right)u[{\mathop{\rm sgn}} ({\alpha_2})(w-l)]+\frac{1}{2}\delta(w-l)}\right]dl. \end{split} \end{equation} \end{small} where the function $sgn(.)$ represents the signum function. To simplify the derivation process, ${\alpha_1}$ and ${\alpha_2}$ are set to positive in the following. \end{document} ```
2条回答
仅仅从你的问题来看,导言区有很多包都是不需要的。 ```tex \documentclass[10pt,twocolumn,twoside]{IEEEtran} \usepackage{amsmath} \begin{document} \begin{small} \begin{equation} \begin{split} &{f_{{\alpha_1}x_{{\rm{THO}}}^{(1)}}}(w)\otimes{f_{{\alpha_2}x_{{\rm{THO}}}^{(2)}}}(w)\!=\!\int_{-\infty}^\infty{{f_{{\alpha _1}x_{{\rm{THO}}}^{(1)}}}(l)}{f_{{\alpha_2}x_{{\rm{THO}}}^{(2)}}}(w-l)dl\\ &= \int_{-\infty }^\infty{\left[ {\frac{1}{{\sqrt {2\pi}{\alpha_1}{\sigma_1}}}\exp \left({\frac{{-{l^2}}}{{2\alpha_1^2\sigma_1^2}}}\right)u[{\mathop{\rm{sgn}}}({\alpha_1})l]+\frac{1}{2}\delta(l)}\right]}\times\\ &\left[{\frac{1}{{\sqrt{2\pi}{\alpha_2}{\sigma_2}}}\exp\left({\frac{{-{{(w-l)}^2}}}{{2\alpha_2^2\sigma_2^2}}}\right)u[{\mathop{\rm sgn}} ({\alpha_2})(w-l)]+\frac{1}{2}\delta(w-l)}\right]dl. \end{split} \end{equation} \end{small}% where the function $sgn(.)$ represents the signum function. To simplify the derivation process, ${\alpha_1}$ and ${\alpha_2}$ are set to positive in the following. \end{document} ```

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