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```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 回答
2
那是换行符带来的空格。把
```tex
\end{small}
```
改成
```tex
\end{small}%
```
即可。
-
好的谢谢，问题已解决 – HanJi 2019-12-01 14:02 回复
0
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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