1

I see there are plenty of questions posted here with similar topics, but I can't find a solution to my question. Below is my latex code for a few equations. How can I force text on the 2nd, 3rd (and so on) lines to appear right below the equals sign? Can I also ask latex to insert a newline at the appropriate place so as to make this possible? enter image description here

\documentclass{article}

\usepackage{amsmath}

\begin{document}

%41
\[
\begin{split}
\frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^R = \omega\langle
\Delta i_{fi1} \rangle_{1}^I +\frac{1}{L_{fi1}}
\Big[\frac{2}{\pi}\Big\{(V_{dcL0})_{0}sin(2 \pi D_{i})\\+(\langle
\Delta v_{dcL} \rangle_{0} sin(2 \pi D_{i}))+(V_{dcL} cos(2 \pi
D_{i})2 \pi \Delta d_{i}) \Big\}-R_{fi} \langle i_{fi}
\rangle_{1}^R-\langle v_{cfi} \rangle_{1}^R \Big]
\end{split}
\]

%42 

\[
\begin{split}
\frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^I = -\omega\langle
\Delta i_{fi1} \rangle_{1}^R +\frac{1}{L_{fi1}}
\Big[\frac{2}{\pi}\Big\{(V_{dcL0}cos(2 \pi D_{i}))+ \\(\langle \Delta
v_{dcL} \rangle_{0} cos(2 \pi D_{i}))-(V_{dcL} sin(2 \pi D_{i})2 \pi
\Delta d_{i}) \Big\}-R_{fi} \langle i_{fi} \rangle_{1}^I-\langle
v_{cfi} \rangle_{1}^I \Big]
\end{split}
\]

% 43
\[
\begin{split}
\frac{d}{dt}\langle \Delta v_{Ci}\rangle_1^IR= \omega\langle\Delta v_{Ci}\rangle_1^I +\frac{1}{C_{fi}}\Big[ \langle\Delta i_{fi1}\rangle_1^R - \frac{\langle\Delta v_{Ci}\rangle_1^R}{R_L1}\Big]
\end{split}
\]


% 44
\[
\frac{d}{dt}\langle \Delta v_{Ci} \rangle_1^I = -\omega\langle\Delta v_{Ci}\rangle_1^R +\frac{1}{C_{fi}}\Big[ \langle\Delta i_{fi1}\rangle_1^I - \frac{\langle\Delta v_{Ci}\rangle_1^I}{R_L1}\Big]
\]

% 45 
\[
\frac{d}{dt}\Big[\Delta X_{SST}\Big]=[A_{SST}][\Delta X_{SST}]+[B_{SST}][\Delta d_{SST}]
+[C_{SST}][\Delta U_{SST}]+[D_{SST}]
\]

%46 
\[
\Big[\Delta X_{SST}\Big]=\Big[X_{rectifier} | X_{DAB} | X_{inverter} \Big]$$ where
\small\small $$\  X_{rectifier} = \Big[\Delta \langle   i_{g}  \rangle  _{1}^R\hspace{0.15cm} \Delta \langle   i_{g}  \rangle  _{1}^I\hspace{0.15cm} \Delta \langle  i_{fr} \rangle  _{1}^R\hspace{0.15cm} \Delta \langle  i_{fr} \rangle  _{1}^I\hspace{0.15cm} \Delta \langle  V_{cfr} \rangle  _{1}^R\hspace{0.15cm} \Delta \langle  V_{cfr} \rangle  _{1}^I\hspace{0.15cm} \Delta \langle  V_{dcH} \rangle  _{0} \Big])\normalsize$$
$$\ X_{DAB} = \Big[\Delta \langle  i_{tp} \rangle  _{1}^R\hspace{0.3cm}   \Delta \langle  i_{tp} \rangle  _{1}^I\hspace{0.3cm}  \Delta \langle  V_{dcL} \rangle  _{0}\Big]$$ and 
$$\ X_{inverter}= \Big[\Delta \langle  i_{fi1} \rangle  _{1}^R\hspace{0.3cm} \Delta \langle  i_{fr1} \rangle  _{1}^I\hspace{0.3cm} \Delta \langle  V_{cfi1} \rangle  _{1}^R\hspace{0.3cm} \Delta \langle  V_{cfi1} \rangle  _{1}^I\Big]
\]




\[
\Big[\small \Delta d_{SST}\Big]=\Big[\Delta d_{r}\hspace{0.3cm}\Delta d_{D1}\hspace{0.3cm}\Delta d_{D2}\hspace{0.3cm}\Delta d_{i}\Big]\eqno{\hbox{(47)}}
\]

\end{document}
2

I need quit a lot a time to rearrange your MWE ... I strongly suggest you that even in editor you wrote equation on the way, that it is easy find 0 and end of lines (\\) i.e. begining of new line of equation. Beside this for trigonometric function I suggest to use \sin or \cos instead of sin and cos.

For aligning equation at some point, for example at equal sign, you should use ampersand (&).

I merge all your equations in one system. for lines, where you not like to have equation numbers, I put \notag before \\. Beside align I also use multlined environment for broke very long equations in more lines. Result is the following:

enter image description here

Code:

\documentclass{article}
    \usepackage{mathtools}

\begin{document}
%41
    \begin{align}
\frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^R 
    & = \begin{multlined}[t]
    \omega\langle \Delta i_{fi1}\rangle_{1}^I +\frac{1}{L_{fi1}}
    \Big[\frac{2}{\pi}\Big\{(V_{dcL0})_{0}\sin(2\pi D_{i})      \\           
    +(\langle\Delta v_{dcL} \rangle_{0}\sin(2 \pi D_{i}))
    +(V_{dcL} \cos(2\pi D_{i}) 2\pi\Delta d_{i}) \Big\} -R_{fi} 
      \langle i_{fi}\rangle_{1}^R-\langle v_{cfi} \rangle_{1}^R \Big]    
      \end{multlined}                \notag \\
%42
\frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^I 
    & = \begin{multlined}[t]
    -\omega\langle\Delta i_{fi1} \rangle_{1}^R + \frac{1}{L_{fi1}}q \\ 
        (\langle \Delta v_{dcL} \rangle_{0}\cos(2 \pi D_{i}))                
    -(V_{dcL}\sin(2 \pi D_{i})2 \pi\Delta d_{i}) \Big\}                       
        -R_{fi} \langle i_{fi} \rangle_{1}^I-\langle
        v_{cfi} \rangle_{1}^I \Big]                                
        \end{multlined}              \notag \\
% 43
\frac{d}{dt}\langle \Delta v_{Ci}\rangle_1^IR
    & = \omega\langle\Delta v_{Ci}\rangle_1^I +\frac{1}{C_{fi}}
        \Big[ \langle\Delta i_{fi1}\rangle_1^R - \frac{\langle\Delta v_{Ci}\rangle_1^R}{R_L1}\Big]                                   \notag \\
% 44
\frac{d}{dt}\langle \Delta v_{Ci} \rangle_1^I 
    & = -\omega\langle\Delta v_{Ci}\rangle_1^R +\frac{1}{C_{fi}}
        \Big[ \langle\Delta i_{fi1}\rangle_1^I - \frac{\langle\Delta v_{Ci}\rangle_1^I}{R_L1}\Big]                                   \notag \\
% 45
\frac{d}{dt}\Big[\Delta X_{SST}\Big]
    & =[A_{SST}][\Delta X_{SST}]+[B_{SST}][\Delta d_{SST}] +[C_{SST}]
        [\Delta U_{SST}]+[D_{SST}]                                      \notag \\
%46
\Big[\Delta X_{SST}\Big]
    &   =\Big[X_{rectifier} | X_{DAB} | X_{inverter}\Big]               \notag 
\intertext{where}
X_{rectifier} 
    &   = \Big[\Delta\langle i_{g} \rangle_{1}^R\hspace{0.15cm}
        \Delta\langle i_{g}  \rangle_{1}^I\
        \Delta\langle i_{fr} \rangle_{1}^R\ 
        \Delta\langle i_{fr} \rangle_{1}^I\ 
        \Delta\langle V_{cfr}\rangle_{1}^R\ 
        \Delta\langle V_{cfr}\rangle_{1}^I\
        \Delta\langle V_{dcH}\rangle_{0}\Big]                           \notag \\
X_{DAB} 
    & = \Big[\Delta\langle i_{tp} \rangle_{1}^R\
             \Delta\langle i_{tp} \rangle_{1}^I\
             \Delta\langle V_{dcL}\rangle_{0}\Big]                      \notag \\
\intertext{and}
X_{inverter}
    & = \Big[\Delta\langle i_{fi1} \rangle_{1}^R\ 
             \Delta\langle i_{fr1} \rangle_{1}^I\
             \Delta\langle V_{cfi1}\rangle_{1}^R\ 
             \Delta\langle V_{cfi1}\rangle_{1}^I\Big]                   \notag \\
%
\Big[\small \Delta d_{SST}\Big]
    & =\Big[\Delta d_{r} \ \Delta d_{D1}\
            \Delta d_{D2}\ \Delta d_{i}\Big]
    \end{align}
\end{document}

Is this what you looking for?

  • You are awesome Zarko - thank you very much for your time and effort. – linuxfan Jun 15 '15 at 0:12
2

I usually use the align environment. It works pretty much the same way as tables, for example

\begin{align}
f(x) &= |x|^2\\
&= x\bar x
\end{align}

All n-th & signs of each line will be placed below each other, and \\ is the line break.

If you don't want numbering, you can usealign*

2

Here is a solution, based on align*, alignat* and flalign*. I grouped some equations in the same environment, which may be debatable, and decided to put the second part of split equations at end of line. I also systematically used the bmatrix environment, setting a proper value of arraycolsep, rather than repetitively inserting the same length manually:

\documentclass{article}
\usepackage{geometry}
\usepackage{mathtools, array}
\usepackage{showframe}

\begin{document}

%41
\begin{alignat*}{2}
  \frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^R & = & \omega \langle\Delta i_{fi1} \rangle_{1}^I + \frac{1}{L_{fi1}} \biggl[\frac{2}{\pi} & \Bigl\{(V_{dcL0})_{0}\sin(2 \pi D_{i})+ (\langle \Delta v_{dcL} \rangle_{0} \sin(2 \pi D_{i})) \\[-1.5ex]
                                                   & & & +(V_{dcL} \cos(2 \pi D_{i})2 \pi \Delta d_{i}) \Bigr\}-R_{fi} \langle i_{fi}
  \rangle_{1}^R-\langle v_{cfi} \rangle_{1}^R \biggr]\\
  \frac{d}{dt}\langle \Delta i_{fi1} \rangle_{1}^I & ={} & -\omega \langle \Delta i_{fi1} \rangle_{1}^R + \frac{1}{L_{fi1}} \biggl[\frac{2}{\pi} & \Bigl\{(V_{dcL0}\cos(2 \pi D_{i}))+(\langle \Delta v_{dcL} \rangle_{0} \cos(2 \pi D_{i})) \\[-1.5ex]
                                                   & & & -(V_{dcL} \sin(2 \pi D_{i})2 \pi \Delta d_{i}) \Bigr\}-R_{fi} \langle i_{fi} \rangle_{1}^I-\langle v_{cfi} \rangle_{1}^I \biggr]
\end{alignat*}
% 43
\begin{align*}
  \frac{d}{dt}\langle \Delta v_{Ci}\rangle_1^IR & = \omega\langle\Delta v_{Ci}\rangle_1^I +\frac{1}{C_{fi}}\biggl[ \langle\Delta i_{fi1}\rangle_1^R - \frac{\langle\Delta v_{Ci}\rangle_1^R}{R_L1}\biggr] \\
  % 44
  \frac{d}{dt}\langle \Delta v_{Ci} \rangle_1^I & = -\omega\langle\Delta v_{Ci}\rangle_1^R +\frac{1}{C_{fi}}\biggl[ \langle\Delta i_{fi1}\rangle_1^I - \frac{\langle\Delta v_{Ci}\rangle_1^I}{R_L1}\biggr] \\
  % 45
  \frac{d}{dt}\bigl[\Delta X_{SST}\bigr] & =[A_{SST}][\Delta X_{SST}]+[B_{SST}][\Delta d_{SST}]
  +[C_{SST}][\Delta U_{SST}]+[D_{SST}]
\end{align*}
%46
\begin{flalign*}
  & & \Big[\Delta X_{SST}\Big] & =\Bigl[X_\mathrm{rectifier}\mid X_{DAB}\mid X_\mathrm{inverter} \Bigr] & & \\[1ex]
    &
    \mathrlap{\setlength\arraycolsep{4pt}\renewcommand\arraystretch{1.5}
      \begin{array}[t]{@{}l@{\quad}l}
        \text{where} & X_\mathrm{rectifier} =
        \begin{bmatrix}\Delta \langle i_{g} \rangle _{1}^R & \Delta \langle i_{g} \rangle _{1}^I & \Delta \langle i_{fr} \rangle _{1}^R & \Delta \langle i_{fr} \rangle _{1}^I & \Delta \langle V_{cfr} \rangle _{1}^R & \Delta \langle V_{cfr} \rangle _{1}^I & \Delta \langle V_{dcH} \rangle _{0}
        \end{bmatrix}\\
        & X_{DAB} = \begin{bmatrix}\Delta \langle i_{tp} \rangle _{1}^R & \Delta \langle i_{tp} \rangle _{1}^I & \Delta \langle V_{dcL} \rangle _{0}\end{bmatrix} \\[2ex]
        \text{and} & X_\mathrm{inverter}= \begin{bmatrix}\Delta \langle i_{fi1} \rangle _{1}^R & \Delta \langle i_{fr1} \rangle _{1}^I & \Delta \langle V_{cfi1} \rangle _{1}^R & \Delta \langle V_{cfi1} \rangle _{1}^I \end{bmatrix}
      \end{array}} & &
  \end{flalign*}

  \[
    \begin{bmatrix} \Delta d_{SST}\Big]=\Big[\Delta d_{r} & \Delta d_{D1} & \Delta d_{D2} & \Delta d_{i}\end{bmatrix}\eqno{\hbox{(47)}}
  \]

\end{document} 

enter image description here

  • Bernard, thanks for the solution. I continue to be amazed at how much enthusiasm latex lovers have to solve other's problems. Cheers! – linuxfan Jun 15 '15 at 4:08

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