LaTeX Forum

\documentclass{article}
\usepackage{mathtools}
\renewcommand{\vec}[1]{\underline{#1}}
\begin{document}
\begin{align*} 
	\underline{T}\cdot \underline{U}_s&=
	\underline{Z}\cdot \underline{T}\cdot i_s\\ 
	\vec{T}\cdot \vec{U}_s&=
	\vec{Z}\cdot \vec{T}\cdot i_s 
\end{align*}
\end{document}

\documentclass[a4paper, 11pt]{scrartcl} 
\usepackage[european]{circuitikz} 
\usepackage{amsmath} 

\begin{document} 
\begin{align*} 
\underline{U}&=\underline{Z}\cdot I\\ 
\underline{T}\cdot \underline{U}_s&=\underline{Z}\cdot \underline{T}\cdot i_s\\ 
\underline{S}\cdot \underline{T}\cdot \underline{U}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}\cdot i_s\\ 
\underline{E}\cdot \underline{U}_s&=\underline{Z}\cdot \underline{E}\cdot i_s\\ 
\underline{Z}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}=\underline{E}\cdot\underline{Z} \\
\underline{Z}\cdot \underline{T}&= 
\begin{pmatrix} 
\underline{Z}_a & 0 & 0 \\ 
 0 & \underline{Z}_b & 0 \\ 
 0 & 0 & \underline{Z}_c \\ 
\end{pmatrix} 
\begin{pmatrix}
 1 & 1 & 1 \\ 
 1 & \underline{a}^2 & \underline{a} \\ 
 1 & \underline{a} & \underline{a}^2 
\end{pmatrix}
\hspace{1cm}\smash{\makebox[0pt][l]{%
\begin{circuitikz}[scale=.4, transform shape,baseline=(current bounding box.center)] 
\draw 
 (0,0) to[R, l=$Z_a$, i=$\underline{I}_a$, o-o] (4,0) 
 (0,-2) to[R, l=$Z_b$, i=$\underline{I}_b$, o-o] (4,-2) 
 (0,-4) to[R, l=$Z_c$, i=$\underline{I}_c$, o-o] (4,-4);
\end{circuitikz} }}
\\ 
&= 
\begin{pmatrix} 
\underline{Z}_a & \underline{Z}_a & \underline{Z}_a \\ 
\underline{Z}_b & \underline{a}^2\underline{Z}_b & \underline{a}\ \underline{Z}_b \\ 
\underline{Z}_c & \underline{a}\ \underline{Z}_c & \underline{a}^2\underline{Z}_c \\ 
\end{pmatrix}\\
\underline{Z}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}=\frac{1}{3} 
\begin{pmatrix} 
 1 & 1 & 1 \\ 
 1 & \underline{a} & \underline{a}^2 \\ 
 1 & \underline{a}^2 & \underline{a} 
\end{pmatrix} 
\begin{pmatrix} 
\underline{Z}_a & \underline{Z}_a & \underline{Z}_a \\ 
\underline{Z}_b & \underline{a}^2\underline{Z}_b & \underline{a}\ \underline{Z}_b \\ 
\underline{Z}_c & \underline{a}\ \underline{Z}_c & \underline{a}^2\underline{Z}_c \\ 
\end{pmatrix} \\ 
&= 
\frac{1}{3} 
\begin{pmatrix} 
 (\underline{Z}_a+\underline{Z}_b+\underline{Z}_c) & (\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) & (\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c) \\ 
\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c & \underline{Z}_a+\underline{Z}_b+\underline{Z}_c) & (\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) \\ 
\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) &  (\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c) & \underline{Z}_a+\underline{Z}_b+\underline{Z}_c) 
\end{pmatrix} 
\end{align*} 
\end{document}

\documentclass[a4paper, 11pt]{scrartcl}
\usepackage[european]{circuitikz}
\usepackage{amsmath}

\begin{document}
\begin{align*}
\underline{U}&=\underline{Z}\cdot I
\\
\underline{T}\cdot \underline{U}_s&=\underline{Z}\cdot \underline{T}\cdot i_s
\\
\underline{S}\cdot \underline{T}\cdot \underline{U}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}\cdot i_s
\\
\underline{E}\cdot \underline{U}_s&=\underline{Z}\cdot \underline{E}\cdot i_s
\\
\underline{Z}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}=\underline{E}\cdot \underline{Z}
\end{align*}
\centering\begin{tabular}{cc}
$\begin{aligned}
\underline{Z}\cdot \underline{T}&=
\begin{pmatrix}
\underline{Z}_a & 0 & 0 \\
0 & \underline{Z}_b & 0 \\
0 & 0 & \underline{Z}_c \\
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 1 \\
1 & \underline{a}^2 & \underline{a} \\
1 & \underline{a} & \underline{a}^2
\end{pmatrix} \\
&=
\begin{pmatrix}
\underline{Z}_a & \underline{Z}_a & \underline{Z}_a \\
\underline{Z}_b & \underline{a}^2\underline{Z}_b & \underline{a}\ \underline{Z}_b \\
\underline{Z}_c & \underline{a}\ \underline{Z}_c & \underline{a}^2\underline{Z}_c \\
\end{pmatrix}
\end{aligned}$
&
\begin{circuitikz}[scale=.4, transform shape]
\draw
(0,0) to[R, l=$Z_a$, i=$\underline{I}_a$, o-o] (4,0)
(0,-2) to[R, l=$Z_b$, i=$\underline{I}_b$, o-o] (4,-2)
(0,-4) to[R, l=$Z_c$, i=$\underline{I}_c$, o-o] (4,-4);
\end{circuitikz}
\end{tabular}
\begin{align*}\underline{Z}_s&=\underline{S}\cdot \underline{Z}\cdot \underline{T}=\frac{1}{3}
\begin{pmatrix}
1 & 1 & 1 \\
1 & \underline{a} & \underline{a}^2 \\
1 & \underline{a}^2 & \underline{a}
\end{pmatrix}
\begin{pmatrix}
\underline{Z}_a & \underline{Z}_a & \underline{Z}_a \\
\underline{Z}_b & \underline{a}^2\underline{Z}_b & \underline{a}\ \underline{Z}_b \\
\underline{Z}_c & \underline{a}\ \underline{Z}_c & \underline{a}^2\underline{Z}_c \\
\end{pmatrix} \\
&=
\frac{1}{3}
\begin{pmatrix}
(\underline{Z}_a+\underline{Z}_b+\underline{Z}_c) & (\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) & (\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c) \\
\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c & \underline{Z}_a+\underline{Z}_b+\underline{Z}_c) & (\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) \\
\underline{Z}_a+\underline{a}^2\underline{Z}_b+\underline{a}\ \underline{Z}_c) &  (\underline{Z}_a+\underline{a}\ \underline{Z}_b+\underline{a}^2\underline{Z}_c) & \underline{Z}_a+\underline{Z}_b+\underline{Z}_c)
\end{pmatrix}
\end{align*}
\end{document}