From cb482776c5891841efe3f77237620868a42c9f60 Mon Sep 17 00:00:00 2001 From: "[SilasElter]" <[SilasElter]> Date: Mon, 27 Jan 2025 14:03:23 +0100 Subject: [PATCH 1/6] Add changes Fig PWM ex05 --- exercise/fig/ex05/Fig_TEST.tex | 23 +++ exercise/fig/ex05/PWM_PFC_example.csv | 121 +++++++++++++++ exercise/fig/ex05/PWM_sawtooth_example.csv | 97 ++++++++++++ exercise/fig/ex05/PWM_triangular_example.csv | 150 +++++++++++++++++++ exercise/main.tex | 2 +- exercise/tex/exercise05.tex | 2 +- lecture/main.tex | 2 +- 7 files changed, 394 insertions(+), 3 deletions(-) create mode 100644 exercise/fig/ex05/Fig_TEST.tex create mode 100644 exercise/fig/ex05/PWM_PFC_example.csv create mode 100644 exercise/fig/ex05/PWM_sawtooth_example.csv create mode 100644 exercise/fig/ex05/PWM_triangular_example.csv diff --git a/exercise/fig/ex05/Fig_TEST.tex b/exercise/fig/ex05/Fig_TEST.tex new file mode 100644 index 0000000..a3c4f41 --- /dev/null +++ b/exercise/fig/ex05/Fig_TEST.tex @@ -0,0 +1,23 @@ +\begin{figure} + \begin{tikzpicture} + \pgfplotsset{table/search path={fig/ex05}} + \begin{groupplot}[group style={group size=1 by 3, xticklabels at = edge bottom, vertical sep=0.75cm}, height=0.375\textheight, width=0.875\textwidth, xmin=0, xmax=pi, grid,clip = false, ymin = -0.1, ymax =1.1, xtick = {0, pi/4, pi/2, 3*pi/4,pi}, xticklabels = {0,$\frac{1}{4}\pi$, $\frac{1}{2}\pi$,$\frac{3}{4}\pi$, $\pi$}, ytick = {-1, 0, 1}, yticklabels = {, 0, 1}] + + % % Top plot: duty cycle and carrier signal + \nextgroupplot[ylabel = {$d(t), c(t)$}, legend pos=south east, legend columns=2] + \addplot[signalred, thick] table[x=wt, y=d, col sep=comma] {PWM_PFC_example.csv}; + \addplot[signalblue, thick] table[x=wt, y=c, col sep=comma] {PWM_PFC_example.csv}; + \legend{$d(t)$, $c(t)$} + + % middle plot: switching signal + \nextgroupplot[ylabel = {$s(t)$}] + \addplot[signalblue, thick] table[x=wt, y=s, col sep=comma] {PWM_PFC_example.csv}; + + % bottom plot: current response + \nextgroupplot[ylabel = {$i_1(t)$}, xlabel={$\omega t$}, ytick = {-1, 0, 0.5, 1}, yticklabels = {}, legend columns=2] + \addplot[signalred, thick] table[x=wt, y=i1, col sep=comma] {PWM_PFC_example.csv}; + \addplot[thick, dashed] table[x=wt, y=i1ref, col sep=comma] {PWM_PFC_example.csv}; + \legend{$i_1(t)$, $i_1^{(1)}(t)$} + \end{groupplot} + \end{tikzpicture} +\end{figure} \ No newline at end of file diff --git a/exercise/fig/ex05/PWM_PFC_example.csv b/exercise/fig/ex05/PWM_PFC_example.csv new file mode 100644 index 0000000..9493721 --- /dev/null +++ b/exercise/fig/ex05/PWM_PFC_example.csv @@ -0,0 +1,121 @@ +wt, s, d, c, i1, i1ref, i1delta +1.000000000000000021e-03,1.000000000000000000e+00,1.095771190744057266e+00,9.872676045526483923e-01,3.804168234329677784e-05,9.356357241542493514e-04,4.737205200063761279e-02 +7.855569411307920802e-02,1.000000000000000000e+00,1.032472879864483994e+00,2.021620889167685675e-04,2.355289335359267780e-02,7.342395509232234330e-02,-1.601415696281104084e-03 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+3.080306446020112610e+00,1.000000000000000000e+00,9.804919942438587999e-01,9.754002878070622273e-01 +3.140592653589793226e+00,1.000000000000000000e+00,9.996816901138162459e-01,1.591549430919059205e-02 diff --git a/exercise/main.tex b/exercise/main.tex index d1418e6..09b7352 100644 --- a/exercise/main.tex +++ b/exercise/main.tex @@ -1,7 +1,7 @@ \documentclass[solution]{../course_template/exerciseClass} \title{Power Electronics} -\includeonly{tex/exercise07} +\includeonly{tex/exercise05} \begin{document} \include{tex/exercise01} diff --git a/exercise/tex/exercise05.tex b/exercise/tex/exercise05.tex index dadbbd2..00977bc 100644 --- a/exercise/tex/exercise05.tex +++ b/exercise/tex/exercise05.tex @@ -472,4 +472,4 @@ I_{\mathrm{D,1-4}} = \sqrt{\frac{\hat{i}^2_{\mathrm{1}}}{2\pi}\frac{\pi}{2}} = \frac{\hat{i}^2_{\mathrm{1}}}{2}. \end{equation} \end{solutionblock} -% \input{fig/ex05/Fig_Current_Courses.tex} +\input{fig/ex05/Fig_TEST.tex} diff --git a/lecture/main.tex b/lecture/main.tex index 7bb4bfa..b1191ec 100644 --- a/lecture/main.tex +++ b/lecture/main.tex @@ -5,7 +5,7 @@ \author{Oliver Wallscheid} \date{} -%\includeonly{tex/temp.tex} % build only selected sections +\includeonly{tex/Lecture06} % build only selected sections \begin{document} From c73321f11931c4f1b07869302776428fdf714b7b Mon Sep 17 00:00:00 2001 From: "[SilasElter]" <[SilasElter]> Date: Mon, 27 Jan 2025 16:01:16 +0100 Subject: [PATCH 2/6] Add changes ex07 task 2 Fig --- .../fig/ex07/Fig_Voltage_U_um_excerpt.tex | 2 +- .../ex07/Fig_graphic_solutions_cos_terms.tex | 30 +++++++++++-------- ...Fig_standardization_to_fudamental_freq.tex | 8 +++-- exercise/main.tex | 4 +-- 4 files changed, 27 insertions(+), 17 deletions(-) diff --git a/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex b/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex index 3091db1..39bd077 100644 --- a/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex +++ b/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex @@ -11,7 +11,7 @@ enlargelimits, axis line style={->}, % Pfeilspitzen an den Achsen xlabel={$\omega t$}, - ylabel={$U_\mathrm{d}$}, + ylabel={$U_\mathrm{1}$}, xmin=0, xmax=13/6*pi, ymin=-1, ymax=1, xtick={0, pi/3, 2*pi/3, pi, 4*pi/3, 5*pi/3, 2*pi}, diff --git a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex index 083c8b7..25c591f 100644 --- a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex +++ b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex @@ -11,23 +11,29 @@ axis lines=middle, xlabel={$\cos(\varphi)$}, ylabel={$\sin(\varphi)$}, + xlabel style={xshift=0.5cm}, width=7cm, height=7cm, major grid style={line width=.2pt,draw=gray!50}, minor grid style={line width=.1pt,draw=gray!20}, xmin=-1.5, xmax=1.5, ymin=-1.5, ymax=1.5, - ] - - \draw[thick, ->] (0,0) -- ({cos(40)}, {sin(40)}) node[above right] {}; - \node at ({0.5*cos(40)}, {0.5*sin(40)}) [below] {$\varphi$}; - - \addplot[ - domain=0:360, - samples=200, - thick, - color=blue, - ] - ({cos(x)}, {sin(x)}); + xtick={-1, 0, 1}, % Manuelle Ticks auf der x-Achse + ytick={-1, 0, 1}, % Manuelle Ticks auf der y-Achse + tick label style={xshift=5pt, yshift=5pt}, % Verschiebt die Beschriftungen nach außen + ] + + % Vektor einzeichnen + \draw[thick, ->] (0,0) -- ({cos(40)}, {sin(40)}) node[above right] {}; + \node at ({0.5*cos(40)}, {0.5*sin(40)}) [below] {$\varphi$}; + + % Kreis zeichnen + \addplot[ + domain=0:360, + samples=200, + thick, + color=blue, + ] + ({cos(x)}, {sin(x)}); \end{axis} \end{tikzpicture} \end{minipage} diff --git a/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex b/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex index 4060230..9d69392 100644 --- a/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex +++ b/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex @@ -3,7 +3,7 @@ \begin{tikzpicture} % Achsen zeichnen \draw[->] (0,0) -- (14,0) node[right] {$k$}; % x-Achse - \draw[->] (0,0) -- (0,5) node[above] {$\frac{\hat{u}_\mathrm{UM,k}}{\hat{u}_\mathrm{UM,1}}$}; % y-Achse + \draw[->] (0,0) -- (0,5) node[above] {$\frac{\hat{u}_\mathrm{2ae,k}}{\hat{u}_\mathrm{2ae,1}}$}; % y-Achse % Ticks und Beschriftungen auf der x-Achse \foreach \x in {1, 5, 7, 11, 13} { @@ -15,7 +15,11 @@ % Ticks und Beschriftung auf der y-Achse \draw (-0.05,4.5) -- (0.05,4.5) node[left] {1}; - + \draw (-0.05,3.6) node[left] {0.8}; + \draw (-0.05,2.7) node[left] {0.6}; + \draw (-0.05,1.8) node[left] {0.4}; + \draw (-0.05,0.9) node[left] {0.2}; + \draw (-0.05,0) node[left] {0}; % Balken zeichnen \foreach \x/\y in {1/4.5, 5/0.9, 7/0.6, 11/0.5, 13/0.4} { \draw[thick] (\x,0) -- (\x,\y); % Balken diff --git a/exercise/main.tex b/exercise/main.tex index 7da18f3..d1418e6 100644 --- a/exercise/main.tex +++ b/exercise/main.tex @@ -1,7 +1,7 @@ -\documentclass[]{../course_template/exerciseClass} +\documentclass[solution]{../course_template/exerciseClass} \title{Power Electronics} -\includeonly{tex/exercise05} +\includeonly{tex/exercise07} \begin{document} \include{tex/exercise01} From acb53dd876af4632aa785e1f8bfc3f98afd76acd Mon Sep 17 00:00:00 2001 From: Wallscheid Date: Mon, 27 Jan 2025 17:34:15 +0100 Subject: [PATCH 3/6] add to lec06 --- lecture/main.ist | 2 +- lecture/tex/Lecture06.tex | 168 ++++++++++++++++++++++++++++++-------- 2 files changed, 136 insertions(+), 34 deletions(-) diff --git a/lecture/main.ist b/lecture/main.ist index 4289e8e..100bbf4 100644 --- a/lecture/main.ist +++ b/lecture/main.ist @@ -1,5 +1,5 @@ % makeindex style file created by the glossaries package -% for document 'main' on 2025-1-26 +% for document 'main' on 2025-1-27 actual '?' encap '|' level '!' diff --git a/lecture/tex/Lecture06.tex b/lecture/tex/Lecture06.tex index 1a3e51e..672cec1 100644 --- a/lecture/tex/Lecture06.tex +++ b/lecture/tex/Lecture06.tex @@ -7,7 +7,7 @@ \section{Transistor-based AC/DC converters} %% Extending AC/DC converter to four quadrant operation %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} - \frametitle{Extending AC/DC converters to four quadrant operation} + \frametitle{Transistor-based AC/DC converters: self-commutated converters} \begin{columns} \begin{column}{0.5\textwidth} @@ -16,12 +16,12 @@ \section{Transistor-based AC/DC converters} \item<1-> Diode-based converters \begin{itemize} \item<1-> Rectification only - \item<1-> Single quadrant operation + \item<1-> No control \end{itemize} \item<2-> Thyristor-based converters \begin{itemize} \item<2-> Rectification and inversion - \item<2-> Two quadrant operation + \item<2-> Limited control / line commutation \end{itemize} \end{itemize} \vspace{1em} @@ -31,7 +31,7 @@ \section{Transistor-based AC/DC converters} \item Transistor-based converters \begin{itemize} \item Rectification and inversion - \item Four quadrant operation + \item Fully controllable / self-commutated \end{itemize} \end{itemize}} \end{column} @@ -47,8 +47,8 @@ \section{Transistor-based AC/DC converters} \fill[shadecolor, opacity=0.3] (0,0) -- (-2,0) -- (-2,2) -- (0,2) -- cycle; \fill[shadecolor, opacity=0.3] (0,0) -- (-2,0) -- (-2,-2) -- (0,-2) -- cycle; \end{scope} - \draw[<->] (-2,0) -- (2,0) node[anchor=west] {$i$}; - \draw[<->] (0,-2) -- (0,2) node[anchor=south] {$u$}; + \draw[<->] (-2,0) -- (2,0) node[anchor=west] {$i_1$}; + \draw[<->] (0,-2) -- (0,2) node[anchor=south] {$u_1$}; \node[anchor=center, align = center] at (1.0,1.0) {$\mathrm{I}$\\$P \geq 0$\\(recitifier)}; \node[anchor=center, align = center] at (-1.0,1.0) {$\mathrm{II}$\\$P \leq 0$\\(inverter)}; \node[anchor=center, align = center] at (-1.0,-1.0) {$\mathrm{III}$\\$P \geq 0$\\(recitifier)}; @@ -86,22 +86,22 @@ \subsection{Single-phase AC/DC bridge converter} \begin{columns} \begin{column}{0.4\textwidth} \onslide<2->{% - Define \hl{switching function}: - \begin{equation} - s_i(t)=\begin{cases} + Define \hl{switching function}:% + \begin{equation}% + s_i(t)=\begin{cases}% +1 & \text{upper position,}\\ -1 & \text{lower position.} - \end{cases} - \label{eq:switching_function_VSI} + \end{cases}% + \label{eq:switching_function_VSI}% \end{equation}}% - \onslide<3->{Output voltage considering a voltage source at the input is: - \begin{equation} - u_2(t)=\underbrace{\frac{1}{2}\left(s_1(t)-s_2(t)\right)}_{s(t)}u_1(t). + \onslide<3->{Output voltage considering a voltage source at the input is:% + \begin{equation}% + u_2(t)=\underbrace{\frac{1}{2}\left(s_1(t)-s_2(t)\right)}_{s(t)}u_1(t).% + \end{equation}}% + \onslide<4->{Input current assuming a current source at the output results in:% + \begin{equation}% + i_1(t)=s(t)i_2(t).% \end{equation}}% - \onslide<4->{Input current assuming a current source at the output results in: - \begin{equation} - i_1(t)=s(t)i_2(t). - \end{equation}} \end{column} \begin{column}{0.6\textwidth} \begin{figure} @@ -303,7 +303,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s^*(t), c(t)$}, legend pos=north east, legend columns=2] @@ -340,7 +340,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s^*(t), c(t)$}, legend pos=north east, legend columns=2, ymin = -1.1, ymax =1.1] @@ -616,7 +616,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s^*(t), c(t)$}, ymin = -1.25, ymax =1.25, legend pos=north east, legend columns=2 ] @@ -653,7 +653,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{columns} \begin{column}{0.45\textwidth} Considering a normalized input reference - $$ s^*(t) = m\sin(\omega t)$$ + $$ s^*(t) = m\sin(\omega t)=\frac{\hat{u}_2^*}{U_1}\sin(\omega t)$$ with the \hl{modulation ratio} $m$ one can distinguish two PWM operation areas: \begin{itemize} \item $m \leq 1$: linear modulation, @@ -682,12 +682,12 @@ \subsection{Single-phase AC/DC bridge converter} yticklabels = {$0$, $0.25$, $0.5$, $0.75$, $1$, $\nicefrac{4}{\pi}$}, ] \addplot[signalblue, domain=0:1, samples=10] {x}; - \addplot[signalblue, domain=1:3, samples=40, smooth] {4/pi + rad(asin(1/x))*(2/pi -8/pi^2)}; + \addplot[signalblue, domain=1:3, samples=40, smooth] {2/pi*(x*rad(asin(1/x))+sqrt(1-1/x^2))}; \draw[dashed] (axis cs:0, 4/pi) -- (axis cs:3, 4/pi); \draw[dash dot] (axis cs:1, 0) -- (axis cs:1, 4/pi); \draw node[fill=white, inner sep = 1pt] at (axis cs:0.5,1) {\small linear mod.}; \draw node[fill=white, inner sep = 1pt] at (axis cs:2,1) {\small overmodulation}; - \draw node[fill=white, inner sep = 1pt, above] at (axis cs:1.5,4/pi) {\small Fundamental freq. mod. ($f_\mathrm{s}=2\cdot\nicefrac{\omega}{2 \pi}$)}; + \draw node[fill=white, inner sep = 1pt, above] at (axis cs:1.5,4/pi) {\small Fundamental freq. mod. ($f_\mathrm{s}=\nicefrac{\omega}{2 \pi}$)}; \end{axis} \end{tikzpicture} \caption{Reference amplitude to output voltage fundamental amplitude} @@ -697,6 +697,84 @@ \subsection{Single-phase AC/DC bridge converter} \end{columns} \end{frame} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% Overmodulation (cont.) %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame} + \frametitle{Overmodulation (cont.)} + \begin{columns} + \begin{column}{0.45\textwidth} + Due to the converter's constraints, the reference voltage is limited to + $$ + s^*_\mathrm{lim}(t)=\begin{cases} + 1 & \text{if } s^*(t) > 1,\\ + s^*(t) & \text{if } -1 \leq s^*(t) \leq 1,\\ + -1 & \text{if } s^*(t) < -1. + \end{cases} + $$ + Hence, from $\omega t_0$ to $\omega t_1$ the converter's output voltage is clipped for $m>1$. With + $$ + m \sin(\omega t_0)\stackrel{!}{=}1 + $$ + one can find + \begin{align} + \omega t_0 &= \arcsin\left(\frac{1}{m}\right). + \label{eq:Overmodulation_t0} + \end{align} + \end{column} + \begin{column}{0.55\textwidth} + \begin{figure} + \begin{tikzpicture} + \def\m{1.2} + \begin{axis}[ + xlabel={$\omega t$}, + ylabel={$s^*(t)$}, + ymin=0, ymax=1.25, + xmin=0, xmax=pi, + width = 0.95\textwidth, + height = 0.6\textheight, + grid, + thick, + clip = false, + ytick = {0, 0.25, 0.5, 0.75, 1.0, 1.25}, + yticklabels = {$0$, $0.25$, $0.5$, $0.75$, $1$, $1.25$}, + xtick = {0, pi/2, pi}, + xticklabels = {$0$,$\frac{1}{2}\pi$,$\pi$}, + ] + \addplot[signalblue, domain=0:pi, samples=50] {min(\m*sin(deg(x)),1)}; + \addplot[signalblue, domain=rad(asin(1/\m)):pi-rad(asin(1/\m)), samples=50, dashed] {\m*sin(deg(x))}; + \draw[<-] (axis cs:pi/3,1.1) to (axis cs:pi/5, 1.1) node[left] {$s^*(t)$}; + \draw[<-] (axis cs:pi/2,0.95) to (axis cs:pi/2, 0.7) node[below] {$s_\mathrm{lim}^*(t)$}; + \draw[dashed] (axis cs:{rad(asin(1/\m))},1) to (axis cs:{rad(asin(1/\m))},0) node[below] {$\omega t_0$}; + \draw[dashed] (axis cs:{pi-rad(asin(1/\m))},1) to (axis cs:{pi-rad(asin(1/\m))},0) node[below] {$\omega t_1$}; + \end{axis} + \end{tikzpicture} + \caption{Exemplary time series between reference and actual voltage in the overmodulation range} + \label{fig:Overmodulation_time_series_example} + \end{figure} + \end{column} + \end{columns} +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% Overmodulation (cont.) %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame} + \frametitle{Overmodulation (cont.)} + To calculate the resulting fundamental output voltage during overmodulation, a Fourier analysis is performed while utilizing the quarter-wave symmetry of the output voltage signal: + \begin{equation} + \begin{split} + \frac{u_2^{(1)}}{U_1} &= \frac{1}{\pi} \int_0^{2\pi}s^*_\mathrm{lim}(\omega \tau)\sin(\omega \tau)\mathrm{d}\omega \tau \\ + &=\frac{4}{\pi}\left( \int_0^{\omega t_0} m\sin^2(\omega \tau)\mathrm{d}\omega \tau + \int_{\omega t_0}^{\frac{\pi}{2}} 1\sin(\omega \tau)\mathrm{d}\omega \tau\right)\\ + &=\frac{4}{\pi}\left[\frac{m}{2}\left(\omega t_0 -\frac{1}{2}\sin(2\omega t_0)\right)+\cos(\omega t_0)\right]. + \end{split} + \end{equation} + Inserting $\omega t_0$ from \eqref{eq:Overmodulation_t0} and applying trigonometric identities yields: + \begin{equation} + \frac{u_2^{(1)}}{U_1} = \frac{2}{\pi}\left[m\arcsin\left(\frac{1}{m}\right)+\sqrt{1-\frac{1}{m^2}}\right]\in\left[1,\frac{4}{\pi}\right] \quad \mbox{for} \quad m \geq 1. + \end{equation} +\end{frame} + %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Fundamental frequency modulation %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% @@ -705,7 +783,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \def\a{0.6*pi} - \begin{groupplot}[group style={group size=1 by 4, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.34\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 4, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.34\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % top middle plot: individual switching signals @@ -886,7 +964,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{tikzpicture} \def\a{0.33*pi} \def\it{0.1*pi} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % switching input \nextgroupplot[ylabel = {$s_i(t)$}] @@ -937,7 +1015,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{tikzpicture} \def\a{0.33*pi} \def\it{0.1*pi} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % switching input \nextgroupplot[ylabel = {$s_i(t)$}] @@ -987,7 +1065,7 @@ \subsection{Single-phase AC/DC bridge converter} \begin{tikzpicture} \def\a{0.33*pi} \def\it{0.1*pi} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{4}$,$\nicefrac{1}{2}$, $\nicefrac{3}{2}$, $1$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % switching input \nextgroupplot[ylabel = {$s_i(t)$}] @@ -1223,7 +1301,7 @@ \subsection{Rectifier operation for single-phase grids} \draw (c1) to [capacitor, *-*, l=$C$, i_=$i_\mathrm{C}(t)$] (c1 |- int2); \draw (out1) to [inductor, v<=$u_{\mathrm{L}}(t)$, voltage = straight, l=$L$, mirror, o-] ++(-2.5,0) to [sinusoidal voltage source, v_=$u_{\mathrm{g}}(t)$] (\tikztostart |- out2) to [short,-o] (out2); \end{circuitikz} - \caption{Single-phase grid rectification: full bidirectional operation possible (e.g., for electrical rail vehicles with a \SI{15}{\kilo\volt}, \SI[parse-numbers = false]{16\frac{2}{3}}{\hertz} grid). Note: converter topology is flipped to align $u_2$ with the AC grid side while $u_1$ is the DC output.} + \caption{Single-phase grid rectification: full bidirectional operation possible (e.g., for electrical rail vehicles with a \SI{15}{\kilo\volt}, \SI[parse-numbers = false]{16\frac{2}{3}}{\hertz} grid). Note: converter topology is flipped to align $u_2$ with the AC grid side while $u_1$ is the DC output. Also known as \hl{active front end (AFE) rectifier}.} \label{fig:recitifier_single_phase_transistor_bridge_converter} \end{figure} \end{frame} @@ -1352,7 +1430,7 @@ \subsection{Rectifier operation for single-phase grids} \begin{figure} \begin{tikzpicture} \def\p{0.2*pi} - \begin{groupplot}[group style={group size=1 by 2, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.4\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 2, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.4\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % top middle plot: individual switching signals @@ -1761,7 +1839,7 @@ \subsection{Three-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s_{i}^*(t), c(t)$}] @@ -1803,7 +1881,7 @@ \subsection{Three-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s_{i}^*(t), c(t)$}] @@ -1845,7 +1923,7 @@ \subsection{Three-phase AC/DC bridge converter} \begin{figure} \begin{tikzpicture} \pgfplotsset{table/search path={fig/lec06}} - \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {0,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] + \begin{groupplot}[group style={group size=1 by 5, xticklabels at = edge bottom, vertical sep=0.25cm}, height=0.31\textheight, width=0.875\textwidth, xmin=0, xmax=2*pi, grid,clip = false, ymin = -1.1, ymax =1.1, xtick = {0, pi/2, pi, 3/2*pi, 2*pi}, xticklabels = {$0$,$\nicefrac{1}{2}\pi$,$\pi$, $\nicefrac{3}{2}\pi$, $2\pi$}, ytick = {-1, 0, 1}, yticklabels = {$-1$, $0$, $1$}] % Top plot: duty cycle reference and carrier signal \nextgroupplot[ylabel = {$s_{i}^*(t), c(t)$}, ymin = -1.2, ymax =1.2] @@ -1876,4 +1954,28 @@ \subsection{Three-phase AC/DC bridge converter} \end{groupplot} \end{tikzpicture} \end{figure} +\end{frame} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% Section summary %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{frame} + \frametitle{Section summary} + This section provided an introduction to transistor-based AC/DC converters. The key takeaways are: + \begin{itemize} + \item<2-> They render themselves (half/full) bridge topologies as already known from the DC/DC converter context. + \item<3-> Can transfer power in both directions and handle all four quadrants on the AC side. + \item<4-> Require modulation strategies to generate the desired output voltage: + \begin{itemize} + \item<4-> High switching frequency PWM (low harmonics, below maximum conv. utilization) or + \item<5-> Low switching frequency fundamental modulation (max. utilization, but high harmonics). + \end{itemize} + \item<6-> The output voltage amplitude and phase angle can be adjusted to achieve arbitrary power factors for grid operation or to supply various loads such as DC or AC motors. + \end{itemize} + \onslide<7->{While this section only covered a very brief overview about these self-commutated converters, the following aspects are, among other, important for practical applications:} + \begin{itemize} + \item<7-> closed-loop control, + \item<8-> Further modulation strategies (e.g., \href{https://en.wikipedia.org/wiki/Space_vector_modulation}{space vector modulation} or \href{https://doi.org/10.1109/TIE.2010.2047824}{optimized pulse pattern}), + \item<9-> converters with a \href{https://en.wikipedia.org/wiki/Power_electronics\#Current_source_inverters}{current source} (instead of voltage source) within the DC link. + \end{itemize} \end{frame} \ No newline at end of file From 991dee617ef49f2f9c7147d78a07b5a8858dd413 Mon Sep 17 00:00:00 2001 From: SevenOfNinePE Date: Mon, 27 Jan 2025 21:26:23 +0100 Subject: [PATCH 4/6] Ex07Task2 Add solution text and update drawings. --- ...cerpt.tex => Fig_Voltage_u2a0_excerpt.tex} | 8 +- .../ex07/Fig_graphic_solutions_cos_terms.tex | 79 ++++++------ ...Fig_standardization_to_fudamental_freq.tex | 2 +- exercise/tex/exercise07.tex | 118 +++++++++++++++--- 4 files changed, 152 insertions(+), 55 deletions(-) rename exercise/fig/ex07/{Fig_Voltage_U_um_excerpt.tex => Fig_Voltage_u2a0_excerpt.tex} (84%) diff --git a/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex b/exercise/fig/ex07/Fig_Voltage_u2a0_excerpt.tex similarity index 84% rename from exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex rename to exercise/fig/ex07/Fig_Voltage_u2a0_excerpt.tex index 39bd077..f6aaa2b 100644 --- a/exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex +++ b/exercise/fig/ex07/Fig_Voltage_u2a0_excerpt.tex @@ -11,13 +11,13 @@ enlargelimits, axis line style={->}, % Pfeilspitzen an den Achsen xlabel={$\omega t$}, - ylabel={$U_\mathrm{1}$}, + ylabel={$u_\mathrm{2a0}(\omega t)/ \SI{}{\volt}$}, xmin=0, xmax=13/6*pi, ymin=-1, ymax=1, xtick={0, pi/3, 2*pi/3, pi, 4*pi/3, 5*pi/3, 2*pi}, xticklabels={0, $\frac{1\pi}{3}$, $\frac{2\pi}{3}$,$\pi$, $\frac{4\pi}{3}$, $\frac{5\pi}{3}$, $2\pi$}, ytick={-2/3, -1/3, 0, 1/3, 2/3}, - yticklabels={$-\frac{2}{3}$, $-\frac{1}{3}$, $0$, $\frac{1}{3}$, $\frac{2}{3}$}, + yticklabels={$-340$, $-170$, $0$, $170$, $340$}, % grid=both, % major grid style={line width=.2pt,draw=gray!50}, % minor grid style={line width=.1pt,draw=gray!20}, @@ -31,6 +31,6 @@ }; \end{axis} \end{tikzpicture} - \caption{Section of the voltage curve $U_\mathrm{UM}$.} - \label{fig:voltage_uum_section} + \caption{Section of the voltage curve $u_\mathrm{2a0}(\mathrm{\omega t})$.} + \label{fig:voltage_u2a0_section} \end{solutionfigure} diff --git a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex index 25c591f..a8d0de0 100644 --- a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex +++ b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex @@ -8,32 +8,39 @@ \centering \begin{tikzpicture} \begin{axis}[ + % x/y range adjustment + xmin=-20, xmax=420, + ymin=-160, ymax=180, + width=7cm, height=7cm, axis lines=middle, xlabel={$\cos(\varphi)$}, ylabel={$\sin(\varphi)$}, - xlabel style={xshift=0.5cm}, - width=7cm, height=7cm, major grid style={line width=.2pt,draw=gray!50}, minor grid style={line width=.1pt,draw=gray!20}, xmin=-1.5, xmax=1.5, ymin=-1.5, ymax=1.5, - xtick={-1, 0, 1}, % Manuelle Ticks auf der x-Achse - ytick={-1, 0, 1}, % Manuelle Ticks auf der y-Achse - tick label style={xshift=5pt, yshift=5pt}, % Verschiebt die Beschriftungen nach außen - ] - - % Vektor einzeichnen - \draw[thick, ->] (0,0) -- ({cos(40)}, {sin(40)}) node[above right] {}; - \node at ({0.5*cos(40)}, {0.5*sin(40)}) [below] {$\varphi$}; - - % Kreis zeichnen - \addplot[ - domain=0:360, - samples=200, - thick, - color=blue, - ] - ({cos(x)}, {sin(x)}); + % Label adjustment + x label style={at={(axis description cs:1,0.5)},anchor=west}, + % x-Ticks + xtick={-1,0,1}, + xticklabels={-1,,1}, + xticklabel style = {anchor=north,shift={(0.25cm,0.1cm)}}, + % y-Ticks + ytick={-1,0,1}, + yticklabels={-1,,1}, + yticklabel style = {anchor=east,shift={(0.1cm,0.2cm)}}, + ] + + \draw[thick, ->] (0,0) -- ({cos(40)}, {sin(40)}) node[above right] {}; + \node at ({0.5*cos(40)}, {0.5*sin(40)}) [below] {$\varphi$}; + + \addplot[ + domain=0:360, + samples=200, + thick, + color=blue, + ] + ({cos(x)}, {sin(x)}); \end{axis} \end{tikzpicture} \end{minipage} @@ -44,10 +51,10 @@ \draw[->] (-2,0) -- (2,0) node[right] {}; \draw[->] (0,-2) -- (0,2) node[above] {}; \node at (-1.5,1.5) {$\cos\left(k \frac{\pi}{2}\right)$}; - \node at (-2,0) [left] {$2,6,\ldots$}; - \node at (2,0) [right] {$0,4,\ldots$}; - \node at (0,2) [above] {$1,5,9,13,\ldots$}; - \node at (0,-2) [below] {$3,7,11,15,\ldots$}; + \node at (-2,0) [left] {$k=2,6,\ldots$}; + \node at (2,0) [right] {$k=0,4,\ldots$}; + \node at (0,2) [above] {$k=1,5,9,13,\ldots$}; + \node at (0,-2) [below] {$k=3,7,11,15,\ldots$}; % Kreuz bei jedem markierten Punkt \foreach \x/\y in {-1.5/0, 1.5/0, 0/1.5, 0/-1.5} { @@ -61,13 +68,13 @@ \begin{tikzpicture} \draw[->] (-2,0) -- (2,0) node[right] {}; \draw[->] (0,-2) -- (0,2) node[above] {}; - \node at (-2.5,1.5) {$\cos\left(k \frac{\pi}{3}\right)$}; - \node at (-1,0.2) [left] {$3,9,15,21\ldots$}; - \node at (1,0.2) [right] {$0,6\ldots$}; - \node at (2,1) [above] {$1,7,13,19\ldots$}; - \node at (-0.8,1.6) [below] {$2,8\ldots$}; - \node at (-0.8,-1) [below] {$4,10\ldots$}; - \node at (2,-1) [below] {$5,11,17,23\ldots$}; + \node at (-2.7,1.8) {$\cos\left(k \frac{\pi}{3}\right)$}; + \node at (-0.3,0.4) [left] {$k=3,9,15,21\ldots$}; + \node at (1,0.4) [right] {$k=0,6\ldots$}; + \node at (2,1) [above] {$k=1,7,13,19\ldots$}; + \node at (-1.2,1.6) [below] {$k=2,8\ldots$}; + \node at (-1.2,-1.1) [below] {$k=4,10\ldots$}; + \node at (2,-1.1) [below] {$k=5,11,17,23\ldots$}; \node at (-0.75,-0.3) [left] {-1}; \node at (1,-0.3) [left] {0.5}; % Kreuz bei jedem markierten Punkt @@ -93,7 +100,9 @@ % Koordinatensystem zeichnen \draw[->] (-2,0) -- (2,0) node[right] {}; \draw[->] (0,-2) -- (0,2) node[above] {}; - \node at (-1.5,1.5) {$\cos\left(k \frac{\pi}{3}\right)+1$}; + % \node at (-1.5,1.5) {$\cos\left(k \frac{\pi}{3}\right)+1$}; + \node at (-1.7,1.8) {$\cos\left(k \frac{\pi}{3}\right)+1$}; + % Beschriftungen an der x-Achse \node at (-0.00009,-0.2) [left] {0}; @@ -102,8 +111,8 @@ } % Beschriftungen an spezifischen Punkten - \node at (1.5,1.1) [above] {$1,7,13,19,\ldots$}; - \node at (1.5,-1.1) [below] {$5,11,17,23,\ldots$}; + \node at (2,1.1) [above] {$k=1,7,13,19,\ldots$}; + \node at (2,-1.1) [below] {$k=5,11,17,23,\ldots$}; % Kreuz bei jedem markierten Punkt \foreach \x/\y in {0/0, 1.5/1, 1.5/-1} { @@ -116,6 +125,6 @@ \draw[thick, color=blue!70!black] (0,0) -- (1.5,1) -- (1.5,-1) -- cycle; \end{tikzpicture} - \caption{Graphical solution of the cos terms.} - \label{fig:Graphical solution of the cos terms} + \caption{Graphical solution of the cos terms within complex plane.} + \label{fig:GraphicalSolutionOfInComplexPlane} \end{solutionfigure} diff --git a/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex b/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex index 9d69392..9c0a8fd 100644 --- a/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex +++ b/exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex @@ -3,7 +3,7 @@ \begin{tikzpicture} % Achsen zeichnen \draw[->] (0,0) -- (14,0) node[right] {$k$}; % x-Achse - \draw[->] (0,0) -- (0,5) node[above] {$\frac{\hat{u}_\mathrm{2ae,k}}{\hat{u}_\mathrm{2ae,1}}$}; % y-Achse + \draw[->] (0,0) -- (0,5) node[above] {$\frac{\hat{u}_\mathrm{2a}^\mathrm{(k)}}{\hat{u}_\mathrm{2a}^\mathrm{(1)}}$}; % y-Achse % Ticks und Beschriftungen auf der x-Achse \foreach \x in {1, 5, 7, 11, 13} { diff --git a/exercise/tex/exercise07.tex b/exercise/tex/exercise07.tex index 39c6200..ca0ba83 100644 --- a/exercise/tex/exercise07.tex +++ b/exercise/tex/exercise07.tex @@ -228,11 +228,10 @@ \end{equation} \end{solutionblock} -\subtask{Decompose the voltage $u_\mathrm{a}(t)$ into a Fourier series and sketch the spectral lines related to the -amplitude of the fu -ndamental signal up to order n=13. Hint: The following applies to the Fourier coefficients of an odd and alternating function: +\subtask{Decompose the voltage $u_\mathrm{2a}(t)$ into a Fourier series and sketch the spectral lines related to the +amplitude of the fundamental signal up to order n=13. Hint: The following applies to the Fourier coefficients of an odd and alternating function: \begin{align*} -b_k = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} f(x)\sin(kx) \mathrm{d}x \quad k =\mathrm{odd} \quad \quad + b_k = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} f(x)\sin(kx) \mathrm{d}x \quad k =\mathrm{odd} \quad \quad \end{align*} \label{sub:DecomposeVoltage} } @@ -241,28 +240,117 @@ \begin{equation} \begin{split} a_\mathrm{k} &= 0 \\ - a_\mathrm{k} &= \frac{4}{\pi} \int_0^{\pi/2} f(x)\sin(x) \mathrm{d}x \quad k=\mathrm{odd} \\ + b_\mathrm{k} &= \frac{4}{\pi} \int_0^{\pi/2} f(x)\sin(x) \mathrm{d}x \quad k=\mathrm{odd} \\ f(x) &= \sum_{k}^{} \left( b_k \sin(kx) \right). - \end{split} + \end{split} \end{equation} The coefficients $b_k$ are the amplitudes of the respective harmonic. The voltage $u_{\mathrm{2a}}(t)$ needs only to be integrated up to $\pi/2$. Only the terms with odd order numbers are taken into account. + Apply this to the current signal is expressed by + \begin{equation} + b_\mathrm{k} = \frac{4}{\pi} \int_0^{\pi/3} \frac{U_{\mathrm{1}}}{3} \sin(kt) \mathrm{d}t + + \frac{4}{\pi} \int_{\pi/3}^{\pi/2} \frac{2U_{\mathrm{1}}}{3} \sin(kt) \mathrm{d}t + \end{equation} + Signal ratio $u_{\mathrm{2a0}}(\omega t)/U_{\mathrm{1}}$ is depicted in \autoref{fig:voltage_u2a0_section} for one periode. + \input{fig/ex07/Fig_Voltage_u2a0_excerpt} \begin{equation} \begin{split} - a_\mathrm{k} &= \frac{4}{\pi} \int_0^{\pi/3} f(x)\sin(x) \mathrm{d}x \quad k=\mathrm{odd} \\ - f(x) &= \sum_{k}^{} \left( b_k \sin(kx) \right). - \end{split} + b_\mathrm{k} &= \frac{4U_{\mathrm{1}}}{3k\pi} \big[-\cos(kt)\mathrm{d}t \big]_0^{\pi/3} + + \frac{4U_{\mathrm{1}}}{3k\pi} \big[-2\cos(kt)\mathrm{d}t \big]_{\pi/3}^{\pi/2} \\ + &= \frac{4U_{\mathrm{1}}}{3k\pi} \left(-\cos(k\frac{\pi}{3})+\cos(0) -2 \cos(k\frac{\pi}{2})+ 2\cos(0) \right) \\ + &= \frac{4U_{\mathrm{1}}}{3k\pi} \left( \cos(k\frac{\pi}{3})+ 1 -2 \cos(k\frac{\pi}{2})\right) \\ + \text{with } &\cos(k\frac{\pi}{3})+1=1.5 \quad \text{for } k=n \cdot 6\pm1 \text{ is odd} \\ + \text{and }&\cos(k\frac{\pi}{2})=0 \quad \text{ for } k=\text{ odd} + \end{split} \end{equation} + The result is displayed in the complex plane in \autoref{fig:voltage_u2a0_section}. - - - \input{fig/ex07/Fig_Voltage_U_um_excerpt} \input{fig/ex07/Fig_graphic_solutions_cos_terms} + + \autoref{fig:voltage_u2a0_section} leads to + \begin{equation} + \hat{u}_\mathrm{2a0,k} = b_\mathrm{k} = \frac{4U_{\mathrm{1}}}{3k\pi} \cdot \frac{3}{2}=\frac{2U_{\mathrm{1}}}{k\pi} + \label{eq:Ex07T2_FundamentelVoltage} + \end{equation} + The amplitudes are depicted in \autoref{fig:NormalizationToTheAmplitude}, \input{fig/ex07/Fig_standardization_to_fudamental_freq.tex} - \input{fig/ex07/Fig_ trigonometric_approach_triangle.tex} -\end{solutionblock} + The relation between fundamental und harmonic amplitude is calculated by + \begin{equation} + \begin{split} + \frac{\hat{u}_\mathrm{2a0,1}}{\hat{u}_\mathrm{2a0,k}} &= \frac{1}{k} \\ + \text{with } &k=n \cdot 6\pm1 \text{ and } n=1,2,3... + \end{split} + \end{equation} + \label{subtask:Ex07T2_FourierSeries} + \end{solutionblock} -\subtask{Based on \autoref{sub:DecomposeVoltage}, calculate the fundamental amplitude $\hat{i}_\mathrm{a}^1$ using a vector diagram and complex alternating current calculations. +\subtask{Based on \autoref{sub:DecomposeVoltage}, calculate the fundamental amplitude $\hat{i}_\mathrm{a}^1$ using a vector diagram and +complex alternating current calculations. From this, determine the total active power converted in the load.} \begin{solutionblock} + According \eqref{eq:Ex07T2_FundamentelVoltage} the fundamental voltage is calculated as: + \begin{equation} + \hat{u}_\mathrm{2a0,1}(t)=\frac{2U_{\mathrm{1}}}{1\pi}=\frac{2\cdot \SI{510}{\volt}}{1\pi}=\SI{324,68}{\volt} + \end{equation} + The amplitude of $u_\mathrm{2ae}(t)$ is given with + \begin{equation} + \hat{u}_\mathrm{2ae}=\sqrt{2} \cdot \SI{220}{\volt}=\SI{311,13}{\volt} + \end{equation} + + A triangle is formed by the inverter voltage $u_\mathrm{2a}^\mathrm{(1)}(t)$, the voltage $u_\mathrm{2ae}(t)$ + and the voltage drop across the inductance $L$. Two sides and one angle are known. By applying the sine theorem results in + \begin{equation} + \begin{split} + \frac{a}{\sin(\alpha)} = &\frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)} \\ + \text{with } a=\SI{324,68}{\volt} &\quad \alpha=\SI{120}{\degree} \quad b=\SI{311,13}{\volt}. + \label{eq:Ex07T2_SinTheorem} + \end{split} + \end{equation} + Solving \eqref{eq:Ex07T2_SinTheorem} with respect to $\beta$ leads to + \begin{equation} + \beta=\arcsin\big(\frac{b}{a}\sin(\alpha)\big) + =\arcsin\big(\frac{\SI{311,13}{\volt}}{\SI{324,68}{\volt}}\sin(\SI{120}{\degree})\big) = \arcsin(0,8298)=\SI{56.1}{\degree}. + \label{eq:Ex07T2_sin_beta} + \end{equation} + Using the result for $\beta$ leads to + \begin{equation} + \gamma=\SI{180}{\degree}-\alpha-\beta = \SI{180}{\degree}-\SI{120}{\degree}-\SI{56.1}{\degree}= \SI{3.9}{\degree} + \label{eq:Ex07T2_sin_beta} + \end{equation} + In \autoref{fig:IllustrationForUsingSineTheorem} the triangle is depicted. + \input{fig/ex07/Fig_ trigonometric_approach_triangle.tex} + In a symmetrical three-phase system, the active power is: + \begin{equation} + P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\phi) + \label{eq:Ex07T2_EffPowergen} + \end{equation} + + $U_{\mathrm{L-L}}$ corresponds to the effective value of the line-to-line voltage and $IU_{\mathrm{L}}$ is the effective value + of the line current and $\phi$ is the phase angle between voltage and current. For this case $U_{\mathrm{L-L}}$ is calculated by + \begin{equation} + U_{\mathrm{L-L}}=\sqrt{3}\frac{\hat{u}_\mathrm{2a}^\mathrm{(1)}}{\sqrt{2}} + = \sqrt{\frac{3}{2}} \frac{2U_\mathrm{1}}{\pi} = \sqrt{3}\sqrt{2}\frac{U_\mathrm{1}}{\pi} + = \sqrt{3}\sqrt{2} \cdot \frac{\SI{510}{\volt}}{\pi}=\SI{397.6}{\volt} + \label{eq:Ex07T2_EffPowervoltage} + \end{equation} + The line current $I_{\mathrm{L}}$ is obtained by + \begin{equation} + I_{\mathrm{L}}=\frac{\hat{i}_\mathrm{2a}^\mathrm{(1)}}{\sqrt{2}} + = \frac{\SI{12.37}{\ampere}}{\sqrt{2}}=\SI{8.75}{\ampere} + \label{eq:Ex07T2_EffPowercurrent} + \end{equation} + The angle $\phi$ results in + \begin{equation} + \begin{split} + &\phi=\SI{30}{\degree}+ \gamma= \SI{30}{\degree}+\SI{3.9}{\degree}= \SI{33.9}{\degree}\phi=\SI{30}{\degree}+ \gamma= \SI{30}{\degree}+\SI{3.9}{\degree}= \SI{33.9}{\degree} \\ + &\cos(\phi)=\cos(\SI{33.9}{\degree})=0.83 + \end{split} + \label{eq:Ex07T2_EffPowerangle} + \end{equation} + Using \eqref{eq:Ex07T2_EffPowervoltage}, \eqref{eq:Ex07T2_EffPowercurrent} and \eqref{eq:Ex07T2_EffPowerangle} + in \eqref{eq:Ex07T2_EffPowergen} leads to + \begin{equation} + P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\phi) + = \sqrt{3} \SI{397.6}{\volt} \cdot \SI{8.75}{\ampere} \cdot 0.83= \SI{5}{\kilo\watt} + \end{equation} \end{solutionblock} From 24a176334c38af144e866704cd84d489a7a4d000 Mon Sep 17 00:00:00 2001 From: Wallscheid Date: Tue, 28 Jan 2025 10:39:23 +0100 Subject: [PATCH 5/6] minor corre and animations to lec06 --- exercise/main.tex | 2 +- lecture/main.ist | 2 +- lecture/tex/Lecture06.tex | 97 +++++++++++++++++++++------------------ 3 files changed, 55 insertions(+), 46 deletions(-) diff --git a/exercise/main.tex b/exercise/main.tex index d1418e6..0cdb76c 100644 --- a/exercise/main.tex +++ b/exercise/main.tex @@ -1,4 +1,4 @@ -\documentclass[solution]{../course_template/exerciseClass} +\documentclass[]{../course_template/exerciseClass} \title{Power Electronics} \includeonly{tex/exercise07} diff --git a/lecture/main.ist b/lecture/main.ist index 100bbf4..1446aa9 100644 --- a/lecture/main.ist +++ b/lecture/main.ist @@ -1,5 +1,5 @@ % makeindex style file created by the glossaries package -% for document 'main' on 2025-1-27 +% for document 'main' on 2025-1-28 actual '?' encap '|' level '!' diff --git a/lecture/tex/Lecture06.tex b/lecture/tex/Lecture06.tex index 672cec1..4a251fa 100644 --- a/lecture/tex/Lecture06.tex +++ b/lecture/tex/Lecture06.tex @@ -712,15 +712,15 @@ \subsection{Single-phase AC/DC bridge converter} -1 & \text{if } s^*(t) < -1. \end{cases} $$ - Hence, from $\omega t_0$ to $\omega t_1$ the converter's output voltage is clipped for $m>1$. With + \onslide<2->{Hence, from $\omega t_0$ to $\omega t_1$ the converter's output voltage is clipped for $m>1$. With $$ m \sin(\omega t_0)\stackrel{!}{=}1 - $$ - one can find + $$}% + \onslide<3->{one can find \begin{align} \omega t_0 &= \arcsin\left(\frac{1}{m}\right). \label{eq:Overmodulation_t0} - \end{align} + \end{align}}% \end{column} \begin{column}{0.55\textwidth} \begin{figure} @@ -749,7 +749,7 @@ \subsection{Single-phase AC/DC bridge converter} \draw[dashed] (axis cs:{pi-rad(asin(1/\m))},1) to (axis cs:{pi-rad(asin(1/\m))},0) node[below] {$\omega t_1$}; \end{axis} \end{tikzpicture} - \caption{Exemplary time series between reference and actual voltage in the overmodulation range} + \caption{Exemplary time series between average reference and actual voltage in the overmodulation range} \label{fig:Overmodulation_time_series_example} \end{figure} \end{column} @@ -761,18 +761,18 @@ \subsection{Single-phase AC/DC bridge converter} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} \frametitle{Overmodulation (cont.)} - To calculate the resulting fundamental output voltage during overmodulation, a Fourier analysis is performed while utilizing the quarter-wave symmetry of the output voltage signal: + \onslide<1->{To calculate the resulting fundamental output voltage during overmodulation, a \hl{Fourier analysis} is performed while utilizing the quarter-wave symmetry of the output voltage signal:}% \begin{equation} \begin{split} - \frac{u_2^{(1)}}{U_1} &= \frac{1}{\pi} \int_0^{2\pi}s^*_\mathrm{lim}(\omega \tau)\sin(\omega \tau)\mathrm{d}\omega \tau \\ - &=\frac{4}{\pi}\left( \int_0^{\omega t_0} m\sin^2(\omega \tau)\mathrm{d}\omega \tau + \int_{\omega t_0}^{\frac{\pi}{2}} 1\sin(\omega \tau)\mathrm{d}\omega \tau\right)\\ - &=\frac{4}{\pi}\left[\frac{m}{2}\left(\omega t_0 -\frac{1}{2}\sin(2\omega t_0)\right)+\cos(\omega t_0)\right]. + \onslide<1->{\frac{u_2^{(1)}}{U_1} &= \frac{1}{\pi} \int_0^{2\pi}s^*_\mathrm{lim}(\omega \tau)\sin(\omega \tau)\mathrm{d}\omega \tau \\}% + \onslide<2->{&=\frac{4}{\pi}\left( \int_0^{\omega t_0} m\sin^2(\omega \tau)\mathrm{d}\omega \tau + \int_{\omega t_0}^{\frac{\pi}{2}} 1\sin(\omega \tau)\mathrm{d}\omega \tau\right)\\}% + \onslide<3->{&=\frac{4}{\pi}\left[\frac{m}{2}\left(\omega t_0 -\frac{1}{2}\sin(2\omega t_0)\right)+\cos(\omega t_0)\right].}% \end{split} \end{equation} - Inserting $\omega t_0$ from \eqref{eq:Overmodulation_t0} and applying trigonometric identities yields: + \onslide<4->{Inserting $\omega t_0$ from \eqref{eq:Overmodulation_t0} and applying trigonometric identities yields: \begin{equation} \frac{u_2^{(1)}}{U_1} = \frac{2}{\pi}\left[m\arcsin\left(\frac{1}{m}\right)+\sqrt{1-\frac{1}{m^2}}\right]\in\left[1,\frac{4}{\pi}\right] \quad \mbox{for} \quad m \geq 1. - \end{equation} + \end{equation}}% \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% @@ -820,9 +820,9 @@ \subsection{Single-phase AC/DC bridge converter} \onslide<2->{The fundamental output voltage amplitude can be derived from the corresponding \hl{Fourier coefficient}} \begin{equation} \begin{split} - \onslide<2->{\hat{u}_2^{(k)} &= \frac{1}{\pi} \int_{\alpha}^{\alpha + \pi} u_2(t)\sin(k (\omega t-\alpha)) \mathrm{d}\omega t}\onslide<3->{ = \frac{2}{\pi} \int_{0}^{\pi/2} U_1 \sin(k \omega t) \mathrm{d}\omega t} \\ - & \onslide<4->{= \frac{2}{\pi} \left[-\frac{U_1}{k} \cos(k \omega t)\right]_{0}^{\pi/2}} \onslide<5->{= \frac{2}{\pi} \left[\frac{U_1}{k} \left(\cos(0) - \cos(k(\pi/2))\right)\right]} \\ - &\onslide<6->{= \frac{4}{\pi} U_1 \frac{1}{k}, \quad k=1,3,5,7,\ldots} + \onslide<2->{\frac{u_2^{(k)}}{U_1} &= \frac{1}{\pi} \int_{\alpha}^{\alpha + \pi} \frac{u_2(t)}{U_1}\sin(k (\omega t-\alpha)) \mathrm{d}\omega t}\onslide<3->{ = \frac{2}{\pi} \int_{0}^{\pi/2} \sin(k \omega t) \mathrm{d}\omega t} \\ + & \onslide<4->{= \frac{2}{\pi} \left[-\frac{1}{k} \cos(k \omega t)\right]_{0}^{\pi/2}} \onslide<5->{= \frac{2}{\pi} \left[\frac{1}{k} \left(\cos(0) - \cos(k\frac{\pi}{2})\right)\right]} \\ + &\onslide<6->{= \frac{4}{\pi} \frac{1}{k}, \quad k=1,3,5,7,\ldots} \end{split} \end{equation} \onslide<7->{The fundamental output voltage amplitude is thus given by $ \hat{u}_2^{(1)} = \nicefrac{4}{\pi} \cdot U_1$ which is fixed due to fundamental frequency modulation while only the phase angle $\alpha$ can be adjusted.} @@ -835,11 +835,11 @@ \subsection{Single-phase AC/DC bridge converter} \frametitle{Blanking / interlocking time} \begin{columns} \begin{column}{0.45\textwidth} - When the $i$-th half bridge is actuated, i.e., changes it switching state, an interlocking / blanking time $t_0$ is introduced to avoid short-circuiting the DC link: + \onslide<2->{When the $i$-th half bridge is actuated, i.e., changes it switching state, an \hl{interlocking / blanking time} $t_0$ is introduced to avoid short-circuiting the DC link:}% \begin{itemize} - \item First: turn off conducting transistor, - \item Second: wait $t_0$ \newline(ensure safe turn off), - \item Third: turn on the other transistor. + \item<2-> First: turn off conducting transistor, + \item<3-> Second: wait $t_0$ \newline(ensure safe turn off), + \item<4-> Third: turn on the other transistor. \end{itemize} \vspace{-0.5cm} \begin{varblock}{Background} @@ -903,6 +903,7 @@ \subsection{Single-phase AC/DC bridge converter} \caption{Upper transistor on} \end{subfigure} % + \onslide<2->{% \begin{subfigure}{0.32\textwidth} \centering \begin{circuitikz}[] @@ -926,6 +927,8 @@ \subsection{Single-phase AC/DC bridge converter} \end{circuitikz} \caption{Lower transistor on} \end{subfigure} + }% + \onslide<3->{% \begin{subfigure}{0.32\textwidth} \centering \begin{circuitikz}[] @@ -951,6 +954,7 @@ \subsection{Single-phase AC/DC bridge converter} \end{circuitikz} \caption{Both transistors off} \end{subfigure} + }% \end{figure} \end{frame} @@ -1117,8 +1121,9 @@ \subsection{Single-phase AC/DC bridge converter} \begin{equation} \Delta u =\overline{u}_2 - U_1s^* = -\mathrm{sgn}(i_2)\frac{t_0}{T_\mathrm{s}} U_1= -\mathrm{sgn}(i_2)t_0 f_\mathrm{s} U_1. \end{equation} + \pause Hence, the error depends on the relative duration of the interlocking time $t_0$ compared to the switching period $T_\mathrm{s}$ which is a device-specific parameter (cf. below). - + \begin{table} \begin{tabular}{c c c} \toprule @@ -1181,6 +1186,7 @@ \subsection{Single-phase AC/DC bridge converter} \end{tikzpicture} \caption{2-level half bridge} \end{subfigure} + \onslide<2->{% \begin{subfigure}{0.32\textwidth} \centering \begin{circuitikz} @@ -1222,6 +1228,8 @@ \subsection{Single-phase AC/DC bridge converter} \end{tikzpicture} \caption{3-level half bridge} \end{subfigure} + }% + \onslide<3->{% \begin{subfigure}{0.32\textwidth} \centering \begin{circuitikz} @@ -1266,6 +1274,7 @@ \subsection{Single-phase AC/DC bridge converter} \end{tikzpicture} \caption{4-level half bridge} \end{subfigure} + }% \end{figure} \end{frame} @@ -1316,11 +1325,11 @@ \subsection{Rectifier operation for single-phase grids} \onslide<1->{Assuming steady state, the grid side input loop from \figref{fig:recitifier_single_phase_transistor_bridge_converter} can be described with \hl{complex phasors}: \begin{equation} \underline{\hat{u}}_2 = \underline{\hat{u}}_\mathrm{g} - \mathrm{j}\omega L \underline{\hat{i}}_2. - \end{equation}} + \end{equation}}% \onslide<2->{The converter's input voltage amplitude is \begin{equation} \hat{u}_2 = \sqrt{\hat{u}^2_\mathrm{g} + \left(\omega L \hat{i}_2\right)^2}. - \end{equation}} + \end{equation}}% \onslide<3->{As the converter boosts the grid voltage towards the DC-link, the following condition must apply: \begin{equation} u_1(t)\approx U_\mathrm{dc} \geq \hat{u}_2 = \sqrt{\hat{u}^2_\mathrm{g} + \left(\omega L \hat{i}_2\right)^2}. @@ -1361,7 +1370,7 @@ \subsection{Rectifier operation for single-phase grids} \end{equation} \onslide<6->{The resulting apparent power $S_2$ is} \begin{equation} - \onslide<6->{S_2 = \sqrt{P^2 + Q_2^2}} \onslide<7->{= \sqrt{P^2 + \left(\omega L I_\mathrm{g}\right)^2}} \onslide<8->{= \sqrt{P^2 + \left(\frac{\omega L}{U_\mathrm{g}^2}P^2\right)^2}}\onslide<9->{ = P\sqrt{1 + \left(\frac{\omega L}{U_\mathrm{g}^2}P\right)^2}.} + \onslide<6->{S_2 = \sqrt{P^2 + Q_2^2}} \onslide<7->{= \sqrt{P^2 + \left(\omega L I^2_\mathrm{g}\right)^2}} \onslide<8->{= \sqrt{P^2 + \left(\frac{\omega L}{U_\mathrm{g}^2}P^2\right)^2}}\onslide<9->{ = P\sqrt{1 + \left(\frac{\omega L}{U_\mathrm{g}^2}P\right)^2}.} \end{equation} \end{frame} @@ -1370,11 +1379,11 @@ \subsection{Rectifier operation for single-phase grids} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} \frametitle{Steady-state operation (cont.)} - \onslide<1->{Neglecting the switching-induced current and voltage ripples, the instantaneous grid power is} + \onslide<1->{Neglecting the switching-induced current and voltage ripples, the \hl{instantaneous grid power} is} \begin{equation} \onslide<1->{p_\mathrm{g}(t) = u_\mathrm{g}(t)i_\mathrm{g}(t) = \hat{u}_\mathrm{g}\hat{i}_\mathrm{g}\cos^2(\omega t)} \onslide<2->{= P + P \cos(2\omega t).} \end{equation} - \onslide<3->{The instantaneous converter power at its AC input is} + \onslide<3->{The \hl{instantaneous converter power at its AC input} is} \begin{equation} \begin{split} \onslide<3->{p_2(t) &= u_2(t)i_2(t)} \onslide<4->{= \left(u_\mathrm{g}(t) + u_\mathrm{L}(t)\right)i_\mathrm{g}(t)} \onslide<5->{= \left(u_\mathrm{g}(t) + L\frac{\mathrm{d}}{\mathrm{d}t}i_\mathrm{g}(t)\right)i_\mathrm{g}(t)}\\ @@ -1382,7 +1391,7 @@ \subsection{Rectifier operation for single-phase grids} & \onslide<7->{= P\left(1 + \cos(2\omega t)\right) + Q_2 \sin(2\omega t)} \onslide<8->{= P + S_2\cos(2\omega t- 2\varphi_2)} \end{split} \end{equation} - \onslide<8->{with $\varphi_2$ being the phase angle between $i_2(t)$ and $u_2(t)$.}\onslide<9->{ Hence, the converter power oscillates at twice the grid frequency with an amplitude of $S_2$. As $S_2>P$ applies, the instantaneous output power gets temporarily negative as a result of the reactive power compensation on the grid input side. } + \onslide<8->{with $\varphi_2$ being the phase angle between $i_2(t)$ and $u_2(t)$.}\onslide<9->{ Hence, the \hl{converter power oscillates at twice the grid frequency} with an amplitude of $S_2$. As $S_2>P$ applies, \hl{the instantaneous output power gets temporarily negative as a result of the reactive power compensation} on the grid input side. } \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% @@ -1394,11 +1403,11 @@ \subsection{Rectifier operation for single-phase grids} \begin{equation} \onslide<1->{i_1(t) = \frac{p_1(t)}{U_\mathrm{dc}}}\onslide<2->{= \frac{p_2(t)}{U_\mathrm{dc}} = \frac{P}{U_\mathrm{dc}}+ \frac{S_2}{U_\mathrm{dc}}\cos(2\omega t - 2\varphi_2).} \end{equation} - \onslide<3->{For a constant load current $$I_0 = \frac{P}{U_\mathrm{dc}},$$ the converter's output current can be rewritten as + \onslide<3->{For a \hl{constant load} current $$I_0 = \frac{P}{U_\mathrm{dc}},$$ the converter's output current can be rewritten as \begin{equation} i_1(t) = I_0 \left(1 + \sqrt{1+\left(\frac{\omega L U_\mathrm{dc}}{U_\mathrm{g}^2}\right)^2}\cos(2\omega t - 2\varphi_2)\right). \end{equation}}% - \onslide<4->{Consequently, the DC-link capacitor current carries the harmonic content: + \onslide<4->{Consequently, the \hl{DC-link capacitor carries the harmonic current content}: \begin{equation} i_\mathrm{C}(t) = i_1(t) - I_0 = I_0 \sqrt{1+\left(\frac{\omega L U_\mathrm{dc}}{U_\mathrm{g}^2}\right)^2}\cos(2\omega t - 2\varphi_2). \label{eq:DC_link_current_single_phase_ACDC} @@ -1410,11 +1419,11 @@ \subsection{Rectifier operation for single-phase grids} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} \frametitle{Steady-state operation (cont.)} - \onslide<1->{Assuming that the voltage ripple of the DC-link capacitor does not significantly affect the output current, the voltage ripple amplitude can be approximated as: + \onslide<1->{Assuming that the voltage ripple of the DC-link capacitor does not significantly affect the output current, the \hl{voltage oscillation amplitude} can be approximated as: \begin{equation} \hat{u}_\mathrm{C} = \hat{u}_1 \approx \frac{\hat{i}_1}{2\omega C} = \frac{I_0}{2C} \sqrt{1+\left(\frac{\omega L U_\mathrm{dc}}{U_\mathrm{g}^2}I_0\right)^2}. \label{eq:DC_link_voltage_single_phase_ACDC}}% - \end{equation} + \end{equation}% \onslide<2->{This relation results from the complex phasor analysis of the capacitor's impedance given the current ripple \eqref{eq:DC_link_current_single_phase_ACDC}.} \onslide<3->{From \eqref{eq:DC_link_voltage_single_phase_ACDC} one can } \begin{itemize} \item<3-> derive the required DC-link capacitance for a given voltage ripple, @@ -1568,7 +1577,7 @@ \subsection{Three-phase AC/DC bridge converter} u_{2\mathrm{bc}}(t) &= \frac{1}{2}\left(s_{\mathrm{b}}(t)-s_{\mathrm{c}}(t)\right)u_1(t),\\ u_{2\mathrm{ca}}(t) &= \frac{1}{2}\left(s_{\mathrm{c}}(t)-s_{\mathrm{a}}(t)\right)u_1(t). \end{split} - \end{equation} + \end{equation}\pause The \hl{line-to-ground voltages} are given by \begin{equation} \begin{split} @@ -1639,24 +1648,24 @@ \subsection{Three-phase AC/DC bridge converter} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} \frametitle{Three-phase converter with symmetric load in star connection (cont.)} - Assuming a star-connected load, the three-phase currents sum up to zero: + Assuming a \hl{star-connected load}, the three-phase currents sum up to zero: \begin{equation} i_{2\mathrm{a}}(t) + i_{2\mathrm{b}}(t) + i_{2\mathrm{c}}(t) = 0. \label{eq:three_phase_current_sum_star} - \end{equation} + \end{equation}\pause If the star point is not connected to ground, $u_{\mathrm{n}0}(t)\neq 0$ may occur leading to a load voltage of \begin{equation} u_{2\mathrm{a}}(t) = u_{2\mathrm{a}0}(t) - u_{\mathrm{n}0}(t), \quad u_{2\mathrm{b}}(t) = u_{2\mathrm{b}0}(t) - u_{\mathrm{n}0}(t), \quad u_{2\mathrm{c}}(t) = u_{2\mathrm{c}0}(t) - u_{\mathrm{n}0}(t). - \end{equation} - To calculate $u_{\mathrm{n}0}(t)$ one can utilize the load equation (assuming an inductive load): + \end{equation}\pause + To calculate $u_{\mathrm{n}0}(t)$ one can utilize the load equation (assuming an \hl{inductive load}): \begin{equation} u_{2i}(t) = L \frac{\mathrm{d}}{\mathrm{d}t} i_{2i}(t) + u_{\mathrm{n}0}(t) - \end{equation} + \end{equation}\pause summing up to \begin{equation} 3u_{\mathrm{n}0}(t) + L \frac{\mathrm{d}}{\mathrm{d}t} \left(i_{2\mathrm{a}}(t) + i_{2\mathrm{b}}(t) + i_{2\mathrm{c}}(t)\right) = u_{2\mathrm{a}0}(t) + u_{2\mathrm{b}0}(t) +u_{2\mathrm{c}0}(t) - \end{equation} - and finally delivering the star-to-ground voltage as + \end{equation}\pause + and finally delivering the \hl{star-to-ground voltage} as \begin{equation} u_{\mathrm{n}0}(t) = \frac{1}{3} \left(u_{2\mathrm{a}0}(t) + u_{2\mathrm{b}0}(t) +u_{2\mathrm{c}0}(t)\right). \end{equation} @@ -1673,7 +1682,7 @@ \subsection{Three-phase AC/DC bridge converter} \centering \begin{tabular}{c c c c c c c c c c c c c c} \toprule - $\mathrm{No.}$ & $s_\mathrm{a}$ & $s_\mathrm{b}$ & $s_\mathrm{c}$ & $\frac{u_{2a0}}{u_1}$ & $\frac{u_{2b0}}{u_1}$ & $\frac{u_{2c0}}{u_1}$ & $\frac{u_{2a}}{u_1}$ & $\frac{u_{2b}}{u_1}$ & $\frac{u_{2c}}{u_1}$ & $\frac{u_{\mathrm{ab}}}{u_1}$ & $\frac{u_{\mathrm{bc}}}{u_1}$ & $\frac{u_{\mathrm{ca}}}{u_1}$ & $\frac{u_{\mathrm{n}0}}{u_1}$ \\ + $\mathrm{No.}$ & $s_\mathrm{a}$ & $s_\mathrm{b}$ & $s_\mathrm{c}$ & $\frac{u_{2\mathrm{a}0}}{u_1}$ & $\frac{u_{2\mathrm{b}0}}{u_1}$ & $\frac{u_{2\mathrm{c}0}}{u_1}$ & $\frac{u_{2\mathrm{a}}}{u_1}$ & $\frac{u_{2\mathrm{b}}}{u_1}$ & $\frac{u_{2\mathrm{c}}}{u_1}$ & $\frac{u_{\mathrm{ab}}}{u_1}$ & $\frac{u_{\mathrm{bc}}}{u_1}$ & $\frac{u_{\mathrm{ca}}}{u_1}$ & $\frac{u_{\mathrm{n}0}}{u_1}$ \\ \midrule $0$ & $-1$ & $-1$ & $-1$ & $-\frac{1}{2}$ & $-\frac{1}{2}$ & $-\frac{1}{2}$ & $0$ & $0$ & $0$ & $0$& $0$& $0$&$-\frac{1}{2}$\\ @@ -1703,7 +1712,7 @@ \subsection{Three-phase AC/DC bridge converter} %% Three-phase fundamental frequency modulation %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame} - \frametitle{Three-phase fundamental frequency modulation} + \frametitle{Three-phase fundamental frequency modulation (aka six-step mode)} \vspace{-0.1cm} \begin{figure} \begin{tikzpicture} @@ -1754,13 +1763,13 @@ \subsection{Three-phase AC/DC bridge converter} \frametitle{Three-phase fundamental frequency modulation (cont.)} From the previous figure and voltage equations, we can summarize the following observations: \begin{itemize} - \item Due to the fundamental frequency modulation, the switching frequency of the inverter is identical to the fundamental frequency: $f_{\mathrm{s}} = \nicefrac{\omega}{2\pi}$. - \item The star-to-ground voltage $u_{\mathrm{n0}}(t)$ shows a rectangular signal pattern with triple fundamental frequency. - \item Consequently, it does not influence the fundamental output voltage, that is, the fundamental components of the line-to-ground voltage $u_{2i0}(t)$ as well as the load voltage $u_{2i}(t)$ are identical: $\hat{u}^{(1)}_{2i0}=\hat{u}^{(1)}_{2i}$. - \end{itemize}\vspace{-0.25cm} + \item<1-> Due to the fundamental frequency modulation, the switching frequency of the inverter is identical to the fundamental frequency: $f_{\mathrm{s}} = \nicefrac{\omega}{2\pi}$. + \item<2-> The star-to-ground voltage $u_{\mathrm{n0}}(t)$ shows a rectangular signal pattern with triple fundamental frequency. + \item<3-> Consequently, it does not influence the fundamental output voltage, that is, the fundamental components of the line-to-ground voltage $u_{2i0}(t)$ as well as the load voltage $u_{2i}(t)$ are identical: $\hat{u}^{(1)}_{2i0}=\hat{u}^{(1)}_{2i}$. + \end{itemize}\vspace{-0.25cm}\onslide<4->{% \begin{varblock}{Note on the star point} The previous analysis assumed a non-connected star point, which comes with certain advantages, e.g., on the rejection of current harmonics. If, however, the star point would be connected, the three-phase converter can be interpreted and analyzed as three independent single-phase converters (each driven by a half bridge). - \end{varblock} + \end{varblock}} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% From ee4726f8944ecc7ec2a9631329a188938a647709 Mon Sep 17 00:00:00 2001 From: SevenOfNinePE Date: Tue, 28 Jan 2025 11:31:00 +0100 Subject: [PATCH 6/6] Update Ex07 Task2 according review result Ex07_Tasks_OW.pdf --- .../ex07/Fig_ThreePhaseInverter_6StepMode.tex | 10 ++-- exercise/tex/exercise07.tex | 55 ++++++++++--------- 2 files changed, 34 insertions(+), 31 deletions(-) diff --git a/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex b/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex index b9540ac..0bbcf9a 100644 --- a/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex +++ b/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex @@ -5,9 +5,9 @@ \begin{center} \begin{circuitikz} % Add voltage U1p - \draw (0,0) coordinate (U1p) to [open, o-o, v = $U_1p\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (Gnd) + \draw (0,0) coordinate (U1p) to [open, o-o, v = $\frac{U_1}{2}\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (Gnd) (Gnd) node[rground, rotate = 270 ](){} ++(0.4,0) - (Gnd) to [open, -o, v = $U_1m\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (U1m) + (Gnd) to [open, -o, v = $\frac{U_1}{2}\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (U1m) % Add current (U1p) to [short, o-, i=$i_1(t)$] ++(2,0) coordinate (jT1c) % Add T1 @@ -53,17 +53,17 @@ % Add u2a inductor (ju2ax) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2ae) % Add u2ae - (ju2ae) to [sV=$u_\mathrm{2ae}$] ++(1.5,0) coordinate (ju2an) + (ju2ae) to [sV=$u_\mathrm{2ai}$] ++(1.5,0) coordinate (ju2an) % Add u2b inductor (ju2b) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2be) % Add u2be - (ju2be) to [sV=$u_\mathrm{2be}$] ++(1.5,0) coordinate (ju2bn) + (ju2be) to [sV=$u_\mathrm{2bi}$] ++(1.5,0) coordinate (ju2bn) % Add connection to u2c inductor (ju2c) to [short,-] ++(0,-2) coordinate (ju2cx) % Add u2a inductor (ju2cx) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2ce) % Add u2ce - (ju2ce) to [sV=$u_\mathrm{2ce}$] ++(1.5,0) coordinate (ju2cn) + (ju2ce) to [sV=$u_\mathrm{2ci}$] ++(1.5,0) coordinate (ju2cn) % Add connection of u2in (ju2an) to [short,-*] (ju2bn) to [short,-] (ju2cn) % Add connection point u2n diff --git a/exercise/tex/exercise07.tex b/exercise/tex/exercise07.tex index ca0ba83..6c5ca08 100644 --- a/exercise/tex/exercise07.tex +++ b/exercise/tex/exercise07.tex @@ -61,41 +61,42 @@ %% Task 2: Three-phase inverter in six-step mode %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\task{Symmetrical 3-phase switching rectifier} +\task{Symmetrical 3-phase rectifier} -An symmetrical switching rectifier in three-phase bridge topology shall supply a -symmetrical three-phase consumer in star connection. The consumer is simulated by an inductance and -a sinusoidal counter voltage per phase. The inverter is operated with a basic switching frequency. -The switching elements are considered as ideal. +A rectifier in three-phase bridge topology shall supply a +symmetrical three-phase load in star connection. The load is represented by an inductance and +a sinusoidal internal (or inner) voltage per phase. The inverter is operated with the fundamental +frequency modulation (also known as six-step mode) and the switching elements are considered as ideal. +The schematic is depicted in \autoref{fig:Fig_ThreePhaseInverter_6StepMode}. -% \input{fig/ex07/Fig_ThreePhaseInverter_6StepMode} \input{fig/ex07/Fig_ThreePhaseInverter_6StepMode} \begin{table}[ht] \centering % Center the table \begin{tabular}{ll} \toprule - Input voltages: & $U_\mathrm{1}=\SI{510}{\volt}$ \quad $U_\mathrm{1p}=U_\mathrm{1m}=U_\mathrm{1}/2$ \\ - Internal voltages: & $u_{\mathrm{2ae}}(t) = \sqrt{2} \cdot \SI{220}{\volt} \cdot \sin(\omega_1t)$ \\ - Circular frequency: & $\omega_1 = \SI{2 \pi \cdot 30}{\frac{1}{\second}}$ \\ - Inductivity per phase: & $L= \SI{10}{\milli \henry}$ \\ - Phase angle between $u_{\mathrm{2ae}}(t)$ and $i_{\mathrm{2ae}^\mathrm{(1)}}(t)$ & $\alpha=\SI{30}{\degree}$ \\ + Input voltages: & $U_\mathrm{1}=\SI{510}{\volt}$ \\ + Internal voltages: & $u_{\mathrm{2ai}}(t) = \sqrt{2} \cdot \SI{220}{\volt} \cdot \sin(\omega_1t)$ \\ + Angular load frequency: & $\omega_1 = \SI{2 \pi \cdot 30}{\frac{1}{\second}}$ \\ + Inductance per phase: & $L= \SI{10}{\milli \henry}$ \\ + Phase angle between $u_{\mathrm{2ai}}(t)$ and $i_{\mathrm{2ai}}^\mathrm{(1)}(t)$ & $\alpha=\SI{30}{\degree}$ \\ \bottomrule \end{tabular} \caption{Parameters of three-phase inverter in six-step mode.} \label{table:ex07_Task2_ParametersOfTheCircuit} \end{table} - -\subtask{Create a table with all possible switching states for basic switching frequency. -Use the following notation: \\ -$(s_\mathrm{a}(t),s_\mathrm{b}(t),s_\mathrm{c}(t))=\begin{cases} +\subtask{Create a table with all possible switching states for fundamental frequency modulation. +Use the following notation: +\begin{align*} + \{ s_\mathrm{a}(t),s_\mathrm{b}(t),s_\mathrm{c}(t) \}=\begin{cases} s_i(t)= +1 & \text{upper position,}\\ s_i(t)= -1 & \text{lower position.} - \end{cases}$\\ + \end{cases} \bigskip -Sketch the switching states in the correct chronological order for minimum one periode. -Calculate and sketch the voltages $u_\mathrm{a0}(t)$, $u_\mathrm{b0}(t)$ and $u_\mathrm{c0}(t)$ +\end{align*} +Sketch the switching states in the correct chronological order for one periode. +Calculate and sketch the voltages $u_\mathrm{2a0}(t)$, $u_\mathrm{2b0}(t)$ and $u_\mathrm{2c0}(t)$ depending on these switching states.} \begin{solutionblock} Each half bridge has got the 2 states '+1' and '-1', which results in $2^3 = 8$ combinations according table \autoref{stable:ex07_Task2_Switchingstates}. @@ -144,9 +145,9 @@ the switching only occurs twice per period. This results in the maximum possible voltage (square wave) at the output. \end{solutionblock} -\subtask{The internal voltages $u_\mathrm{ea}(t)$, $u_\mathrm{eb}(t)$ and $u_\mathrm{ec}(t)$ are a symmetrical voltage system, -i.e. the following is always applicable: $u_\mathrm{ea}(t)+u_\mathrm{eb}(t)+u_\mathrm{ec}(t)=0V$. -Show that this equation is also applicable for the voltages $u_\mathrm{a}(t)$, $u_\mathrm{b}(t)$ and $u_\mathrm{c}(t)$ under the same conditions. +\subtask{The internal voltages $u_\mathrm{2ai}(t)$, $u_\mathrm{2bi}(t)$ and $u_\mathrm{2ci}(t)$ are a symmetrical voltage system, +i.e., the following is always applicable: $u_\mathrm{2ai}(t)+u_\mathrm{2bi}(t)+u_\mathrm{2ci}(t)=\SI{0}{\volt}$. +Show that this equation is also applicable for the voltages $u_\mathrm{2a}(t)$, $u_\mathrm{2b}(t)$ and $u_\mathrm{2c}(t)$ under the same conditions. } \begin{solutionblock} In the case of a symmetrical three-phase consumer where the current sum at the consumer star point is zero, @@ -229,11 +230,13 @@ \end{solutionblock} \subtask{Decompose the voltage $u_\mathrm{2a}(t)$ into a Fourier series and sketch the spectral lines related to the -amplitude of the fundamental signal up to order n=13. Hint: The following applies to the Fourier coefficients of an odd and alternating function: +amplitude of the fundamental signal up to order $k=13$. \\ +Hint: \begin{align*} - b_k = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} f(x)\sin(kx) \mathrm{d}x \quad k =\mathrm{odd} \quad \quad + b_k = \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} f(x)\sin(kx) \mathrm{d}x \quad k =\mathrm{odd}. \end{align*} \label{sub:DecomposeVoltage} +The formula above applies to the Fourier coefficients of an odd and alternating function. } \begin{solutionblock} In the case of odd and alternating functions corresponding to $f(x)=-f(x+\pi)$ the Fourier coefficients are: @@ -284,9 +287,9 @@ \label{subtask:Ex07T2_FourierSeries} \end{solutionblock} -\subtask{Based on \autoref{sub:DecomposeVoltage}, calculate the fundamental amplitude $\hat{i}_\mathrm{a}^1$ using a vector diagram and +\subtask{Based on \autoref{sub:DecomposeVoltage}, calculate the fundamental amplitude $\hat{i}_\mathrm{a}^\mathrm{(1)}$ using a vector diagram and complex alternating current calculations. -From this, determine the total active power converted in the load.} +From this, determine the total active power fed to the load.} \begin{solutionblock} According \eqref{eq:Ex07T2_FundamentelVoltage} the fundamental voltage is calculated as: \begin{equation} @@ -353,4 +356,4 @@ P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\phi) = \sqrt{3} \SI{397.6}{\volt} \cdot \SI{8.75}{\ampere} \cdot 0.83= \SI{5}{\kilo\watt} \end{equation} -\end{solutionblock} +\end{solutionblock} \ No newline at end of file