From e44b28941c63ab376e98902e7c28f74b779f6cdf Mon Sep 17 00:00:00 2001 From: Ali Date: Wed, 22 Jan 2025 00:29:09 +0100 Subject: [PATCH 1/3] Units and alpha greater than pi/2 --- exercise/fig/ex06/Solution_subtask_6_1_2.tex | 10 +++++----- exercise/tex/exercise06.tex | 6 +++--- 2 files changed, 8 insertions(+), 8 deletions(-) diff --git a/exercise/fig/ex06/Solution_subtask_6_1_2.tex b/exercise/fig/ex06/Solution_subtask_6_1_2.tex index c92a689..edbe4ed 100644 --- a/exercise/fig/ex06/Solution_subtask_6_1_2.tex +++ b/exercise/fig/ex06/Solution_subtask_6_1_2.tex @@ -11,16 +11,16 @@ \begin{axis}[ % x/y range adjustment xmin=0, xmax=180, - ymin=0, ymax=1, + ymin=-1, ymax=1, samples=500, axis y line=center, - axis x line=bottom, + axis x line=middle, extra y ticks=0, % Label text xlabel={$\alpha / \text{rad}$}, ylabel={$\frac{ U_\mathrm{2}(\alpha)}{U_\mathrm{2}(\alpha=0)}$}, % Label adjustment - x label style={at={(axis description cs:1,-0.1)},anchor=south west}, + x label style={at={(axis description cs:1,0.4)},anchor=south west}, y label style={at={(axis description cs:-.05,.97)},anchor=south,yshift=0.2cm}, width=0.6\textwidth, height=0.3\textwidth, @@ -29,8 +29,8 @@ xticklabels={0,0.871,$\frac{\pi}{2}$,$\pi$}, xticklabel style = {anchor=north}, % y-Ticks - ytick={0,0.5,0.644,1}, - yticklabels={0,0.5,0.644,1}, + ytick={-1,-0.5,0,0.5,0.644,1}, + yticklabels={-1,,0,,0.644,1}, yticklabel style = {anchor=east}, % Grid layout grid, diff --git a/exercise/tex/exercise06.tex b/exercise/tex/exercise06.tex index 185cdea..34612cc 100644 --- a/exercise/tex/exercise06.tex +++ b/exercise/tex/exercise06.tex @@ -34,7 +34,7 @@ \end{equation} where \begin{equation} - \hat{u}_\mathrm{1} = \sqrt{2}U_\mathrm{1} = \sqrt{2} \cdot 230 \approx \SI{325.27}{\volt}. + \hat{u}_\mathrm{1} = \sqrt{2}U_\mathrm{1} = \sqrt{2} \cdot \SI{230}{\volt} \approx \SI{325.27}{\volt}. \label{sub6.1.1:eq:voltage_amplitude} \end{equation} Hence, @@ -76,7 +76,7 @@ period is equal to $\frac{2\pi}{3}$. Thus, $a_1$ ($\hat{i}^\mathrm{(1)}_\mathrm{1a}$) can be calculated as: \begin{equation} - a_1 = \hat{i}^\mathrm{(1)}_\mathrm{1a} = \frac{2}{\pi}\int_{0}^{\frac{\pi}{3}}I_\mathrm{2}\cos(\omega t) d\omega t = \left[\frac{2}{\pi} \sin(\omega t)\right]^{\frac{\pi}{3}}_{0} = \frac{1}{\pi}I_\mathrm{2}\sqrt{3}. + a_1 = \hat{i}^\mathrm{(1)}_\mathrm{1a} = \frac{2}{\pi}\int_{0}^{\frac{\pi}{3}}I_\mathrm{2}\cos(\omega t) \mathrm{d}\omega t = \left[\frac{2}{\pi} \sin(\omega t)\right]^{\frac{\pi}{3}}_{0} = \frac{1}{\pi}I_\mathrm{2}\sqrt{3}. \label{sub6.1.4:eq:calculate_a1} \end{equation} @@ -101,7 +101,7 @@ \begin{solutionblock} The fundamental reactive power drawn for one full period $(2\pi)$ can be calculated using \begin{equation} - Q^\mathrm{(1)} = 3 U_1 I^\mathrm{(1)}_1 \sin(\alpha) = 3 \cdot 230 \cdot 13.5 \cdot \sin(0.871) \approx \SI{7.12}{\kilo \volt \ampere}. + Q^\mathrm{(1)} = 3 U_1 I^\mathrm{(1)}_1 \sin(\alpha) = 3 \cdot \SI{230}{\volt} \cdot \SI{13.5}{\ampere} \cdot \sin(0.871) \approx \SI{7.12}{\kilo \volt \ampere}. \end{equation} \end{solutionblock} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% From 5df4f1982e0c1cd94123db47a744de2458e3f607 Mon Sep 17 00:00:00 2001 From: "[SilasElter]" <[SilasElter]> Date: Wed, 22 Jan 2025 15:07:17 +0100 Subject: [PATCH 2/3] Add Solutionfigures ex07 task2 --- .../ex07/Fig_graphic_solutions_cos_terms.tex | 178 +++++++++--------- 1 file changed, 86 insertions(+), 92 deletions(-) diff --git a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex index c813237..e583461 100644 --- a/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex +++ b/exercise/fig/ex07/Fig_graphic_solutions_cos_terms.tex @@ -4,103 +4,97 @@ \begin{figure}[htb] \centering - \begin{tikzpicture} - \begin{axis}[ - axis lines=middle, % Achsen wie im Koordinatensystem - xlabel={$\cos(\varphi)$}, - ylabel={$\sin(\varphi)$}, - width=7cm, height=7cm, - grid=both, - 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, -0.5, 0, 0.5, 1}, - % ytick={-1, -0.5, 0, 0.5, 1}, - ] - \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, % 360° für den ganzen Kreis - samples=200, % Anzahl der Punkte - thick, - color=blue, - ] - ({cos(x)}, {sin(x)}); % Kreisparameter (cos, sin) - - \end{axis} - \end{tikzpicture} - - \hspace{1cm} % Abstand zwischen den beiden Diagrammen - - \begin{tikzpicture} - % Achsen - \draw[->, thick] (-2,0) -- (2,0) node[right] {$(x)$}; - \draw[->, thick] (0,-2) -- (0,2) node[above] {$(y)$}; + \begin{minipage}[t]{0.45\textwidth} + \centering + \begin{tikzpicture} + \begin{axis}[ + axis lines=middle, + xlabel={$\cos(\varphi)$}, + ylabel={$\sin(\varphi)$}, + 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] {$\varphi$}; - % Punkte auf den Achsen - \foreach \x in {-2,2} { - \draw[thick] (\x,0) node[below] {\x} -- (\x,0.2); - } - \foreach \y in {-1,1} { - \draw[thick] (0,\y) -- (0.2,\y) node[right] {\y}; - } - - % Werte cos(k*pi/2) - \node at (1.5,1.5) {$\cos\left(k \frac{\pi}{2}\right)$}; - \node[below right] at (0.1,0) {1,5,9,...}; - \end{tikzpicture} - \caption{Switch-on behavior and switch-off behavior of $i_{\mathrm{C}}(t)$.} - \label{fig:Switch-on behavior and switch-off behavior of} + \addplot[ + domain=0:360, + samples=200, + thick, + color=blue, + ] + ({cos(x)}, {sin(x)}); + \end{axis} + \end{tikzpicture} + \end{minipage} + \hspace{0.5cm} + \begin{minipage}[t]{0.45\textwidth} + \centering + \begin{tikzpicture} + \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] {$3,7,11,15,\ldots$}; + % Kreuz bei jedem markierten Punkt + \foreach \x/\y in {-1.5/0, 1.5/0, 0/1.5, 0/-1.5} { + \draw[thick] + (\x,\y) +(-0.1,0.1) -- +(0.1,-0.1) % Diagonale des Kreuzes + +(-0.1,-0.1) -- +(0.1,0.1); % Andere Diagonale des Kreuzes + } + \end{tikzpicture} + \end{minipage} + \begin{tikzpicture} - \begin{axis}[ - width=7cm, height=4.5cm, - grid=both, - major grid style={line width=.2pt,draw=gray!50}, - minor grid style={line width=.1pt,draw=gray!20}, - xlabel={$t$ / ns}, - ylabel={$u_{\mathrm{CE}}(t)$ / V}, - xmin=0, xmax=150, - ymin=0, ymax=800, - xtick={0, 50, 100, 150}, - ytick={0,200, 400, 600, 800}, - ] - % Einschaltverhalten graph - \addplot[ - thick, - mark=none, - color=black, - ] coordinates { - (0,600) (100, 600) (100, 0) - }; - \end{axis} + \draw[->] (-2,0) -- (2,0) node[right] {}; + \draw[->] (0,-2) -- (0,2) node[above] {}; + \node at (-1.5,1.5) {$\cos\left(k \frac{\pi}{3}\right)$}; + \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] {$3,7,11,15,\ldots$}; + + % Kreuz bei jedem markierten Punkt + \foreach \x/\y in {-1.5/0, 1.5/0, 0/1.5, 0/-1.5} { + \draw[thick] + (\x,\y) +(-0.1,0.1) -- +(0.1,-0.1) % Diagonale des Kreuzes + +(-0.1,-0.1) -- +(0.1,0.1); % Andere Diagonale des Kreuzes + } \end{tikzpicture} \hspace{1cm} % Abstand zwischen den beiden Diagrammen \begin{tikzpicture} - \begin{axis}[ - width=7cm, height=4.5cm, - grid=both, - major grid style={line width=.2pt,draw=gray!50}, - minor grid style={line width=.1pt,draw=gray!20}, - xlabel={$t$ / ns}, - ylabel={$u_{\mathrm{CE}}(t)$ / V}, - xmin=0, xmax=600, - ymin=0, ymax=1000, - xtick={0,200, 400, 600}, - ytick={0,200, 400, 600,800, 1000}, - ] - % Ausschaltverhalten graph - \addplot[ - thick, - mark=none, - color=black, - ] coordinates { - (50,0) (200, 900) (250, 600) (600, 600) - }; - \end{axis} + % Koordinatensystem zeichnen + \draw[->] (-2,0) -- (2,0) node[right] {}; + \draw[->] (0,-2) -- (0,2) node[above] {}; + \node at (-1.5,1.5) {$\cos\left(k \frac{\pi}{3}\right)+1$}; + + % Beschriftungen an der x-Achse + \foreach \x in {0, 1, 1.5} { + \node at (\x, 0) [below] {\x}; + } + + % Beschriftungen an spezifischen Punkten + \node at (1.5,1.1) [above] {$1,7,13,19,\ldots$}; + \node at (1.5,-1.1) [below] {$5,11,17,23,\ldots$}; + + % Kreuz bei jedem markierten Punkt + \foreach \x/\y in {0/0, 1.5/1, 1.5/-1} { + \draw[thick] + (\x,\y) +(-0.1,0.1) -- +(0.1,-0.1) % Diagonale des Kreuzes + +(-0.1,-0.1) -- +(0.1,0.1); % Andere Diagonale des Kreuzes + } + + % Verbindungslinien zwischen den Punkten + \draw[thick, color=blue!70!black] + (0,0) -- (1.5,1) -- (1.5,-1) -- cycle; \end{tikzpicture} - \caption{Switch-on behavior and switch-off behavior of $u_{\mathrm{CE}}(t)$.} - \label{fig:Switch-on behavior and switch-off behavior of voltage} + \caption{Graphical solution of the cos terms.} + \label{fig:Graphical solution of the cos terms} \end{figure} From a938f5449615ae2010c82cf3fb223e091314f900 Mon Sep 17 00:00:00 2001 From: SevenOfNinePE Date: Wed, 22 Jan 2025 18:42:31 +0100 Subject: [PATCH 3/3] Ex07 Task2: Add remaining subtask and update Fig_ThreePhaseInverter_6StepMode --- .../ex07/Fig_ThreePhaseInverter_6StepMode.tex | 141 +++++++++++------- exercise/tex/exercise07.tex | 44 +++++- 2 files changed, 125 insertions(+), 60 deletions(-) diff --git a/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex b/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex index 68cda43..e679b6a 100644 --- a/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex +++ b/exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex @@ -4,64 +4,91 @@ \begin{figure}[htb] \begin{center} \begin{circuitikz} - \def\vd{1.5cm} % vertical distance AC sources - \def\hd{1.5cm} % horizontal distance diode bridge - \def\h1d{5.0cm} % horizontal position first diode string - % Base point for voltage supplies - \coordinate (orig) at (0,0); - % Voltage sources and neutral connection - \draw - % draw the neutral connection - (0,0) to [short, -*] ++(0,-1.5) to [short] ++(0,-1.5) - % draw first phase ua - (0,0) to [sinusoidal voltage source, v^<=$u_{1\mathrm{a}}(t)$] ++(1.5, 0) to [short, i=$i_{1\mathrm{a}}(t)$]++(0.75,0) -- ++(0.25,0) coordinate (A) - % draw second phase ub - (0,-1*\vd) to [sinusoidal voltage source, v^<=$u_{1\mathrm{b}}(t)$] ++(1.5, 0) to [short, i=$i_{1\mathrm{b}}(t)$]++(0.75,0) -- ++(0.25,0) coordinate (B) - % draw third phase uc - (0,-2*\vd) to [sinusoidal voltage source, v^<=$u_{1\mathrm{c}}(t)$] ++(1.5,0) to [short, i=$i_{1\mathrm{c}}(t)$]++(0.75,0) -- ++(0.25,0) coordinate (C) - %thyristor bridge - % Add thyristor T1 - (\h1d,0) to [thyristor, l=$T_1$, name=D1] ++(0,1.25) coordinate (D1top) - % Add thyristor T2 - (\h1d,-4.25) coordinate (D2bot) to [thyristor, l=$T_2$, name=D2] ++(0,1.25) to [short] (\h1d, 0) - % Add connection to junction A - (\h1d, 0) to [short, *-] (A) - % Add thyristor T3 - (\h1d+\hd,0) to [thyristor, l=$T_3$, name=D3] ++(0,1.25) coordinate (D3top) - % Add thyristor T4 - (\h1d+\hd,-4.25) coordinate (D4bot) to [thyristor, l=$T_4$, name=D4] ++(0,1.25) to [short] (\h1d+\hd, 0) - % Add thyristor T5 - (\h1d+2*\hd,0) to [thyristor, l=$T_5$, name=D5] ++(0,1.25) coordinate (D5top) - % Add thyristor T6 - (\h1d+2*\hd,-4.25) coordinate (D6bot) to [thyristor, l=$T_6$, name=D6] ++(0,1.25) to [short] (\h1d+2*\hd, 0) - % Add connection to junction B - (B -| D3) to [crossing, *-, mirror] ++(-2*\hd,0) -- (B) - % Add connection to junction C - (C -| D5) to [short, *-] ++(-\hd/2,0) to [crossing, mirror] ++(-\hd,0) to [crossing, mirror] ++(-\hd,0) -- (C) - % Add wire T1-T3-T5 - (D1top) to [short, -*] (D3top) to [short, -*] (D5top) to [short, -] ++(0.5,0) coordinate (jL1) - % Add inductor L and motor current - (jL1) to [L, l=$L$, name = L] ++(2,0) to [short,i=$\overline{i}_\mathrm{mot}$] ++(0.5,0) coordinate (jL2) - % Add DC-motor and motor voltage - (jL2) ++ (0,-3) node[elmech](motor){M} - (jL2) to (motor.north) - (motor.bottom) to (D6bot -| \tikztostart) to (D6bot) - % (jL2) to [R, l=$R$, name = R, v_>=$\overline{u}_\mathrm{mot}$, voltage = straight] (D6bot -| \tikztostart) to (D6bot) + % 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) + (Gnd) to [short,o-o] ++(1,0) + (Gnd) to [open, -o, v = $U_1m\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (U1m) + % Add current + (U1p) to [short, o-, i=$i_1(t)$] ++(2,0) coordinate (jT1c) + % Add T1 + (jT1c) to [Tnpn, n=T1, invert, bodydiode] ++(0,-2) coordinate (jT1e) + % Add connection to u2a + (jT1e) to [short, *-] ++(1,0) to [crossing] ++(2,0) to [crossing] ++(2,0) to [short,-] ++(1,0) coordinate (ju2a) + % Add junction to T2 + (jT1e) to [short] ++(0,-1) coordinate (jT2c) + % Add T2 + (jT2c) to [Tnpn, n=T2, invert, bodydiode] ++(0,-2) coordinate (jT2e) + % Add connection to T3 + (jT1c) to [short, *-] ++(2,0) coordinate (jT3c) + % Add T3 + (jT3c) to [Tnpn, n=T3, invert, bodydiode] ++(0,-2) coordinate (jT3e) + % Add junction to ju2b + (jT3e) to [short] ++(0,-0.5) coordinate (jmu2b) + % Add connection to u1b + (jmu2b) to [short,*-] ++(1,0) to [crossing] ++(2,0) to [short,-] ++(1,0) coordinate (ju2b) + % Add junction to T4 + (jmu2b) to [short] ++(0,-0.5) coordinate (jT4c) + % Add T4 + (jT4c) to [Tnpn, n=T4, invert, bodydiode] ++(0,-2) coordinate (jT4e) + % Add connection to T5 + (jT3c) to [short, *-] ++(2,0) coordinate (jT5c) + % Add T5 + (jT5c) to [Tnpn, n=T5, invert, bodydiode] ++(0,-2) coordinate (jT5e) + % Add junction to T6 + (jT5e) to [short] ++(0,-1) coordinate (jT6c) + % Add T6 + (jT6c) to [Tnpn, n=T6, invert, bodydiode] ++(0,-2) coordinate (jT6e) + % Add connection to T4 + (jT6e) to [short, -*] (jT4e) + % Add connection to T2 + (jT4e) to [short, -*] (jT2e) + % Add connection to U1m + (jT2e) to [short, -] (U1m) + % Add connection to u1c + (jT6c) to [short,*-] ++(2,0) coordinate (ju2c) + % Add connection to u2a inductor + (ju2a) to [short,-] ++(0,2) coordinate (ju2ax) + % Add u2a inductor + (ju2ax) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2ae) + % Add u2ae + (ju2ae) to [sV=$u_\mathrm{1ae}$] ++(1.5,0) coordinate (ju2an) + % Add u2b inductor + (ju2b) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2be) + % Add u2be + (ju2be) to [sV=$u_\mathrm{1be}$] ++(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=$u_\mathrm{1ce}$] ++(1.5,0) coordinate (ju2cn) + % Add connection of u2in + (ju2an) to [short,-*] (ju2bn) to [short,-] (ju2cn); + + + % Add component name of transistors + \draw let \p1 = (T1.B) in node[anchor=east] at (\x1,\y1) {$T_1$}; + \draw let \p1 = (T2.B) in node[anchor=east] at (\x1,\y1) {$T_2$}; + \draw let \p1 = (T3.B) in node[anchor=east] at (\x1,\y1) {$T_3$}; + \draw let \p1 = (T4.B) in node[anchor=east] at (\x1,\y1) {$T_4$}; + \draw let \p1 = (T5.B) in node[anchor=east] at (\x1,\y1) {$T_5$}; + \draw let \p1 = (T6.B) in node[anchor=east] at (\x1,\y1) {$T_6$}; + % Add current arrows i2a, i2b and i2c + \draw (jT1e) ++(1,0) node[currarrow](i2a){} + (i2a) node[anchor=north,color=black]{$i_\mathrm{2a}(t)$} + (jmu2b) ++(1,0) node[currarrow](i2b){} + (i2b) node[anchor=north,color=black]{$i_\mathrm{2b}(t)$} + (jT6c) ++(1,0) node[currarrow](i2c){} + (i2c) node[anchor=north,color=black]{$i_\mathrm{2c}(t)$} + % Add voltage arrows u2an, u2bn and u2cn + (ju2ax) ++(0,-0.8) to [open,v^=$u_\mathrm{2a}(t)$,voltage = straight] ++(3.8,0) + (ju2b) ++(0,-0.8) to [open,v^=$u_\mathrm{2b}(t)$,voltage = straight] ++(3.8,0) + (ju2cx) ++(0,-0.8) to [open,v^=$u_\mathrm{2c}(t)$,voltage = straight] ++(3.8,0) + % Add voltage arrows u2ab and u2bc + (ju2ax) ++(0.2,0) to [open,v^=$u_\mathrm{2ab}(t)$,voltage = straight] ++(0,-2.5) + (ju2b) ++(0.2,0) to [open,v^=$u_\mathrm{2bc}(t)$,voltage = straight] ++(0,-2.5); + - % Add wire T2-T3-T6 - (D2bot) to [short, -*] (D4bot) to [short, -*] (D6bot) - % Add voltage arrow u2(t) between Dtop and Dbot - (jL1) to [open, v^>=$\hspace{0.5cm}u_2(t)$, voltage = straight] (D6bot-|jL1) - % Add voltage arrow u2+n(t) between Dtop and neutral - (D1top) ++(-0.2,0) to [open, v_>=$u_\mathrm{2,p}(t)$, voltage = straight] ++(-5.5,0) - % Add voltage arrow u2-n(t) between Dbot and neutral - (D2bot) ++(-0.2,0) to [open, v_>=$u_\mathrm{2,m}(t)$, voltage = straight] ++(-5.5,0) - % Add voltage arrow between AC source a and b - (A) to [open, v^>=$\hspace{0.75cm}u_{1\mathrm{ab}}(t)$, voltage = straight] (B) - % Add voltage arrow between AC source b and c - (B) to [open, v^>=$\hspace{0.75cm}u_{1\mathrm{bc}}(t)$, voltage = straight] (C) - % Add voltage arrow between AC source a and c - (-0.5,-2*\vd) to [open, v^>=$u_{1\mathrm{ca}}(t)\hspace{0.75cm}$, voltage = straight] (-0.5,0); \end{circuitikz} \end{center} \caption{Three-phase inverter in six-step mode.} diff --git a/exercise/tex/exercise07.tex b/exercise/tex/exercise07.tex index 2d95368..5dbce2c 100644 --- a/exercise/tex/exercise07.tex +++ b/exercise/tex/exercise07.tex @@ -21,10 +21,48 @@ a sinusoidal counter voltage per phase. The inverter is operated with a basic frequency clock. The switching elements are considered as ideal. +% \input{fig/ex07/Fig_ThreePhaseInverter_6StepMode} \input{fig/ex07/Fig_ThreePhaseInverter_6StepMode} -\subtask{Aufgabe 4} + +\subtask{Create a table with all possible switching states for basic frequency clocking. +Use the following notation: \\ +$(s_\mathrm{a}(t),s_\mathrm{b}(t),s_\mathrm{c}(t))=\begin{cases} + s_i(t)= +1 & \text{upper position,}\\ + s_i(t)= -1 & \text{lower position.} + \end{cases}$\\ +Sketch the switching states in the correct chronological order for minimum one periode. +Calculate and sketch the voltages $u_\mathrm{a,0}(t)$, $u_\mathrm{b,0}(t)$ and $u_\mathrm{c,0}(t)$ depending on these switching states. +} +\begin{solutionblock} +\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 always applies: $u_\mathrm{ea}(t)+u_\mathrm{eb}(t)+u_\mathrm{ec}(t)=0V$. +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. +} +\begin{solutionblock} +\end{solutionblock} + +\subtask{Calculate and sketch the voltages $u_\mathrm{ab}(t)$, $u_\mathrm{bc}(t)$, $u_\mathrm{a}(t)$ and $u_\mathrm{a,0}(t)$ +depending on these switching states.} +\begin{solutionblock} +\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 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 +\end{align*} +\label{sub:DecomposeVoltage} +} +\begin{solutionblock} + \input{fig/ex07/Fig_Voltage_U_um_excerpt} + \input{fig/ex07/Fig_graphic_solutions_cos_terms} +\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. +From this, determine the total active power converted in the load.} \begin{solutionblock} -\input{fig/ex07/Fig_Voltage_U_um_excerpt} -\input{fig/ex07/Fig_graphic_solutions_cos_terms} \end{solutionblock} \ No newline at end of file