Merge branch 'main' of https://github.com/IAS-Uni-Siegen/PE_course

exercise/fig/ex07/Fig_Voltage_U_um_excerpt.tex → 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}

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}

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} {

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}