From 991dee617ef49f2f9c7147d78a07b5a8858dd413 Mon Sep 17 00:00:00 2001 From: SevenOfNinePE Date: Mon, 27 Jan 2025 21:26:23 +0100 Subject: [PATCH] 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}