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ET2_Uebung_BEI/ET2_L_B12_A6.tex

ET2_L_B12_A6.tex 4.3KB
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  1. \section{Rechteckspannung}

  2. Gegeben ist eine periodische Rechteckspannung mit der Periodendauer von $10\,\milli\second$.\\

  3. Berechnen Sie den Effektivwert, wenn der arithmetische Mittelwert gleich Null ist.\\

  4. \begin{align*}

  5. 	\begin{tikzpicture}[scale=0.5]

  6. 		\begin{scope}[>=latex,thick] 			

  7. 			\draw [->](0,-2) -- (10.75,-2) node [right] {$t\,[\milli\second]$};

  8. 			\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};

  9. 			\draw [<->,blue, very thick] (-.5,0)--(-.5,5) node at (0,2.5)[right]{$5\,\volt$};

  10. 			\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)

  11. 						--(10,5)--(10.5,5);				

  12. 				\foreach \x in {0,1,...,10}

  13. 					\draw (\x,-2) -- (\x,-2.2) node[anchor=north] {$\x$};

  14. 				\foreach \y in {0,1,...,5}

  15. 					\draw (0,\y) -- (-0.2,\y);% node[anchor=east] {$\y$};

  16. 		\end{scope}

  17. 	\end{tikzpicture}

  18. \end{align*}

  19. \ifthenelse{\equal{\toPrint}{Lösung}}{%

  20. \begin{align}

  21. \intertext{Formeln:}

  22. \overline{u}&=\frac{1}{T}\cdot \int_{t=0}^{T}{u(t)\cdot dt}&\text{Arithmetischer Mittelwert}\\

  23. U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&

  24. \end{align}

  25. \begin{align*}

  26. \intertext{Berechnung:}

  27. \intertext{a) Arithmetischer Mittelwert $\overline{u}=0$}

  28. &\text{$\Rightarrow$ Fläche ober- und unterhalb der Nulllinie muß gleich sein!}\\

  29. &\text{$\Rightarrow$ Wo ist die Nulllinie?}

  30. \end{align*}

  31. \begin{align*}

  32. 	\begin{tikzpicture}[scale=0.5]

  33. 		\begin{scope}[>=latex,thick] 			

  34. 			\draw [->](0,0) -- (10.75,0) node [right] {$t\,[\milli\second]$};

  35. 			\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};

  36. 			\fill [black!15!] (0,3.5)rectangle(7,5) (7,3.5)rectangle(10,0);

  37. 			\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)

  38. 						--(10,5)--(10.5,5);	

  39. 			\draw [blue,very thick] (0,3.5)--(10,3.5) node [right]{NULL};	

  40. 			\draw [->,blue,very thick] (3.5,3.5)--(3.5,5) node at(3.5,4.25)[right]{$u_1$};			 

  41. 			\draw [<-,blue,very thick] (8.5,0)--(8.5,3.5) node  at(8.5,1.75)[right]{$u_2$};

  42. 				\foreach \x in {0,1,...,10}

  43. 					\draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$};

  44. 				\foreach \y in {-3.5,-2.5,...,1.5}

  45. 					\draw (0,\y+3.5) -- (-0.2,\y+3.5) node[anchor=east] {$\y$};

  46. 		\end{scope}

  47. 	\end{tikzpicture}

  48. \end{align*}

  49. \enlargethispage{1cm}

  50. \begin{align*}

  51. &u_1+(-u_2)=5\,\volt\rightarrow u_2=-(5\,\volt-u_1)\\

  52. &\text{Fläche:}\\

  53. &\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\

  54. &\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\

  55. &u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\

  56. &u_1\cdot 7\,\milli\second-(5\,\volt-U_1)\cdot 3\,\milli\second=0\\

  57. &(7\,\milli\second+3\,\milli\second)\cdot u_1=15\,\volt\cdot \milli\second\\

  58. &u_1=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uuline{1{,}5\,\volt}\\

  59. &u_2=-(5\,\volt-1{,}5\,\volt)=\uuline{-3{,}5\,\volt}

  60. \intertext{Alternativ mit Beträgen}

  61. &|u_1|+|u_2|=5\,\volt\rightarrow |u_2|=5\,\volt-|u_1|\\

  62. &\text{Fläche:}\\

  63. &\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\

  64. &\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\

  65. &u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\

  66. &u_1\cdot 7\,\milli\second-(5\,\volt-|u_1|)\cdot 3\,\milli\second=0\\

  67. &(7\,\milli\second+3\,\milli\second)\cdot |u_1|=15\,\volt\cdot \milli\second\\

  68. &|u_1|=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uline{1{,}5\,\volt}\qquad

  69. |u_2|=5\,\volt-1{,}5\,\volt)=\uline{3{,}5\,\volt}\\

  70. &\text{Da $u_1$ positives Vorzeichen in der Skizze hat, muss $u_2$ ein negatives Vorzeichen erhalten.}\\

  71. \Rightarrow& u_1=\uuline{1{,}5\,\volt} \qquad u_2=\uuline{-3{,}5\,\volt}

  72. \intertext{b) Effektivwert}

  73. U^2&=U^2_{\textrm{eff}}=\frac{1}{T}\int{\left(u(t)\right)^2\cdot dt}\\

  74. &=\frac{1}{10\,\milli\second}\left(\int_{0}^{7\,\milli\second}{(1{,}5\,\volt)^2\cdot dt}+\int_{7\,\milli\second}^{10\,\milli\second}{(-3{,}5\,\volt)^2\cdot dt}\right)\\

  75. &=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot \big[t\big]_{0}^{7\,\milli\second}+12{,}25\,\volt^2\cdot \big[t\big]_{7\,\milli\second}^{10\,\milli\second}\right)\\

  76. &=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot 7\,\milli\second+12{,}25\,\volt^2\cdot (10\,\milli\second-7\,\milli\second)\right)=5{,}25\,\volt^2\\

  77. U&=\sqrt{5{,}25\,\volt^2}=\uuline{2{,}29\,\volt}

  78. \end{align*}

  79. \clearpage

  80. }{}%