Grundlagen Elektrotechnik 2 (GET2)19 VierpoleProf. Dr. Jörg Vollrath18 Bode Diagramm |
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Länge: 1:22:04 |
0:0:0 Evaluierung 0:0:0 Differenzverstärker 0:2:0 Eingangs und Ausgangswiderstand |
Widerstandsmatrix Z |
Reihen-Parallel-Matrix H |
\( \left( \begin{array}{r} \underline{U}_1 \\ \underline{U}_2 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{Z}_{11} & \underline{Z}_{12} \\ \underline{Z}_{12} & \underline{Z}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{I}_1 \\ \underline{I}_2 \\ \end{array} \right) \) | \( \left( \begin{array}{r} \underline{U}_1 \\ \underline{I}_1 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{H}_{11} & \underline{H}_{12} \\ \underline{H}_{12} & \underline{H}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{I}_1 \\ \underline{U}_2 \\ \end{array} \right) \) |
Leitwertmatrix Z |
Parallel-Reihen-Matrix P/G |
\( \left( \begin{array}{r} \underline{I}_1 \\ \underline{I}_2 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{Y}_{11} & \underline{Y}_{12} \\ \underline{Y}_{12} & \underline{Y}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{U}_1 \\ \underline{U}_2 \\ \end{array} \right) \) | \( \left( \begin{array}{r} \underline{I}_1 \\ \underline{U}_2 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{P}_{11} & \underline{P}_{12} \\ \underline{P}_{12} & \underline{P}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{U}_1 \\ \underline{I}_2 \\ \end{array} \right) \) |
Kettenmatrix A |
Inverse Kettenmatrix B |
\( \left( \begin{array}{r} \underline{U}_1 \\ \underline{I}_1 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{A}_{11} & \underline{A}_{12} \\ \underline{A}_{12} & \underline{A}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{U}_2 \\ - \underline{I}_2 \\ \end{array} \right) \) | \( \left( \begin{array}{r} \underline{U}_2 \\ \underline{I}_2 \\ \end{array} \right) = \left( \begin{array}{rr} \underline{B}_{11} & \underline{B}_{12} \\ \underline{B}_{12} & \underline{B}_{22} \\ \end{array} \right) \left( \begin{array}{r} \underline{U}_1 \\ - \underline{I}_1 \\ \end{array} \right) \) |
Maschengleichung: \( \underline{U}_2 = R_2 \cdot \underline{I}_1 + R_2 \cdot \underline{I}_2 \) \( \underline{U}_1 = (R_1 + R_2) \cdot \underline{I}_1 + R_2 \cdot \underline{I}_2 \) | ![]() |
\( \left( \begin{array}{r} \underline{U}_1 \\ \underline{U}_2 \\
\end{array} \right)
= \left( \begin{array}{rr} \underline{Z}_{11} & \underline{Z}_{12} \\
\underline{Z}_{12} & \underline{Z}_{22} \\
\end{array} \right)
\left( \begin{array}{r} \underline{I}_1 \\ \underline{I}_2 \\
\end{array} \right) \)
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![]() \( \underline{Z}_{11} = \left. \frac{\underline{U}_{1}}{\underline{I}_{1}} \right\vert_{\underline{I}_2 = 0} \) \( \underline{Z}_{21} = \left. \frac{\underline{U}_{2}}{\underline{I}_{1}} \right\vert_{\underline{I}_2 = 0} \) |
\( \left( \begin{array}{r} \underline{U}_1 \\ \underline{U}_2 \\
\end{array} \right)
= \left( \begin{array}{rr} \underline{Z}_{11} & \underline{Z}_{12} \\
\underline{Z}_{12} & \underline{Z}_{22} \\
\end{array} \right)
\left( \begin{array}{r} \underline{I}_1 \\ \underline{I}_2 \\
\end{array} \right) \)
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![]() \( \underline{Z}_{12} = \left. \frac{\underline{U}_{1}}{\underline{I}_{2}} \right\vert_{\underline{I}_1 = 0} \) \( \underline{Z}_{22} = \left. \frac{\underline{U}_{2}}{\underline{I}_{2}} \right\vert_{\underline{I}_1 = 0} \) |
R1=100Ω, R2=200Ω Bestimmen Sie die Z Parameter. Das Zweitor wird an eine ideale Stromquelle von 10mA angeschlossen. Berechnen Sie die Spannung U2. Am Ausgang wird ein Strom von 2mA entnommen. Berechnen Sie Eingangs- und Ausgangsspannung. |
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R1=100Ω, C1=45nF Bestimmen Sie die Z Parameter. Das Zweitor wird an eine ideale Stromquelle von 10mA und 20kHz angeschlossen. Berechnen Sie die Spannung U2. |
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