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electrical_engineering_1:circuits_under_different_frequencies [2023/03/27 09:47]
mexleadmin 		electrical_engineering_1:circuits_under_different_frequencies [2023/09/19 23:37] (aktuell)
mexleadmin
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-====== 7. Networks at variable frequency ======+====== 7 Networks at variable frequency ======
  
 Further content can be found at this [[https://www.electronics-tutorials.ws/accircuits/series-circuit.html|Tutorial]] or that [[https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-rlc-natural-response-intuition|Tutorial]] Further content can be found at this [[https://www.electronics-tutorials.ws/accircuits/series-circuit.html|Tutorial]] or that [[https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-rlc-natural-response-intuition|Tutorial]]
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   * the amplitude response: $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}$ and   * the amplitude response: $A = \frac {\omega L}{\sqrt{R^2 + (\omega L)^2}}$ and
-  * the phase response: $\Delta\varphi_{u} = arctan \frac{R}{\omega L} = \frac{\pi}{2} - \arctan \frac{\omega L}{R}$+  * the phase response: $\Delta\varphi_{u} = \arctan \frac{R}{\omega L} = \frac{\pi}{2} - \arctan \frac{\omega L}{R}$
  
 The main focus should first be on the amplitude response. Its frequency response can be derived from the equation in various ways. The main focus should first be on the amplitude response. Its frequency response can be derived from the equation in various ways.
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 This can also be derived from understanding the components:  This can also be derived from understanding the components: 
   * At small frequencies, the current in the coil and thus the magnetic field changes only slowly. So only a negligibly small reverse voltage is induced. The coil acts like a short circuit at low frequencies.    * At small frequencies, the current in the coil and thus the magnetic field changes only slowly. So only a negligibly small reverse voltage is induced. The coil acts like a short circuit at low frequencies. 
-  * At higher frequencies, the current generated by $U_I$ through the coil changes faster, the induced voltage $U_i = - {\rm d}I / {\rm d}t$ becomes large. \\ As a result, the coil inhibits the current flow and a voltage drops across the coil.  +  * At higher frequencies, the current generated by $U_I$ through the coil changes faster, the induced voltage $U_{\rm i} = - {\rm d}I / {\rm d}t$ becomes large. \\ As a result, the coil inhibits the current flow and a voltage drops across the coil.  
-  * If the frequency becomes very high, only a negligible current flows through the coil - and hence through the resistor. The voltage drop at $R$ thus approaches zero and the output voltage $U_O$ tends towards $U_I$.+  * If the frequency becomes very high, only a negligible current flows through the coil - and hence through the resistor. The voltage drop at $R$ thus approaches zero and the output voltage $U_\rm O$ tends towards $U_\rm I$.
  
 The transfer function can also be decomposed into amplitude response and frequency response. \\  The transfer function can also be decomposed into amplitude response and frequency response. \\ 
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 <WRAP centeralign> <WRAP centeralign>
-$\large{\underline{A} = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}}  +\begin{align*}  
-                      = \frac {\omega L}    {\sqrt{R^2 +    (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)}}$ +\large{\underline{A}  = \frac {\underline{U}_{\rm O}^\phantom{O}}{\underline{U}_{\rm I}^\phantom{O}}  
-$ \quad  \quad \vphantom{\HUGE{I \\ I}} \large{\xrightarrow{\text{normalization}}} \vphantom{\HUGE{I \\ I}} \quad \quad \quad $ +                      = \frac {\omega L}    {\sqrt{R^2 +    (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)}} 
-$\large{\underline{A}_{norm}  + \quad  \quad \vphantom{\HUGE{I \\ I}} \large{\xrightarrow{\text{normalization}}} \vphantom{\HUGE{I \\ I}} \quad \quad \quad  
-                      = \frac {\omega L / R}{\sqrt{1  + (\omega L / R)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)} } $  +\large{\underline{A}_{norm}  
-$\large{              = \frac {x}           {\sqrt{1  + x^2             }}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan x \right)} }$+                      = \frac {\omega L / R}{\sqrt{1  + (\omega L / R)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)} }  
 +\large{               = \frac {x}           {\sqrt{1  + x^2             }}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan x \right)} } 
 +\end{align*} 
 </WRAP> </WRAP>
  
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 \begin{align*}  \begin{align*} 
 \vphantom{\HUGE{I }} \\ \vphantom{\HUGE{I }} \\
-\underline{A}_{\rm norm} = \frac{x}{\sqrt{1 + x^2}}    \cdot {\rm e}^{{\rm e}\left(\frac{\pi}{2} - arctan x \right)} +\underline{A}_{\rm norm} = \frac{x}{\sqrt{1 + x^2}}    \cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - arctan x \right)} 
                          = \frac{U_{\rm O}}{U_{\rm I}} \cdot {\rm e}^{{\rm j}\varphi}                           = \frac{U_{\rm O}}{U_{\rm I}} \cdot {\rm e}^{{\rm j}\varphi} 
 \end{align*} \end{align*}
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 \begin{align*}  \begin{align*} 
-R            &= \omega L \\ +R               &= \omega L \\ 
-\omega_{c}  &= \frac{R}{L} \\ +\omega _{\rm c} &= \frac{R}{L} \\ 
-2 \pi f_{c} &= \frac{R}{L} \quad \rightarrow \quad \boxed{f_{\rm c} = \frac{R}{2 \pi \cdot L}} \end{align*}+2 \pi f_{\rm c} &= \frac{R}{L} \quad \rightarrow \quad \boxed{f_{\rm c} = \frac{R}{2 \pi \cdot L}} \end{align*}
  
 ==== 7.2.2 RL Low Pass ==== ==== 7.2.2 RL Low Pass ====
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 \begin{align*}  \begin{align*} 
-\underline{A}_{\rm norm} = \frac {1}{\sqrt{1 + (\omega L / R)^2}}\cdot {\rm e}^{-{\rm j} \; arctan \frac{\omega L}{R} } +\underline{A}_{\rm norm} = \frac {1}{\sqrt{1 + (\omega L / R)^2}}\cdot {\rm e}^{-{\rm j} \; \arctan \frac{\omega L}{R} } 
 \end{align*} \end{align*}
  
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 \begin{align*}  \begin{align*} 
-\underline{A}_{\rm norm} = \frac {\omega RC}{\sqrt{1 + (\omega RC)^2}}\cdot {\rm e}^{\frac{\pi}{2}-{\rm j} \; arctan (\omega RC) } +\underline{A}_{\rm norm} = \frac {\omega RC}{\sqrt{1 + (\omega RC)^2}}\cdot {\rm e}^{\frac{\pi}{2}-{\rm j} \; \arctan (\omega RC) } 
 \end{align*} \end{align*}