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circuit_design:5_filter_circuits_i [2021/12/06 13:22]
tfischer
circuit_design:5_filter_circuits_i [2023/07/17 11:59] (aktuell)
mexleadmin
Zeile 3: Zeile 3:
 <callout> <callout>
  
-  * Also recommended for basic circuits II is the nice introduction and textbook found [[https://wiki.mexle.org/circuit_design/3_opamp_basic_circuits_i|here]].+  * Also recommended for basic circuits II is the nice introduction and textbook found [[https://link.springer.com/book/10.1007%2F978-3-319-28127-8|here]].
  
 </callout> </callout>
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 === Introductory example === === Introductory example ===
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l5BOJyVKtALADgOxjAGwCsyATBkRjgMw4jXUhEgaNECmAtPgFACGIUlizgh9LBlEiRYZt2Zh48dEuokciDJFKQi5HAQJRwSyDwDu4yflJWQBSWYDGdh4OH3HTWKd9wwXHSc1NDqYKTIYBgOCHiMMP4W7iJuQjLhUDwAToKQ6bZp4BmQJnDZdtQFHpXGimWWhTWFbmaW1BL0tu2SLTwA5sngGJKFYAbGZgBKIAhG4SnWYiW2WkwT6EQ8OnTdUnZjRpIAqgD6APJgPFqM1JAjeSzYgji2JYonGCeQJ+ERnzjQLC6IiESA0AgaIgSL4+SBgL4nUg-b7UE6nC48MZvYYzOY41K5UolZgwS4AZXA4xsQ2sxWMADM+AAbADO7GMpCSByKBQe1LM13ABGWD0KpBexk+n2+vwQnwQ0FiRDBWiIUIMpEMMPg8O+SM+qNOAEkADosgAUADsAPYASxZAE8AJRbMG4vazIUlY4nI08ABuNI9cyW9hYyhJE1CrroRGFezjb3GPrOnIG0WsOIzlKMJVaTHjNUTnkylmzNWzvRyxYrbhqb3g5V21Oz-KJSUrklbxR4LmLbk7UdMsnQYCwRBeFFIYCQlFuapUcKSxfmBZKvQErbEK7EMjkI7qShAnASSjHE9ImocBlwWCEXkP9X2GWbSyukEY2BFC2WEreWA+BEMEUJEsEwEh-DyJBWAA084QRJExxOQ1zk5QVvyDcVXiFQCZUvOUTiIWE-DPbV-AQn4CGQtFfR4IA noborder}} </WRAP>+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5BOJyVKtALADgOwEY8A2AVmQCYNiMcwdwJiQMGBTAWgICgBDEMrLCDz9wWDEJGC8jDozzx4INjAWVCkLMIIJCWMGURQhCyJwDuo8QTKWQhcaYDGt+3wF2HIYrBO+4edjoMaGJ1LGIcHDIcLEgMDEgSdH9zN0FXfilhKE4AJz4NIWzMoptIYzg823008DKK0wsSmpLXRuqbMDEPHIBzWrx4gZxCI1MAJQkpER0poxsErzH0Yk4DOi6rEU2+IiYQAFUAfQB5PE4EiDBdAsFiYqijeSOMI8gjgA89SCIjnGgsGQIghaAgMCQwNIyG8fD8jhwjtDXu9nmAYapUJiUIQ3kcADoAZ04eBw5RJo1cswyBQaSxg5wAyrdSkIRizyuUAGbcAA2BJY8zWkDoBiyNlFu1G4mOAElOAA3VmjYTTVVGKWKRgclZCuihMkifWS-bHE5kTj9cFWIZWpVjVJGmpGtqpW01W0u-KOhauGpk+BVHbWJiuYP+yoWD3iW1hzjOZ3R1xtZIKaRKMDQQjqewYaQEKhIbAp84WI0qrzqHqmXgxw2V8tSGRp+QmFNwa7Z4VUYjChBUTwtgMWIPZIMiUyXLxDCWk8TReq-ZGfTQ-HFYTCkfwaQtddH+eF4REvXG-NEqeAYLFXnHvQmpckgWYPl2TvAIcWFMDXPiPMmvJdfPIvzPPSB5HkugxHGesDELid5AA noborder}} </WRAP>
  
-Various applications work in harsh environments, there a clear digital signal becomes a noisy signal at the receiver (e.g. sensors in the engine compartment or in industrial environments, satelite communication). In the simulation above, the left scope  shows the original signal. The second scope shows the noisy signal.+Various applications work in harsh environments, where a clear digital signal becomes a noisy signal at the receiver (e.g. sensors in the engine compartment or industrial environments, satellite communication). In the simulation above, the left scope shows the original signal. The second scope shows the noisy signal.
  
-One possibility to process such noisy signals is the use of filters. Filters have already been described in [[:electrical_engineering_1:circuits_under_different_frequencies|Electrical Engineering 1]]. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequent optimize the use of the signal, active filters can be applied. These are often built up by operational amplifiers.+One possibility to process such noisy signals is the use of filters. Filters have already been described in [[:electrical_engineering_1:circuits_under_different_frequencies|Electrical Engineering 1]]. The classical $R C$-filters are passive. This means that although the voltage value is filtered, the output current of the filter is always lower than the current measured at the input. To enable better filtering and subsequently optimize the use of the signal, active filters can be applied. These are often built up by operational amplifiers.
  
 Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{O1}$ after the first filter stage already shows significantly less noise. In the signal $U_{O2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply. Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{O1}$ after the first filter stage already shows significantly less noise. In the signal $U_{O2}$, even less noise is visible, but the rising and falling edges are also no longer displayed sharply.
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 After this lesson, you should: After this lesson, you should:
  
-  - Be able to apply the superpostition method to operational amplifier circuits. +  - Be able to apply the superposition method to operational amplifier circuits. 
-  - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantages and disadvantages).+  - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, advantagesand disadvantages).
   - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.   - Know what the circuit and transfer function of a voltage-to-current converter and current-to-voltage converter look like.
-  - Be able to name applications for the inverse-radiator, voltage-to-current converter, and current-to-voltage converter.+  - Be able to name applications for the inverse integrator, voltage-to-current converter, and current-to-voltage converter.
  
 </callout></WRAP> </callout></WRAP>
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 ===== 5.0 Representation of numerical values ===== ===== 5.0 Representation of numerical values =====
  
-In order to be able to analyse filter circuits, various possibilities for representing the numerical values should be explained beforehand.+To be able to analyze filter circuits, various possibilities for representing the numerical values should be explained beforehand.
  
 <WRAP> Decibels (dB's) for Engineers - A Tutorial <WRAP> Decibels (dB's) for Engineers - A Tutorial
Zeile 48: Zeile 48:
 </WRAP> </WRAP>
  
-==== 5.0.1 The dB measure ====+==== 5.0.1 The dB scale ====
  
-The decibel measure is an auxiliary unit of measure that facilitates handling with ratios (e.g. $U_2/U_1$). These ratios are called **level**  in engineering. The level makes it possible to refer to a reference quantity. In electronic circuitry, the decibel is used as a dimensionless unit for current or voltage ratios. In the future, this will be particularly interesting for the amplification $A_V = \frac{U_O}{U_I}$ and factors.+The Decibel scale is an auxiliary unit of measure that facilitates handling with ratios (e.g. $U_2/U_1$). These ratios are called **level** in engineering (in German: //Pegel//). The level makes it possible to refer to a reference quantity. In electronic circuitry, the decibel is used as a dimensionless unit for current or voltage ratios. In the future, this will be particularly interesting for the amplification $A_{\rm V} = \frac{U_\rm O}{U_\rm I}$ and factors.
  
 The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation: The conversion to a level in $dB$ is defined for current or voltage ratios by the following equation:
  
-$\boxed{A_V^{dB}=20 dB \cdot log_{10}\left(\frac{U_2}{U_1}\right)=20 dB \cdot log_{10}A_V}$(nbsp)(nbsp)(nbsp)(nbsp) resp. (nbsp)(nbsp)(nbsp)(nbsp) $A_C^{dB}=20 dB \cdot log_{10}\left(\frac{I_2}{I_1}\right)$+$\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB\cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 {\rm dB\cdot \log_{10} A_\rm V}$(nbsp)(nbsp)(nbsp)(nbsp)  resp. (nbsp)(nbsp)(nbsp)(nbsp) $A_{\rm C}^{\rm dB}=20 ~{\rm dB\cdot \log_{10}\left(\frac{I_2}{I_1}\right)$
  
 \\ \\
-=== Technical level in db ===+=== Technical level in dB ===
 <WRAP> <WRAP>
  
-^ Name                            ^ Symbol  ^ Formula                            ^ Reference value for 0dB         +^ Name                          ^ Symbol      ^ Formula                                      ^ Reference value for 0dB       
-| [[wpde>Voltage level]]         | $dBV$   | $20dB \cdot log_{10}(V/V_{ref})$  | $0dBV  \widehat{=} 1V$        | +| [[wpde>Voltage level]]        | $\rm dBV$   | $20{\rm dB} \cdot \log_{10}(V/V_{\rm ref})$  | $\rm 0dBV  \widehat{=} 1V$        | 
-| [[wpde>Leistungspegel]]         | $dBm$   | $10dB \cdot log_{10}(P/P_{ref})$  | $0dBm  \widehat{=} 1mW$       | +| [[wp>Power level]]            | $\rm dBm$   | $10{\rm dB} \cdot \log_{10}(P/P_{\rm ref})$  | $\rm 0dBm  \widehat{=} 1mW$       | 
-| [[wpde>Power level]]         | $dBW$   | $10dB \cdot log_{10}(P/P_{ref})$  | $0dBW  \widehat{=} 1W$        | +| [[wp>Power level]]            | $\rm dBW$   | $10{\rm dB} \cdot \log_{10}(P/P_{\rm ref})$  | $\rm 0dBW  \widehat{=} 1W$        | 
-| [[wpde>dBFS|Full scale level]]  | $dBFS$  | $20dB \cdot log_{10}(V/V_{max})$  | $0dBFS  \widehat{=} V_{max}$ +| [[wp>dBFS|Full scale level]]  | $\rm dBFS$  | $20{\rm dB} \cdot \log_{10}(V/V_{\rm max})$  | $\rm 0dBFS  \widehat{=} V_{max}$ 
-| [[wpde>Sound pressure level]]       | $dBA$   | $20dB \cdot log_{10}(p/p_{ref})$  | $0dBA  \widehat{=} 20\mu Pa |+| [[wp>Sound pressure level]]   | $\rm dBA$   | $20{\rm dB} \cdot \log_{10}(p/p_{\rm ref})$  | $\rm 0dBA  \widehat{=} 20 µPa   |
 </WRAP> </WRAP>
  
-Note that this equation changes somewhat for __power__  quantities, i.e. ratios of $P$. If $P \sim U^2$ or $U \sim P^\frac{1}{2}$ is consideredthen we get: +Note that this equation changes somewhat for __power__  quantities, i.e. ratios of $P$. \\ 
-\\ $A_P^{dB}=20 dB \cdot log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$\\ (nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp)$ =\color{blue}{20 dB \cdot \frac{1}{2}} \cdot log_{10}\left(\frac{P_2}{P_1}\right)  =\color{blue}{10 dB} \cdot log_{10}\left(\frac{P_2}{P_1}\right) $+Since $P \sim U^2$ or $U \sim P^\frac{1}{2}$, we get: \\  
 +$A_P^{\rm dB}=             20 ~{\rm dB\cdot                    \log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$\\  
 +(nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp) 
 +           =\color{blue}{20 ~{\rm dB\cdot \frac{1}{2}} \cdot \log_{10}\left(\frac{P_2}{P_1}\right)  =\color{blue}{10 {\rm dB}} \cdot \log_{10}\left(\frac{P_2}{P_1}\right) $
  
-The table above shows various dB-levels which are frequently used in engineering. In the following, only the voltage level is used and indicated with the symbol $dB$.+The table above shows various dB-levels which are frequently used in engineering. In the following, only the voltage level is used and indicated with the symbol $\rm dB$.
 === Simple examples of voltage levels in dB === === Simple examples of voltage levels in dB ===
  
 <WRAP> <WRAP>
-^linear factor^level [$dB$]| +^ linear factor  ^ level [$\rm dB$]  
-|$\times 0.01$|$-40dB$| +|$\times 0.01$   |$-40 ~{\rm dB}   
-|$\times 0.1$|$-20dB$| +|$\times 0.1$    |$-20 ~{\rm dB}   
-|$\times 1$|$0dB$| +|$\times 1$      |$  0 ~{\rm dB}   
-|$\times 2$|$\approx +6dB$| +|$\times 2$      |$\approx +6~{\rm dB}$| 
-|$\times 10$|$+20dB$| +|$\times 10$     |$+20 ~{\rm dB}   
-|$\times 100$|$+40dB$|+|$\times 100$    |$+40 ~{\rm dB}   |
 </WRAP> </WRAP>
  
-By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{dB}$ in $dB$. Occasionally, the superscript $\boxed{}^{dB}$ is omitted below, in which case the level is denoted by the unit after the numerical value.+By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{\rm dB}$ in $\rm dB$.  
 +Occasionally, the superscript $\boxed{}^{\rm dB}$ is omitted below. Instead, the level is denoted by the unit after the numerical value.
  
 Examples: Examples:
  
-  - For $A_V= \color{green}{1} $ we get $ A_V^{dB}(\color{green}{1}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{1})} \quad = 20 dB \cdot 0 \quad \ \boldsymbol{= 0 dB}$ \\ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$. \\  \\ +  - For $A_{\rm V}= \color{green}{1} $ we get <WRAP>  
-  - For $A_V= \color{green}{0,01} $ we get $ A_V^{dB}(\color{green}{0,01}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{0,01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{0,01})} = 20 dB \cdot (-2) \boldsymbol{= -40 dB}$ \\ Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, so even $\color{blue}{x}=-2$. \\  \\ +A_{\rm V}^{\rm dB}(\color{green}{1}) = 20 {\rm dB\cdot \underbrace{\log_{10}\left(\color{green}{1}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{1})} \quad = 20 {\rm dB\cdot 0 \quad \ \boldsymbol{= 0 {\rm dB}}$ \\  
-  - For $A_V= \color{green}{2} $, we get $ A_V^{dB}(\color{green}{2}) = 20 dB \cdot \underbrace{log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_V^{dB}(\color{green}{2})} \quad\approx 20 dB \cdot 0.30103 \boldsymbol{\approx 6 dB}$ \\ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. \\  \\+Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$. </WRAP> \\  \\ 
 +  - For $A_{\rm V}= \color{green}{0.01} $ we get <WRAP>  
 +A_{\rm V}^{\rm dB}(\color{green}{0.01}) = 20 {\rm dB\cdot \underbrace{\log_{10}\left(\color{green}{0.01}\right)}_\color{blue}{x} \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{0.01})} = 20 {\rm dB\cdot (-2) \boldsymbol{= -40 {\rm dB}}$ \\  
 +Here $\color{blue}{x}$ has the value, that gives $10^\color{blue}{x} = \color{green}{0.01}$, so even $\color{blue}{x}=-2$. </WRAP>   \\  \\ 
 +  - For $A_{\rm V}= \color{green}{2} $, we get <WRAP>  
 +A_{\rm V}^{\rm dB}(\color{green}{2}) = 20 {\rm dB\cdot \underbrace{\log_{10}\left(\color{green}{2}\right)}_\color{blue}{x} \qquad \qquad \rightarrow \boldsymbol{A_{\rm V}^{\rm dB}(\color{green}{2})} \quad\approx 20 {\rm dB\cdot 0.30103 \boldsymbol{\approx 6 {\rm dB}}$ \\  
 +Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. </WRAP> \\  \\
  
 === Use of the dB measure === === Use of the dB measure ===
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 The decibel offers some advantages, which are used in the filter elements considered below: The decibel offers some advantages, which are used in the filter elements considered below:
  
-  * **handier numerical values**: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_V = 10000000 \rightarrow A_V^{dB}= 140dB$. This also results in less "zero counting"+  * **handier numerical values**: If very large or very small linear values are needed, the number of the resulting level has fewer digits. Example: $A_{\rm V} = 10000000 \rightarrow A_{\rm V}^{\rm dB}= 140{\rm dB}$. This also results in less "zero counting"
-  * **Relationship to Sensory Perceptions**: Sensory perceptions such as brightness and loudness have an almost exponential effect. That is, any tenfold increase in the underlying physical quantity (number of photons or sound pressure) does not have ten times the effectbut seems to have an additive effect. +  * **Relationship to Sensory Perceptions**: Sensory perceptions such as brightness and loudness have an almost exponential effect. That is, any tenfold increase in the underlying physical quantity (number of photons or sound pressure) does not have ten times the effect but seems to have an additive effect. 
-  * **easier math**: The logarithm in the defining equation turns any multiplication of linear factors into an addition of levels: $A_V^{dB}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1 \cdot A_2) = 20dB \cdot log_{10}(A_1) + 20dB \cdot log_{10}(A_2) = A_V^{dB}(A_1) + A_V^{dB}(A_2)$+  * **easier math**: The logarithm in the defining equation turns any multiplication of linear factors into addition of levels: $A_{\rm V}^{\rm dB}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1 \cdot A_2) = 20~{\rm dB} \cdot \log_{10}(A_1) + 20~{\rm dB} \cdot \log_{10}(A_2) = A_{\rm V}^{\rm dB}(A_1) + A_{\rm V}^{\rm dB}(A_2)$
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <imgcaption pic0| Example: filters in series>   <imgcaption pic0| Example: filters in series>  
 </imgcaption> </imgcaption>
-\\ {{drawio>Beispiel_Filter_in_Reihe}}+\\ {{drawio>Beispiel_Filter_in_Reihe.svg}}
 </panel></WRAP> </panel></WRAP>
  
-Especially the last point of the calculation should be considered again. In <imgref pic0> several amplifiers connected in series can be seen with exemplary voltage gain values. \\ The total gain here is the product of the individual gains: $A_{V,ges}=\prod A_i = A_1 \cdot A_2 \cdot A_3$. \\ The determination of the total gain was rather laborious before the times of the pocket calculator due to the multiplications. For the levels, the result is an addition: $A_{V,ges}^{dB}=\sum A_i^{dB} = A_1^{dB} + A_2^{dB} + A_3^{dB}$. \\ Here this would be: $A_{V,ges}^{dB}=\sum A^{dB} = 88dB + (-58dB) + 14dB 44dB$.+Especially the last point of the calculation should be considered again. In <imgref pic0> several amplifiers connected in series can be seen with exemplary voltage gain values. \\  
 +The total gain here is the product of the individual gains: $A_{\rm V, eq}=\prod A_{i} = A_1 \cdot A_2 \cdot A_3$. \\  
 +The determination of the total gain was rather laborious before the time of the pocket calculator due to the multiplications.  
 +For the levels, the result is an addition: $A_{\rm V,eq}^{\rm dB}=\sum A_i^{\rm dB} = A_1^{\rm dB} + A_2^{\rm dB} + A_3^{\rm dB}$. \\  
 +Here this would be: $A_{\rm V,eq}^{\rm dB} = \sum A^{\rm dB} = 88~{\rm dB} + (-58~{\rm dB}) + 14~{\rm dB} 44~{\rm dB}$.
  
 <WRAP column 100%> <panel type="danger" title="Remember: dB measure"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> <WRAP column 100%> <panel type="danger" title="Remember: dB measure"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
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 For current and voltage levels: For current and voltage levels:
  
-  - A linear factor of $\color{green}{\times 10}$ results in level $+ 20dB$. +  - A linear factor of $\color{green}{\times 10}$ results in level $+ 20~{\rm dB}$. 
-  - A linear factor of $\color{green}{\times 2}$ results in a level of $+ 6dB$. +  - A linear factor of $\color{green}{\times 2} $ results in a level of $+ 6~{\rm dB}$. 
-  - The linear value $A_V = 1$ corresponds to $0 dB$.+  - The linear value $A_{\rm V} = 1$ corresponds to $0 ~{\rm dB}$.
  
 For systems connected in series, to determine the amplification For systems connected in series, to determine the amplification
  
-  - multiply the linear measure $A_V$ and +  - multiply the linear measure $A_\rm V$ and 
-  - add the level $A_V^{dB}$.+  - add the level $A_{\rm V}^{\rm dB}$.
  
 </WRAP></WRAP></panel> </WRAP> </WRAP></WRAP></panel> </WRAP>
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 === More difficult examples of voltage levels in dB=== === More difficult examples of voltage levels in dB===
  
-With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20dB$ and $\color{green}{\times 2} \rightarrow + 6dB$ the linear values can easily be determined from a level in $dB$ without a calculator.+With this knowledge, the interpolation points $\color{green}{\times 10} \rightarrow + 20~{\rm dB}$ and $\color{green}{\times 2} \rightarrow + 6~{\rm dB}$ the linear values can easily be determined from a level in $dB$ without a calculator.
  
 Examples: Examples:
-  - $A_V^{dB}=58dB$ \\ mit Stützstellen: $A_V^{dB}=58dB 40dB 18dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB$ \\ This becomes linear to  $ \qquad \qquad \qquad \qquad \qquad \  A_V = 10^\color{blue}{2} \qquad  \cdot \qquad  2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$ \\ \\ +  - <WRAP> $A_{\rm V}^{\rm dB}=58~{\rm dB}$ \\ with interpolation points 
-  - $A_V^{dB}=56dB$ \\ with interpolation points: $A_V^{dB}=56dB 80dB 24dB = \color{blue}{4}\cdot 20dB + \color{magenta}{-4}\cdot 6dB$ \\ This becomes linear $ \qquad \qquad \qquad \qquad \qquad \ A_V = 10^\color{blue}{4} \qquad \cdot \qquad  2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ \\  oder $A_V^{dB} = 20dB 36dB \rightarrow A_V= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ \\ \\ +$A_{\rm V}^{dB}=58~{\rm dB} 40~{\rm dB} 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\  
-  - $A_V^{dB}=55dB$ \\ with interpolation points: $A_V^{dB}=56dB 40dB 18dB 3dB = \color{blue}{2}\cdot 20dB + \color{magenta}{3}\cdot 6dB  + \color{teal}{-\frac{1}{2}}\cdot 6dB$ \\ This becomes linear to $ \qquad \qquad \qquad \qquad \qquad \qquad A_V = 10^\color{blue}{2} \qquad \cdot \qquad  2^\color{magenta}{3}  \qquad \cdot \qquad  2^\color{teal}{-\frac{1}{2}} \approx 100 \cdot 8 \cdot 0,707  = 560$ +This becomes linear   $ \qquad \qquad \qquad \qquad \qquad \qquad \quad \  A_{\rm V} = 10^\color{blue}{2} \qquad  \cdot \qquad  2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$ </WRAP>\\ \\ 
 +  - <WRAP>$A_{\rm V}^{\rm dB}=56~{\rm dB}$ \\ with interpolation points:  
 +$A_{\rm V}^{dB}=56~{\rm dB} 80~{\rm dB} 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$ \\  
 +This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \quad\ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad  2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ \\  or alternatively  $qquad \qquad \qquad \qquad \qquad \qquad A_{\rm V}^{\rm dB} = 20~{\rm dB} 36~{\rm dB} \rightarrow A_{\rm V}= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ </WRAP>\\ \\ 
 +  - <WRAP>$A_{\rm V}^{\rm dB}=55~{\rm dB}$ \\ with interpolation points:  
 +$A_{\rm V}^{dB}=56~{\rm dB} 40~{\rm dB} 18~{\rm dB} 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}  + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\  
 +This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad A_{\rm V} = 10^\color{blue}{2} \qquad \cdot \qquad  2^\color{magenta}{3}  \qquad \cdot \qquad  2^\color{teal}{-\frac{1}{2}} \approx 100 \cdot 8 \cdot 0.707  = 560$ </WRAP>
  
 \\  \\ 
-The value $-3dB$ will still be used in the following examples.+The value $-3~{\rm dB}$ will still be used in the following examples.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <imgcaption picA| Representation of complex numbers> <imgcaption picA| Representation of complex numbers>
 </imgcaption> </imgcaption>
-\\ {{drawio>Darstellung_von_komplexen_Zahlen}}+\\ {{drawio>Darstellung_von_komplexen_Zahlen.svg}}
 </panel></WRAP> </panel></WRAP>
  
  
-The aim of the floor diagram is to show the transmission behaviour of systems clearly and concisely.+The Bode diagram aims to show the transmission behavior of systems clearly and concisely.
  
 === Preliminary consideration: complex numbers === === Preliminary consideration: complex numbers ===
  
-A complex number can always be reduced to two real number values. For the exact definition of these numerical values there are different possibilities (<imgref picA>):+A complex number can always be reduced to two real number values. For the exact definition of these numerical valuesthere are different possibilities (<imgref picA>):
  
-  - Definition over real part $\Re(\underline{A}_V)=A_V \cdot cos(\varphi)$ and imaginary part $\Im(\underline{A}_V)=A_V \cdot sin(\varphi)$ in $\underline{A}_V= \Re(\underline{A}_V) + j \cdot \Im(\underline{A}_V)$ +  - Definition over real part $\Re(\underline{A}_{\rm V})=A_{\rm V} \cdot \cos(\varphi)$ and imaginary part $\Im(\underline{A}_{\rm V})=A_{\rm V} \cdot \sin(\varphi)$ in $\underline{A}_{\rm V}= \Re(\underline{A}_{\rm V}) + {\rm j\cdot \Im(\underline{A}_{\rm V})$ 
-  - Definition over magnitude $A_V = |\underline{A}_V|$ and phase $\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right)$ in $\underline{A}_V=A_V \cdot e^{j \varphi}$+  - Definition over absolute value $A_{\rm V} = |\underline{A}_{\rm V}|$ and phase $\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right)$ in $\underline{A}_{\rm V}=A_{\rm V} \cdot {\rm e}^{{\rm j\varphi}$
  
-The 2nd definition is more appropriate when considering frequency-dependent voltage gainsince it allows the "time shift" (phase) to be separated from the gain.+The 2nd definition is more appropriate when considering frequency-dependent voltage gain since it allows the "time shift" (phase) to be separated from the gain.
  
 ~~PAGEBREAK~~ ~~PAGEBREAK~~
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 <WRAP>{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+44+0.5+50%0A%25+1+8761050.782000184%0Ac+240+160+240+48+0+1e-9+0%0Ar+112+48+240+48+0+16000%0AO+240+48+352+48+1%0Ag+240+160+240+192+0%0A170+112+48+80+48+3+20+1000+50+0.1%0Ax+326+77+343+80+0+24+U%0Ax+339+86+357+89+0+24+O%0Ax+120+88+126+91+0+24+I%0Ax+101+79+118+82+0+24+U%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A noborder}} </WRAP> <WRAP>{{url>https://www.falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+44+0.5+50%0A%25+1+8761050.782000184%0Ac+240+160+240+48+0+1e-9+0%0Ar+112+48+240+48+0+16000%0AO+240+48+352+48+1%0Ag+240+160+240+192+0%0A170+112+48+80+48+3+20+1000+50+0.1%0Ax+326+77+343+80+0+24+U%0Ax+339+86+357+89+0+24+O%0Ax+120+88+126+91+0+24+I%0Ax+101+79+118+82+0+24+U%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A noborder}} </WRAP>
  
-To better understand the frequency dependence of the voltage gain, it can be plotted as $|A_V(f)|$ as a function of frequency. It is useful to represent the voltage gain as level $|A_V^{dB}(f)|$. The simulation above shows the $|A_V^{dB}(f)|$ curve for a lowpass filter in the lower half of the image. The curve starts at $0dB$ on the left and drops to about $-45dB$. If the mouse is dragged over the diagram, the gain value in $dB$ can be displayed for each frequency. Clicking on the curve displays current flow and voltage ratios. Only clicking on the leftmost frequency range results in a situation where $U_O$ assumes noticeably high voltages (visible via the color of the line).+To better understand the frequency dependence of the voltage gain, it can be plotted as a function of frequency $f$ as $|A_{\rm V}(f)|$. It is useful to represent the voltage gain as level $|A_{\rm V}^{\rm dB}(f)|$. The simulation above shows the $|A_{\rm V}^{\rm dB}(f)|$ curve for a low pass filter in the lower half of the image. The curve starts at $0~{\rm dB}$ on the left and drops to about $-45~{\rm dB}$. If the mouse is dragged over the diagram, the gain value in ${\rm dB}$ can be displayed for each frequency. Clicking on the curve displays current flow and voltage ratios. Only clicking on the leftmost frequency range results in a situation where $U_{\rm O}$ assumes noticeably high voltages (visible via the color of the line).
  
-In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve. However, if the frequency axis is also logarithmized, a different picture results. This is done via ''Options''  >> ''Linear Scale''. Now the initially flat course of the voltage gain for low frequencies and the drop for higher frequencies becomes very clear. Also the cut-off frequency can be read at the point of the "bending".+In this type of display, it is difficult to determine values such as the cut-off frequency from the frequency curve. However, if the frequency axis is also logarithmized, a different picture results. This is done via ''Options''  >> ''Linear Scale''. Now the initially flat course of the voltage gain for low frequencies and the drop for higher frequencies becomes very clear. Alsothe cut-off frequency can be read at the point of the "bending".
  
 The phase can be made visible via ''Options''  >> ''Show Phase''. The phase can be made visible via ''Options''  >> ''Show Phase''.
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 <imgcaption picB| Principle image of the Bode diagram> <imgcaption picB| Principle image of the Bode diagram>
 </imgcaption> </imgcaption>
-\\ {{drawio>Prinzipbild_des_Bodediagramms}}+\\ {{drawio>Prinzipbild_des_Bodediagramms.svg}}
 </panel></WRAP> </panel></WRAP>
  
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 A jump in frequency by a factor of $\times 10$ is called a **decade**  (abbreviated Dec.).  The amplitude response of various functions will be briefly discussed ($\mathcal{C}$ is an arbitrary frequency-independent factor): A jump in frequency by a factor of $\times 10$ is called a **decade**  (abbreviated Dec.).  The amplitude response of various functions will be briefly discussed ($\mathcal{C}$ is an arbitrary frequency-independent factor):
  
-  - $|A_V(f)| = \mathcal{C} \cdot f$: If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain. This results in an increase of $+20dB$ per decade.    - $|A_V(f)| = \mathcal{C} / f$: If a function is considered that is reciprocal to $f$, then a tenfold increase in frequency results in a decrease in voltage gain to one-tenth. This results in a drop of $-20dB$ per decade (cf. <imgref picB> at high frequencies).    - $|A_V(f)| = \mathcal{C} / f^n$: Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/10^n$th. This results in a drop of $-20dB \cdot n$ per decade.+  - $|A_{\rm V}(f)| = \mathcal{C} \cdot f$: \\ If a function is considered that increases linearly with $f$, then a tenfold increase in frequency results in a tenfold increase in voltage gain. \\ This results in an increase of $+20~{\rm dB}$ per decade.     
 +  - $|A_{\rm V}(f)| = \mathcal{C} / f$:  \\ If a function is considered that is reciprocal to $f$, then a tenfold increase in frequency results in a decrease in voltage gain to one-tenth. \\ This results in a drop of $-20~{\rm dB}$ per decade (cf. <imgref picB> at high frequencies).      
 +  - $|A_{\rm V}(f)| = \mathcal{C} / f^n$: \\ Considering a function reciprocal to $f^n$, a tenfold increase in frequency results in a drop in voltage gain to one $1/10^n$th. \\ This results in a drop of $-20~{\rm dB} \cdot n$ per decade.
  
-As an alternative to the actual course, $|A_V(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments.+As an alternative to the actual course, $|A_{\rm V}(f)|$ and $\varphi(f)$ are occasionally represented idealized with straight line segments.
  
 ~~PAGEBREAK~~ <WRAP column 100%> <panel type="danger" title="Remember: Bode diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> ~~PAGEBREAK~~ <WRAP column 100%> <panel type="danger" title="Remember: Bode diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
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   - Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation)   - Phase response: linear phase over logarithmic frequency (i.e. single logarithmic representation)
  
-This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$.  In particular, this is true for $A(\omega) \sim \omega$, i.e., a slope of +20dB/decade, and for $A(\omega) \sim \frac{1}{\omega}$, i.e., a slope of -20dB/decade </WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~+This gives a straight line in the amplitude response for functions of the form $A(\omega) \sim \omega^n$.   
 +In particular, this is true for $A(\omega) \sim \omega$, i.e., a slope of $~\rm 20dB/decade$, and for $A(\omega) \sim \frac{1}{\omega}$, i.e., a slope of $-20~\rm dB/decade$  
 +</WRAP></WRAP></panel> </WRAP> ~~PAGEBREAK~~
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <imgcaption pic1| Filter circuits> <imgcaption pic1| Filter circuits>
 </imgcaption> </imgcaption>
-\\ {{drawio>allgemeine_Filterschaltungen}}+\\ {{drawio>allgemeine_Filterschaltungen.svg}}
 </panel></WRAP> </panel></WRAP>
  
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 In electronics, inductors are rarely used for this purpose. This has several reasons: In electronics, inductors are rarely used for this purpose. This has several reasons:
  
-  - Inductors are possible in integrated circuitsbut are somewhat more difficult to calculate as such an element.+  - Inductors are possible in integrated circuits but are somewhat more difficult to calculate as such an element.
   - Inductors require a current source as current storage. The internal resistance results in a continuous power loss.   - Inductors require a current source as current storage. The internal resistance results in a continuous power loss.
  
Zeile 223: Zeile 247:
 The following basic circuit is a modified, [[:circuit_design:3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]] (<imgref pic1>) in which one or both ohmic resistors are replaced by (complex-valued) impedances. The following basic circuit is a modified, [[:circuit_design:3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]] (<imgref pic1>) in which one or both ohmic resistors are replaced by (complex-valued) impedances.
  
-For first circuit only the part between output voltage $U_O$ and virtual ground should be replaced by a capacitor (<imgref pic2>).+For the first circuitonly the part between output voltage $U_{\rm O}$ and virtual ground should be replaced by a capacitor (<imgref pic2>).
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 ==== 5.1.1 Circuit analysis with differential equations ==== ==== 5.1.1 Circuit analysis with differential equations ====
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.00009999999999999999+0.32112705431535615+57+5+50%0Ag+96+224+96+240+0%0Aw+336+112+336+160+0%0Aw+192+112+192+144+0%0Aa+192+160+336+160+8+15+-15+1000000+-0.00007695554889407606+0+100000%0Ac+192+112+336+112+0+0.0000058+-7.695631844956501%0AO+336+160+400+160+0%0Av+96+224+96+160+0+2+40+2+0+3.141592653589793+0.5%0Av+96+160+96+112+0+2+80+1+0+0+0.5%0Aw+96+112+128+112+0%0Ar+128+112+192+112+0+1000%0Ag+192+176+192+192+0%0A403+304+16+432+80+0+5_4_0_12290_8.523156660430175_0.0001_1_1%0A403+16+16+144+80+0+8_4_0_12290_3_0.0001_0_2_8_3%0Ax+194+97+210+100+4+24+K%0Ax+209+105+222+108+4+24+1%0A noborder}} </WRAP>+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l5BOJyWoVaBmATGM2B2SAVgBZMxjNiA2SkYghhyBgUwFo8AoAcxAQ0Q2bKQFDRrKdwDuITJiF5s8xeBrS5YBCuXgd4UmMjcAhvt0bVSqwA5wxEF0dh48KB9dvuAYwvh8awCVKVg3RjRIhEdpAHkgsCtSOHUpKG4AN3FhUWzEtJVk4Q9MaDBSSh0aKmJbBAJojGJMvKtBYI8VW1YINJhmuXa9fHs9EwAncGxRwO1dQJ74Pn8wAiUDOY8TZMx5SDFEkHIutOIAfVIzyDOAD1tMSHwLrDBbN4IH2oRXS5h4MDOXDOAP6gIQtm4O3U0PKYm6HlsFyut3uj2wZ0wVzCj0BAPRILOiMxf3gpCikRoV24N30Ynqwlc4BSYlyAGlqcJEEzHCJdJB7CyDtwgA noborder}} </WRAP>
  
 The first active filter circuit can be seen in the simulation above. The superimposed square-wave voltage sources result in a step function as input voltage. This generates a current via the RC element. If we now look at the output voltage in comparison to the input voltage, we can see that: The first active filter circuit can be seen in the simulation above. The superimposed square-wave voltage sources result in a step function as input voltage. This generates a current via the RC element. If we now look at the output voltage in comparison to the input voltage, we can see that:
  
-  - for each constant input value $U_I \neq 0$ an output value with a fixed slope results and +  - for each constant input value $U_{\rm I} \neq 0$ an output value with a fixed slope results and 
-  - for each positive input value $U_I > 0$ a negative slope results, for a negative input value a positive slope results.+  - for each positive input value $U_{\rm I}    0$ a negative slope results, and for a negative input value a positive slope results.
  
-The circuit thus created is called **inverse integrator** or inverting integrator.+The circuit thus created is called an **inverse integrator** or inverting integrator.
  
-If you look at the circuit, you can see that the node $K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly.+If you look at the circuit, you can see that the node $\rm K_1$ is at virtual ground. With a constant input voltage, the input current is therefore constant and defined only by the resistance. Thus, the capacitor charges with a constant current; the charge increases linearly. The voltage across the capacitor therefore also increases linearly.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <imgcaption pic2| Inverse Integrator> <imgcaption pic2| Inverse Integrator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Umkehrintegrator}}+\\ {{drawio>Umkehrintegrator.svg}}
 </panel></WRAP> </panel></WRAP>
  
-Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in <imgref pic2> for this purpose. The **transfer function $U_O = f(U_I)$**  is now to be determined.+Just as in the basic circuits, the relationship between the output value and the input value is now to be determined mathematically. The meshes are drawn in <imgref pic2> for this purpose. The **transfer function $U_{\rm O} = f(U_{\rm I})$**  is now to be determined.
  
-$A_V = ? \quad -> \quad U_O = f(U_I) $+$A_{\rm V} = ? \quad \rightarrow \quad U_{\rm O} = f(U_{\rm I}) $
  
 === given equations === === given equations ===
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 Given the following equations: Given the following equations:
  
-|I.|Basic equation|$U_O A_D \cdot U_D$| +|I.    |Basic equation| $                   U_{\rm O} A_{\rm D} \cdot U_{\rm D}                                 
-|II.|mesh 1|$ -U_I+U_R-U_D=0 $| +|II.   |Mesh        | $-U_{\rm I} + U_R - U_{\rm D} = 0$                                                          
-|III.|mesh 2|$U_D+U_C+U_O=0$| +|III.  |Mesh        | $ U_{\rm D} + U_C + U_{\rm O} = 0$                                                          
-|IV.|Mesh|$I_R=I_C$| +|IV.   |Node          | $ I_R                         = I_C$                                                        
-|V.|Capacity C|$C= { Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C dt+ Q_0(t_0)) $| +|V.    |Capacity C    | $C   = {   Q \over U_C } = { 1 \over U_C }\cdot(\int_{t_0}^{t_1} I_C {\rm d}t+ Q_0(t_0)) $  
-|VI.|Resistance R|$R = { U_R \over I_R }$|+|VI.   |Resistance R  | $R   = { U_R \over I_R }$                                                                   |
  
-=== Derivation ===+=== Calculation ===
  
-The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[rechnung_umkehrintegrator|alternative representation]]):{{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2024x128#/ 1024,108 left noborder}}+The calculation is performed once in detail here (clicking on the right arrow "►" leads to the next step, [[rechnung_umkehrintegrator|alternative representation]]): \\  
 +{{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_umkehrintegrator?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <imgcaption pic3| Example of a signal-time curve of the Inverting Integrator> <imgcaption pic3| Example of a signal-time curve of the Inverting Integrator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Signal_Zeit_Verlauf_des_Umkehrintegrator}}+\\  {{drawio>Signal_Zeit_Verlauf_des_Umkehrintegrator.svg}}
 </panel></WRAP> </panel></WRAP>
  
-By means of an example, the signal-time curve at the inverting integrator shall be explained.   +Using an example, the signal-time curve at the inverting integrator shall be explained.  
- +
-  - Let $R=5 c\Omega$, $C=1 \mu F$, and the input voltage waveform $U_I$ shown in <imgref pic3> be given. +
- +
-  - We are looking for the output voltage $U_O$. +
  
 +  - Let $R=5 ~\rm k\Omega$, $C=1 ~\rm µF$, and the input voltage waveform $U_{\rm I}$ shown in <imgref pic3> be given.
 +  - We are looking for the output voltage $U_{\rm O}$. 
  
 Solution:\\   Solution:\\  
   - Over the given values of $R$ and $C$, the time constant $\tau$ is determined.   - Over the given values of $R$ and $C$, the time constant $\tau$ is determined.
 +  - With the inverse integrator the input value is integrated and inverted. For the given course of the input voltage, the calculation of interpolation points is sufficient.
 +  - With the formula derived in 5.1.1 $U_{\rm O}$ can be composed section by section:
  
-  - With the inverse integrator the input value is integrated and inverted. For the given course of the input voltage the calculation of interpolation points is sufficient. +The calculation is performed once in detail here (clicking on the right arrow "►" leads to the next step, [[rechnung_signalzeitverlauf_umkehrintegrator|alternative representation]]): \\ {{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_signalzeitverlauf_umkehrintegrator?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
-  - With the formula derived in 5.1.$U_O$ can be composed section by section:+
  
-The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[rechnung_signalzeitverlauf_umkehrintegrator|alternative representation]]): {{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_signalzeitverlauf_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=1624x128#/ 624,108 left noborder}} 
  
 <WRAP column 100%> <panel type="danger" title="Notice: signal-time curve of the inverting integrator"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> <WRAP column 100%> <panel type="danger" title="Notice: signal-time curve of the inverting integrator"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
  
-If a constant input voltage $U_I$ is applied to the inverting integrator, the output voltage $U_O$ just equals $-U_I$ after the time constant $\tau$ (see <imgref pic3>, light gray arrow).+If a constant input voltage $U_{\rm I}$ is applied to the inverting integrator, the output voltage $U_{\rm O}$ just equals $-U_{\rm I}$ after the time constant $\tau$ (see <imgref pic3>, light gray arrow).
  
 </WRAP></WRAP></panel> </WRAP> </WRAP></WRAP></panel> </WRAP>
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-==== 5.1.3 Determination of amount and phase ====+==== 5.1.3 Determination of absolute value and phase ====
  
-In order to be able to determine magnitude and phase, purely sinusoidal input and output variables are to be considered first.  The following function is used as input voltage $U_I$:+To be able to determine absolute value and phase, purely sinusoidal input and output variables are to be considered first.  The following function is used as input voltage $U_{\rm I}$:
  
-U_I(t)= \hat{U}_I \cdot sin(\omega \cdot t)$+U_{\rm I}(t)= \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$
  
-This definition of the input voltage can now be substituted into the above equation for $U_O$:+This definition of the input voltage can now be substituted into the above equation for $U_{\rm O}$:
  
-The calculation is performed once in detail here (clicking on right arrow "►" leads to the next step, [[rechnung_betragundphase_umkehrintegrator|alternative representation]]): {{url>https://wiki.mexle.org/_export/revealjs/elektronische_schaltungstechnik/rechnung_betragundphase_umkehrintegrator?theme=dokuwiki&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0&size=2524x128#/ 1024,108 left noborder}}+The calculation is performed once in detail here (clicking on the right arrow "►" leads to the next step, [[rechnung_betragundphase_umkehrintegrator|alternative representation]]): \\ {{url>https://wiki.mexle.org/_export/revealjs/circuit_design/rechnung_betragundphase_umkehrintegrator?theme=white&fade=fade&controls=1&show_progress_bar=1&build_all_lists=1&show_image_borders=0&horizontal_slide_level=2&enlarge_vertical_slide_headers=0#/ 400,300}}
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
 <imgcaption pic4| Sketch of the bottom diagram from the inverting integrator> <imgcaption pic4| Sketch of the bottom diagram from the inverting integrator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Skizze_des_Bodediagramms_vom_Umkehrintegrator}}+\\ {{drawio>Skizze_des_Bodediagramms_vom_Umkehrintegrator.svg}}
 </panel></WRAP> </panel></WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
  
-The **amount**  $|A_V|$ is given by the amplitude ratio of $\hat{U}_O \over \hat{U}_I$: $$|A_V|={\hat{U}_O \over \hat{U}_I} = {1 \over {\omega \cdot R\cdot C}} $$+The **absolute value**  $|A_{\rm V}|$ is given by the amplitude ratio of $\hat{U}_{\rm O} \over \hat{U}_{\rm I}$: $$|A_{\rm V}|={\hat{U}_{\rm O} \over \hat{U}_{\rm I}} = {1 \over {\omega \cdot R\cdot C}} $$
  
-The **phase**  can be determined from the "time offset" of the peak input voltage $U_I = \hat{U}_I \cdot sin(\omega \cdot t)$ and output voltage $U_O = { {\hat{U}_I } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$ can be determined. The phase is given by considering the trigonometric functions and the sign:+The **phase**  can be determined from the "time offset" of the peak input voltage $U_{\rm I} = \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$ and output voltage $U_{\rm O} = { {\hat{U}_{\rm I} } \over {\omega \cdot R\cdot C} } \cdot cos(\omega \cdot t)$ can be determined. The phase is given by considering the trigonometric functions and the sign:
  
-$U_I = + \hat{U}_I \cdot sin(\omega \cdot t)$\\ +$U_{\rm I} = + \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$\\ 
-$U_O = + \hat{U}_O \cdot cos(\omega \cdot t) = + \hat{U}_O \cdot sin(\omega \cdot t + 90°)$\\+$U_{\rm O} = + \hat{U}_{\rm O} \cdot \cos(\omega \cdot t) = + \hat{U}_{\rm O} \cdot \sin(\omega \cdot t + 90°)$\\
 $ \rightarrow  \varphi = 90°$ $ \rightarrow  \varphi = 90°$
 \\ \\
-=== Consideration of extreme values ===+=== Consideration of extreme frequencies ===
  
-In order to be able to sketch the course in the bottom diagram, the **behavior of the transfer function $U_O=f(U_I)$ in the extreme cases for low (${\omega}\rightarrow 0$) and high frequencies (${\omega}\rightarrow \infty$)**  shall be considered. For magnitude $|A_V|$ and phase $\varphi$ we obtain:+To be able to sketch the course in the bottom diagram, the **behavior of the transfer function $U_{\rm O}=f(U_{\rm I})$ in the extreme cases for low (${\omega}\rightarrow 0$) and high frequencies (${\omega}\rightarrow \infty$)**  shall be considered. For absolute value $|A_{\rm V}|$ and phase $\varphi$ we obtain:
  
-$ |A_V({\omega}\rightarrow 0 \ \; )| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$ \\ +$ |A_{\rm V}({\omega}\rightarrow 0 \ \; )  | \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow 0}\quad \infty$ \\ 
-$ |A_V({\omega}\rightarrow\infty)| \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow\infty}\quad 0$ \\+$ |A_{\rm V}({\omega}\rightarrow \infty)   | \quad=\quad{1 \over {\color{blue}{\omega} \cdot R\cdot C}} \quad\xrightarrow{\color{blue}{\omega}\rightarrow\infty}\quad 0$ \\ \\ 
 $\varphi = +90° \qquad \forall \ \omega$ $\varphi = +90° \qquad \forall \ \omega$
  
-From these boundary conditions the frequency response can already be sketched, see <imgref pic4>.+From these boundary conditionsthe frequency response can already be sketched, see <imgref pic4>.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-==== 5.1.4 Circuit analysis with complex calculation ====+==== 5.1.4 Circuit Analysis with complex Calculation ====
  
-In the previous chapters it could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the inverting integrator. Now the complex calculation will be considered as a method which simplifies the analysis of such circuits. For the complex calculation, the resistances and capacitances are replaced by complex impedances:+In the previous chaptersit could be seen that the circuit analysis with differential equations is already very tough and computationally intensive for a simple circuit like the inverting integrator. Now the complex calculation will be considered as a method that simplifies the analysis of such circuits. For the complex calculation, the resistances and capacitances are replaced by complex impedances:
  
 $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\
-$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ dt+ Q_0(t_0)) \qquad  \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ mit $\quad \underline{Z}_C= \frac{1}{j \cdot \omega \cdot C}$ \\+$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad  \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ with $\quad \underline{Z}_C= \frac{1}{{\rm j\cdot \omega \cdot C}$ \\
  
 However, this consideration can only be implemented under certain boundary conditions: However, this consideration can only be implemented under certain boundary conditions:
  
-  - **sinusoidal quantities**: complex current or voltage pointers (cf. [[:elektrotechnik_2:wechselstromtechnik|ET2 Wechselstromtechnik]]) can only represent sinusoidal quantities. +  - **sinusoidal quantities**: complex current or voltage pointers (see [[:electrical_engineering_1:introduction_in_alternating_current_technology|Introduction in AC circuits]]) can only represent sinusoidal quantities. 
-  - **Swinged state**: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since switch-on. This takes out disturbances that are generated by switching on.+  - **Steady state**: The systems of equations consider only sinusoidal oscillations that have already existed for infinite time. This corresponds to a long time since the switch-on. This takes out disturbances that are generated by switching on.
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
 <imgcaption pic5| Circuit of the inverting integrator with complex impedances> <imgcaption pic5| Circuit of the inverting integrator with complex impedances>
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_des_Umkehrintegrators_mit_komplexen_Impedanzen}}+\\ {{drawio>Schaltung_des_Umkehrintegrators_mit_komplexen_Impedanzen.svg}}
 </panel></WRAP> </panel></WRAP>
  
 This can now be used to calculate the circuit (<imgref pic5>): This can now be used to calculate the circuit (<imgref pic5>):
  
-$\underline{Z}_1=R$   $\underline{Z}_2=\frac{1}{j \cdot \omega \cdot C} = \frac{-j}{\omega \cdot C}$+$\underline{Z}_1=R$   
 + 
 +$\underline{Z}_2=\frac{1}{{\rm j\cdot \omega \cdot C} = \frac{\rm -j}{\omega \cdot C}$
  
 From the basic circuit of the [[:circuit_design:3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]], its voltage gain is known: From the basic circuit of the [[:circuit_design:3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]], its voltage gain is known:
  
-$A_V = \frac{U_O}{U_I}=-\frac{R_2}{R_1}$+$A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=-\frac{R_2}{R_1}$
  
 This results in the complex: This results in the complex:
  
-$\underline{A}_V = \frac{\underline{U}_O}{\underline{U}_I}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{j}{\omega \cdot R \cdot C} $+$\underline{A}_{\rm V} = \frac{\underline{U}_{\rm O}}{\underline{U}_{\rm I}}=-\frac{\underline{Z}_2}{\underline{Z}_1} = \frac{\rm j}{\omega \cdot R \cdot C} $
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== Plausibility check via extreme value consideration ===+=== Plausibility check via extreme frequency consideration ===
  
 ^From the formula^From the circuit| ^From the formula^From the circuit|
-|$\underline{A}_V \xrightarrow{\omega \rightarrow 0} \infty$|For $\omega \rightarrow 0$, the capacitor becomes a high impedance resistor \\ The opamp must output an output voltage of $\underline{U}_D \rightarrow 0$ for $\underline{U}_O \rightarrow \infty$.  +|$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow 0} \infty$  <WRAP> 
-|$\underline{A}_V \xrightarrow{\omega \rightarrow \infty} 0$|For $\omega \rightarrow \infty$ the capacitor becomes a short circuit \ The operational amplifier must output an output voltage of $\underline{U}_D \rightarrow 0$ for $\underline{U}_O \rightarrow 0$.|+For $\omega \rightarrow 0$, the capacitor behaves like a high impedance resistor \\  
 +Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of  
 +$\underline{U}_{\rm O} \rightarrow \infty$. </WRAP> 
 +|$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow \infty} 0$  <WRAP> 
 +For $\omega \rightarrow \infty$ the capacitor behaves like a short circuit \\  
 +Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of  
 +$\underline{U}_{\rm O} \rightarrow 0$. </WRAP>|
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== magnitude and phase ===+=== Absolute Value and Phase ===
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
 <imgcaption pic6| Course of the arc tangent> <imgcaption pic6| Course of the arc tangent>
 </imgcaption> </imgcaption>
-\\ {{drawio>Verlauf_Arcustangens}}+\\ {{drawio>Verlauf_Arcustangens.svg}}
 </panel></WRAP> </panel></WRAP>
  
  
-The amount $A_V$ is given by:+The absolute value $A_{\rm V}$ is given by:
  
-$|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$+$|\underline{A}_{\rm V}|= \frac{1}{\omega \cdot R \cdot C} \sim \frac{1}{f}$
  
-Specifically, for a magnitude of $|\underline{A}_V(0dB)|$ at $0dB$, the result is:+Specifically, for a absolute value of $|\underline{A}_{\rm V}(0~{\rm dB})|$ at $0~{\rm dB}$, the result is:
  
-$|\underline{A}_V(0dB)|\overset{!}{=} 1 \widehat{=} 0dB \rightarrow \omega(0dB) = \frac{1}{R \cdot C}$+$|\underline{A}_{\rm V}(0~{\rm dB})|\overset{!}{=} 1 \widehat{=} 0~{\rm dB} \rightarrow \omega(0~{\rm dB}) = \frac{1}{R \cdot C}$
  
 The phase $\varphi$ is calculated via The phase $\varphi$ is calculated via
  
-$\varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = arctan \left( \infty \right) = +90°$+$\varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = \arctan \left( \frac{\omega \cdot R \cdot C}{0} \right) = \arctan \left( \infty \right) = +90°$
  
 The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (<imgref pic6>) for $x \rightarrow \infty$. The phase $\varphi=+90°$ for $arctan(x)|_{x\rightarrow \infty}$ is also evident from the course of the arctangent (<imgref pic6>) for $x \rightarrow \infty$.
Zeile 409: Zeile 440:
 <imgcaption pic7| Bode diagram of the inverting integrator> <imgcaption pic7| Bode diagram of the inverting integrator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Bodediagramm_des_Umkehrintegrators}}+\\ {{drawio>Bodediagramm_des_Umkehrintegrators.svg}}
 </panel></WRAP> </panel></WRAP>
  
-The frequency response is to be illustrated by means of an example.+The frequency response is to be illustrated using an example.
  
-  - Let $R=1 k\Omega$, $C=16 nF$ be given.+  - Let $R=1 ~\rm k\Omega$, $C=16 ~\rm nF$ be given.
   - We are looking for the Bode diagram   - We are looking for the Bode diagram
  
 Solution Solution
  
-  - Determining the time constant: \\ $\tau = R \cdot C = 16 \mu s$ \\ \\ +  - Determining the time constant: \\ $\tau = R \cdot C = 16 ~\rm µs$ \\ \\ 
-  - Determining the frequency $f$ für $|\underline{A}_V(0dB)|$: $\omega(0dB)= \frac{1}{\tau} = 2\pi \cdot f(0dB)$ \\ This gives $f(0dB)$ via: \\ $f(0dB)=\frac{1}{2\pi} \cdot \frac{1}{16 \mu s} \approx 10kHz$ \\ \\ +  - Determining the frequency $f$ for $|\underline{A}_{\rm V}(0~{\rm dB})|$: $\omega(0~{\rm dB})= \frac{1}{\tau} = 2\pi \cdot f(0~{\rm dB})$ \\ This gives $f(0~{\rm dB})$ via: \\ $f(0~{\rm dB})=\frac{1}{2\pi} \cdot \frac{1}{16 ~\rm µs} \approx 10 ~\rm kHz$ \\ \\ 
-  - Consideration of the slope: \\ $|\underline{A}_V|= \frac{1}{\omega \cdot R \cdot C}  \sim \frac{1}{f}$ \\ From this, a tenfold increase in $f$ results in one-tenth the amount $|\underline{A}_V|$, i.e., a slope of $-20dB$ per decade \\ \\+  - Consideration of the slope: \\ $|\underline{A}_{\rm V}|= \frac{1}{\omega \cdot R \cdot C}  \sim \frac{1}{f}$ \\ From this, a tenfold increase in $f$ results in one-tenth the absolute value $|\underline{A}_{\rm V}|$, i.e., a slope of $-20~{\rm dB}$ per decade \\ \\
   - From this information, the full Bode diagram can be determined (<imgref pic7>)   - From this information, the full Bode diagram can be determined (<imgref pic7>)
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.2 Lowpass =====+===== 5.2 Low Pass Filter =====
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic7_1| Circuit of the active low-pass filter>+<imgcaption pic7_1| Circuit of the active Low Pass Filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_des_aktiven_Tiefpasses}}+\\ {{drawio>Schaltung_des_aktiven_Tiefpasses.svg}}
 </panel></WRAP> </panel></WRAP>
  
-Another circuit can be derived from the inverting integrator. For this purpose, the hitherto purely capacitive value of the impedance between the output side and virtual ground is to be supplemented by a resistive component. This circuit can be seen in <imgref pic7_1>. In the following, this circuit+Another circuit can be derived from the inverting integrator. For this purpose, the hitherto purely capacitive value of the impedance between the output side and the virtual ground is to be supplemented by a resistive component. This circuit can be seen in <imgref pic7_1>. In the following, this circuit
  
   * first be considered practically with a simulation,   * first be considered practically with a simulation,
-  * then a picture of the system's effect will be drawn up without detailed calculation, and finally * +  * then a picture of the system's effect will be drawn up without detailed calculation, and finally 
-  * be checked by a circuit analysis with complex calculation.+  * be checked by a circuit analysis with complex calculations.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== Lowpass in simulation ===+=== Low Pass Filter in Simulation ===
  
 <WRAP>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+208+208+208+256+0%0Aw+352+112+352+192+0%0Aw+208+112+208+176+0%0Aa+208+192+352+192+4+15+-15+100000000%0Ac+288+112+336+112+0+1.5915000000000002e-7+0.16559840149986407%0Ar+112+112+208+112+0+10000%0AO+352+192+416+192+0%0A170+112+112+64+112+2+20+4000+5+0.1%0Ar+288+64+336+64+0+1000%0Aw+352+64+352+112+0%0Aw+208+64+208+112+0%0AB+224+32+336+144+0+Box%0As+288+64+224+64+0+0+false%0As+288+112+224+112+0+0+false%0Aw+208+64+224+64+0%0Aw+208+112+224+112+0%0Aw+336+112+352+112+0%0Aw+336+64+352+64+0%0Ax+229+53+251+56+0+18+S1%0Ax+229+98+251+101+0+18+S2%0Ao+0+32+0+34+10+0.0125+0+-1%0A noborder}} </WRAP> <WRAP>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+208+208+208+256+0%0Aw+352+112+352+192+0%0Aw+208+112+208+176+0%0Aa+208+192+352+192+4+15+-15+100000000%0Ac+288+112+336+112+0+1.5915000000000002e-7+0.16559840149986407%0Ar+112+112+208+112+0+10000%0AO+352+192+416+192+0%0A170+112+112+64+112+2+20+4000+5+0.1%0Ar+288+64+336+64+0+1000%0Aw+352+64+352+112+0%0Aw+208+64+208+112+0%0AB+224+32+336+144+0+Box%0As+288+64+224+64+0+0+false%0As+288+112+224+112+0+0+false%0Aw+208+64+224+64+0%0Aw+208+112+224+112+0%0Aw+336+112+352+112+0%0Aw+336+64+352+64+0%0Ax+229+53+251+56+0+18+S1%0Ax+229+98+251+101+0+18+S2%0Ao+0+32+0+34+10+0.0125+0+-1%0A noborder}} </WRAP>
  
-In the simulation above, the circuit from <imgref pic7_1> is shown again. In addition, two switches $S1$ and $S2$ are built into the circuit by which the various feedback paths can be disabled:+In the simulation above, the circuit from <imgref pic7_1> is shown again. In addition, two switches $\rm S1$ and $\rm S2$ are built into the circuit by which the various feedback paths can be disabled:
  
-  - If only switch $S1$ is closed, the circuit is an inverting amplifier. +  - If only switch $\rm S1$ is closed, the circuit is an inverting amplifier. 
-  - If only switch $S2$ is closed, the circuit is an inverting integrator.+  - If only switch $\rm S2$ is closed, the circuit is an inverting integrator.
  
-In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in $dB$, or the phase in degrees.+In the simulation, the Bode diagram is sketched below. By clicking on the Bode diagram, the distribution of the current in the circuit corresponding to the frequency is displayed and - in addition to the Bode diagram - also the gain in ${\rm dB}$, or the phase in degrees.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 461: Zeile 492:
   - Now both switches should be closed.   - Now both switches should be closed.
       - In which frequency ranges does the inverting amplifier or the inverting integrator work approximately?       - In which frequency ranges does the inverting amplifier or the inverting integrator work approximately?
-      - Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1k\Omega$ resistor and the capacitor? What is the value of the gain and phase here?+      - Consider the distribution of the currents in the circuit at different frequencies. At what frequency does the current divide approximately equally between the $1~\rm k\Omega$ resistor and the capacitor? What is the value of the gain and phase here?
       - After reading through the following analyses, the gain and phase at the "kink point" can be determined. Do these deviate from your measurement?       - After reading through the following analyses, the gain and phase at the "kink point" can be determined. Do these deviate from your measurement?
  
Zeile 468: Zeile 499:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-==== 5.2.1 Consideration without detailed account ====+==== 5.2.1 First Consideration ====
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== Extreme value consideration ===+=== Extreme frequency consideration ===
  
-From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of magnitude of voltage gain and phase can be analyzed.+From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of the absolute value of voltage gain and phase can be analyzed.
  
   - $\omega \rightarrow 0$:   - $\omega \rightarrow 0$:
Zeile 480: Zeile 511:
       - Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$       - Thus the impedance of the capacitor is greater than that of the resistor: $R_2 \ll |\underline{X}_C|$
       - Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially       - Thus, in the parallel circuit of $R_2$ and $C$, $R_2$ acts essentially
-      - Thus the circuit behaves like an inverting amplifier: $|\underline{A}_V|=|-\frac{R_2}{R_1}|$+      - Thus the circuit behaves like an inverting amplifier: $|\underline{A}_{\rm V}|=|-\frac{R_2}{R_1}|$
       - For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier.       - For the inverting amplifier, a sinusoidal input signal results in a negated sinusoidal signal. This corresponds to a phase of $\varphi=\pm 180°$. The sign is not defined by the inverting amplifier.
   - $\omega \rightarrow \infty$:   - $\omega \rightarrow \infty$:
Zeile 486: Zeile 517:
       - Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$       - Thus the impedance of the capacitor is smaller than that of the resistor: $R_2 \gg |\underline{X}_C|$
       - Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts       - Thus, in the parallel circuit of $R_2$ and $C$, essentially $C$ acts
-      - Thus the circuit behaves like a inverting integrator: $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$ and $\varphi=+ 90°$+      - Thus the circuit behaves like a inverting integrator: $|\underline{A}_{\rm V}|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$ and $\varphi=+ 90°$
  
-From this it can be seen that+From thisit can be seen that
  
   * for low frequencies a constant gain is expected and   * for low frequencies a constant gain is expected and
-  * for high frequencies a drop as known from the inverting integrator.+  * for high frequencies a drop is known from the inverting integrator.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 497: Zeile 528:
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic10| Expected bottom diagram of low-pass filter>+<imgcaption pic10| Expected bottom diagram of Low Pass Filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>erwartetes_Bodediagramm_des_Tiefpass_Filter}}+\\ {{drawio>erwartetes_Bodediagramm_des_Tiefpass_Filter.svg}}
 </panel></WRAP> </panel></WRAP>
  
-floor diagram can be estimated from the extreme value consideration.+Bode diagram can be estimated from the extreme frequency consideration.
  
 **Frequency Response:** **Frequency Response:**
  
-  * For low frequencies, the filter behaves like an inverting amplifier with $|\underline{A}_V|=|-\frac{R_2}{R_1}|$ +  * For low frequencies, the filter behaves like an inverting amplifier with     $|\underline{A}_{\rm V}\xrightarrow{f \rightarrow 0}      |-\frac{R_2}{R_1}                   |$ 
-  * For higher frequencies, the filter behaves like an inverting integrator with $|\underline{A}_V|=|-\frac{1}{\omega \cdot R_1 \cdot C}|$+  * For higher frequencies, the filter behaves like an inverting integrator with $|\underline{A}_{\rm V}\xrightarrow{f \rightarrow \infty} |-\frac{1}{\omega \cdot R_1 \cdot C}|$
   * There is a frequency where both situations seem to occur simultaneously   * There is a frequency where both situations seem to occur simultaneously
  
Zeile 518: Zeile 549:
 For the intermediate area, there must be a transition between the two extremal situations. For the intermediate area, there must be a transition between the two extremal situations.
  
-One problem still seems to be that for the inverting amplifier it is not clear whether the phase is now $+180°$ or $-180°$. In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ dt$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just $90°$ phase shift from inverting amplifier to high frequencies at low frequencies.+One problem still seems to be that for the inverting amplifierit is not clear whether the phase is now $+180°$ or $-180°$. In the mathematical consideration of the inverting integrator, it turned out that for a capacitor an integration step ($U=\frac{1}{C} \int I_C \ {\rm d}t$) must be performed. In the inverting amplifier, no integration step was necessary. Thus, a sinusoidal input signal is shifted by 90° at most. So there must be just $90°$ phase shift from the inverting amplifier to high frequencies at low frequencies.
  
-From this knowledge, we get an expected floor diagram as seen in <imgref pic10>.+From this knowledge, we get an expected Bode diagram as seen in <imgref pic10>.
  
 <WRAP column 100%> <panel type="danger" title="Notice: filters in the Bode diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> <WRAP column 100%> <panel type="danger" title="Notice: filters in the Bode diagram"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
Zeile 526: Zeile 557:
 The following rules apply to filters: The following rules apply to filters:
  
-  * for each energy store present, the order of the filter increases by 1. +  * for each energy storage element in the circuit, the order of the filter increases by 1. 
-  * for each energy store present ($C$, $L$) there is an amplitude change of $-20dB/dec$ +  * for each energy storage element in the circuit ($C$, $L$) there is an amplitude change of $-20~\rm dB/dec$ 
-  * each energy store present ($C$, $L$) results in a phase change of $-90°$+  * each energy storage element in the circuit ($C$, $L$) results in a phase change of $-90°$
   * The phase response is monotonically decreasing.   * The phase response is monotonically decreasing.
  
Zeile 535: Zeile 566:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== RC element and cutoff frequency ===+=== RC element and cut-off frequency ===
  
-In the circuit, the parallel circuit $R_2$ and $C$ behaves like a passive RC element. That is, it affects the frequency response. At a certain frequency, the circuit behaves just in such a way that the current runs half over $R_2$ and $C$, thus acting "half" like an inverting amplifier and "half" like inverting integrator. This frequency is the cutoff frequency $f_{Gr}$:+In the circuit, the parallel circuit $R_2$ and $C$ behave like a passive RC element. That is, it affects the frequency response. At a certain frequency, the circuit behaves just in such a way that the current runs half over $R_2$ and $C$, thus acting "half" like an inverting amplifier and "half" like an inverting integrator. This frequency is the cut-off frequency $f_{c}$:
  
 For this: \\  For this: \\ 
 $|\underline{X}_C|=R_2$ \\ $|\underline{X}_C|=R_2$ \\
-$\frac{1}{\omega_{Gr} \cdot C}=R_2 \rightarrow \omega_{Gr} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{Gr}$\\ +$\frac{1}{\omega_{c} \cdot C}=R_2 \rightarrow \omega_{c} = \frac{1}{R_2 \cdot C} = 2\pi \cdot f_{c}$\\ 
-$\boxed{f_{Gr} = \frac{1}{2\pi \cdot R_2 \cdot C}} $\\+$\boxed{f_{c} = \frac{1}{2\pi \cdot R_2 \cdot C}} $\\
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 548: Zeile 579:
 ==== 5.2.2 Circuit Analysis with Complex Calculus ==== ==== 5.2.2 Circuit Analysis with Complex Calculus ====
  
-Now the circuit is to be analyzed again by means of complex calculation. The impedances are again understood as complex numbers. The starting point is again the voltage gain of the (complex) inverting amplifier. The impedances from <imgref pic7_1> are taken into account:+Now the circuit is to be analyzed again using complex calculations. The impedances are again understood as complex numbers. The starting point is again the voltage gain of the (complex) inverting amplifier. The impedances from <imgref pic7_1> are taken into account:
  
-$\underline{A}_V=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{j \cdot \omega \cdot C}}{R_2 + \frac{1}{j \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ +$\underline{A}_{\rm V}=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{{\rm j\cdot \omega \cdot C}}{R_2 + \frac{1}{{\rm j\cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ 
-$\boxed{\underline{A}_V= - \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}}$+$\boxed{\underline{A}_{\rm V}= - \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j\omega \cdot R_2 \cdot C}}$
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== Calculation of magnitude and phase ===+=== Calculation of Absolute Value and Phase ===
  
-For the calculation of the amount $A_V$ a "trick" can be used. In principle, the amount can always be determined by multiplication with the conjugate complex value. But here it is easier to calculate the amount of the fraction via the amount of numerator and amount of denominator:+For the calculation of the absolute value $A_{\rm V}$ a "trick" can be used. In principle, the absolute value can always be determined by multiplication with the conjugate complex value. But here it is easier to calculate the absolute value of the fraction via the absolute value of the numerator and the absolute value of the denominator:
  
-$|\underline{A}_V| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot  \frac{|\mathcal{b}|}{|\mathcal{c}|}$+$|\underline{A}_{\rm V}| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot  \frac{|\mathcal{b}|}{|\mathcal{c}|}$
  
-This results in for the amount:+This results in the absolute value:
  
-$\boxed{|\underline{A}_V| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$+$\boxed{|\underline{A}_{\rm V}| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega \cdot R_2 \cdot C)^2}}}$
  
-For the phase $\varphi$ real value $\Re(\underline{A}_V)$ and imaginary value $\Im(\underline{A}_V)$ must be determined by multiplication with the conjugate complex value.+For the phase$\varphi$ real value $\Re(\underline{A}_{\rm V})$ and imaginary value $\Im(\underline{A}_{\rm V})$ must be determined by multiplication with the conjugate complex value.
  
-$\varphi = arctan(\frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)})$+$\varphi = \arctan(\frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})})$
  
 But here, too, there is a "trick": But here, too, there is a "trick":
  
-$\underline{A}_V= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}} \cdot \frac{1 - j \omega \cdot R_2 \cdot C}{\color{blue}{1 - j \omega \cdot R_2 \cdot C}}$+$\underline{A}_{\rm V}= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j\omega \cdot R_2 \cdot C}} \cdot \frac{1 - {\rm j\omega \cdot R_2 \cdot C}{\color{blue}{1 - {\rm j\omega \cdot R_2 \cdot C}}$
  
-After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$because all factors of the constant are real:+After all, it is just the conjugate complex value that is multiplied to get a real denominator. Thus the blue marked part is a real constant $\mathcal{C}$ because all factors of the constant are real:
  
-$\underline{A}_V= \color{blue}{\mathcal{C}} \cdot (1 - j \omega \cdot R_2 \cdot C)$+$\underline{A}_{\rm V}= \color{blue}{\mathcal{C}} \cdot (1 - {\rm j\omega \cdot R_2 \cdot C)$
  
-Thus, the phase $\varphi = arctan\left(\frac{\color{teal}{\Im(\underline{A}_V)}}{\color{brown}{\Re(\underline{A}_V)}}\right)$ is obtained as.+Thus, the phase $\varphi = \arctan\left(\frac{\color{teal}{\Im(\underline{A}_{\rm V})}}{\color{brown}{\Re(\underline{A}_{\rm V})}}\right)$ is obtained as.
  
-$\underline{A}_V= \mathcal{C} \cdot (\color{brown}{1} + j \cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$+$\underline{A}_{\rm V}= \mathcal{C} \cdot (\color{brown}{1} + {\rm j\cdot (\color{teal}{-\omega \cdot R_2 \cdot C}))$
  
-$\boxed{\varphi = arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$+$\boxed{\varphi = \arctan\left(\frac{\color{teal}{-\omega \cdot R_2 \cdot C}}{\color{brown}{1}}\right)}$
  
 \\ \\
-=== External value assessment ===+=== Consideration of extreme frequencies ===
  
-For the __amount__  we get +For the __absolute value__  we get 
-  - for $\omega \rightarrow 0$:  \\ $|\underline{A}_V| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 }}$, (nbsp) (nbsp) since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \gg 1$ \\  So the magnitude of the gain goes towards $|\underline{A}_V| = \frac{R_2}{R_1}$. \\  The effect is similar to the inverting amplifier\\ \\  +  - for $\omega \rightarrow 0$:  \\ $|\underline{A}_{\rm V}| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 }}$, (nbsp) (nbsp) since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \gg 1$ \\  So the absolute value of the gain tends towards $|\underline{A}_{\rm V}| = \frac{R_2}{R_1}$. \\  The effect is similar to the inverting amplifier\\ \\  
-  - at $\omega \rightarrow \infty$: \\ $|\underline{A}_V| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{\color{blue}{(\omega \cdot R_2 \cdot C)^2}}} $, (nbsp) (nbsp) da $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \ll 1$  \\  So the amount of gain goes towards $|\underline{A}_V| = \frac{1}{\color{blue}{\omega \cdot} R_1 \color{blue}{\cdot C}}$. \\  The action resembles the inverse integrator+  - at $\omega \rightarrow \infty$: \\ $|\underline{A}_{\rm V}| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + \color{blue}{(\omega \cdot R_2 \cdot C)^2}}} \rightarrow  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{\color{blue}{(\omega \cdot R_2 \cdot C)^2}}} $, (nbsp) (nbsp) since $\color{blue}{(\omega \cdot R_2 \cdot C)^2} \ll 1$  \\  So the absolute value of gain tends towards $|\underline{A}_{\rm V}| = \frac{1}{\color{blue}{\omega \cdot} R_1 \color{blue}{\cdot C}}$. \\  This  resembles the inverse integrator
  
 <WRAP> <WRAP>
Zeile 601: Zeile 632:
 </panel> </WRAP> </panel> </WRAP>
  
-For finding the phase $\color{red}{\varphi} = arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (<imgref pic20>). For the extremal values $\omega$ of results in:+For finding the phase $\color{red}{\varphi} = \arctan\left(\color{teal}{-\omega \cdot R_2 \cdot C}\right)$ it helps to look at the arctangent in the diagram. To do this, the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is plotted against the phase $\color{red}{\varphi}$ (<imgref pic20>). For the extremal values $\omega$ of results in:
  
   - at $\omega \rightarrow 0$: the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$.   - at $\omega \rightarrow 0$: the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ goes towards $-0$.
   - at $\omega \rightarrow \infty$: $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$.   - at $\omega \rightarrow \infty$: $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ argument goes towards $-\infty$.
  
-In the diagram the argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuousbecause also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the __upper branch__: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in <imgref pic20>. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram.+In the diagramthe argument $Arg$ is changeable by the slider in the upper left corner. The course in the diagram must be continuous because also the argument $Arg = \color{teal}{-\omega \cdot R_2 \cdot C}$ is continuous between $-0$ and $-\infty$. This is only possible on the __upper branch__: the point $Arg \rightarrow -0$ then just corresponds to approaching the y-axis (here $\color{red}{\varphi}$-axis) from the left in <imgref pic20>. The point $Arg \rightarrow -\infty$ corresponds to the path to the left in the diagram.
  
 This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$. This results in a progression of the phase $\color{red}{\varphi}$ from $\color{red}{\varphi}(\omega \rightarrow -0)=\pi = 180°$ to $\color{red}{\varphi}(\omega \rightarrow -\infty)=\frac{\pi}{2} = 90°$.
Zeile 612: Zeile 643:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-=== Calculation of the cutoff frequency ===+=== Calculation of the cut-off frequency ===
  
 The cut-off frequency can also be understood as the transition from the inverting amplifier to the inverting integrator. In the Bode diagram (<imgref pic10>), the cut-off frequency can be found at the intersection of the straight lines for the inverting amplifier and the inverting integrator. The cut-off frequency can also be understood as the transition from the inverting amplifier to the inverting integrator. In the Bode diagram (<imgref pic10>), the cut-off frequency can be found at the intersection of the straight lines for the inverting amplifier and the inverting integrator.
  
-Thus, for the cut-off frequency $f_{Gr}$ we get+Thus, for the cut-off frequency $f_{c}$ we get
  
-$\frac{R_2}{R_1} = \frac{1}{\omega_{Gr} R_1 \cdot C}$ \\ +$\frac{R_2}{R_1} = \frac{1}{\omega_{c} R_1 \cdot C}$ \\ 
-$\omega_{Gr} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{Gr}$+$\omega_{c} = \frac{1}{ R_2 \cdot C} = 2 \pi \cdot f_{c}$
  
-At the cut-off frequency, the result is an amount of:+At the cut-off frequency, the result is an absolute value of:
  
-$|\underline{A}_{V,Gr}| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega_{Gr} \cdot R_2 \cdot C)^2}}=  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C)^2}} =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + 1^2}} =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{2}}$+$|\underline{A}_{V,c}| =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\omega_{c} \cdot R_2 \cdot C)^2}}=  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + (\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C)^2}} =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{1^2 + 1^2}} =  \frac{R_2}{R_1}\cdot \frac{1}{\sqrt{2}}$
  
-$\boxed{|\underline{A}_{V,Gr}| =  \frac{1}{2} \sqrt{2} \cdot \frac{R_2}{R_1} = -3dB + |\underline{A}^{dB}_V(\omega \rightarrow 0)|}$+$\boxed{|\underline{A}_{V,c}| =  \frac{1}{2} \sqrt{2} \cdot \frac{R_2}{R_1} = -3~{\rm dB} + |\underline{A}^{\rm dB}_{\rm V}(\omega \rightarrow 0)|}$
  
-The phase at the cutoff frequency is:+The phase at the cut-off frequency is:
  
-$\varphi_{Gr} = arctan\left(-\omega_{Gr} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$ +$\varphi_{c} = \arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = \arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = \arctan\left(-1 \right)$ 
  
-$\boxed{\varphi_{Gr} = \frac{3}{4} \pi =135°}$+$\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$
  
-Because of the $-3dB$ attenuation of the low-frequency gain at the cutoff frequency, it is also called the **$-3dBcutoff frequency**.+Because of the $-3~{\rm dB}$ attenuation of the low-frequency gain at the cut-off frequency, it is also called the **$-3~{\rm dB}cut-off frequency**.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 641: Zeile 672:
 <imgcaption pic11_1|Circuit of the inverting Differentiator> <imgcaption pic11_1|Circuit of the inverting Differentiator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_des_Umkehrdifferentiators}}+\\ {{drawio>Schaltung_des_Umkehrdifferentiators.svg}}
 </panel></WRAP> </panel></WRAP>
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+0.000049999999999999996+1.9265835257097934+41+5+43%0Aa+288+128+384+128+8+15+-15+1000000+-0.000039999600019248555+0+100000%0Aw+384+112+384+80+0%0Ac+208+112+160+112+0+5.000000000000001e-7+-3.42003999865076%0Aw+288+80+288+112+0%0Ar+384+80+288+80+0+10000%0Ag+288+144+288+176+0%0AR+128+112+96+112+0+3+40+5+0+0+0.5%0A207+384+128+432+128+4+U_A%0A403+320+208+464+272+0+7_8_0_12294_4.008796818347498_0.0001_0_2_7_3_U%5CsA%0A207+128+112+128+144+4+U_E%0Aw+384+112+384+128+0%0A403+112+208+256+272+0+9_8_0_12294_4.9840000009474466_0.0001_0_2_9_3_U%5CsE%0Aw+128+112+160+112+0%0Aw+272+112+288+112+0%0Ar+208+112+272+112+0+10%0A noborder}} </WRAP>+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l5AWAnC1b0DZzSQJgwKwAcAzAbgQOyRKVIkIgIQFMkgECmAtGGAFABDELiJFwokCSKMwk8WFa9WYePBDcYakmgzwweBEQImo4NZH4B3KTPBzbjIpCj8AxiMgKHYPfdxmBLAWIXBgPJQaJNAIuHA6KESEkJRYWpACNqLiziJi-q4ATo4gudmlLi6q8PwA5nkKCIzlYKlmlgBKEt4BSFh8AS7sCC6slVDQBPxxkdKykggkAXLijACqAPoA8vwj7EsucasYzZSDIJQbRBuQG3K4SAg3wRnP+hu8G7gblyQbmzsZt0CitwE0mCBNgBJawlAZwySWPYFI4iAhYXBnMxIK43O64B5PdKvYlgD5k744v7Q2Gg+G+KoOSxZLHwlpM-jFVFs1lM8z8IA noborder}} </WRAP>
  
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Zeile 651: Zeile 682:
 <imgcaption pic11| Bode diagram of the inverting differentiator> <imgcaption pic11| Bode diagram of the inverting differentiator>
 </imgcaption> </imgcaption>
-\\ {{drawio>Bode_Diagramm_des_Umkehrdifferentiators}}+\\ {{drawio>Bode_Diagramm_des_Umkehrdifferentiators.svg}}
 </panel></WRAP> </panel></WRAP>
  
 In <imgref pic11_1> an inverting differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped. In <imgref pic11_1> an inverting differentiator is shown. Compared to the integrator here just the resistor and the capacitor are swapped.
  
-In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is resulting at the output. The derivative at the reversal points ("peaks") of the signal cannot be determined (see [[wpde>Differentiable_function|Differentiability of the sum function]]). This leads to problems in the calculation during the simulation and can be seen as overshoot or "deflection" at $U_O$. To reduce this, a small resistor (relative to the feedback resistor) is inserted after the capacitor.+In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is resulting in the output. The derivative at the reversal points ("peaks") of the signal cannot be determined (see [[wpde>Differentiable_function|Differentiability of the sum function]]). This leads to problems in the calculation during the simulation and can be seen as overshoot or "deflection" at $U_{\rm O}$. To reduce this, a small resistor (relative to the feedback resistor) is inserted after the capacitor.
  
  
-In the following, only the results will be discussed without derivation.+In the following, only the results will be discussed without calculation.
  
 Circuit analysis via differential equation yields: Circuit analysis via differential equation yields:
  
-$\boxed{U_O = - R \cdot C \frac{d}{dt}U_I}$+$\boxed{U_{\rm O} = - R \cdot C \frac{\rm d}{{\rm d}t}U_{\rm I}}$
  
 With complex calculation, the transfer function becomes: With complex calculation, the transfer function becomes:
  
-$\boxed{\underline{A}_V=-j \cdot \omega \cdot R \cdot C}$+$\boxed{\underline{A}_{\rm V}=-{\rm j\cdot \omega \cdot R \cdot C}$
  
 From this, the Bode diagram shown in <imgref pic11> can be determined. From this, the Bode diagram shown in <imgref pic11> can be determined.
Zeile 675: Zeile 706:
 <panel type="info" title="Exercise 5.3.1 Inverting Differentiator"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 5.3.1 Inverting Differentiator"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 <WRAP> <WRAP>
-{{drawio>Vorgabe_Zeitverlauf_Umkehrintegrator}}</WRAP>+{{drawio>Vorgabe_Zeitverlauf_Umkehrintegrator.svg}} 
 +</WRAP>
  
-For the inverting differentiator shown in <imgref pic11_1>, derive the complex voltage gain, and its magnitude and phase using complex calculus as shown for the [[#inverting_integrator|inverting integrator]]. In doing so, implement the following steps:+For the inverting differentiator shown in <imgref pic11_1>, derive the complex voltage gain, and its absolute value and phase using complex calculus as shown for the [[#inverting_integrator|inverting integrator]]. In doing so, implement the following steps:
  
-  - Circuit analysis by means of differential equation +  - Circuit analysis using differential equation 
-  - Determination of magnitude and phase from differential equation (incl. consideration of extreme cases) +  - Determination of absolute value and phase from differential equation (incl. consideration of extreme cases) 
-  - Example of a signal-time-curve with$R = 10 k\Omega$ and $C = 2µF$ and $U_I$ as shown in the diagram +  - Example of a signal-time-curve with $R = 10 ~\rm k\Omega$ and $C = 2 ~\rm µF$ and $U_{\rm I}as shown in the diagram. 
-  - Circuit analysis by means of complex calculation +  - Circuit analysis using a complex calculation 
-  - Consideration of magnitude and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$ +  - Consideration of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$ 
-  - Frequency response (Bode plot) for circuit with$R = 10 k\Omega$ and $C = 16nF$+  - Frequency response (Bode plot) for circuit with $R = 10 ~\rm k\Omega$ and $C = 16 ~\rm nF$.
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 690: Zeile 722:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.4 High Pass =====+===== 5.4 High Pass Filter =====
  
-<WRAP left>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+240+176+240+224+0%0Aw+384+80+384+160+0%0Aw+240+80+240+144+0%0Aa+240+160+384+160+4+15+-15+10000000000%0Ac+144+80+208+80+0+1.5915e-9+0.16559840149986407%0Ar+80+80+144+80+0+10000%0AO+384+160+448+160+0%0A170+80+80+32+80+2+2+400000+0.005+0.1%0Ar+256+80+368+80+0+1000%0AB+64+48+224+160+0+Box%0As+144+112+80+112+0+1+false%0As+208+112+144+112+0+1+false%0Aw+240+80+256+80+0%0Aw+368+80+384+80+0%0Ax+78+147+100+150+0+18+S1%0Ax+149+149+171+152+0+18+S2%0Aw+144+80+144+112+0%0Aw+80+80+80+112+0%0Aw+208+80+208+112+0%0Aw+208+80+240+80+0%0Ao+0+32+0+34+10+0.0125+0+-1%0A 600,500 noborder}} </WRAP>+A high pass filter can be created from the inverting differentiator, if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (<imgref pic12_1>)The simulation above shows this high passA reverse integrator forms from this with switch $\rm S1$ closed and switch $\rm S2$ openWhen the switch is invertedan inverting amplifier is formed. In the simulation, clicking on a frequency point again shows the distribution of the currents.
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic12_1| Circuit of the high-pass filter>+<imgcaption pic12_1| Circuit of the High Pass Filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_des_Hochpass_Filter}}+\\ {{drawio>Schaltung_des_Hochpass_Filter.svg}}
 </panel></WRAP> </panel></WRAP>
  
-~~PAGEBREAK~~ ~~CLEARFIX~~+With complex calculation this results in: $\boxed{\underline{A}_{\rm V} = - \frac{R_2}{R_1} \cdot \frac{{\rm j} \cdot \omega \cdot R_1 \cdot C}{1 + {\rm j} \cdot \omega \cdot R_1 \cdot C}} $ 
 + 
 +From this, the Bode diagram shown in <imgref pic12> can be determined.
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic12| Bode diagram of the high-pass filter>+<imgcaption pic12| Bode diagram of the High Pass Filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Bode_Diagramm_des_Hochpass_Filter}}+\\ {{drawio>Bode_Diagramm_des_Hochpass_Filter.svg}}
 </panel></WRAP> </panel></WRAP>
  
-A high-pass filter can be created from the inverting differentiator, if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (<imgref pic12_1>). The simulation above shows this high pass. A reverse integrator forms from this with switch $S1$ closed and switch $S2$ open. When the switch is inverted, an inverting amplifier is formed. In the simulation, clicking on a frequency point again shows the distribution of the currents.+~~PAGEBREAK~~ ~~CLEARFIX~~
  
-With complex calculation this results in: $\boxed{\underline{A}_V = - \frac{R_2}{R_1} \cdot \frac{j \cdot \omega \cdot R_1 \cdot C}{1 + j \cdot \omega \cdot R_1 \cdot C}} $+<WRAP>{{url>https://www.falstad.com/afilter/circuitjs.html?running=false&cct=$+1+0.000005+5+63+5+50%0A%25+4+984968.4014609919%0Ag+240+176+240+224+0%0Aw+384+80+384+160+0%0Aw+240+80+240+144+0%0Aa+240+160+384+160+4+15+-15+10000000000%0Ac+144+80+208+80+0+1.5915e-9+0.16559840149986407%0Ar+80+80+144+80+0+10000%0AO+384+160+448+160+0%0A170+80+80+32+80+2+2+400000+0.005+0.1%0Ar+256+80+368+80+0+1000%0AB+64+48+224+160+0+Box%0As+144+112+80+112+0+1+false%0As+208+112+144+112+0+1+false%0Aw+240+80+256+80+0%0Aw+368+80+384+80+0%0Ax+78+147+100+150+0+18+S1%0Ax+149+149+171+152+0+18+S2%0Aw+144+80+144+112+0%0Aw+80+80+80+112+0%0Aw+208+80+208+112+0%0Aw+208+80+240+80+0%0Ao+0+32+0+34+10+0.0125+0+-1%0A 600,500 noborder}} </WRAP>
  
-From this, the Bode diagram shown in <imgref pic12> can be determined. 
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 718: Zeile 751:
 <panel type="info" title="Exercise 5.4.1 1st order high pass"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 5.4.1 1st order high pass"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-In [[#lowpass|previous chapter]] the gain $A_V$ of the 1st order low pass was derived based on its circuit. In the same way now the gain for a high pass (cf. <imgref pic12_1>) shall be derived.+In [[#lowpass|previous chapter]] the gain $A_{\rm V}$ of the 1st order low pass filter was derived based on its circuit. In the same waynow the gain for a high pass filter (cf. <imgref pic12_1>) shall be derived.
  
-  - Behavior of magnitude and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$.+  - Behavior of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$.
   - Expected Bode diagram   - Expected Bode diagram
-  - RC element and cutoff frequency+  - RC element and cut-off frequency
   - Circuit analysis with complex calculation   - Circuit analysis with complex calculation
-  - Calculation of magnitude and phase+  - Calculation of absolute value and phase
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 730: Zeile 763:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.5 Overview high pass / low pass =====+===== 5.5 Overview high pass Filter / low pass Filter =====
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic13| Overview high pass / low pass>+<imgcaption pic13| Overview high pass Filter / low pass Filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Übersicht_Hochpass_Tiefpass}}+\\ {{drawio>Übersicht_Hochpass_Tiefpass.svg}}
 </panel></WRAP> </panel></WRAP>
  
Zeile 742: Zeile 775:
 ====== Exercises ====== ====== Exercises ======
  
-<panel type="info" title="Exercise 5.0.1. converting linear factors to dB"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.0.1. Converting linear Factors to dB"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points 20 dB ≙ factor 10 and 6 dB ≙ factor 2.+Derive the linear factor for the following levels in dB. State how this factor can be determined in each case using the interpolation points $20 ~{\rm dB}$ ≙ factor 10 and $~{\rm dB}$ ≙ factor 2.
  
 Solve without a calculator  (note: $\sqrt{2}\approx 1.414$, ${1\over\sqrt{2}}\approx 0.707$). Solve without a calculator  (note: $\sqrt{2}\approx 1.414$, ${1\over\sqrt{2}}\approx 0.707$).
  
-As an example, the derivation is sketched for the value $10 dB$.+As an example, the calculation is sketched for the value $10 ~{\rm dB}$.
  
-^level^over interpolation points in dB^over interpolation points linear^linear factor| +^level^over interpolation points in $\rm dB$^over interpolation points linear^linear factor| 
-|$10 dB$|$5 \cdot 6 dB - 20 dB$|$2^5 \cdot {1\over 10}$|$3,2$| +|$10  ~{\rm dB}$|$5 \cdot 6 ~{\rm dB- 20 ~{\rm dB}$|$2^5 \cdot {1\over 10}$|$3,2$| 
-|$2 dB$| | | | +|$2   ~{\rm dB}$| | | | 
-|$4 dB$| | | | +|$4   ~{\rm dB}$| | | | 
-|$6 dB$| | | | +|$6   ~{\rm dB}$| | | | 
-|$8 dB$| | | | +|$8   ~{\rm dB}$| | | | 
-|$12 dB$| | | | +|$12  ~{\rm dB}$| | | | 
-|$14 dB$| | | | +|$14  ~{\rm dB}$| | | | 
-|$16 dB$| | | | +|$16  ~{\rm dB}$| | | | 
-|$18 dB$| | | | +|$18  ~{\rm dB}$| | | | 
-|$15 dB$| | | | +|$15  ~{\rm dB}$| | | | 
-|$79 dB$| | | | +|$79  ~{\rm dB}$| | | | 
-|$128 dB$| | | |+|$128 ~{\rm dB}$| | | |
  
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.0.2. Series of Amplifiers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +The following image shows a series of amplifiers, which shall be analyzed.
 +
 +<WRAP>
 +{{drawio>SeriesOfAmps.svg}}
 +</WRAP>
 +
 +The amplification factors are:
 +  * $A_{\rm V1} =80$
 +  * $A_{\rm V2} =0.0125$
 +  * $A_{\rm V3} =250'000$
 +
 +Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). 
 +
 +<button size="xs" type="link" collapse="Loesung_5_0_2_1_SolutionSteps">{{icon>eye}} Solution steps</button>
 +<collapse id="Loesung_5_0_2_1_SolutionSteps" collapsed="true"> 
 +
 +  - Rearrange the given linear factors as exponents of $2$ and $10$, e.g. $2^6 \cdot 10^7$
 +  - Use the exponent values to transfer it into $\rm dB$, in this example: $6 \cdot 6~{\rm dB} + 7 \cdot 20~{\rm dB} $
 +  - Calculate the $\rm dB$ value, in this example: $36~{\rm dB} + 140~{\rm dB} = 176~{\rm dB}$
 +</collapse>
 +
 +<button size="xs" type="link" collapse="Loesung_5_0_2_1_Solution">{{icon>eye}} Solution </button>
 +<collapse id="Loesung_5_0_2_1_Solution" collapsed="true"> 
 +^  Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^  transfer into $\rm dB$  $\rightarrow$^  result in $\rm dB$  ^
 +|  $A_{\rm V1}$   | $80$       | $8    \cdot 10$        | $2^3    \cdot 10^1$          | $ 3 \cdot 6~{\rm dB} +    \cdot 20~{\rm dB} $     | $ 38~{\rm dB}$         |
 +|  $A_{\rm V2}$   | $0.0125$   | $0.125\cdot 0.1$       | $2^{-3} \cdot 10^{-1}$       | $-3 \cdot 6~{\rm dB} + (-1) \cdot 20~{\rm dB} $     | $-38~{\rm dB}$         |
 +|  $A_{\rm V3}$   | $250'000$  | $0.25 \cdot 1'000'000$ | $2^{-2} \cdot 10^{6}$        | $-2 \cdot 6~{\rm dB} +    \cdot 20~{\rm dB} $     | $108~{\rm dB}$         |
 +</collapse>
 +
 +<button size="xs" type="link" collapse="Loesung_5_0_2_1_Result">{{icon>eye}} Result</button>
 +<collapse id="Loesung_5_0_2_1_Result" collapsed="true"> 
 + $A_{\rm V} = 250'000$ \\
 + $A_{\rm V}^{\rm dB} = 108~{\rm dB}$
 +</collapse>
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
 <panel type="info" title="Exercise 5.1.1 Analysis of Circuits in Tina TI"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 5.1.1 Analysis of Circuits in Tina TI"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +The following circuit shall be given with $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ and a sinusoidal input voltage $U_{\rm I} = 1 ~\rm V $ with $f = 1 ~\rm kHz$ 
  
 <WRAP> <WRAP>
-{{drawio>Umkehrintegrator}}</WRAP>+{{drawio>Umkehrintegrator.svg}} 
 +</WRAP>
  
-Let the circuit shown opposite with $R= 10 k\Omega$, $C = 1.6 uF$ and a sinusoidal input voltage $U_I = 1 V $ with $f = 1 kHz$ be given. As described in the course, the Bode diagram can be displayed in Tina TI via Analysis > AC Analysis > AC Transfer Characteristic. In the following, frequencies from 100 Hz to 1 GHz are relevant.+As described in the course, the Bode diagram can be displayed in Tina TI via ''Analysis'' ''AC Analysis'' ''AC Transfer Characteristic''\\ In the following, frequencies from 100 Hz to 1 GHz are relevant.
  
   - Simulate this circuit in Tina TI \\ - with an ideal operational amplifier \\ - using the operational amplifiers uA776, LM301A and LM318.   - Simulate this circuit in Tina TI \\ - with an ideal operational amplifier \\ - using the operational amplifiers uA776, LM301A and LM318.
       - Attach the Bode diagram.       - Attach the Bode diagram.
-      - Briefly describe the differences in the amplitude response of the gain $A_V$. +      - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$. 
-  - What happens if instead of $R= 10 k\Omega$, $C = 1.6 uF$ the same time constant is implemented with $R= 10 M\Omega$, $C = 1.6 nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier. +  - What happens if instead of $R= 10 ~\rm k\Omega$, $C = 1.6 ~\rm µF$ the same time constant is implemented with $R= 10 ~\rm M\Omega$, $C = 1.6 ~\rm nF$? Verify this by comparing the Bode plot of the LM318 operational amplifier. 
-  - Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 k\Omega$ and use the LM318 op-amp.+  - Simulate an inverting amplifier in Tina TI. To do this, replace the capacitor in the circuit with a resistor $R_2 = 10 ~\rm k\Omega$ and use the LM318 op-amp.
       - Attach the Bode diagram.       - Attach the Bode diagram.
       - What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup?       - What should the Bode diagram (amplitude and phase response) look like for an ideal inverting amplifier? What deviation do you notice in the real setup?
-      - Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of 1dB)? Use zoom and/or cursor to determine. +      - Up to what frequency can the circuit be operated as an inverting amplifier (maximum deviation of $1~{\rm dB}$)? Use zoom and/or cursor to determine. 
-  - Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to ground. The inverting input should have the above sinusoidal input voltage.+  - Simulate an open-loop operational amplifier LM318 (i.e. without feedback network). The non-inverting input should be connected to the ground. The inverting input should have the above sinusoidal input voltage.
       - Attach the Bode diagram.       - Attach the Bode diagram.
       - What is the cut-off frequency?       - What is the cut-off frequency?
-      - How many dB per decade does the amplitude response drop at high frequencies?+      - How many $\rm dBper decade does the amplitude response drop at high frequencies?
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>