Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung Letzte Überarbeitung Beide Seiten der Revision | ||
circuit_design:5_filter_circuits_i [2021/12/06 12:20] tfischer |
circuit_design:5_filter_circuits_i [2023/03/28 16:10] mexleadmin |
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< | < | ||
- | * 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]]. |
</ | </ | ||
Zeile 11: | Zeile 11: | ||
=== Introductory example === | === Introductory example === | ||
- | < | + | < |
- | In various | + | Various |
- | One possibility to process such signals is the use of filters. Filters have already been described in [[:elektrotechnik_2:netzwerke_bei_veraenderlicher_frequenz|Electrical | + | 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 |
- | Two low-pass filters are shown in the simulation. These attenuate the high-frequency components in the signal. The signal $U_{A1}$ after the first filter stage already shows significantly less noise. In the signal $U_{A2}$, 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. |
- | With the switch (left in the simulation) a frequency-variable test signal ([[https://de.wikipedia.org/ | + | With the switch (left in the simulation) a frequency-variable test signal ([[https://en.wikipedia.org/ |
This chapter will explain the basics of active high and low pass filters from operational amplifiers. | This chapter will explain the basics of active high and low pass filters from operational amplifiers. | ||
Zeile 31: | Zeile 31: | ||
After this lesson, you should: | After this lesson, you should: | ||
- | - Be able to apply the superpostition | + | - Be able to apply the superposition |
- | - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, | + | - Know how differential amplifiers and instrumentation amplifiers differ (circuit, applications, |
- 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 |
</ | </ | ||
Zeile 40: | Zeile 40: | ||
===== 5.0 Representation of numerical values ===== | ===== 5.0 Representation of numerical values ===== | ||
- | In order to be able to analyse | + | To be able to analyze |
< | < | ||
Zeile 48: | Zeile 48: | ||
</ | </ | ||
- | ==== 5.0.1 The dB measure | + | ==== 5.0.1 The dB scale ==== |
- | The decibel measure | + | 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 |
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 |
\\ | \\ | ||
- | === Technical level in db === | + | === Technical level in dB === |
< | < | ||
- | ^ Name ^ Symbol | + | ^ Name ^ Symbol |
- | | [[wpde> | + | | [[wpde> |
- | | [[wpde>Leistungspegel]] | + | | [[wp>Power level]] | $\rm dBm$ | $10{\rm dB} \cdot \log_{10}(P/ |
- | | [[wpde>Power level]] | + | | [[wp>Power level]] |
- | | [[wpde> | + | | [[wp> |
- | | [[wpde>Sound pressure level]] | + | | [[wp>Sound pressure level]] |
</ | </ | ||
- | Note that this equation changes somewhat for __power__ | + | Note that this equation changes somewhat for __power__ |
- | \\ $A_P^{dB}=20 dB \cdot log_{10}\left(\frac{P_2^\frac{1}{2}}{P_1^\frac{1}{2}}\right)$\\ (nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp)$ =\color{blue}{20 dB \cdot \frac{1}{2}} \cdot log_{10}\left(\frac{P_2}{P_1}\right) | + | Since $P \sim U^2$ or $U \sim P^\frac{1}{2}$, |
+ | $A_P^{\rm dB}= | ||
+ | (nbsp)(nbsp)(nbsp)(nbsp)(nbsp)(nbsp) | ||
+ | $ =\color{blue}{20 | ||
- | 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 === | ||
< | < | ||
- | ^linear factor^level [$dB$]| | + | ^ linear factor |
- | |$\times 0.01$|$-40dB$| | + | |$\times 0.01$ |
- | |$\times 0.1$|$-20dB$| | + | |$\times 0.1$ |$-20 ~{\rm dB}$ | |
- | |$\times 1$|$0dB$| | + | |$\times 1$ |$ 0 ~{\rm dB}$ | |
- | |$\times 2$|$\approx +6dB$| | + | |$\times 2$ |$\approx +6~{\rm dB}$| |
- | |$\times 10$|$+20dB$| | + | |$\times 10$ |
- | |$\times 100$|$+40dB$| | + | |$\times 100$ |$+40 ~{\rm dB}$ | |
</ | </ | ||
- | By this equation, various linear factors and ratios $A = \boxed{}_2 / \boxed{}_1$ can be converted into a level $A^{dB}$ in $dB$. Occasionally, | + | 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, | ||
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 $10^\color{blue}{x} = \color{green}{1}$, | + | - For $A_{\rm V}= \color{green}{1} $ we get < |
- | - 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 $10^\color{blue}{x} = \color{green}{0.01}$, | + | $ 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 $10^\color{blue}{x} = \color{green}{1}$, |
+ | - For $A_{\rm V}= \color{green}{0.01} $ we get < | ||
+ | $ 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 $10^\color{blue}{x} = \color{green}{0.01}$, | ||
+ | - For $A_{\rm V}= \color{green}{2} $, we get < | ||
+ | $ 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, $\color{blue}{x}\approx 0.30103$. | ||
=== Use of the dB measure === | === Use of the dB measure === | ||
Zeile 96: | Zeile 106: | ||
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 | + | * **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 |
- | * **Relationship to Sensory Perceptions**: | + | * **Relationship to Sensory Perceptions**: |
- | * **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~~ | ||
Zeile 105: | Zeile 115: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | 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 | + | 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 | ||
+ | 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=" | <WRAP column 100%> <panel type=" | ||
Zeile 114: | Zeile 128: | ||
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 $A_{\rm V}^{\rm dB}$. |
</ | </ | ||
Zeile 130: | Zeile 144: | ||
=== 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 | + | - < |
- | - $A_V^{dB}=56dB$ \\ with interpolation points: $A_V^{dB}=56dB = 80dB - 24dB = \color{blue}{4}\cdot | + | $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot |
- | - $A_V^{dB}=55dB$ \\ with interpolation points: $A_V^{dB}=56dB = 40dB + 18dB - 3dB = \color{blue}{2}\cdot | + | This becomes linear |
+ | - < | ||
+ | $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot | ||
+ | This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \quad\ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad | ||
+ | - < | ||
+ | $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot | ||
+ | This becomes linear $ \qquad \qquad | ||
\\ | \\ | ||
- | 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~~ | ||
Zeile 147: | Zeile 167: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | The aim of the floor diagram | + | The Bode diagram |
=== Preliminary consideration: | === Preliminary consideration: | ||
- | 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 values, there 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 | + | - 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 gain, since 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~~ | ||
- | === path to the Bode diagram | + | === Developing |
- | < | + | < |
- | 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 | + | To better understand the frequency dependence of the voltage gain, it can be plotted |
- | 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, | + | 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, |
The phase can be made visible via '' | The phase can be made visible via '' | ||
Zeile 180: | Zeile 200: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
Zeile 190: | Zeile 210: | ||
A jump in frequency by a factor of $\times 10$ is called a **decade** | A jump in frequency by a factor of $\times 10$ is called a **decade** | ||
- | - $|A_V(f)| = \mathcal{C} \cdot f$: 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. | + | - $|A_{\rm V}(f)| = \mathcal{C} \cdot f$: \\ 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. |
+ | | ||
+ | | ||
- | 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=" | ~~PAGEBREAK~~ <WRAP column 100%> <panel type=" | ||
Zeile 201: | Zeile 223: | ||
- 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 $A(\omega) \sim \omega^n$. | + | 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 $A(\omega) \sim \frac{1}{\omega}$, | ||
+ | </ | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 5.1 Reverse integrator | + | ===== 5.1 Inverse Integrator |
< | < | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
Zeile 216: | Zeile 240: | ||
In electronics, | In electronics, | ||
- | - Inductors are possible in integrated circuits, but 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, [[: | The following basic circuit is a modified, [[: | ||
- | For first circuit only the part between output voltage $U_A$ and virtual ground should be replaced by a capacitor (<imgref pic2>). | + | For the first circuit, only the part between output voltage $U_{\rm O}$ and virtual ground should be replaced by a capacitor (<imgref pic2>). |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
Zeile 229: | Zeile 253: | ||
==== 5.1.1 Circuit analysis with differential equations ==== | ==== 5.1.1 Circuit analysis with differential equations ==== | ||
- | < | + | < |
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, |
- | The circuit thus created is called | + | The circuit thus created is called |
- | 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~~ | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | 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) $ | + | $A_{\rm V} = ? \quad \rightarrow |
=== given equations === | === given equations === | ||
Zeile 256: | Zeile 280: | ||
Given the following equations: | Given the following equations: | ||
- | |I.|Basic equation|$U_A = 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 $| | + | |II. |
- | |III.|mesh 2|$U_D+U_C+U_O=0$| | + | |III. |Mesh 2 | $ U_{\rm D} + U_C + U_{\rm O} = 0$ | |
- | |IV.|Mesh|$I_R=I_C$| | + | |IV. |
- | |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 | $C |
- | |VI.|Resistance R|$R = { U_R \over I_R }$| | + | |VI. |
- | === Derivation | + | === Calculation |
- | The calculation is performed once in detail here (clicking on right arrow " | + | The calculation is performed once in detail here (clicking on the right arrow " |
+ | {{url> | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
\\ | \\ | ||
- | ==== 5.1.2 Signal-Zeit-Verlauf | + | ==== 5.1.2 Signal-Time Curve ==== |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | By means of an example, the signal-time curve at the reverse | + | Using an example, the signal-time curve at the inverting |
- | + | ||
- | - Let $R=5 c\Omega$, $C=1 \mu F$, and the input voltage waveform $U_E$ 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 |
- | - With the formula derived in 5.1.1 $U_O$ can be composed section by section: | + | |
- | The calculation is performed once in detail here (clicking on right arrow " | ||
- | <WRAP column 100%> <panel type=" | + | <WRAP column 100%> <panel type=" |
- | If a constant input voltage $U_I$ is applied to the reverse | + | If a constant input voltage $U_{\rm I}$ is applied to the inverting |
</ | </ | ||
Zeile 301: | Zeile 324: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ==== 5.1.3 Determination of amount | + | ==== 5.1.3 Determination of absolute value and phase ==== |
- | In order to be able to determine | + | To be able to determine |
- | $ 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 " | + | The calculation is performed once in detail here (clicking on the right arrow " |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
~~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** | + | The **phase** |
- | $U_I = + \hat{U}_I \cdot sin(\omega \cdot t)$\\ | + | $U_{\rm I} = + \hat{U}_{\rm I} \cdot \sin(\omega \cdot t)$\\ |
- | $U_O = + \hat{U}_O \cdot cos(\omega \cdot t) = + \hat{U}_O \cdot sin(\omega \cdot t + 90°)$\\ | + | $U_{\rm O} = + \hat{U}_{\rm O} \cdot \cos(\omega \cdot t) = + \hat{U}_{\rm O} \cdot \sin(\omega \cdot t + 90°)$\\ |
$ \rightarrow | $ \rightarrow | ||
\\ | \\ | ||
- | === Consideration of extreme | + | === Consideration of extreme |
- | 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 (${\omega}\rightarrow \infty$)** | + | 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 (${\omega}\rightarrow \infty$)** |
- | $ |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) |
$\varphi = +90° \qquad \forall \ \omega$ | $\varphi = +90° \qquad \forall \ \omega$ | ||
- | From these boundary conditions the frequency response can already be sketched, see <imgref pic4>. | + | From these boundary conditions, the frequency response can already be sketched, see <imgref pic4>. |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ==== 5.1.4 Circuit | + | ==== 5.1.4 Circuit |
- | 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 reverse | + | 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 |
$U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ | $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ | ||
- | $U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ dt+ Q_0(t_0)) \qquad | + | $U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad |
However, this consideration can only be implemented under certain boundary conditions: | However, this consideration can only be implemented under certain boundary conditions: | ||
- | - **sinusoidal quantities**: | + | - **sinusoidal quantities**: |
- | - **Swinged | + | - **Steady |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
This can now be used to calculate the circuit (<imgref pic5>): | This can now be used to calculate the circuit (<imgref pic5>): | ||
- | $\underline{Z}_1=R$ | + | $\underline{Z}_1=R$ |
+ | |||
+ | $\underline{Z}_2=\frac{1}{{\rm j} \cdot \omega \cdot C} = \frac{\rm -j}{\omega \cdot C}$ | ||
From the basic circuit of the [[: | From the basic circuit of the [[: | ||
- | $A_V = \frac{U_A}{U_E}=-\frac{R_2}{R_1}$ | + | $A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}=-\frac{R_2}{R_1}$ |
This results in the complex: | This results in the complex: | ||
- | $\underline{A}_V = \frac{\underline{U}_A}{\underline{U}_E}=-\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 | + | === Plausibility check via extreme |
^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 | + | |$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow 0} \infty$ |
- | |$\underline{A}_V \xrightarrow{\omega \rightarrow \infty} 0$|For $\omega \rightarrow \infty$ the capacitor | + | For $\omega \rightarrow 0$, the capacitor |
+ | Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of | ||
+ | $\underline{U}_{\rm O} \rightarrow \infty$. | ||
+ | |$\underline{A}_{\rm V} \xrightarrow{\omega \rightarrow \infty} 0$ | < | ||
+ | For $\omega \rightarrow \infty$ the capacitor | ||
+ | Given $\underline{U}_{\rm D} \rightarrow 0$, the operational amplifier must output an voltage of | ||
+ | $\underline{U}_{\rm O} \rightarrow 0$. </ | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | === magnitude | + | === Absolute Value and Phase === |
< | < | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | The amount | + | 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, | + | Specifically, |
- | $|\underline{A}_V(0dB)|\overset{!}{=} 1 \widehat{=} | + | $|\underline{A}_{\rm V}(0~{\rm dB})|\overset{!}{=} 1 \widehat{=} |
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 407: | Zeile 438: | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | The frequency response is to be illustrated | + | The frequency response is to be illustrated |
- | - 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 | + | - 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 |
- | - 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 | + | - 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, | - From this information, | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 5.2 Lowpass | + | ===== 5.2 Low Pass Filter |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | Another circuit can be derived from the reverse | + | Another circuit can be derived from the inverting |
* first be considered practically with a simulation, | * first be considered practically with a simulation, | ||
- | * then a picture of the system' | + | * then a picture of the system' |
- | * be checked by a circuit analysis with complex | + | * be checked by a circuit analysis with complex |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | === Lowpass | + | === Low Pass Filter |
< | < | ||
- | 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 460: | Zeile 491: | ||
- Now change the circuit to an inverting integrator and read off the gain and phase there as well. | - Now change the circuit to an inverting integrator and read off the gain and phase there as well. | ||
- Now both switches should be closed. | - Now both switches should be closed. | ||
- | - In which frequency ranges does the inverting amplifier or the reverse | + | - In which frequency ranges does the inverting amplifier or the inverting |
- | - 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 | + | ==== 5.2.1 First Consideration ==== |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | === Extreme | + | === Extreme |
- | From the observation of the capacitor for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$, the variation of magnitude | + | 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 reverse | + | - Thus the circuit behaves like a inverting |
- | From this it can be seen that | + | From this, it 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 reverse | + | * for high frequencies a drop is known from the inverting |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
Zeile 497: | Zeile 528: | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | A floor diagram can be estimated from the extreme | + | A Bode diagram can be estimated from the extreme |
**Frequency Response:** | **Frequency Response:** | ||
- | * For low frequencies, | + | * For low frequencies, |
- | * For higher frequencies, | + | * For higher frequencies, |
* 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 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 ($U=\frac{1}{C} \int I_C \ {\rm d}t$) 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 a $90°$ phase shift from the inverting amplifier to high frequencies at low frequencies. |
- | From this knowledge, we get an expected | + | From this knowledge, we get an expected |
<WRAP column 100%> <panel type=" | <WRAP column 100%> <panel type=" | ||
Zeile 526: | Zeile 557: | ||
The following rules apply to filters: | The following rules apply to filters: | ||
- | * for each energy | + | * for each energy |
- | * for each energy | + | * for each energy |
- | * each energy | + | * each energy |
* The phase response is monotonically decreasing. | * The phase response is monotonically decreasing. | ||
Zeile 535: | Zeile 566: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | === RC element and cutoff | + | === RC element and cut-off |
- | In the circuit, the parallel circuit $R_2$ and $C$ behaves | + | In the circuit, the parallel circuit $R_2$ and $C$ behave |
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 | + | Now the circuit is to be analyzed again using complex |
- | $\underline{A}_V=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{j \cdot \omega \cdot C}}{R_2 + \frac{1}{j \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ | + | $\underline{A}_{\rm V}=\frac{-\underline{Z}_2}{\underline{Z}_1}=\frac{-R_2||C}{R_1}=\frac{-\frac{R_2 \cdot \frac{1}{{\rm j} \cdot \omega \cdot C}}{R_2 + \frac{1}{{\rm j} \cdot \omega \cdot C}}}{R_1}=\frac{-R_2}{R_1 \cdot R_2 \cdot \omega \cdot C + R_1}$ \\ |
- | $\boxed{\underline{A}_V= - \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}}$ | + | $\boxed{\underline{A}_{\rm V}= - \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}}$ |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | === Calculation of magnitude | + | === Calculation of Absolute Value and Phase === |
- | For the calculation of the amount | + | For the calculation of the absolute value $A_{\rm V}$ a " |
- | $|\underline{A}_V| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{|\mathcal{c}|}$ | + | $|\underline{A}_{\rm V}| = |\mathcal{a} \cdot \frac{\mathcal{b}}{\mathcal{c}}| = |\mathcal{a}| \cdot \frac{|\mathcal{b}|}{|\mathcal{c}|}$ |
- | 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 " | But here, too, there is a " | ||
- | $\underline{A}_V= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + j \omega \cdot R_2 \cdot C}} \cdot \frac{1 - j \omega \cdot R_2 \cdot C}{\color{blue}{1 - j \omega \cdot R_2 \cdot C}}$ | + | $\underline{A}_{\rm V}= \color{blue}{- \frac{R_2}{R_1}\cdot \frac{1}{1 + {\rm j} \omega \cdot R_2 \cdot C}} \cdot \frac{1 - {\rm j} \omega \cdot R_2 \cdot C}{\color{blue}{1 - {\rm j} \omega \cdot R_2 \cdot C}}$ |
- | 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__ | + | For the __absolute value__ |
- | - 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 | + | - 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 |
- | - 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 | + | - 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 |
< | < | ||
Zeile 601: | Zeile 632: | ||
</ | </ | ||
- | 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 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. | + | In the diagram, the 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 | + | === Calculation of the cut-off |
- | The cut-off frequency can also be understood as the transition from the inverting amplifier to the reverse | + | The cut-off frequency can also be understood as the transition from the inverting amplifier to the inverting |
- | 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 | + | At the cut-off frequency, the result is an absolute value of: |
- | $|\underline{A}_{V, | + | $|\underline{A}_{V, |
- | $\boxed{|\underline{A}_{V, | + | $\boxed{|\underline{A}_{V, |
- | The phase at the cutoff | + | The phase at the cut-off |
- | $\varphi_{Gr} = arctan\left(-\omega_{Gr} \cdot R_2 \cdot C\right) = arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = arctan\left(-1 \right)$ | + | $\varphi_{c} = \arctan\left(-\omega_{c} \cdot R_2 \cdot C\right) = \arctan\left(-\frac{1}{ R_2 \cdot C} \cdot R_2 \cdot C\right) = \arctan\left(-1 \right)$ |
- | $\boxed{\varphi_{Gr} = \frac{3}{4} \pi =135°}$ | + | $\boxed{\varphi_{c} = \frac{3}{4} \pi =135°}$ |
- | Because of the $-3dB$ attenuation of the low-frequency gain at the cutoff | + | Because of the $-3~{\rm dB}$ attenuation of the low-frequency gain at the cut-off |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 5.3 Reverse Differential | + | ===== 5.3 Inverting Differentiator |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | < | + | < |
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | In <imgref pic11_1> an inverse | + | In <imgref pic11_1> an inverting |
- | In the simulation next to it, the effect of the circuit can be seen: the derivative of the inverted input signal is output at the output. The derivative at the reversal points (" | + | 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 (" |
- | In the following, only the results will be discussed without | + | In the following, only the results will be discussed without |
Circuit analysis via differential equation yields: | Circuit analysis via differential equation yields: | ||
- | $\boxed{U_A = - R \cdot C \frac{d}{dt}U_E}$ | + | $\boxed{U_{\rm O} = - R \cdot C \frac{\rm d}{{\rm d}t}U_{\rm I}}$ |
With complex calculation, | With complex calculation, | ||
- | $\boxed{\underline{A}_V=-j \cdot \omega \cdot R \cdot C}$ | + | $\boxed{\underline{A}_{\rm V}=-{\rm j} \cdot \omega \cdot R \cdot C}$ |
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 673: | Zeile 704: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | <panel type=" | + | <panel type=" |
< | < | ||
- | {{drawio> | + | {{drawio> |
+ | </ | ||
- | For the inverse | + | For the inverting |
- | - Circuit analysis | + | - Circuit analysis |
- | - Determination of magnitude | + | - 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 $U_E$ 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 | + | - Circuit analysis |
- | - Consideration of magnitude | + | - 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 a circuit with $R = 10 ~\rm k\Omega$ and $C = 16 ~\rm nF$. |
</ | </ | ||
Zeile 690: | Zeile 722: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 5.4 High Pass ===== | + | ===== 5.4 High Pass Filter |
- | <WRAP left>{{url> | + | A high pass filter can be created from the inverting differentiator, |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | ~~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. | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | A high-pass filter can be created from the reversal differentiator if the purely capacitive impedance for $Z_1$ is suitably extended via a resistive component (<imgref pic12_1> | + | ~~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}} $ | + | < |
- | From this, the Bode diagram shown in <imgref pic12> can be determined. | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
Zeile 718: | Zeile 751: | ||
<panel type=" | <panel type=" | ||
- | In [[# | + | In [[# |
- | - Behavior of magnitude | + | - Behavior of absolute value and phase for $\omega \rightarrow 0$ and $\omega \rightarrow \infty$. |
- Expected Bode diagram | - Expected Bode diagram | ||
- | - RC element and cutoff | + | - RC element and cut-off |
- Circuit analysis with complex calculation | - Circuit analysis with complex calculation | ||
- | - Calculation of magnitude | + | - Calculation of absolute value and phase |
</ | </ | ||
Zeile 730: | Zeile 763: | ||
~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
- | ===== 5.5 Overview high pass / low pass ===== | + | ===== 5.5 Overview high pass Filter |
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
Zeile 742: | Zeile 775: | ||
====== Exercises ====== | ====== Exercises ====== | ||
- | <panel type=" | + | <panel type=" |
- | 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 |
Solve without a calculator | Solve without a calculator | ||
- | As an example, the derivation | + | As an example, the calculation |
- | ^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, | + | |$10 |
- | |$2 dB$| | | | | + | |$2 ~{\rm dB}$| | | | |
- | |$4 dB$| | | | | + | |$4 ~{\rm dB}$| | | | |
- | |$6 dB$| | | | | + | |$6 ~{\rm dB}$| | | | |
- | |$8 dB$| | | | | + | |$8 ~{\rm dB}$| | | | |
- | |$12 dB$| | | | | + | |$12 |
- | |$14 dB$| | | | | + | |$14 |
- | |$16 dB$| | | | | + | |$16 |
- | |$18 dB$| | | | | + | |$18 |
- | |$15 dB$| | | | | + | |$15 |
- | |$79 dB$| | | | | + | |$79 |
- | |$128 dB$| | | | | + | |$128 ~{\rm dB}$| | | | |
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | |||
+ | The following image shows a series of amplifiers, which shall be analyzed. | ||
+ | |||
+ | < | ||
+ | {{drawio> | ||
+ | </ | ||
+ | |||
+ | The amplification factors are: | ||
+ | * $A_{\rm V1} =80$ | ||
+ | * $A_{\rm V2} =0.0125$ | ||
+ | * $A_{\rm V3} =250' | ||
+ | |||
+ | Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). | ||
+ | |||
+ | <button size=" | ||
+ | < | ||
+ | |||
+ | - 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: $36~{\rm dB} + 140~{\rm dB} = 176~{\rm dB}$ | ||
+ | </ | ||
+ | |||
+ | <button size=" | ||
+ | < | ||
+ | ^ Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^ | ||
+ | | $A_{\rm V1}$ | $80$ | $8 \cdot 10$ | $2^3 \cdot 10^1$ | $ 3 \cdot 6~{\rm dB} + | ||
+ | | $A_{\rm V2}$ | $0.0125$ | ||
+ | | $A_{\rm V3}$ | $250' | ||
+ | </ | ||
+ | |||
+ | <button size=" | ||
+ | < | ||
+ | | ||
+ | | ||
+ | </ | ||
</ | </ | ||
<panel type=" | <panel type=" | ||
+ | |||
+ | 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 $f = 1 ~\rm kHz$ | ||
< | < | ||
- | {{drawio> | + | {{drawio> |
+ | </ | ||
- | Let the circuit shown opposite with $R= 10 k\Omega$, $C = 1.6 uF$ and a sinusoidal input voltage $U_E = 1 V $ with $f = 1 kHz$ be given. | + | As described in the course, the Bode diagram can be displayed in Tina TI via '' |
- 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 $\rm dB$ per decade does the amplitude response drop at high frequencies? |
</ | </ |