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circuit_design:5_filter_circuits_i [2023/03/28 15:02]
mexleadmin 		circuit_design:5_filter_circuits_i [2023/07/17 11:59] (aktuell)
mexleadmin
Zeile 11: Zeile 11:
 === 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, 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. 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.
Zeile 54: Zeile 54:
 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:
  
-$\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 dB \cdot \log_{10} A_\rm V}$(nbsp)(nbsp)(nbsp)(nbsp) {\rm resp.} (nbsp)(nbsp)(nbsp)(nbsp) $A_{\rm C}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{I_2}{I_1}\right)$+$\boxed{A_{\rm V}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{U_2}{U_1}\right)=20 {\rm dB} \cdot \log_{10} A_\rm V}$(nbsp)(nbsp)(nbsp)(nbsp)  resp. (nbsp)(nbsp)(nbsp)(nbsp) $A_{\rm C}^{\rm dB}=20 ~{\rm dB} \cdot \log_{10}\left(\frac{I_2}{I_1}\right)$
  
 \\ \\
Zeile 93: Zeile 93:
  
   - For $A_{\rm V}= \color{green}{1} $ we get <WRAP>    - For $A_{\rm V}= \color{green}{1} $ we get <WRAP> 
-$ 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}}$ \\ +$ 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}}$ \\ 
 Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{1}$, so just $\color{blue}{x}=0$. </WRAP> \\  \\ 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>    - 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}}$ \\ +$ 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>   \\  \\ 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>    - 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}}$ \\ +$ 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> \\  \\ Here $\color{blue}{x}$ has the value that gives $10^\color{blue}{x} = \color{green}{2}$. Thus, $\color{blue}{x}\approx 0.30103$. </WRAP> \\  \\
  
Zeile 149: Zeile 149:
   - <WRAP> $A_{\rm V}^{\rm dB}=58~{\rm dB}$ \\ with interpolation points:    - <WRAP> $A_{\rm V}^{\rm dB}=58~{\rm dB}$ \\ with interpolation points: 
 $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\  $A_{\rm V}^{dB}=58~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}$ \\ 
-This becomes linear   $ \qquad \qquad \qquad \qquad \qquad \qquad \  A_{\rm V} = 10^\color{blue}{2} \qquad  \cdot \qquad  2^\color{magenta}{3} \qquad = 100 \cdot 8 = 800$ </WRAP>\\ \\+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:    - <WRAP>$A_{\rm V}^{\rm dB}=56~{\rm dB}$ \\ with interpolation points: 
 $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$ \\  $A_{\rm V}^{dB}=56~{\rm dB} = 80~{\rm dB} - 24~{\rm dB} = \color{blue}{4}\cdot 20~{\rm dB} + \color{magenta}{-4}\cdot 6~{\rm dB}$ \\ 
-This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad  2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ \\  oder $A_{\rm V}^{\rm dB} = 20~{\rm dB} + 36~{\rm dB} \rightarrow A_{\rm V}= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ </WRAP>\\ \\+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:    - <WRAP>$A_{\rm V}^{\rm dB}=55~{\rm dB}$ \\ with interpolation points: 
 $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}  + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\  $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}  + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\ 
-This becomes linear $ \qquad \qquad \qquad \qquad \qquad \qquad \qquad  \ 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>+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>
  
 \\  \\ 
Zeile 223: 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 $A(\omega) \sim \omega^n$.  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~~
Zeile 365: Zeile 367:
  
 $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\ $U_R=R\cdot I \qquad \qquad \qquad \qquad \qquad \rightarrow \underline{U}_R = R \cdot \underline{I}$ \\
-$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad  \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ with $\quad \underline{Z}_C= \frac{1}{j \cdot \omega \cdot C}$ \\+$U_C={ 1 \over C }\cdot(\int_{t_0}^{t_1} I_C \ {\rm d}t+ Q_0(t_0)) \qquad  \rightarrow \underline{U}_C = \underline{Z}_C \cdot \underline{I} \quad$ 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.
   - **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.   - **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.
  
Zeile 475: Zeile 477:
 <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 $\rm S1$ is closed, the circuit is an inverting amplifier.   - If only switch $\rm S1$ is closed, the circuit is an inverting amplifier.
Zeile 673: Zeile 675:
 </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_O%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_I%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>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 722: Zeile 724:
 ===== 5.4 High Pass Filter ===== ===== 5.4 High Pass Filter =====
  
-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.+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 $\rm S1$ closed and switch $\rm S2$ 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.
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
Zeile 761: Zeile 763:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.5 Overview high pass filter / low pass filter =====+===== 5.5 Overview high pass Filter / low pass Filter =====
  
 <WRAP><panel type="default">  <WRAP><panel type="default"> 
-<imgcaption pic13| Overview high pass filter/ low pass filter>+<imgcaption pic13| Overview high pass Filter / low pass Filter>
 </imgcaption> </imgcaption>
 \\ {{drawio>Übersicht_Hochpass_Tiefpass.svg}} \\ {{drawio>Übersicht_Hochpass_Tiefpass.svg}}
Zeile 806: Zeile 808:
  
 The amplification factors are: The amplification factors are:
-  * $A_{V1} =80$ +  * $A_{\rm V1} =80$ 
-  * $A_{V2} =0.0125$ +  * $A_{\rm V2} =0.0125$ 
-  * $A_{V3} =250'000$+  * $A_{\rm V3} =250'000$
  
 Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$).  Calculate manually the resulting total amplification $A_{\rm V}$ as a linear factor and in $\rm dB$ ($A_{\rm V}^{\rm dB}$). 
Zeile 823: Zeile 825:
 <collapse id="Loesung_5_0_2_1_Solution" collapsed="true">  <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 $\rm dB$  ^ ^  Amp. Name $\rightarrow$ ^ lin. Factor $\rightarrow$^ re-arrange $\rightarrow$ ^ as exponents of $2$ and $10$ $\rightarrow$^  transfer into $\rm dB$  $\rightarrow$^  result in $\rm dB$  ^
-|  $A_{V1}$   | $80$       | $8    \cdot 10$        | $2^3    \cdot 10^1$          | $ 3 \cdot 6~{\rm dB} +   1  \cdot 20~{\rm dB} $     | $ 38~{\rm dB}$         | +|  $A_{\rm V1}$   | $80$       | $8    \cdot 10$        | $2^3    \cdot 10^1$          | $ 3 \cdot 6~{\rm dB} +   1  \cdot 20~{\rm dB} $     | $ 38~{\rm dB}$         | 
-|  $A_{V2}$   | $0.0125$   | $0.125\cdot 0.1$       | $2^{-3} \cdot 10^{-1}$       | $-3 \cdot 6~{\rm dB} + (-1) \cdot 20~{\rm dB} $     | $-38~{\rm dB}$         | +|  $A_{\rm V2}$   | $0.0125$   | $0.125\cdot 0.1$       | $2^{-3} \cdot 10^{-1}$       | $-3 \cdot 6~{\rm dB} + (-1) \cdot 20~{\rm dB} $     | $-38~{\rm dB}$         | 
-|  $A_{V3}$   | $250'000$  | $0.25 \cdot 1'000'000$ | $2^{-2} \cdot 10^{6}$        | $-2 \cdot 6~{\rm dB} +   6  \cdot 20~{\rm dB} $     | $108~{\rm dB}$         |+|  $A_{\rm V3}$   | $250'000$  | $0.25 \cdot 1'000'000$ | $2^{-2} \cdot 10^{6}$        | $-2 \cdot 6~{\rm dB} +   6  \cdot 20~{\rm dB} $     | $108~{\rm dB}$         |
 </collapse> </collapse>
  
Zeile 849: Zeile 851:
       - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$.       - Briefly describe the differences in the amplitude response of the gain $A_{\rm V}$.
   - 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.   - 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 $R_2 = 10 \rm k\Omega$ 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?