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circuit_design:5_filter_circuits_i [2023/03/28 16:00]
mexleadmin 		circuit_design:5_filter_circuits_i [2023/03/28 16:10]
mexleadmin
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 152: Zeile 152:
   - <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 \quad\ A_{\rm V} = 10^\color{blue}{4} \qquad \cdot \qquad  2^\color{magenta}{-4} \qquad = 10000 \cdot \frac{1}{16} = 625$ \\  or $A_{\rm V}^{\rm dB} = 20~{\rm dB} + 36~{\rm dB} \rightarrow A_{\rm V}= 10^\color{blue}{1} \cdot 2^\color{magenta}{6} = 10 \cdot 64 = 640$ </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: 
 $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}  + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\  $A_{\rm V}^{dB}=56~{\rm dB} = 40~{\rm dB} + 18~{\rm dB} - 3~{\rm dB} = \color{blue}{2}\cdot 20~{\rm dB} + \color{magenta}{3}\cdot 6~{\rm dB}  + \color{teal}{-\frac{1}{2}}\cdot 6~{\rm dB}$ \\ 
Zeile 367: 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:
Zeile 477: 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 675: 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 724: 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 763: 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}}