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circuit_design:6_filter_circuits_ii [2021/12/05 19:28]
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-====== 6. Filter Circuits II - Higher Order Filters ======+====== 6 Filter Circuits II - Higher Order Filters ======
  
 ===== 6.1 Bandpass filter ===== ===== 6.1 Bandpass filter =====
  
-<WRAP right><panel type="default"> +<WRAP><panel type="default"> 
 <imgcaption pic0| Example: WLAN channels> <imgcaption pic0| Example: WLAN channels>
 </imgcaption> </imgcaption>
-\\ {{drawio>WLAN_Kanäle}}+\\ {{drawio>WLAN_Kanäle.svg}}
 </panel></WRAP> </panel></WRAP>
  
-When analyzing different signals, only a part of the entire frequency spectrum is desired. In <imgref pic0>, the channels of the WLAN standard 802.11 are shown as an example; these are used alternately for data transmission. Another example arises with vibration spectra of a motor in a machine, which contains not only the vibrations (usable for diagnostics), but also interference from other machine parts. Other examples are cabled data transmission or [[https://en.wikipedia.org/wiki/Electroencephalography|bands of brain waves]].+When analyzing different signals, only a part of the entire frequency spectrum is desired. In <imgref pic0>, the channels of the WLAN standard 802.11 are shown as an example; these are used alternately for data transmission. Another example arises with the vibration spectra of a motor in a machine, which contains not only the vibrations (usable for diagnostics) but also interference from other machine parts. Other examples are cabled data transmission or [[https://en.wikipedia.org/wiki/Electroencephalography|bands of brain waves]].
  
-To separate the desired frequencies, a filter can be used which only passes a given band between two frequencies (//frequency band//). This is possible with a **bandpass filter**.+To separate the desired frequencies, a filter can be used that only passes a given band between two frequencies (frequency band). This is possible with a **bandpass filter**.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-<WRAP right><panel type="default"> +<WRAP><panel type="default"> 
 <imgcaption pic1b| Tolerance scheme for a bandpass filter> <imgcaption pic1b| Tolerance scheme for a bandpass filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Toleranzschema_Bandpassfilter}}+\\ {{drawio>Toleranzschema_Bandpassfilter.svg}}
 </panel></WRAP> </panel></WRAP>
  
Zeile 24: Zeile 24:
  
 The range between the two frequencies is called the **passband**, or bandwidth. The range between the two frequencies is called the **passband**, or bandwidth.
-Outside the passband, the gain drops off. A real filter cannot attenuate infinitely.+Outside the passband, the gain drops off. A real filter can not attenuate infinitely.
 Also, there are various ideal filters where outside the passband, gain does not approach zero, but just falls below a threshold. Also, there are various ideal filters where outside the passband, gain does not approach zero, but just falls below a threshold.
 Often the sloping region is called the **transition region** and the region below the threshold is called the **blocking region**. Often the sloping region is called the **transition region** and the region below the threshold is called the **blocking region**.
-The threshold itself is called **blocking area**. In <imgref pic1b>, the ranges are drawn. However, the terms are not clearly defined; in various textbooks, the transition region is already called the blocking region.+The threshold itself is called the **blocking area**. In <imgref pic1b>, the ranges are drawn. However, the terms are not clearly defined; in various textbooks, the transition region is already called the blocking region.
  
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-<WRAP right><panel type="default">  +<WRAP><panel type="default">  
-<imgcaption pic1| block diagram bandpass>+<imgcaption pic1| block diagram of a bandpass>
 </imgcaption> </imgcaption>
-\\ {{drawio>Blockschaltbild_Bandpass}}+\\ {{drawio>Blockschaltbild_Bandpass.svg}}
 </panel></WRAP> </panel></WRAP>
  
 === Assembling the bandpass filter === === Assembling the bandpass filter ===
  
-This filter can be composed over by basic low-pass and high-pass filters. If the signal is first filtered through a low-pass filter and then through a high-pass filter, the desired filter is created. The order of the filters can be reversed. <imgref pic1> shows this in the block diagram - where in (1) is a commonly used and in (2) with the circuit symbols to be used according to EN 60617. Thus the transfer function $\underline{A}_{BP}$ of the bandpass filter simply results from the transfer function of the lowpass and highpass filters $\underline{A}_{TP}$ and $\underline{A}_{HP}$, since the signal passes through the filter stages one after the other:+This filter can be composed of basic low-pass and high-pass filters. If the signal is first filtered through a low-pass filter and then through a high-pass filter, the desired filter is created. The order of the filters can be reversed. <imgref pic1> shows this in the block diagram - where (1) is a commonly used and (2) with the circuit symbols to be used according to EN 60617. Thus the transfer function $\underline{A}_{\rm BP}$ of the bandpass filter simply results from the transfer function of the lowpass and highpass filters $\underline{A}_{\rm LP}$ and $\underline{A}_{\rm HP}$, since the signal passes through the filter stages one after the other:
  
-$$\underline{A}_{BP}= {{\underline{U}_A}\over{\underline{U}_E}} = {{\underline{U}_A}\over{\underline{U}_1}} \cdot {{\underline{U}_1}\over{\underline{U}_E}} = \underline{A}_{TP} \cdot \underline{A}_{HP}$$+$$\underline{A}_{\rm BP}= {{\underline{U}_{\rm O}}\over{\underline{U}_{\rm I}}} = {{\underline{U}_{\rm O}}\over{\underline{U}_1}} \cdot {{\underline{U}_1}\over{\underline{U}_{\rm I}}} = \underline{A}_{\rm LP} \cdot \underline{A}_{\rm HP}$$
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
- 
-<WRAP right><panel type="default">  
-<imgcaption pic2| Amplitude response Bandpass> 
-</imgcaption> 
-\\ {{drawio>Amplitudengang_Bandpass}} 
-</panel></WRAP> 
  
 === Amplitude response of the bandpass filter === === Amplitude response of the bandpass filter ===
  
-In <imgref pic2>, the amplitude response of the bandpass filter can be seen. Since in the amplitude response the transfer function is represented in $dB$ ($\underline{A}^{dB}$), multiplying the transfer functions of the low-pass and high-pass filters $\underline{A}_{TP}$ and $\underline{A}_{HP}$ results in an addition of $\underline{A}_{TP}^{dB}$ and $\underline{A}_{HP}^{dB}$. In the amplitude response, we can see that it results in a $20 dB/dec$ change twice: once at $f_{Gr,HP}$ and once at $f_{Gr,TP}$. So the filter has an order of 2. 
  
-Important: The cutoff frequency of the low-pass filter $f_{Gr,TP}$ must be larger than the cutoff frequency of the high-pass filter $f_{Gr,HP}$ (see <imgref pic2>).+In <imgref pic2>, the amplitude response of the bandpass filter can be seen. Since in the amplitude response, the transfer function is represented in $\rm dB$ ($\underline{A}^{\rm dB}$), multiplying the transfer functions of the low-pass and high-pass filters $\underline{A}_{\rm LP}$ and $\underline{A}_{\rm HP}$ results in an addition of $\underline{A}_{\rm LP}^{\rm dB}$ and $\underline{A}_{\rm HP}^{\rm dB}$. In the amplitude response, we can see that it results in a $20 ~\rm dB/dec$ change twice: once at $f_{\rm c, HP}$ and once at $f_{\rm c, LP}$. So the filter has an order of 2. 
 + 
 +<WRAP><panel type="default">  
 +<imgcaption pic2| Amplitude response of a Bandpass> 
 +</imgcaption> 
 +\\ {{drawio>Amplitudengang_Bandpass.svg}} 
 +</panel></WRAP> 
 + 
 +Important: The cutoff frequency of the low-pass filter $f_{\rm c, LP}$ must be larger than the cutoff frequency of the high-pass filter $f_{\rm c, HP}$ (see <imgref pic2>).
  
 But what does the frequency response look like? This is to be derived in the following. But what does the frequency response look like? This is to be derived in the following.
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-<WRAP right><panel type="default"> +<WRAP><panel type="default"> 
 <imgcaption pic3| Circuit of the bandpass filter based on the inverting amplifier > <imgcaption pic3| Circuit of the bandpass filter based on the inverting amplifier >
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_Bandpassfilter_invertierender_Verstärker}}+\\ {{drawio>Schaltung_Bandpassfilter_invertierender_Verstärker.svg}}
 </panel></WRAP> </panel></WRAP>
  
Zeile 75: Zeile 76:
 The extremal value consideration yields:  The extremal value consideration yields: 
   * for $ \boldsymbol{\omega \rightarrow 0} $:\\ The magnitude of the impedance of the capacitances becomes large \\ and thus $|\underline{X}_{C_1}| \gg R_1$ , as well as $|\underline{X}_{C_2}| \gg R_2$ \\ Thus $\underline{X}_{C_1}$ prevails at $\underline{Z}_1$ and  $\underline{R}_2$ bei $\underline{Z}_2$. \\ $\rightarrow$ **A reverse differentiator results at low frequencies.**   * for $ \boldsymbol{\omega \rightarrow 0} $:\\ The magnitude of the impedance of the capacitances becomes large \\ and thus $|\underline{X}_{C_1}| \gg R_1$ , as well as $|\underline{X}_{C_2}| \gg R_2$ \\ Thus $\underline{X}_{C_1}$ prevails at $\underline{Z}_1$ and  $\underline{R}_2$ bei $\underline{Z}_2$. \\ $\rightarrow$ **A reverse differentiator results at low frequencies.**
-  * for $ \boldsymbol{\omega \rightarrow \infty} $:\\ The magnitude of the impedance of the capacitances becomes small and thus $|\underline{X}_{C_1}| \ll R_1$ ,  as well as $|\underline{X}_{C_2}| \ll R_2$ \\ Thus $\underline{R}_1$ predominates at $\underline{Z}_1$ and  $\underline{X}_{C_2}$ predominates at $\underline{Z}_2$. \\  $\rightarrow$ **A reverse integrator results at high frequencies.**+  * for $ \boldsymbol{\omega \rightarrow \infty} $:\\ The magnitude of the impedance of the capacitances becomes small and thus $|\underline{X}_{C_1}| \ll R_1$,  as well as $|\underline{X}_{C_2}| \ll R_2$ \\ Thus $\underline{R}_1$ predominates at $\underline{Z}_1$ and  $\underline{X}_{C_2}$ predominates at $\underline{Z}_2$. \\  $\rightarrow$ **A reverse integrator results at high frequencies.**
  
 \\  \\ 
Zeile 83: Zeile 84:
 The transfer function is again to be derived from a complex-valued inverting amplifier: The transfer function is again to be derived from a complex-valued inverting amplifier:
  
-$\underline{A}_V = {{\underline{U}_A}\over{\underline{U}_E}} = - {{\underline{Z}_2}\over{\underline{Z}_1}} = - {\underline{Z}_2}\cdot {1\over{\underline{Z}_1}} =  - \Large{{{R_2\cdot {1\over{j\omega C_2}}}\over{{R_2 + {1\over{j\omega C_2}}}}}}\cdot{1 \over{{R_1 + {1\over{j\omega C_1}}}}}=  - \Large{{{R_2}\over{{j\omega C_2 R_2 + 1}}}}\cdot{j\omega C_1 \over{{j\omega C_1 R_1 + 1}}} \Bigg| {{\cdot R_1}\over{\cdot R_1}}$+$\underline{A}_{\rm V} = {{\underline{U}_{\rm O}}\over{\underline{U}_{\rm I}}} = - {{\underline{Z}_2}\over{\underline{Z}_1}} = - {\underline{Z}_2}\cdot {1\over{\underline{Z}_1}} =  - \Large{{{R_2\cdot {1\over{{\rm j}\omega C_2}}}\over{{R_2 + {1\over{{\rm j}\omega C_2}}}}}}\cdot{1 \over{{R_1 + {1\over{{\rm j}\omega C_1}}}}}=  - \Large{{{R_2}\over{{{\rm j}\omega C_2 R_2 + 1}}}}\cdot{{\rm j}\omega C_1 \over{{{\rm j}\omega C_1 R_1 + 1}}} \Bigg| {{\cdot R_1}\over{\cdot R_1}}$
  
-$\boxed{\underline{A}_V = - \color{blue}{R_2 \over R_1 } \cdot \large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}} \cdot \large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ j\omega \cdot C_1 R_1}}}$+$\boxed{\underline{A}_{\rm V} = - \color{blue}{R_2 \over R_1 } \cdot \large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}} \cdot \large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ {\rm j}\omega \cdot C_1 R_1}}}$
  
 \\ \\
-Clever reshaping yields an interesting result of the following parts:+Better reshaping yields an interesting result of the following parts:
   - $- \color{blue}{R_2 \over R_1 }$: This corresponds to a [[3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]]   - $- \color{blue}{R_2 \over R_1 }$: This corresponds to a [[3_opamp_basic_circuits_i#inverting_amplifier|inverting amplifier]]
-  - $\large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}}$: This corresponds to a [[5_filter circuits_i#lowpass|lowpass 1st order]] with a cutoff frequency of $\color{teal}{\omega_{Gr, TP}= {1 \over {C_2 R_2}}}$ +  - $\large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}}$: This corresponds to a [[5_filter circuits_i#lowpass|lowpass 1st order]] with a cutoff frequency of $\color{teal}{\omega_{\rm c, LP}= {1 \over {C_2 R_2}}}$ 
-  - $\large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ j\omega \cdot C_1 R_1}}$ This corresponds to a [[5_filter circuits_i#highpass|highpass 1st order]] with a cutoff frequency of $\color{brown}{\omega_{Gr, HP}= {1 \over {C_1 R_1}}}$+  - $\large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ {\rm j}\omega \cdot C_1 R_1}}$ This corresponds to a [[5_filter circuits_i#highpass|highpass 1st order]] with a cutoff frequency of $\color{brown}{\omega_{\rm c, HP}= {1 \over {C_1 R_1}}}$
  
 \\ \\
 This results in a function via the extremal value consideration: This results in a function via the extremal value consideration:
  
-  * for $ \boldsymbol{\omega \rightarrow 0     } $:\\ $\underline{A}_V = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ \color{black}{\underbrace{\color{teal}{j\omega \cdot C_2 R_2}}_{\rightarrow 0}}}} \cdot \Large\color{brown}{{j\omega \cdot C_1 R_1} \over {1+ \color{black}{\underbrace{\color{brown}{j\omega \cdot C_1 R_1}}_{\rightarrow 0}}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over 1} \cdot \Large\color{brown}{{j\omega \cdot C_1 R_1} \over 1} \rightarrow - \color{brown}{\normalsize{j\omega \cdot C_1 \color{black}{R_2}}}$ \\ The equation is the same as that of a reverse differentiator \\ \\ +  * for $ \boldsymbol{\omega \rightarrow 0     } $:\\ $\underline{A}_{\rm V} = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ \color{black}{\underbrace{\color{teal}{{\rm j}\omega \cdot C_2 R_2}}_{\rightarrow 0}}}} \cdot \Large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ \color{black}{\underbrace{\color{brown}{{\rm j}\omega \cdot C_1 R_1}}_{\rightarrow 0}}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over 1} \cdot \Large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over 1} \rightarrow - \color{brown}{\normalsize{{\rm j}\omega \cdot C_1 \color{black}{R_2}}}$ \\ The equation is the same as that of a reverse differentiator \\ \\ 
-  * for $ \omega \rightarrow \infty $:\\ $\underline{A}_V = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ j\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{{j\omega \cdot C_1 R_1} \over \color{brown}{1+ {j\omega \cdot C_1 R_1}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over {j\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{1 \over 1} \rightarrow - \color{teal}{1 \over {j\omega \cdot C_2 \color{black}{R_1}}}$ \\ The equation is equivalent to that of an inverse integrator+  * for $ \omega \rightarrow \infty $:\\ $\underline{A}_{\rm V} = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over \color{brown}{1+ {{\rm j}\omega \cdot C_1 R_1}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over {{\rm j}\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{1 \over 1} \rightarrow - \color{teal}{1 \over {{\rm j}\omega \cdot C_2 \color{black}{R_1}}}$ \\ The equation is equivalent to that of an inverse integrator
  
 \\ \\
 == Determination of magnitude and phase from complex-valued observation == == Determination of magnitude and phase from complex-valued observation ==
  
-For the magnitude $|\underline{A}_V|$ of the transfer function, the following hint can be used: $|a\cdot b\cdot c| = |a| \cdot |b| \cdot |c| $. \\ +For the magnitude $|\underline{A}_{\rm V}|$ of the transfer function, the following hint can be used: $|a\cdot b\cdot c| = |a| \cdot |b| \cdot |c| $. \\ 
-Thus, for the magnitude $|\underline{A}_V|$, we get: \\ +Thus, for the magnitude $|\underline{A}_{\rm V}|$, we get: \\ 
-$       |\underline{A}_V| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 C_2^2 R_2^2}} \cdot \large{{\omega \cdot C_1 R_1} \over \sqrt{1+ \omega^2 C_1^2 R_1^2}} $ +$       |\underline{A}_{\rm V}| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 C_2^2 R_2^2}} \cdot \large{{\omega \cdot C_1 R_1} \over \sqrt{1+ \omega^2 C_1^2 R_1^2}} $ 
-$\xrightarrow{\color{teal}{\omega_{Gr, TP}}, \ \ \color{brown}{\omega_{Gr, HP}}}$  +$\xrightarrow{\color{teal}{\omega_{\rm c, LP}}, \ \ \color{brown}{\omega_{\rm c, HP}}}$  
-$\boxed{|\underline{A}_V| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 / \color{teal}{\omega_{Gr, TP}^2}}} \cdot \large{{\omega / \color{brown}{\omega_{Gr, HP}}} \over \sqrt{1+ \omega^2 \color{brown} / {\omega_{Gr, HP}}^2}}}$+$\boxed{|\underline{A}_{\rm V}| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 / \color{teal}{\omega_{\rm c, LP}^2}}} \cdot \large{{\omega / \color{brown}{\omega_{\rm c, HP}}} \over \sqrt{1+ \omega^2 \color{brown}{\omega_{\rm c, HP}^2}}}}$
  
  
-For the phase $\varphi$ must be conjugate complexly extended again. \\ +The phase $\varphi$ must be conjugate complexly extended again. \\ 
-At first this produces an unwieldy equation - but a real-valued constant can be separated from it.+At first, this produces an unwieldy equation - but a real-valued constant can be separated from it.
  
-$\underline{A}_V = - \color{blue}\large{R_2 \over R_1 } $ +$\underline{A}_{\rm V} = - \color{blue}\large{R_2 \over R_1 } $ 
-$\cdot \large\color{teal }{ 1 \over \color{lightgray}{\boxed{\color{teal }{\small{1+ j\omega \cdot C_2 R_2}}}}}$ +$\cdot \large\color{teal }{ 1 \over \color{lightgray}{\boxed{\color{teal }{\small{1+ {\rm j}\omega \cdot C_2 R_2}}}}}$ 
-$\cdot \large\color{teal }{{1- j\omega \cdot C_2 R_2} \over \color{lightgray}{\boxed{\color{teal }{\small{1- j\omega \cdot C_2 R_2}}}}}$ +$\cdot \large\color{teal }{{1- {\rm j}\omega \cdot C_2 R_2} \over \color{lightgray}{\boxed{\color{teal }{\small{1- {\rm j}\omega \cdot C_2 R_2}}}}}$ 
-$\cdot \large\color{brown}{{ j\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1+ j\omega \cdot C_1 R_1}}}}}$ +$\cdot \large\color{brown}{{ {\rm j}\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1+ {\rm j}\omega \cdot C_1 R_1}}}}}$ 
-$\cdot \large\color{brown}{{1- j\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1- j\omega \cdot C_1 R_1}}}}}$+$\cdot \large\color{brown}{{1- {\rm j}\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1- {\rm j}\omega \cdot C_1 R_1}}}}}$
  
-$\underline{A}_V = \quad \quad \mathcal{C} \quad \quad \quad \quad$ +$\underline{A}_{\rm V} = \quad \quad \mathcal{C} \quad \quad \quad \quad$ 
-$ \cdot \color{teal }{(1- j\omega \cdot C_2 R_2)}$ +$ \cdot \color{teal }{(1- {\rm j}\omega \cdot C_2 R_2)}$ 
-$\ \cdot \color{brown}{ j\omega \cdot C_1 R_1 }$ +$\ \cdot \color{brown}{ {\rm j}\omega \cdot C_1 R_1 }$ 
-$\ \cdot \ \color{brown}{(1- j\omega \cdot C_1 R_1)}$+$\ \cdot \ \color{brown}{(1- {\rm j}\omega \cdot C_1 R_1)}$
  
-$\underline{A}_V = \quad \quad \mathcal{C} \quad \quad \quad$ +$\underline{A}_{\rm V} = \quad \quad \mathcal{C} \quad \quad \quad$ 
-$ \cdot (j + \omega R_2 C_2 + \omega R_1 C_1 - j \omega R_1 C_1 \omega R_2 C_2)$+$ \cdot ({\rm j} + \omega R_2 C_2 + \omega R_1 C_1 - {\rm j} \omega R_1 C_1 \omega R_2 C_2)$
  
-From this equation it is easy to read the proportions for real part $\Re(\underline{A}_V)$ and imaginary part $\Im(\underline{A}_V)$. \\+From this equation, it is easy to read the proportions for real part $\Re(\underline{A}_{\rm V})$ and imaginary part $\Im(\underline{A}_{\rm V})$. \\
 This gives for the phase $\varphi$ : This gives for the phase $\varphi$ :
  
-$       \varphi = arctan \left( \frac{\Im(\underline{A}_V)}{\Re(\underline{A}_V)} \right) = arctan \left( \frac{1 - \omega R_1 C_1 \omega R_2 C_2}{\omega R_2 C_2 + \omega R_1 C_1} \right)$ +$       \varphi = \arctan \left( \frac{\Im(\underline{A}_{\rm V})}{\Re(\underline{A}_{\rm V})} \right) = \arctan \left( \frac{1 - \omega R_1 C_1 \omega R_2 C_2}{\omega R_2 C_2 + \omega R_1 C_1} \right)$ 
-$\xrightarrow{\color{teal}{\omega_{Gr, TP}}, \ \ \color{brown}{\omega_{Gr, HP}}}$  +$\xrightarrow{\color{teal}{\omega_{\rm c, LP}}, \ \ \color{brown}{\omega_{\rm c, HP}}}$  
-$\boxed{\varphi = arctan \left( \frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right)}$+$\boxed{\varphi = \arctan \left( \frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right)}$
  
  
Zeile 137: Zeile 138:
 The formal for the phase $\varphi$ says The formal for the phase $\varphi$ says
 The extremal consideration can now be carried out for some salient frequencies: The extremal consideration can now be carried out for some salient frequencies:
-  * for $ \boldsymbol{\omega \rightarrow 0} $:\\ $\varphi(0) = arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow arctan \left( \frac{\mathcal{C}_1 - "0"}{"0"} \right) = arctan \left( "+\infty" \right)$ \\ \\ +  * for $ \boldsymbol{\omega \rightarrow 0} $:\\ $\varphi(0) = \arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow \arctan \left( \frac{\mathcal{C}_1 - "0"}{"0"} \right) = \arctan \left( "+\infty" \right)$ \\ \\ 
-  * for $ \boldsymbol{\omega \rightarrow \infty} $: \\ $\varphi(\infty) = arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow arctan \left( \frac{\mathcal{C}_1 - "\infty"^2}{"\infty"} \right) = arctan \left( "-\infty" \right)$ \\ \\ +  * for $ \boldsymbol{\omega \rightarrow \infty} $: \\ $\varphi(\infty) = \arctan \left( \frac{\mathcal{C}_1 - \omega^2}{\omega \mathcal{C}_2} \right) \rightarrow \arctan \left( \frac{\mathcal{C}_1 - "\infty"^2}{"\infty"} \right) = \arctan \left( "-\infty" \right)$ \\ \\ 
-  * for a **(circular) frequency** $\boldsymbol{\omega= \omega_0}$ **for which the argument of the** $\boldsymbol{arctan}$** function becomes zero**. \\ Thus the phase: \\ $\varphi(\omega_0) = arctan \left( 0 \right)$. \\ The corresponding frequency is given by: \\ $\large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} = 0 \quad\rightarrow\quad \omega_0^2 = \color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} \quad\rightarrow\quad \omega_0 =  \large\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}}$ \\ \\ +  * for a **(circular) frequency** $\boldsymbol{\omega= \omega_0}$ **for which the argument of the** $\boldsymbol{\arctan}$** function becomes zero**. \\ Thus the phase: \\ $\varphi(\omega_0) = \arctan \left( 0 \right)$. \\ The corresponding frequency is given by: \\ $\large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} = 0 \quad\rightarrow\quad \omega_0^2 = \color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} \quad\rightarrow\quad \omega_0 =  \large\sqrt{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}}}$ \\ \\ 
-  * for the cutoff frequency of the high pass filter $\boldsymbol{\omega = \color{brown}{\omega_{Gr, HP} = {1 \over {R_1 C_1}}}}$. \\ For this, if the passband is sufficiently large, $\color{brown}{\omega_{Gr, HP}} \ll \color{teal}\omega_{Gr, TP}$ can be assumed. \\ Thus we get: \\ $\varphi(\color{brown}{\omega_{Gr, HP}}) = arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \color{brown}{\omega_{Gr, HP}}^2 }{\color{brown}{\omega_{Gr, HP}} (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right)      =     arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} - \color{brown}{\omega_{Gr, HP}} }{ (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) \xrightarrow{\color{brown}{\omega_{Gr, HP}} \ll \color{teal}{\omega_{Gr, TP}}} \varphi(\color{brown}{\omega_{Gr, HP}}) = arctan (1)$ \\ \\ +  * for the cutoff frequency of the high pass filter $\boldsymbol{\omega = \color{brown}{\omega_{\rm c, HP} = {1 \over {R_1 C_1}}}}$. \\ For this, if the passband is sufficiently large, $\color{brown}{\omega_{\rm c, HP}} \ll \color{teal}\omega_{\rm c, LP}$ can be assumed. \\ Thus we get: \\ $\varphi(\color{brown}{\omega_{\rm c, HP}}) = \arctan \left( \large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \color{brown}{\omega_{\rm c, HP}}^2 }{\color{brown}{\omega_{\rm c, HP}} (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right)      =     \arctan \left( \large\frac{\color{teal}{\omega_{\rm c, LP}} - \color{brown}{\omega_{\rm c, HP}} }{ (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) \xrightarrow{\color{brown}{\omega_{\rm c, HP}} \ll \color{teal}{\omega_{\rm c, LP}}} \varphi(\color{brown}{\omega_{\rm c, HP}}) = \arctan (1)$ \\ \\ 
-  * for the cutoff frequency of the lowpass filter $\boldsymbol{\omega = \color{teal}{\omega_{Gr, TP} = {1 \over {R_2 C_2}}}}$. \\ For this, if the passband is sufficiently large, $\color{brown}{\omega_{Gr, HP}} \gg \color{teal}{\omega_{Gr, TP}}$ can be assumed. \\ Thus we have: \\ $\varphi(\color{teal}{\omega_{Gr, TP}}) = arctan \left( \large\frac{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}} - \color{teal}{\omega_{Gr, TP}}^2 }{\color{teal}{\omega_{Gr, TP}} (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right)      =     arctan \left( \large\frac{ \color{brown}{\omega_{Gr, HP}} - \color{teal}{\omega_{Gr, TP}}}{ (\color{teal}{\omega_{Gr, TP}}+\color{brown}{\omega_{Gr, HP}})} \right) \xrightarrow{\color{brown}{\omega_{Gr, HP}} \gg \color{teal}{\omega_{Gr, TP}}} \varphi(\color{teal}{\omega_{Gr, TP}}) = arctan (-1)$ +  * for the cutoff frequency of the lowpass filter $\boldsymbol{\omega = \color{teal}{\omega_{\rm c, LP} = {1 \over {R_2 C_2}}}}$. \\ For this, if the passband is sufficiently large, $\color{brown}{\omega_{\rm c, HP}} \gg \color{teal}{\omega_{\rm c, LP}}$ can be assumed. \\ Thus we have: \\ $\varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan \left( \large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}^2 }{\color{teal}{\omega_{\rm c, LP}} (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right)      =     \arctan \left( \large\frac{ \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}}{ (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) \xrightarrow{\color{brown}{\omega_{\rm c, HP}} \gg \color{teal}{\omega_{\rm c, LP}}} \varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan (-1)$ 
  
 <WRAP right><panel type="default"> <WRAP right><panel type="default">
 <imgcaption pic7| Arc tangent for bandpass> <imgcaption pic7| Arc tangent for bandpass>
 </imgcaption> </imgcaption>
-\\ {{drawio>ArcusTangens_Bandpass}}+\\ {{drawio>ArcusTangens_Bandpass.svg}}
 </panel></WRAP> </panel></WRAP>
  
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 This results in the following for individual points: This results in the following for individual points:
  
-^ $\boldsymbol{\omega}\quad$ |  $\rightarrow 0$    |  $\color{brown}{\omega_{Gr, HP}}$  |  $\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}}$ |  $\color{teal}{\omega_{Gr, TP}}$  |  $\rightarrow \infty$ | +^ $\boldsymbol{\omega}\quad$ |  $\rightarrow 0$    |  $\color{brown}{\omega_{\rm c, HP}}$  |  $\sqrt{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}}}$ |  $\color{teal}{\omega_{\rm c, LP}}$  |  $\rightarrow \infty$ | 
-^ $\boldsymbol{\varphi}$ |  $arctan \left( "+\infty" \right)$    |  $arctan ( +1)$  |  $arctan ( 0)$  |  $arctan ( -1)$  |  $arctan \left( "-\infty" \right)$  | +^ $\boldsymbol{\varphi}$ |  $\arctan \left( "+\infty" \right)$    |  $\arctan ( +1)$  |  $\arctan ( 0)$  |  $\arctan ( -1)$  |  $\arctan \left( "-\infty" \right)$  | 
 ^ |  $+90°$    |  $+45°$  |  $0°$  |  $-45°$  |  $-90°$  |  ^ |  $+90°$    |  $+45°$  |  $0°$  |  $-45°$  |  $-90°$  | 
  
-The results also seem plausible with the course of the arc tangent (<fc #ff0000>red curve</fc> in <imgref pic7>): for low frequencies the argument of the arc tangent goes towards $+\infty$ and thus the phase $\varphi$ seems to go towards $+90°$, for high frequencies towards $-90°$.+The results also seem plausible with the course of the arc tangent (<fc #ff0000>red curve</fc> in <imgref pic7>): for low frequencies, the argument of the arc tangent goes towards $+\infty$ and thus the phase $\varphi$ seems to go towards $+90°$, for high frequencies towards $-90°$.
  
-<fs x-large>ABER:</fs> Looking at the phase progression in the simulation below, it shows more of a progression that goes along with the black line.+<fs x-large>BUT:</fs> Looking at the phase progression in the simulation below, it shows more of a progression that goes along with the black line.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 164: Zeile 165:
 <panel type="info" title="Exercise 6.1.1 Bandpass based on the inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.1.1 Bandpass based on the inverting amplifier"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-  - Consider again the [[#uebertragungsfunktion1| transfer function]] and find the complex gain for $\omega_0 = \large\sqrt{\color{teal}{\omega_{Gr, TP}} \color{brown}{\omega_{Gr, HP}}}$. \\ Is this value positive (= no phase shift) or negative (= phase shift by $\pm 180°$)?+  - Consider again the [[#uebertragungsfunktion1| transfer function]] and find the complex gain for $\omega_0 = \large\sqrt{\color{teal}{\omega_{\rm c, LP}} \cdot  \color{brown}{\omega_{\rm c, HP}}}$. \\ Is this value positive (= no phase shift) or negative (= phase shift by $\pm 180°$)?
   - Consider the circuit in the simulation below at the following points:   - Consider the circuit in the simulation below at the following points:
-    - Increase of +20dB/dec at low frequencies.+    - Increase of $+20~\rm dB/Dec$ at low frequencies.
     - Middle of the passband ("plateau")     - Middle of the passband ("plateau")
-    - Drop of -20dB/dec at high frequencies \\ <WRAP outdent> <WRAP outdent> Which capacitor behaves like a short circuit at each point? \\ Knowing the behavior of the capacitors: What equivalent circuit describes the system in the forward region? </WRAP> </WRAP>+    - Drop of $-20 ~\rm dB/Dec$ at high frequencies \\ <WRAP outdent> <WRAP outdent> Which capacitor behaves like a short circuit at each point? \\ Knowing the behavior of the capacitors: What equivalent circuit describes the system in the forward region? </WRAP> </WRAP>
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<WRAP right>{{url>https://falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+33623165.424224265%0Ac+256+128+304+128+0+0.000006799999999999999+0%0Ar+192+128+256+128+0+33%0AO+400+144+464+144+0%0Ag+304+160+304+192+0%0A170+192+128+160+128+3+20+1000+5+0.1%0Aa+304+144+400+144+0+15+-15+100000000%0Ar+304+80+400+80+0+100%0Ac+304+32+400+32+0+6.8000000000000005e-9+0%0Aw+400+32+400+80+0%0Aw+400+80+400+144+0%0Aw+304+128+304+80+0%0Aw+304+80+304+32+0%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A 700,400 noborder}}+<WRAP>{{url>https://falstad.com/afilter/circuitjs.html?cct=$+1+0.000005+5+50+5+50%0A%25+4+33623165.424224265%0Ac+256+128+304+128+0+0.000006799999999999999+0%0Ar+192+128+256+128+0+33%0AO+400+144+464+144+0%0Ag+304+160+304+192+0%0A170+192+128+160+128+3+20+1000+5+0.1%0Aa+304+144+400+144+0+15+-15+100000000%0Ar+304+80+400+80+0+100%0Ac+304+32+400+32+0+6.8000000000000005e-9+0%0Aw+400+32+400+80+0%0Aw+400+80+400+144+0%0Aw+304+128+304+80+0%0Aw+304+80+304+32+0%0Ao+4+16+0+34+5+0.00009765625+0+-1+in%0Ao+2+16+0+34+2.5+0.00009765625+1+-1+out%0A noborder}}
 </WRAP> </WRAP>
  
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== 6.1.2 Multi-Feedback Bandpass ==== ==== 6.1.2 Multi-Feedback Bandpass ====
-<WRAP right><panel type="default"> +<WRAP><panel type="default"> 
 <imgcaption pic7| Circuit of the multi-feedback bandpass filter> <imgcaption pic7| Circuit of the multi-feedback bandpass filter>
 </imgcaption> </imgcaption>
-\\ {{drawio>Schaltung_MultiFeedbackBandpassFilter}}+\\ {{drawio>Schaltung_MultiFeedbackBandpassFilter.svg}}
 </panel></WRAP> </panel></WRAP>
  
  
-<WRAP right><panel type="default"> +<WRAP><panel type="default"> 
 <imgcaption pic6| bottom diagram bandpass> <imgcaption pic6| bottom diagram bandpass>
 </imgcaption> </imgcaption>
-\\ {{drawio>Bodediagramm_Bandpass}}+\\ {{drawio>Bodediagramm_Bandpass.svg}}
 </panel></WRAP> </panel></WRAP>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
-====== 6.2 Tape lock ======+====== 6.2 Band-Reject Filter ======
  
-Electrical Engineering 2 and Electrical Engineering Lab have already given insights into oscillating circuits. Within these circuits, at certain frequencies come up swinging motions, which can take up energies of system+In Electrical Engineering 1 already oscillating circuits have been investigated. Within these circuits, at certain frequencies come up swinging motions, which can take up the energy of the system.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
 <WRAP onlyprint> <WRAP onlyprint>
-{{url>https://www.geogebra.org/material/iframe/id/zhvkeaa8/width/1000/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 1000,700 noborder}}+{{url>https://www.geogebra.org/material/iframe/id/zhvkeaa8/width/1000/height/700/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false noborder}}
  
-From page [[https://www.geogebra.org/m/zhvkeaa8|www.geogebra.org/m/zhvkeaa8]], author: Tim Fischer.+From the page [[https://www.geogebra.org/m/zhvkeaa8|www.geogebra.org/m/zhvkeaa8]], author: Tim Fischer.
 </WRAP> </WRAP>
  
Zeile 210: Zeile 211:
 Example: Evaluation of an infrared sensor: Example: Evaluation of an infrared sensor:
   * Nodes are missing in the circuit from the manufacturer --> correct circuit is to be drawn.   * Nodes are missing in the circuit from the manufacturer --> correct circuit is to be drawn.
-  * to which basic circuits do OPV 1 and 2 correspond? What filtenn do both correspond to?+  * to which basic circuits do OPV 1 and 2 correspond? What filter do both correspond to?
 {{elektronische_schaltungstechnik:murata_beispiel_opv_schaltung.jpg?600}} {{elektronische_schaltungstechnik:murata_beispiel_opv_schaltung.jpg?600}}