====== 6 Filter Circuits II - Higher Order Filters ====== ===== 6.1 Bandpass filter ===== \\ {{drawio>WLAN_Kanäle.svg}} When analyzing different signals, only a part of the entire frequency spectrum is desired. In , 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 that only passes a given band between two frequencies (frequency band). This is possible with a **bandpass filter**. ~~PAGEBREAK~~ ~~CLEARFIX~~ \\ {{drawio>Toleranzschema_Bandpassfilter.svg}} === Frequency response ranges === The range between the two frequencies is called the **passband**, or bandwidth. 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. 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 the **blocking area**. In , 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~~ \\ {{drawio>Blockschaltbild_Bandpass.svg}} === Assembling the bandpass filter === 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. 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: $$\underline{A}_{\rm BP}= {{\underline{U}_{\rm O}}\over{\underline{U}_{\rm I}}} = {{\underline{U}_{\rm O}}\over{\underline{U}_1}} \cdot {{\underline{U}_1}\over{\underline{U}_{\rm I}}} = \underline{A}_{\rm LP} \cdot \underline{A}_{\rm HP}$$ ~~PAGEBREAK~~ ~~CLEARFIX~~ === Amplitude response of the bandpass filter === In , 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. \\ {{drawio>Amplitudengang_Bandpass.svg}} 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 ). But what does the frequency response look like? This is to be derived in the following. ~~PAGEBREAK~~ ~~CLEARFIX~~ \\ {{drawio>Schaltung_Bandpassfilter_invertierender_Verstärker.svg}} ==== 6.1.1 Bandpass based on the inverting amplifier ==== == Realization == From [[5_filter_circuits_i#lowpass|chapter 5]], the circuits of highpass and lowpass filters are known. From this, the circuit shown in can be derived. This will be considered in some detail. 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 \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.** \\ == complex-valued consideration of the transfer function == The transfer function is again to be derived from a complex-valued inverting amplifier: $\underline{A}_{\rm V} = {{\underline{U}_{\rm O}}\over{\underline{U}_{\rm I}}} = - {{\underline{Z}_2}\over{\underline{Z}_1}} = - {\underline{Z}_2}\cdot {1\over{\underline{Z}_1}} = - \Large{{{R_2\cdot {1\over{{\rm j}\omega C_2}}}\over{{R_2 + {1\over{{\rm j}\omega C_2}}}}}}\cdot{1 \over{{R_1 + {1\over{{\rm j}\omega C_1}}}}}= - \Large{{{R_2}\over{{{\rm j}\omega C_2 R_2 + 1}}}}\cdot{{\rm j}\omega C_1 \over{{{\rm j}\omega C_1 R_1 + 1}}} \Bigg| {{\cdot R_1}\over{\cdot R_1}}$ $\boxed{\underline{A}_{\rm V} = - \color{blue}{R_2 \over R_1 } \cdot \large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}} \cdot \large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ {\rm j}\omega \cdot C_1 R_1}}}$ \\ 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]] - $\large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}}$: This corresponds to a [[5_filter circuits_i#lowpass|lowpass 1st order]] with a cutoff frequency of $\color{teal}{\omega_{\rm c, LP}= {1 \over {C_2 R_2}}}$ - $\large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ {\rm j}\omega \cdot C_1 R_1}}$ 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: * 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 $:\\ $\underline{A}_{\rm V} = - \Large{R_2 \over R_1 } \cdot \Large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over \color{brown}{1+ {{\rm j}\omega \cdot C_1 R_1}}} \rightarrow - {R_2 \over R_1 } \cdot \color{teal}{ 1 \over {{\rm j}\omega \cdot C_2 R_2}} \cdot \Large\color{brown}{1 \over 1} \rightarrow - \color{teal}{1 \over {{\rm j}\omega \cdot C_2 \color{black}{R_1}}}$ \\ The equation is equivalent to that of an inverse integrator \\ == Determination of magnitude and phase from complex-valued observation == 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 $|\underline{A}_{\rm V}|$, we get: \\ $ |\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}} $ $\xrightarrow{\color{teal}{\omega_{\rm c, LP}}, \ \ \color{brown}{\omega_{\rm c, HP}}}$ $\boxed{|\underline{A}_{\rm V}| = {R_2 \over R_1 } \cdot \large{1 \over \sqrt{1+ \omega^2 / \color{teal}{\omega_{\rm c, LP}^2}}} \cdot \large{{\omega / \color{brown}{\omega_{\rm c, HP}}} \over \sqrt{1+ \omega^2 \color{brown}{\omega_{\rm c, HP}^2}}}}$ 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. $\underline{A}_{\rm V} = - \color{blue}\large{R_2 \over R_1 } $ $\cdot \large\color{teal }{ 1 \over \color{lightgray}{\boxed{\color{teal }{\small{1+ {\rm j}\omega \cdot C_2 R_2}}}}}$ $\cdot \large\color{teal }{{1- {\rm j}\omega \cdot C_2 R_2} \over \color{lightgray}{\boxed{\color{teal }{\small{1- {\rm j}\omega \cdot C_2 R_2}}}}}$ $\cdot \large\color{brown}{{ {\rm j}\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1+ {\rm j}\omega \cdot C_1 R_1}}}}}$ $\cdot \large\color{brown}{{1- {\rm j}\omega \cdot C_1 R_1} \over \color{ pink }{\boxed{\color{brown}{\small{1- {\rm j}\omega \cdot C_1 R_1}}}}}$ $\underline{A}_{\rm V} = \quad \quad \mathcal{C} \quad \quad \quad \quad$ $ \cdot \color{teal }{(1- {\rm j}\omega \cdot C_2 R_2)}$ $\ \cdot \color{brown}{ {\rm j}\omega \cdot C_1 R_1 }$ $\ \cdot \ \color{brown}{(1- {\rm j}\omega \cdot C_1 R_1)}$ $\underline{A}_{\rm V} = \quad \quad \mathcal{C} \quad \quad \quad$ $ \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 $\Re(\underline{A}_{\rm V})$ and imaginary part $\Im(\underline{A}_{\rm V})$. \\ This gives for the phase $\varphi$ : $ \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)$ $\xrightarrow{\color{teal}{\omega_{\rm c, LP}}, \ \ \color{brown}{\omega_{\rm c, HP}}}$ $\boxed{\varphi = \arctan \left( \frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right)}$ \\ The formal for the phase $\varphi$ says 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 \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: \\ $\large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \omega^2 }{\omega (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} = 0 \quad\rightarrow\quad \omega_0^2 = \color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} \quad\rightarrow\quad \omega_0 = \large\sqrt{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}}}$ \\ \\ * for the cutoff frequency of the high pass filter $\boldsymbol{\omega = \color{brown}{\omega_{\rm c, HP} = {1 \over {R_1 C_1}}}}$. \\ 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 $\boldsymbol{\omega = \color{teal}{\omega_{\rm c, LP} = {1 \over {R_2 C_2}}}}$. \\ 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: \\ $\varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan \left( \large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}^2 }{\color{teal}{\omega_{\rm c, LP}} (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) = \arctan \left( \large\frac{ \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}}{ (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) \xrightarrow{\color{brown}{\omega_{\rm c, HP}} \gg \color{teal}{\omega_{\rm c, LP}}} \varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan (-1)$ \\ {{drawio>ArcusTangens_Bandpass.svg}} \\ This results in the following for individual points: ^ $\boldsymbol{\omega}\quad$ | $\rightarrow 0$ | $\color{brown}{\omega_{\rm c, HP}}$ | $\sqrt{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}}}$ | $\color{teal}{\omega_{\rm c, LP}}$ | $\rightarrow \infty$ | ^ $\boldsymbol{\varphi}$ | $\arctan \left( "+\infty" \right)$ | $\arctan ( +1)$ | $\arctan ( 0)$ | $\arctan ( -1)$ | $\arctan \left( "-\infty" \right)$ | ^ | $+90°$ | $+45°$ | $0°$ | $-45°$ | $-90°$ | The results also seem plausible with the course of the arc tangent (red curve in ): 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 $-90°$. BUT: Looking at the phase progression in the simulation below, it shows more of a progression that goes along with the black line. ~~PAGEBREAK~~ ~~CLEARFIX~~ {{fa>pencil?32}} - 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: - Increase of $+20~\rm dB/Dec$ at low frequencies. - Middle of the passband ("plateau") - Drop of $-20 ~\rm dB/Dec$ at high frequencies \\ 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? {{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}} This can be used to determine the floor diagram. ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== 6.1.2 Multi-Feedback Bandpass ==== \\ {{drawio>Schaltung_MultiFeedbackBandpassFilter.svg}} \\ {{drawio>Bodediagramm_Bandpass.svg}} ~~PAGEBREAK~~ ~~CLEARFIX~~ ====== 6.2 Band-Reject Filter ====== 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~~ {{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 the page [[https://www.geogebra.org/m/zhvkeaa8|www.geogebra.org/m/zhvkeaa8]], author: Tim Fischer. ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Homework ==== Example: Evaluation of an infrared sensor: * 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 filter do both correspond to? {{elektronische_schaltungstechnik:murata_beispiel_opv_schaltung.jpg?600}} ====== References ====== --> References to the media used # ^ Element ^ License ^ Link ^ | : Superposition of sinusoidal oscillations | Public Domain | https://en.wikipedia.org/wiki/File:Fourier_transform_time_and_frequency_domains.gif | | : Superposition of sinusoidal oscillations | [[https://creativecommons.org/licenses/by-sa/3.0/deed.en|CC-BY SA 3.0]] | https://en.wikipedia.org/wiki/File:Fourier_series_square_wave_circles_animation.gif | | : Superposition of sinusoidal oscillations | [[https://creativecommons.org/licenses/by-sa/4.0/deed.en|CC-BY SA 4.0]] | https://commons.wikimedia.org/wiki/File:Example_of_Fourier_Convergence.gif | <--