6 Filter Circuits II - Higher Order Filters

Abb. 1: Example: WLAN channels

When analyzing different signals, only a part of the entire frequency spectrum is desired. In Abbildung 1, 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 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.

Abb. 2: Tolerance scheme for a bandpass filter

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 Abbildung 2, the ranges are drawn. However, the terms are not clearly defined; in various textbooks, the transition region is already called the blocking region.

Abb. 3: block diagram of a bandpass

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. Abbildung 3 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:

$$\underline{A}_{\rm BP}= {{\underline{U}_{\rm O}}\over{\underline{U}_{\rm I}}} = {{\underline{U}_{\rm O}}\over{\underline{U}_1}} \cdot {{\underline{U}_1}\over{\underline{U}_{\rm I}}} = \underline{A}_{\rm LP} \cdot \underline{A}_{\rm HP}$$

Amplitude response of the bandpass filter

In Abbildung 4, 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.

Abb. 4: Amplitude response of a Bandpass

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 Abbildung 4).

But what does the frequency response look like? This is to be derived in the following.

Abb. 5: Circuit of the bandpass filter based on the inverting amplifier


From chapter 5, the circuits of highpass and lowpass filters are known. From this, the circuit shown in Abbildung 5 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:

  1. $- \color{blue}{R_2 \over R_1 }$: This corresponds to a inverting amplifier
  2. $\large\color{teal}{1 \over {1+ {\rm j}\omega \cdot C_2 R_2}}$: This corresponds to a lowpass 1st order with a cutoff frequency of $\color{teal}{\omega_{\rm c, LP}= {1 \over {C_2 R_2}}}$
  3. $\large\color{brown}{{{\rm j}\omega \cdot C_1 R_1} \over {1+ {\rm j}\omega \cdot C_1 R_1}}$ This corresponds to a 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:
    $\varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan \left( \large\frac{\color{teal}{\omega_{\rm c, LP}} \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}^2 }{\color{teal}{\omega_{\rm c, LP}} (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) = \arctan \left( \large\frac{ \color{brown}{\omega_{\rm c, HP}} - \color{teal}{\omega_{\rm c, LP}}}{ (\color{teal}{\omega_{\rm c, LP}}+\color{brown}{\omega_{\rm c, HP}})} \right) \xrightarrow{\color{brown}{\omega_{\rm c, HP}} \gg \color{teal}{\omega_{\rm c, LP}}} \varphi(\color{teal}{\omega_{\rm c, LP}}) = \arctan (-1)$

Abb. 6: Arc tangent for bandpass

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 Abbildung 6): 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.

Exercise 6.1.1 Bandpass based on the inverting amplifier

  1. Consider again the 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 $\pm 180°$)?
  2. Consider the circuit in the simulation below at the following points:
    1. Increase of $+20~\rm dB/Dec$ at low frequencies.
    2. Middle of the passband („plateau“)
    3. 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?

This can be used to determine the floor diagram.

Abb. 6: Circuit of the multi-feedback bandpass filter

Abb. 8: bottom diagram bandpass

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.

From the page www.geogebra.org/m/zhvkeaa8, author: Tim Fischer.

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?



References to the media used