Hi Simo, with reference to your post#17 I like to ask you where is the "square root formula" coming from? I have severe doubts about it´s relevance.
More than that, how could such a formula help?
From the beginning, the wanted overall cut-off frequency is given - and how can you find the parameters of the particular stages?
I must confess, I have problems, in general, to follow your approach.

As I have mentioned in my former post, the classical steps in filter design are as follows:
1.) Usage of filter tables to derive the normalized pole parameters for the required number of stages;
2.) Denormalization for the reqired cut-off frequency resulting in fixed figures (like wc1=2.775 1/sec and Qp1=0.743).
(Note that these steps are independent on any filter topology!).
3.) Selection of an appropriate filter topology
4.) Calculating parts values for the different stages - based on the particular pole parameters.
___________________
Finally, there is somethibg I forgot to mention in my last post:
The normalized pole data for the two stages I have given are NOT for a filter cut-off at the -3dB points. Instead, they are for a cut-off based on the delay characteristic of the Bessel filter. This is quite normal for a Bessel response because this low pass approximation is selected primarily because of its group delay properties. Therefore, in most cases the pass band is defined in the time domain.
That means: values for Wp and Qp are normalized to wo=1/tau,o=wp/sqrt(3). (tau,o: group delay at w=0).
For each 2nd order Bessel filter section the 3-dB cut-off is at wc=0.786*wp.

 

I have the feeling that these examples are not very realistic.
It is self-evident that two lowpass filters in series with cut-offs at 10 kHz and 20 kHz, respectively, result in an overall cut-off below 10 kHz. But such an estimate does not help for designing a 4th order Bessel filter, which needs only a 10 % difference between the pole frequencies of both stages. More than that, you shouldn`t forget that such a series connection changes the characteristic of the magnitude response.



In contrast, both pole frequencies and both pole quality factors must be chosen so that the RESULT has a Thomson-Bessel characteristic.

I never said ANYTHING about the characteristic of the response. Please keep in mind the specific context in which I was making those comments. It was stated, "So your U1 stage has fc=30.2 kHz, your U2 stage has fc=33.8 kHz. I think you take fc of U2 and subtract fc of U1 from it. I got overall fc of 3.643 kHz."

To which I pointed out that this makes no sense because the circuit was a cascade of two low pass filters and therefore the cutoff frequency would be primarily determined by the lower of the two cutoff frequencies and that there was no mechanism for the cutoff frequency to be the difference of the two, which would place it significantly below the lower cutoff frequency. I was making a sanity check observation. That observation was independent of whether it's supposed to be a Bessel, a Butterworth, or a Bob's Bubble filter. Basically, I'm asserting that they give you enough information to recognize when a claimed cutoff frequency is simply not reasonable and that there is no point even talking about the characteristic of the response if you know up front that the calculated cutoff frequency is clearly wrong.

But that sanity check observation is only valid IF knowledge of the two separate cutoff frequencies tells you SOMETHING about the cutoff frequency of the composite filter. I'm not saying they give you enough information to determine exactly what the cutoff frequency is, only that they give you enough information to establish a set of bounds within which that cutoff frequency must be.

But you are saying that knowledge the separate cutoff frequences are meaningless and tells you absolutely nothing about the composite filter.

I think what we have is more a case of mismatched contexts than incompatible claims. Correct me if I am wrong here. When you say that they are meaningless and tell you nothing, you are thinking only in terms of whether they tell you about the detailed characteristic of the response and the exact cutoff frequency. If so, I agree.

 

Hi WBahn - OK OK, no problem.
Perhaps a kind of misunderstanding - perhaps caused by language problems.
I propose to concentrate on the problems described by the OP to find the desired 4th order Bessel filter.
In this context - what do you think about the square-root formula as mentioned by Jony130 as well as the questioner ?

 

Hi WBahn - OK OK, no problem.
Perhaps a kind of misunderstanding - perhaps caused by language problems.
I propose to concentrate on the problems described by the OP to find the desired 4th order Bessel filter.
In this context - what do you think about the square-root formula as mentioned by Jony130 as well as the questioner ?

I'm not sure. I think that it would produce a pretty reasonable estimate of the combined cutoff frequency, but I would be a bit surprised if it is exact. My thinking there is that I would expect the combined cutoff frequency for two cascaded first order filters to be lower than the combined cutoff frequency for two cascaded second order filters with the same cutoff frequencies. But I would expect them to be reasonably close to each other.

I started to play around with the generic transfer functions this weekend but had to get back to real work. Maybe over Thanksgiving I can look at it again.

Filter theory is certainly not something I am strong in.

 

.......My thinking there is that I would expect the combined cutoff frequency for two cascaded first order filters....

In this context, I remember the following:

When we try to measure the rise time (ts) of a signal using an oscilloscope we see on the screen a superposition of the rise time to be measured and the internal rise time (tos) of the oscilloscope (caused by its limited bandwidth).
And it can be shown that - under the assumption that both effects are of first order - the screen shows with good accuracy the value
tscr~sqrt(ts^2 + tos^2).

When we express the rise times by the corresponding bandwidths of the first order systems we arrive exactly on the sqrt formulas as mentioned by Jony130. It is clear that the biggest rise time dominates in this approximation.
Furthermore, it can be shown that the approximation error is maximum for equal times (about 5%) and gets smaller if both rise times are different.

That means: The mentioned sqrt-formula for combining two cut-off frequencies works with good accuracy for first-order systems only.

PS: I am still waiting for Jony`s answer regarding his source of this formula).

 

Hi dears, nice to read you again.

Hi Simo, with reference to your post#17 I like to ask you where is the "square root formula" coming from? (...)
More than that, how could such a formula help?

I copied it from John130's post. I trusted to use it but I was really tired and exausted yesterday, I am not really sure about it right now.

If I am not wrong for an active filter with order \(n > 2\) the cutoff frequency is determined mainly by the cutoff frequency of the first stage.
Correct me if I am wrong please.

I must confess, I have problems, in general, to follow your approach.

I understand and this is because I am confused for first about the Bessel design, and I do not understand how to relate the design of the book with the cutoff frequency -3dB given by the simulation of the example's circuit, also because the -3dB frequency is not given as design parameter in the example in contrast to the delay, as the Bessel filter is delay filter.

The textbook's example says:

"We are required to design a filter having at most 10 % of deviation in delay at \(\omega\) 2.5 rad/s, and at most of 1 dB of loss at frequency 1.0 rad/s. The filter is to provide a \({100\,\mu s}\) of delay."

The filter order is determined after looking a graph for the "Bessel - Thompson" frequency response bode plot for a normalized frequency axis from 0 to 10, and also looking a table for normalized value of dB losses at a given normalized frequency \(\omega\). Finally the determined filter order is \(n = 4\).

Given the delay

\(D = 100\,\mu s\)

From the poles pairs for the normalized unity delay:

\(
p_p = a \pm \text{j}b\\
p_p1 = -2.1038 \pm \text{j}2.6574\\
p_p2 = -2.8962 \pm \text{j}0.8672\\
\)

The poles frequency and quality factor are obtained as follows:

\(
\omega = \sqrt(a^2 + b^2)\\
Q = {1 \over 2a}\omega_0\\
\)

Giving the following values:

\(
\omega_{p_p1} = 3.39\\
Q_1 = 0.8055
\omega__{p_p2} = 3.023
Q_2 = 0.5219
\)

From each normalized 2nd order stage:

the components for each cell are denormalizated from a scaling factor \(k_\text{r}\):

\(
k_\text{r} = 10000\\
k_\text{f} = \omega_0\\
D = 100\,\mu s \\
\\
R_1 = R_2 = k_\text{r} \,(\Omega)\\
C_1 = 2Q \left ( {D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
C_2 = \left ( {1 \over 2Q}\right ) \left ({D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
\)

Filters stage are correctly placed from lower \(Q\) to higher \(Q\) resulting the circuit:

And from the simulation I posted, the resulting frequency for -3db of attenuation is 3.3 kHz.

Now, I do not understand where these 3.3 kHz come from, also because, for the first stage:

\({1 \over 2\pi \sqrt{10000^2\,\Omega \times 3.45\,\text{nF} \times 3.17\,\text{nF}}} \approx 4812.61\,\text{Hz}\).

This is my problem. I just want to understand and learn.

Thank you again.
Simon

 

Hello simo.

Sorry for my earlier mistakes. I forgot to include 2pi when I was doing cutoff frequency calculations.

One thing that I noticed. The problem does not require you to find cutoff frequency ahead of time, they don't say that you need to design a filter with cutoff frequency X. All the calculations are done in terms of other data quantities, like Q. So maybe there is no way to calculate cutoff frequency. Maybe the best you can do is calculate values of resistors and capacitors, build the circuit in simulation/lab, and once it is running, you can see what the cutoff frequency is.

 

I copied it from John130's post. I trusted to use it but I was really tired and exausted yesterday, I am not really sure about it right now.

I am sure that it is not correct. Hopefully, Jony130 can give us some information about his source.

If I am not wrong for an active filter with order \(n > 2\) the cutoff frequency is determined mainly by the cutoff frequency of the first stage.
Correct me if I am wrong please.

Simo, in my post #20 I have mentioned a 4th order Bessel lowpass example with two pole frequencies which differ by about 10% only.
Do you still believe that the lower one has much more influence than the other one? No - that is not the case. It´s not so simple..
Instead, the superposition of both pole frequencies together with the particular pole Q values result in the overall filter response.

At the moment, I do not know what your problem really is.
Therefore my question: Do you intend to design a Bessel filter based on certain requirements - either in the frequency or in the time domain?
If yes - what are these requirements?

 

So maybe there is no way to calculate cutoff frequency. Maybe the best you can do is calculate values of resistors and capacitors, build the circuit in simulation/lab, and once it is running, you can see what the cutoff frequency is.

Filter design is challenging task - however, at the same time it is a straightforward procedure consisting of several logical steps.
Thus, of course it is possible to find/calculate the data for each 2nd order stage - based on the requirements given for a 4th order Bessel response.
And, of course, it is possible to calculate all the parts values to realize such a filter for the various different filter topologies.

 

Hi LvW,

Instead, the superposition of both pole frequencies together with the particular pole Q values result in the overall filter response.

Step by step the concepts are becoming more clear and thank you for clarifying these concepts.

Honestly I am quiet confused. But why do I feel confused?

Because the example of the book takes as main parameter the delay and the 10 % of deviation of it at certain frequency.
This example makes me understand that the Bessel filter is a kind of filter that it is needed when a properly delay is required with a 10 % of deviation at a certain frequency, while I always read about Bessel filter as a filter to be designed for the 3db cutoff frequency as main parameter..

The same occurs as the example posted by Jony130 in #8. That filter have a cutoff frequency of 1 kHz (main requirement) and a constant group delay (not mentioned by the author) of 300 μs (I checked it through a quickly LTspice simulation) until the cutoff frequency of 1 kHz.
In this example the main parameter seems to be the cutoff frequency, in contrast to the textbook's example.

At the moment, I do not know what your problem really is.
Therefore my question: Do you intend to design a Bessel filter based on certain requirements - either in the frequency or in the time domain?
If yes - what are these requirements?

Yes, I want to to understand and learn how to design a Bessel filter based on the constant group delay (as the textbook does) either for the cutoff frequency. Perhaps, LvW, the key to it is in your post #9.

I do not have fixed requirements, but an example based on the both parameters (delay and frequency) will make the concepts clearer to me.

Again, thank you all for your help and for your time.

Filter design is challenging task

I strongly agree.

 

LvW I am reading and studying better your posts.

I am finally realizing that pheraps my approach is too wrong. This means that for a true Bessel approximation it is wrong looking for a -3dB frequency as main requirement.

 

I find this formula in a textbook about analog circuit design in section about how to find "equivalent" ω when we connects the first order filters is "series". And individual poles are not far away (less than one decade).

 

Hi Simo,
your problem touches the advanced filter theory. Therefore, it is not easy to answer your questions within the frame of this forum. Nevertheless, let me try it.

1.) It is the task of lowpass filters to separate the pass band from the stop band - using one of the classical approximations to the ideal brick wall filter (Butterworth, Chebyshev, Cauer,...).
Therefore, for all of these transfer characteristics (approximations) it makes sense to define the end of the pass band in the frequency domain (mostly at the -3dB point, but not always - depending on particular requirements).

2.) There is another low pass approximation (Thomson-Bessel), which also has the task to attenuate "higher" frequencies - however with one additional and important requirement:

Phase shift within the pass band as linear as possible (group delay nearly constant). Because of this, for many applications (but not always) the end of the pass band is defined in the time domain in form of an acceptable deviation from the requirement of a constant phase slope (constant group delay). The price to be paid for this optimized time behaviour is a relatively bad attenuation characteristic.

3.) Summary: For low pass filters with a Thomson-Bessel characteristic the end of the pass band can be defined either in the frequency domain (e.g. -3 dB) or in the time domain (deviation from the delay tau(w=0)=tau,g0). In some cases, both specifications are used at the same time and the filter order is determined by the most critical requirement.

4.) Filter practice for all low pass characteristics (frequency domain specification): The required order to fulfill the respective attenuation requirements can be determined using a corresponding formula. Then, the pole data for each 2nd-order section (Wp and Qp) can be found in corresponding (normalized) tables. The actual values (wp and Qp) are used to calculate the various parts values.

5.) Filter practice for Thomson-Bessel (time domain specification): If the pass band width is specified in the time domain (delay deviation from nominal) the required order can be found using curves which are available for the various filter orders. In addition, tables for standard deviations (1%, 5%, 10%) are used to determine the corresponding pole data for each section. Thus, all the parts values can be calculated.

EDIT: It is interesting to note that for a 2nd order Thomson-Bessel lowpass the group delay deviation from its nominal value (at w=0) at the 3-dB cut-off frequency is about 20%.
_____________

Do not hesitate to ask further questions.
If necessary I can elaborate an example.

 

I find this formula in a textbook about analog circuit design in section about how to find "equivalent" ω when we connects the first order filters is "series". And individual poles are not far away (less than one decade).

Yes - that´s what I have expected. It applies for FIRST-order systems only and NOT for second or higher orders. See my post#25.
Each simple example (two Butterworth 2nd order stages) reveals that this square-root formula cannot be used.

 

Hi LvW,

Sorry for this late reply but I have been busy during the last two days.
I agree with all the points you listed.

I will really appreciate if you can elaborate an example.

After this, I have just two questions:

1) Referring to my example I have the following normalized values for poles frequency:

\(
\omega_1 = 3.023\\
\omega_1 = 3.389\\
\)

And in #9 you have correctly written:

*Stage 1: Wp=3.023 ; Qp=0.5219
*Stage 2: Wp=3.389 : Qp=0.8055
(...)
Please note that Wp is the normalized angular pole frequeny - normalized to the desired angular cut-off frequency wo of the whole filter.
Thus, Wp=wp/wo.
* That means: All you have to do is to use denormalization wp=Wp*wo and to design both 2nd oder stages for the respective wp and Qp values.

As so, if I could determine the components values this way, for a cutoff frequency \(\omega_{0}\) and for a normalized unit delay \(D = 1\,\text{s}\):

\(C_{\text{old}} \cdot \left ( {1 \over k_{\text{r}}\omega{_\text{p}}\omega_{0}} \right ) = C_{\text{new}}\)

Correct?

2) When you talk about the square root formula for a n order filter, do you refer to this:

\(f_{\text{c}1} = {1 \over 2\pi\sqrt{ R{_1}R{_2}C_1C_2}}\)

or this:

\(f_{\text{c}t}=\frac{1}{\sqrt{\frac{1}{Fc1^2}+\frac{1}{Fc2^2}}}\)

Thank you again.
Simo

 

Hi Simo,
to answer your last question at first: Of course, the last formula, which applies to 1st order systems only. The first expression is a classical 2nd-order term.

As to the relation between Cold and Cnew - I am not sure if it makes sense to jump into another example which somebody else has created. At least, I have no real motivation to do this.
I prefer to present a new example - however, I am not certain if I will manage it for tomorrow. At latest on Friday - OK?

 

Hi LvW,

I prefer to present a new example - however, I am not certain if I will manage it for tomorrow. At latest on Friday - OK?

Of course you have the right to take your time.
Thank you in advance.

Regards,
Simo

 

Hi Simo, please see the pdf attachement for the two announced examples for calculating Bessel low pass filters.

As you will see, designing filters with Thomson-Bessel response is a bit more time-consuming than for other approximations (because in most cases filter requirements are given inthe time domain).

* In the text, I have mentioned tables and curves for deriving filter parameters. These tables/curves can be found in many specialized textbooks on filter theory and/or internet contributions.

* In some of these tables filter coefficients a, b are given instead of corresponding pole data (Wp, Qp)
You easily can convert from a,b to Wp,Qp through comparison of the corresponding denominators of both (normalized) transfer functions.

Regards
LvW

 

Hi LvW,

First of all, THANKS.

This little text have clarified all the doubts I practically had..
Relating the normalized poles frequency to the group delay and so on.

I manipulated the data provided by the example of the textbook with the the way the concepts have been exposed by your PDF and I finally made the values to coincide and I feel more secure.

I mean, solving for \(\omega_{\text{p}}\) than solving for \(D\) etc.

What I was needing was a practical example exposed in a different and perhaps more "practically" way, like these two examples posted on this text.

I saved it with my favorites Applications Notes and several PDFs.
I think I am going to read and use it, of course along with the textbook, for some several examples through my study about Bessel - Thompson approximation.

I also have to thank you for your patient.
Through this thread I realized that you are a good educator and I feel happy because I understood and learned something new.

\(\text{Thanks}\,^3\)

Kind Regards,
Simon