bessel filter as a differetiator question - Page 1
Forum section: electronics / projects

author: john23
author_rank: Frequent Contributor
timestamp: March 08, 2025, 10:38:53 am
content: Hello, A circuit below presents bessel filter which is acting as differentiator .few question:1. How the denominator of a transfer function is defining the group delay?2.given the configuration below how did they know to choose the following opamp configuration to implement the transfer function?3.My pluses are smeered ,how do I make them sharper?Ltspice file is attached.Thanks.

author: mawyatt
author_rank: Super Contributor
timestamp: March 08, 2025, 01:31:28 pm
content: 1) Group Delay is defined as the dPhase/dFrequency and for the Bessel is maximally flat at the origin (F=0), the Butterworth is maximally flat amplitude at 0. To calculate the Group Delay you'll need to evaluate the Transfer Function Phase and take thederivatederivative of this wrt frequency. In the circuit C1 creates a High Pass, and C2 and C3 create the Low Pass effect, thus the result at frequency 0 is the output amplitude is also 0, and at infinite frequency the output is also 0 (Low Pass effect).2) The circuit topology looks similar to the classic Multiple-Feedback type.3) Increase the Low Pass Corner Frequency, thus "speeding up" the pulse edges.Best,

author: john23
author_rank: Frequent Contributor
timestamp: March 08, 2025, 02:17:13 pm
content: Hello , few questions:1.Is there a mathemacal example witch extracts group delay from a transfer function of a bessel filter ?2.Is there some document on "classic Multiple-Feedback type" filter3.given the denominator what is the theory which allowed to quickly see the group delay?https://www.eevblog.com/forum/index.php?action=dlattach;topic=455275.0;attach=2517953;image

author: mawyatt
author_rank: Super Contributor
timestamp: March 08, 2025, 02:48:29 pm
content: 1) and 2) Google should be able to provide answers to all your questions.3) This is the normalized S Domain Transfer Function Denominator for a 3rd Order Bessel LPF, for example the Butterworth is S^3 + 2S^2 + 2S +1.Get a copy of Zverev Handbook of Filter Synthesis, this has lots of details wrt to filters, transfer functions and mathematical relationships.BTW the significance of Group Delay is this is the time delay that all frequency components of the input waveform experience as they travel thru the filter/system. Thus one can visualize how a filter/system can "distort" the final waveform outcome if the various frequency components that make up the waveform experience a different time delay as they travel thru the filter/system. ISI is an example of this effect on digital pulse waveforms.Best

author: john23
author_rank: Frequent Contributor
timestamp: March 08, 2025, 05:29:59 pm
content: Hello,I can build a filter with group delay of 0.5 sec and 120nano second .I was told that to get a good differentiator I need a constant group delay so what are the advantages of having larger group delay In my “filter” (differtiator )Thanks .

author: john23
author_rank: Frequent Contributor
timestamp: March 08, 2025, 06:43:05 pm
content: UPDATE:Hello,I have rising edge of 1n and falling edge of 1n second these edges are separates by 2usec at the input at the presented simulation, the filter I built roughly does differntiation a little.Turning a pulse into two spikes.I want sharp peaks with the same amplitude.What group delay should I use for this case?

author: mawyatt
author_rank: Super Contributor
timestamp: March 08, 2025, 08:05:29 pm
content: For an edge pulse just use a RC Differentiator (High Pass). Something like PW/8 Time constant or shorter. So for 2us PW, then an RC of 1K and 220pF should work.Best

author: john23
author_rank: Frequent Contributor
timestamp: March 09, 2025, 07:28:04 pm
content: what is the theory for building the polynomial llike I showed.Is there some gppd example which presents the process?Thanks.

author: Benta
author_rank: Super Contributor
timestamp: March 09, 2025, 09:32:48 pm
content: Try this:https://www.eevblog.com/forum/projects/analog-filters-a-compilation-of-standard-transfer-functions/

author: john23
author_rank: Frequent Contributor
timestamp: March 10, 2025, 07:37:26 pm
content: Hello ,The following guidance showing in the quotes I am trying to understand.As I see It each 1 nano second rise and fall of the pulse has the same spectral data values.each frequency tone needs to subjected to the same group delay to be preserved on the output of the pulse.So we have rising edge (data 1) pause by 0.5usec falling edge (data 2) the filter gets "activated" twice at two time points.We will have two pulses on the output .(simetrical because group delay is constant)why pulse width is (ideally) twice the group delay?my pulse is just a delay between two events(DC spectral data is the pulse width) why the delay which each spectral component feels has to do with distance between the rising and falling edge?What will be the difference if the time distance between rising and falling edge will be much larger?My intution tells me that we are going to get two spikes with the pulse width appart. Thats it.Where am I going wrong?Thanks."If the group delay is perfectly constant, the resulting pulse will be symmetrical. For a symmetrical pulse, the peak is in the middle, at a time delayed by the group delay. So, the pulse width is (ideally) twice the group delay. A finite-order filter isn't perfect, but Bessel is pretty good: even this third-order filter yields a reasonably symmetric pulse."

author: gf
author_rank: Super Contributor
timestamp: March 11, 2025, 10:26:18 am
content: It is correct that a linear phase filter (-> constant group delay) has a symmetric impulse response. But remember that the "impulse response" of a filter is defined as the response to a single Dirac delta impulse. It is not the response toyourimpulse. The filter's response to your impulse is the convolution of your impulse with the filter's impulse response.The response of a filter to a Heaviside step is the integral of its impulse response. A single rectangular impulse can be considered a sum of two time-delayed Heaviside step functions (one step upwards and a second step down) and the filter's response to the rectangular impulse is the sum of the two step responses.Quote from: john23 on March 10, 2025, 07:37:26 pmYour intuition is not wrong. Just remember that the impulse response (and step response) of an IIR filter is infinite. If the width of your impulse is too short, the response to the rising edge may not have settled sufficiently by the time the falling edge occurs, and a non-negligible portion of the response to the rising edge may still overlap (and add up) with the response to the falling edge.

author: john23
author_rank: Frequent Contributor
timestamp: March 11, 2025, 07:46:58 pm
content: Hello,This is the key of what I dont understand.How can I know given ther rising edge of the pulse home much time it needs to sttle?Thanks."if the width of your impulse is too short, the response to the rising edge may not have settled sufficiently by the time the falling edge occurs"

author: gf
author_rank: Super Contributor
timestamp: March 13, 2025, 03:26:27 pm
content: Given that the transfer function of your filter is (b1*s)/(s³+a2*s²+a1*s+a0), its step response is the inverse Laplace transform of b1/(s³+a2*s²+a1*s+a0). The result of the transform is a function of b1 and the poles, and for a stable system it is a sum of decaying exponential functions and exponentially decaying sinusoids. I may be wrong, but I don't think it is possible to calculate analytically the time at which this sum has decayed to a certain level (-> settling time). You can do it numerically, of course, or graphically, if you plot the function.Btw, the transfer function of an (ideal) differentiator is s, therefore your filter (b1*s)/(s³+a2*s²+a1*s+a0) is not a differentiator, but it is effectively a differentiator concatenated with a 3rd order lowpass b1/(s³+a2*s²+a1*s+a0). The width of the impulse (i.e. the width of the step response of your filter) is basically determined by the cut-off frequency of the lowpass. If you require narrower impulses you need to increase the lowpass cut-off frequency. The reason for combining a differentiator with a low pass is to reduce noise sensitivity, since an ideal differentiator has a gain proportional to frequency (-> transfer function = s). Furthermore, an ideal opamp differentiator circuit (R1=R3=0, C2=C3=0) may not be stable because the opamp and the components are not ideal.

author: mawyatt
author_rank: Super Contributor
timestamp: March 13, 2025, 04:15:57 pm
content: Quote from: gf on March 13, 2025, 03:26:27 pmNice analysis!!This is why we suggested the simplest (component-wise) Differentiator as a RC, which has a TF of s/(s+1/RC). Letting the RC Time Constant be less than 1/8 Pulse Widths allows the edges to be sensed by the Differential without any significant interaction.Best