Cutoff frequency of Bessel response S&K.

Discussion in 'Homework Help' started by simo_x, Nov 16, 2013.

1. simo_x Thread Starter Member

Dec 23, 2010
200
6
Hi dears,

I am studying the Bessel - Thompson frequency response and I have a doubt about the Sallen & Key realization.

The textbook I use is "Analog Filter Design" by M. E. Van Valkenburg and the Bessel response is covered in chapter 10 "Delay Filter".

There is a design example for a low pass active filter in this chapter (example 10.1), and I do not understand how to determine the cutoff frequency from the denormalization formula exposed:

${D \over k_\text{r}k_\text{f}}$

The filter is designed to provide a delay

$D = 100\,\mu s$

(while the cutoff frequency is not mentioned).
The order of the filter is $n = 4$ and the complex conjugate poles pairs are:


\begin{align}
&& 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\\
\end{align}

And


\begin{align}
&& \omega_0 = \sqrt(a^2 + b^2)\\
&& Q = {1 \over 2a}\omega_0\\
\end{align}

For a Sallen & Key configuration:

The components for each cell are denormalizated from a scaling factor $k_\text{r}$:


\begin{align}
&& 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})\\
\end{align}

And the final circuits is:

Of course I tried to simulate it in LTspice:

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104.
And I can see through AC analysis that the cutoff frequency $f_c$ at -3 dB is $\approx 3,25\,\text{kHz}$

As we can see from the attached Bode plot image.

At this point it is not clear to me how to relate the cutoff frequency $f_c$ with the denormalization formula:

${D \over k_\text{r}k_\text{f}}$

Could please someone help me to understand it?
Thank you in advance.

simo_x

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• -3dB_bodeplot.png
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Last edited: Nov 16, 2013
2. shteii01 AAC Fanatic!

Feb 19, 2010
4,425
700
I was googling, the second order has a formula for cutoff frequency=1/sqrt(R1*R2*C1*C2).

I forgot to include the 2pi. Sorry.

Last edited: Nov 18, 2013
3. Jony130 AAC Fanatic!

Feb 17, 2009
4,495
1,265
There must be something wrong with you sim. When I ran your simulation file I get -3db at 33.8KHz

• 1.PNG
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4. simo_x Thread Starter Member

Dec 23, 2010
200
6
Hi Jony130,

Thank you for reporting this mistake.

This is because the I did not save the .asc file before posting it here and the resistors value was 1k instead of 10k.

This is the correct .asc file:

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104.
However, from the Sallen and Key configuration:

$f_{\text{c}} = {1 \over 2\pi\sqrt{ R{_1}R{_2}C_1C_2}} = {1 \over 2\pi\sqrt{10\,\text{k}\Omega \times 10\,\text{k}\Omega \times 3.45\,\text{nF} \times 3.17\,\text{nF}}} \approx 4,813\,\text{kHz}$

And the result is not as shown on the Bode plot.

However thanks for the link for the 4th order Bessel with $f_c \approx 1\,\text{kHz}$.

It reports:

$R_1, R_2 = {a_1C_2 \pm \sqrt{a_1^2C_2^2 - 4b_1C_1C_2} \over 2\omega_0C_1C_2}$

I saw this formula on many books including Texas Instruments "Op Amp for Everyone" Chapter 16 page 16.
But my question is related to the denormalization formula:

${D \over k_\text{r}k_\text{f}}$

However I think that maybe the best thing to do is to relate the two formulas:

\begin{align}
&& C_{\text{old}} \left ( {D \over k_\text{r}k_\text{f}} \right ) = C_{\text{new}}
&& \omega_{\text{c}} = {1 \over \sqrt{R{_1}R{_2}C_1C_2}}
\end{align}

I need to rest now, it is very late here.
Tomorrow I will care about it.

Thanks
Simo

Last edited: Nov 16, 2013
5. WBahn Moderator

Mar 31, 2012
23,082
6,936
This doesn't make a lot of sense to me.

The circuit shown by the OP is simply two cascaded second-order low pass filters. The opamp, configured as a voltage follower, should pretty effectively decouple the two stages. The cutoff frequency should be primarily dictated by whichever stage has the lowest cutoff frequency. At that frequency and below the other stage should be within its passband. So I don't see any mechanism whereby the cutoff frequency would somehow be the difference of the two, thus placing it well below the cutoff frequency of either stage.

6. shteii01 AAC Fanatic!

Feb 19, 2010
4,425
700
I forgot to include 2pi again.

Last edited: Nov 18, 2013
7. WBahn Moderator

Mar 31, 2012
23,082
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How are you getting your cutoff frequencies?

If you are just multiplying the two resistances and the two capacitances together and then taking 1/sqrt(that) then you aren't finding the cutoff frequency. For lack of a better term that is the center frequency.

Also, since you aren't tracking your units, you are off by a factor of 6.28.

1/sqrt(R1R2C1C2) has units of 1/sec, not cycles/sec. You need to multiply by 1cycle/2∏radians to get Hz.

Of course, no one believes me that tracking units has any value whatsoever.

8. Jony130 AAC Fanatic!

Feb 17, 2009
4,495
1,265
The active filter form this blog
http://analogelectronix.blogspot.com/2010/08/sallen-key-bessel-filter-2nd-4th-order.html

$f_{\text{c}1} = {1 \over 2\pi\sqrt{ R{_1}R{_2}C_1C_2}} {\approx 1.4307\.\text{kHz}$

And Fc for U4 stage

$f_{\text{c}2} = {1 \over 2\pi\sqrt{ R{_3}R{_4}C_3C_4}} {\approx 1.603\.\text{kHz}$

And combine and the net frequency is equal to

$\Large f_{\text{c}t}=\frac{1}{\sqrt{\frac{1}{Fc1^2}+\frac{1}{Fc2^2}}} = 1.067KHz$

Also in the art of electronics we can find a table in which we can find Fc for a given filter stage.
Or we can use this site
http://www.analog.com/designtools/en/filterwizard/#/type

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9. LvW Active Member

Jun 13, 2013
674
100
Simo, I was absent for 2 days - therefore, my late response.
May I repeat some fundamentals?

In case you want to design a 4th order Bessel filter as a cascade of two second-order filters, the filter theory tells you that none of both stages has a Bessel characteristic.
(As another example: Two 2nd order Butterworth filters do NOT result in a 4th order Butterworth filter).
Instead, particular pole parameters for both stages have to be found by a rather time comsuming calculation. It is not true, that both pole frequencies are to be subtracted (as mentioned in a post before). And it is also not correct that the lowest cut-off would determine the overall cut-off frequency.
Instead, because the derivation of the characteristic values for each stage is rather complicated you should use filter tables which are given in each relevant textbook or in appropriate articles. These tables give normalized pole data for each stage which are used to design each stage separately.
In the following, I give you the values for a 4th order Bessel filter:

*Stage 1: Wp=3.023 ; Qp=0.5219
*Stage 2: Wp=3.389 : Qp=0.8055

(As can be seen by inspecting the Qp values, one stage is "below" Bessel and the other one is "above" Bessel, because a 2nd order Bessel response has a pole quality factor Qp=0.5773)
* 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.
I suppose, all formulas relating parts values to the mentioned pole parameters for 2nd order stages are known to you.
* Please give notice, if you need further information.

Last edited: Nov 17, 2013
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10. shteii01 AAC Fanatic!

Feb 19, 2010
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Ah. Thank you. Yes, I lost 2pi while calculating cutoff frequencies.

Feb 19, 2010
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Thank you.

12. WBahn Moderator

Mar 31, 2012
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Note that I didn't say that the lowest cutoff frequency determines the overall cutoff frequency, but rather that it is primarily determined by the lowest cutoff frequency. This is along the same lines as saying that the equivalent resistance of two parallel resistors is primarily determined by the smaller of the two resistors. That's not to say that the larger resistor plays no role, but the overall resistance is bounded by 50%*Rs and Rs where Rs is the smaller of the two resistors.

A similar thing applies here. If Fcs is the lower of the two cutoff frequencies, then the overall cutoff frequency is bounded by and 70.7%Fcs and Fcs, so the lower cutoff frequency of either stage creates an even tighter bound than the smaller of two parallel resistors does.

13. WBahn Moderator

Mar 31, 2012
23,082
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We all make mistakes -- and this one is a pretty easy one to make. But the takeaway is that if you track your units religiously, then you will almost never make this mistake and not catch it.

14. LvW Active Member

Jun 13, 2013
674
100
My only intention was to make clear that the particular (separate) cut-off frequencies of the two stages plays no role at all !!
They are meaningless. They cannot be combined - neither as a difference nor by a square root. That is the outcome of filter theory.

I repeat the simple example of two identical 2nd order Bessel stages connected in series:
When each 3dB-cut-off is at f=fo, the 4th order filter (cascade of both stages) will have at f=fo an attenuation of -6 dB.
And most important: The response will not be Thomson-Bessel anymore.

15. LvW Active Member

Jun 13, 2013
674
100
Jony130, are you sure that the square root formulas you have mentioned really can be applied with sufficient accuracy for 2nd oder circuits ?
Can you give a reference?

16. WBahn Moderator

Mar 31, 2012
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I don't agree that the particular cut-off frequences are meaningless and play no role at all. For example, I have two second-order low pass filters that are cascaded. One has a 3dB cutoff frequency of 10kHz and the other has a 3dB cutoff frequency of 20kHz. Which of the following is a possible cutoff frequency for the combined cascaded filter?

A) 1kHz
B) 9kHz
C) 15kHz
D) 40kHz

If what you are saying is true, then all of the above choices are possible cutoff frequencies for the combined filter since the individual cutoff frequencies are meaningless and play no role. Yet I would argue that three of the four options can be eliminated based on the assertion that the individual cutoff frequencies DO play a role and DO convey meaning.

17. simo_x Thread Starter Member

Dec 23, 2010
200
6
Hello LvW and first of all thank you for your reply.

Yes, you are absolutely right about the normalized values for $\omega{_0}$ and $Q$, and in fact, the values for capacitors are obtained from that.

****************************************************************

I reprise for a moment my doubt about relating

$D \over k{_\text{r}}\omega{_0}$

and

$\omega{_c}$

As reported in the textbook. Following that solving process, with $R_1 = R_2$ we have:


\begin{align}
&C_1 = 2Q \left ({D \over k{_\text{r}}\omega{_0}} \right )\\
&C1 = {1 \over 2Q} \left ({D \over k{_\text{r}}\omega{_0}} \right )\\
&f_c = {1 \over 2\pi R \sqrt{C{_1}C{_2}}}\\
\\
&C{_1}C{_2} = \left ({D \over k{_\text{r}}\omega{_0}} \right )^2\\
\\
&f_c = \frac{1}{2\pi R\,\frac{D}{k{_\text{r}} \omega{_0}}} = {\omega_0 \over 2\pi D}\qquad(**)
\end{align}

Where $\omega{_0}$ is the normalized pole frequency given by

$\sqrt(a^2 + b^2)$

As mentioned in the post #1, while the overall cutoff frequency is given by the formula:

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

As said earlier..

So, for the equation marked as (**), we just need to solve for

$D = {\omega{_0} \over 2\pi f{_c}} = {\omega{_0} \over \omega{_c}}$

And then solve for the capacitors..
But the delay have to be the same for each 2nd order cell and does not have to change.. Maybe it is better to solve some more math..

****************************************************************

LvW, if I am not wrong, you suggest to solve the circuit for the cutoff frequency and not for the delay, right? Do I am correct?

Again, thank to you all for your interest.
Simo

Last edited: Nov 17, 2013
18. WBahn Moderator

Mar 31, 2012
23,082
6,936
If D is a delay with units of time, then it cannot also be equal to the ratio of two frequencies, as that is dimensionless. Are you sure that D isn't some normalized delay parameter?

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19. simo_x Thread Starter Member

Dec 23, 2010
200
6
Ouch. You are right.

Thinking it better, the scaling factor:

$k_\text{r} = 10000$

Then, if in the example $D = 100\,\mu s = 1e-4 s$ is a normalized delay,

$k_\text{r} = {\text{required delay} \over \text{normalized delay}} = {1 \over 1e-4} = 10000$

Following this, the scaling factor (so the components value) is straight related to the delay D.
Do am I right?

But again, what about the overall 3 dB frequency?

Last edited: Nov 17, 2013
20. LvW Active Member

Jun 13, 2013
674
100

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.

Last edited: Nov 18, 2013