# Universal state-variable filter

Discussion in 'General Electronics Chat' started by atferrari, Mar 14, 2014.

1. ### atferrari Thread Starter AAC Fanatic!

Jan 6, 2004
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Universal state-variable filters

I've lost the magazine where these two circuits were shown. I recall that in both, the formulas to calculate Q and band pass gain were simple relations between two resistors.

Could anyone tell them for A and B?

Gracias for any help.

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2. ### LvW Active Member

Jun 13, 2013
674
100
Hi atferrari, perhaps the following helps.

For both circuits the transfer functions are
H(s)=N(s)/D(s).
Of course, for each circuit each output has the same denominator D(s).

1) Circuit A:
D(s)=R1*R2*R6*R8+s*C2*R1*R2*R4*R5*R7+s^2*C1*C2*R1*R3*R4*R5*R6*R8
N(s) Lowpass: -R2*R5*R6*R8
N(s) Bandpass: +s*C2*R2*R4*R5*R6*R8
N(s) Highpass: +s^2*C1*C2*R2*R3*R4*R5*R6*R8

2) Circuit B:
D(s)=R1*R2*R5*R6+s*C2*R1*R4*R5*R7*R8+s^2*C1*C2*R1*R2*R3*R4*R6*R8
N(s) Lowpass: +R5*R6*R7*R8
N(s) Bandpass: -s*C2*R4*R5*R6*R7*R8
N(s) Highpass: +s^2*C1*C2*R3*R4*R5*R6*R7*R8
_____________________________________________

From these equations all relevant data can be derived if you re-arrange the denominator with the aim to compare D(s) with the nominal form:

D(s)=wp^2 + s*wp/Qp + s^2

For example, the bandpass gain is found by setting s=jwp. Hence, the left most and the right most terms cancel each other. (wp=pole frequency, identical to bandpass center frequency)

Last edited: Mar 14, 2014
3. ### atferrari Thread Starter AAC Fanatic!

Jan 6, 2004
2,663
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Hi LvW,

Filters are a stumbling block to me.

Sincerely, I am still expecting to be lucky in finding the direct answer to my question. The article showed, as I said, the relation between two especific resistors.

I am focused in experimenting with a sensitive microphone and tried to get this sorted out without to delve into the theory and equations.

4. ### crutschow Expert

Mar 14, 2008
13,477
3,362
Perhaps this will help.

atferrari likes this.
5. ### LvW Active Member

Jun 13, 2013
674
100
I`m not sure if it can be helpful since none of the circuits as given in post#1 are identical to the KHN topology.

atferrari - it´s only basic math, nothing else.
Divide N(s) and D(s) by the expression which appears in D(s) as a factor of s^2.
Then, you can compare element by element with the denominator of the general formula I have given.
From this comparison you can find the expressions for wp (center frequency) as well as for Qp.
Qp is the wanted quality factor.

6. ### LvW Active Member

Jun 13, 2013
674
100
atferrari, here comes an additional hint:
When you have found the expressions for wp and Qp you can, of course, simplify the expressions because you have two unknowns and much more components.
In any case, you can set C1=C2=C and you also can select some resistors to be equal (and, hence, cancel each other).
However, you must take care that wp and Qp still can be determined independent on each other.

7. ### crutschow Expert

Mar 14, 2008
13,477
3,362
I gave him that reference since I thought it would be a similar filter that could do what he needs and has cookbook formulas for corner frequency and Q that are easy to use.

Edit: I'm rather with atferrari, I look for easy formulas. I'm not much on wading through equations with a couple dozen variables to beat it into a format that is usable, even if it is just "basic math". You appear to be at home with those type of manipulations and that's great. But I often mess up somewhere along the many steps of the conversion and end up with a formula in error. So if I find a cookbook formula, I go with it.

Last edited: Mar 14, 2014
8. ### LvW Active Member

Jun 13, 2013
674
100
Hi crutschow, I understand and respect your position - however, there are some cases where it is not possible (or not appropriate) to have only one single "easy cookbook formula".
In the present example, we have 10 parts to fix only two filter parameters (wp and Qp). That means, we have several degrees of freedom and we are free to select some values. And it makes sense - for simple formulas - to select some values to be equal, but this is not necessary - and very often not wise.
Hence, several different sets of formulas are possible - and that is a great advantage because we can adapt our selection to "environmental conditions" (lumped parts yes/no, tunability, tolerances),....

I was of the opinion, that it is not a great job to divide numerator and denominator by the expression (case A)
R1*R3*R4*R5*R6*R8*C1*C2,
and to compare the resulting denominator with the general form (for deriving wp and Qp expressions).
OK - I did it and came to the following result (case A only):

If we select R2=R3=R4=R5=R and C1=C2=C we arrive at
wp=1/RC
Qp=R6*R8/(R*R7)=R6/R
(in case we also set R7=R8)
Ao=R6*R8/(R1*R7)=R6/R1 (for R7=R8).
(Ao is the bandpass gain at w=wp; it is found simply by dividing the bandpass numerator by the s-term in the middle of the denominator).

Comment: Perhaps it makes sense and it seems to be simple to set R2=R3=R4=R5. However, in reality it may cause problems because of tolerances. Therefore, it may be advantageous to see how each of the various parts influences the pole parameter (based on the formula without simplification). Thus, it could be possible to cancel the tolerance of one part with proper adjustment of another part.

Conclusion: It seems to be advantageous to have one simple cookbook formula, however this approach also has severe disadvantages.

Simple example: If we do not want to adjust or to tune the pole frequency with C1=C2=C we have to vary 4 (four!) resistors at the same time (based on the simplified cookbook formula).
This seems to be not the best solution.
More than that, for the most simple design formulas (with R7=R8) there is no independent adjustment of Qp and Ao.
Hence, it is very important to know the complete expression (without simplification) - and to have the freedom to decide which single element should be used for adjustment.
Good filter design is sometimes not as easy as it seems to be (based on cookbook recommendations) .
I know what I am speaking of. I have some experience in filter design.

Regards
LvW

@atferrari: I suppose you are able to perform similar mathematcal manipulations for the circuit case B. If you have problems I can provide further help.

In case you are interested in bandpass functions only (ref. your first post), you should know that there are many special bandpass circuits with one opamp only.

Last edited: Mar 15, 2014
9. ### atferrari Thread Starter AAC Fanatic!

Jan 6, 2004
2,663
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I expect part of the weekend to be for me and not for my job. I will do what you suggest and will return with the outcome.

Gracias.

10. ### atferrari Thread Starter AAC Fanatic!

Jan 6, 2004
2,663
784
Hola again LvW,

Last night I started to do the maths but found myself doing something empty of sense (or it is meaning?) to me. After all I have no idea how / why you came to the expresions posted.

Instead, I went to read the AN 779 from National Semiconductors. Even then, I cannot see the relationship between both topologies and the equations posted.

Since I hate people asking in these forums explanantions of a whole subject, please suggest any simple text I could read to see how you came to them.

Sorry but ending learning nothing after using your/my time is not wise.

If you think this is way over my head, you are surely right. Wish I was more versed in all this.

11. ### LvW Active Member

Jun 13, 2013
674
100
Hola atferrari,

I have prepared a one-sheet document which contains all the steps necessary to derive the pole data for circuit A from the given transfer function (see pdf attachement).
This method can be, of course, also applied to find the corresponding data for circuit B.
* Circuit A is a second-order version of the FLF topology
* Circuit B is a second-order version of the so called "Leapfrog topology".
* For example, see chapter 5.6 in