Universal state-variable filters

[atferrari]

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.

 

[MrAl]

Hi,

Do you need them for these exact circuits? If so we can develop the formulas ourselves as i could not find these exact circuits online. I did find other circuits however.

What makes these circuits interesting is that all the non inverting terminals go to ground. Not all SV filters have that.

 

[chemelec]

Here is a Better Drawn Schematic.

 

[MrAl]

Hello again,

Oh yes much nicer drawing, and that makes it easier to look at and read. That's a different circuit though right?

atferrari:
I took a closer look at the second circuit exactly as drawn in post #1, and it appears that the band pass output goes like this...

wc=sqrt(R5/(C1*C2*R3*R4*R8))
Ac=-R6/R1

where
wc is the center angular frequency for the band pass, and
Ac is the center gain for the band pass

The center frequency is therefore:
Fc=wc/(2*pi)

Check these in a circuit simulator to make sure they are right. You said the gain involves just two resistors so it looks right so far but it should be checked.

 

[chemelec]

Not Sure Where you got the one you have or if it works?
Very Poor Drawing!

But My Circuit Is Correct.
The Two Frequency Pots MUST be GANGED Together so each give the Same Resistance.
Or for a Single Frequency, these must be Equil Value Resistors.

I Use this Circuit in This Project on my site.
http://chemelec.com/Projects/Parabolic-Mic/Parabolic-Mic.htm

Have FUN......Gary

 

[RCinFLA]

Also find info under "biquad filter"

https://mysite.du.edu/~etuttle/electron/elect65.htm

 

[atferrari]

Hi,
Do you need them for these exact circuits? If so we can develop the formulas ourselves as i could not find these exact circuits online.

I have sketched the distribution of components with them in, ....and then I lost the magazine. Article by Ron Tipton or Lipton (most probably the first) in Poptronics, some years ago.

If you could help I would like to have the formulas developed for them. Gracias for any help.

What makes these circuits interesting is that all the non inverting terminals go to ground. Not all SV filters have that.

I found that contributing for a simpler layout but I am sure there is a more electronic merit than that!

 

[atferrari]

Not Sure Where you got the one you have or if it works?
Very Poor Drawing!

As I said to MrAl, they come from an article in Poptronics magazine. A quite detailed one.

Gary, not a drawing at all. Just a sketch, which, to understand who goes where, resulted much better.

Bragging now, I am sure I can beat any of your drawings when I fire Corel Draw. But not now. Hardware first!

 

[chemelec]

Anybody can make it look more PRETTY.
I just Draw the schematics, so they are VERY EASILY PRINTED and Very Clear.

I Draw my schematics with "2 Bit wide Lines" and in "True Black and White".
This Creates a "Very Small File" that Does NOT Degrade or go Blurly like I see on many Schematics.
(like many Color or Gray Scale picture do.)

 

[MrAl]

I have sketched the distribution of components with them in, ....and then I lost the magazine. Article by Ron Tipton or Lipton (most probably the first) in Poptronics, some years ago.

If you could help I would like to have the formulas developed for them. Gracias for any help.




I found that contributing for a simpler layout but I am sure there is a more electronic merit than that!

Hi again,


Ok here are the results for the second circuit in the first post...

wc=sqrt(R5/(C1*C2*R3*R4*R8))
Ac=-R6/R1
Q=R2*R6*sqrt((C1*R3)/(C2*R4*R5*R8))/R7

where
wc is the center angular frequency and center frequency in Hertz is fc=wc/(2*pi), and
Ac is the center frequency gain, it's negative which simply indicates the gain is R6/R1 but the phase is 180 degrees, and
Q is the circuit Q.

Note that certain simplifications come about by making some resistors equal and the two caps equal. This leads to a less general solution however, and some response shapes may not be possible.

These results look correct but you should check out these as well as any other formulas you find in a circuit simulator to verify. I checked for a very limited data set myself. You really must check results anyway because for a given set of component values although the output of interest (the BP output in this case) may come out just fine the intermediate op amp outputs may come out to be out of range of the op amp's output capability (for example higher than the power supply in the application) which would mean the construction does not work in the real life application even though the math works out perfect.

I'll try to get to the first circuit next as soon as i can unless someone else finds it on the web first.

 

[MrAl]

Hello again,

For the first circuit (A) in the first post the BP output results are as follows:

wc=sqrt(R2/(C1*C2*R3*R4*R5))
Ac=-R8/R1
Q=sqrt(C1*R3)*R6*R8/(sqrt(C2*R2*R4*R5)*R7)

where
wc is the center angular frequency and fc=wc/2/pi, and
Ac is the center frequency gain, and
Q is the circuit Q.

 

[atferrari]

Gracias MrAl,

I appreciate your help that never fails.

I will build the circuits in the next days. Now I know what to expect.

 

[MrAl]

Gracias MrAl,

I appreciate your help that never fails.

I will build the circuits in the next days. Now I know what to expect.

Hello again,

You are very welcome.

If you like you can also do a simulation and see what's up. With real life op amps or good simulation models for op amps we will see some differences if the frequency is close to the limit of the op amp for the given gain required to do the filter function, but hopefully not too much difference.