The Electrician's friend here. Here is what I get from LTspice:View attachment 162733 View attachment 162734

For the schematic I used a hierarchical subcircuit for each slider section. I think you had a typo for one of the resistor values, so I changed it to achieve equal EQ for all sections. (You're welcome.)

WOW
Thanks alot man. I'm really new to this spice thing so there's alot for me to learn about this still (I ha no idea you could even model the sliders in spice like that)
Gonna take tonight to go through everything and understand what you did...im seeing if statements next to the filter stages so i'm guessing you modelled them as purely theoretical :O ....which would leave it up to me to determine what op-amps to use to get a close enough response to the theoretical model. I'd have to play around with components in the lab to determine this most likely I guess.

I'm doing a nodal analysis of the complete circuit, with the nodal equations in matrix format. Thank goodness for modern mathematical software. The amount of number crunching would not be possible without it. My laptop takes about 5 seconds to solve the matrix numerically, and a few seconds to plot the response. A symbolic solution is possible, but it takes about 15 minutes to do it!!!

I may add a couple more filters, but as it stands now I think the circuit is behaving as expected.

Is the response you show with alternate filters set for boost, cut, boost, cut, boost, cut, etc.? I don't know why you get the irregular response; mine seems perfectly regular if, for example all the bands are set for the same amount of boost or cut.

Here's how I set things up; I'm not showing the plotting code (the easy part):

View attachment 162710

Funny thing is I was in the lab trying to understand this today and my friends saw it and they recognized it was an admittance matrix (They branched off into power systems...I branched into computer systems ...i'd spend some time this week trying to learn this method because it looks pretty cool )

I'm recalling a problem I had with this analysis which may be a clue to your difficulty. When I started the analysis of this circuit I only had a single filter, and everything was working ok.

When I added the second filter, Mathematica (which I'm using to do the number crunching) reported that my admittance matrix was singular. It took me a while to find the problem. I was using standard double precision floating point arithmetic. This means using about 17 digit floating point arithmetic. It turns out that the circuit is large enough that some of the internal calculations were over/under flowing the arithmetic. Mathematica allows for arithmetic other than floating point. The other method is rational arithmetic--all numeric quantities are represented as fractions with integer numerator and denominator. This is essentially "infinite" precision arithmetic; there's no roundoff error. When I switched to this method, the problem of singularity of the admittance matrix went away.

I was getting this problem with the addition of the second filter. Just imagine what would happen with 15 filters! I don't know if LTSpice can do rational arithmetic, but it's possible that it may be having problems with its floating problem arithmetic when solving such a large problem.

I guess this explains all the errors I was having before
I'd never have figured that out...I was starting to doubt the design.


My next phase of the project based on the requirements would be to replace all potentiometers with digital pots and implement a digital interface to control the boosting and cutting of the various bands using a micro-controller(The part i've been looking forward too). Oh with a VU meter in the digital interface as well.


Will post my designs/approach for this soon, as well as for the system's power supply.

 

WOW! Thanks a lot man. I'm really new to this spice thing so there's a lot for me to learn about this still (I ha no idea you could even model the sliders in LTspice like that). Gonna take tonight to go through everything and understand what you did. . . I'm seeing IF statements next to the filter stages so I'm guessing you modeled them as purely theoretical :O . . . which would leave it up to me to determine what op-amps to use to get a close enough response to the theoretical model. I'd have to play around with components in the lab to determine this most likely I guess.

As I already stated, the sliders are hierarchical subcircuits. If you double-click on a slider you will see the LTspice schematic of the slider with the details of its opamp circuit and the potentiometer on its output. Since these are all the same except for component values it is easier, more compact and more readable to use the same graphical subcircuit for each one with the values as parameters. I made the subcircuit symbol in the form of a slider control because it is the best (IMO) representation of the idea of your circuit.

I set up the simulation to run 15 times over with the stepped parameter "n" controlling which individual slider gets pushed to 80 percent boost (the "m" parameter) via the IF statements (check LTspice Help in the b-source section for documentation for IF statement syntax). The rest of the sliders (the non-active ones) at each step of n remain at 50 percent. I also increased the simulation points to 300 per decade to generate smoother output curves. The simulation runs fine with either the normal solver or the alternate, higher accuracy, solver.

 

Hello again,

An interesting version of the transfer function of the circuit with a single stage is:
(s^2*C*L*R22*R23+s*L*R22*R23+R22*R23-s*G*L*R23+s^2*C*L*R23+s
*L*R23+R23+s^2*C*L*R22+s*L*R22+R22)/(s^2*C*L*R22*R23+s*L*R22*R23+R22*
R23+s^2*C*L*R23+s*L*R23+R23-s*G*L*R22+s^2*C*L*R22+s*L*R22+R22)

where we use an LC filter instead of an RC op am filter section and the response is the same when L and C are chosen such that:
C=C1
L=(C1*R1*R2*R3)/(R2+R1)
with the stipulation that R3=2*R1 as in all the filters of the circuit.

The above is in the neat form (all constants except 's'):
(s^2*Z+s*Y-s*G*V+K)/(s^2*Z+s*Y-s*G*U+K)

where we can see that the pot setting affects the damping ratio and that's what changes the effect for a given stage.
Note: U is the upper pot section resistance, V is the lower pot section resistance.
We can also see that the gain G=R6/R5 has a similar effect except it controls both cut and boost.
Z,Y,K all constants that depend on the circuit component values.

 

A friend of mine who is an LTSpice expert mentioned that when LTSpice is having arithmetic problems, there is an alternate solver which can be chosen in the LTSpice control panel. He may get on here to contribute.

Hi,

It's just that he had a lot of pot settings that made it look bad because he did not know how to interpret the solution. With simpler pot settings (many set to 50k and 50k) the solution looks much more comprehensible.

 

As I already stated, the sliders are hierarchical subcircuits. If you double-click on a slider you will see the LTspice schematic of the slider with the details of its opamp circuit and the potentiometer on its output. Since these are all the same except for component values it is easier, more compact and more readable to use the same graphical subcircuit for each one with the values as parameters. I made the subcircuit symbol in the form of a slider control because it is the best (IMO) representation of the idea of your circuit.

I set up the simulation to run 15 times over with the stepped parameter "n" controlling which individual slider gets pushed to 80 percent boost (the "m" parameter) via the IF statements (check LTspice Help in the b-source section for documentation for IF statement syntax). The rest of the sliders (the non-active ones) at each step of n remain at 50 percent. I also increased the simulation points to 300 per decade to generate smoother output curves. The simulation runs fine with either the normal solver or the alternate, higher accuracy, solver.

Oh okay I think I understood the syntax from having a look at it, Its basically (from what I gather) an if-then-else
So it plots for one band boosted (with the others at no boost/cut), then the second band boosted (with the others at no boost/cut) etc until the last band is boosted.

That's what the then 0.5 else 0.8 of the if statement does. One thing i've noticed, if i set m to be 0.5 so that it plots for all pots at 0% boost/cut I get a flat response @ -6dB...and in the higher frequencies I get some ripple. Shouldn't it be 0dB for all pots centered? If it flat lines at -6dB I believe the audio signal will be attenuated at -6dB across all bands (well most) until it hits the ripple in the higher frequencies, so some higher frequencies would be boosted with reference to the -6dB flat line.

Setting m to 1 I see that the maximum boost is about 7dB if I remember correctly, and setting m to 0 the maximum cut was about -20dB (I think it should be +/- 12 dB based on the 39k resistor for R6)

Hello again,

An interesting version of the transfer function of the circuit with a single stage is:
(s^2*C*L*R22*R23+s*L*R22*R23+R22*R23-s*G*L*R23+s^2*C*L*R23+s
*L*R23+R23+s^2*C*L*R22+s*L*R22+R22)/(s^2*C*L*R22*R23+s*L*R22*R23+R22*
R23+s^2*C*L*R23+s*L*R23+R23-s*G*L*R22+s^2*C*L*R22+s*L*R22+R22)

where we use an LC filter instead of an RC op am filter section and the response is the same when L and C are chosen such that:
C=C1
L=(C1*R1*R2*R3)/(R2+R1)
with the stipulation that R3=2*R1 as in all the filters of the circuit.

The above is in the neat form (all constants except 's'):
(s^2*Z+s*Y-s*G*V+K)/(s^2*Z+s*Y-s*G*U+K)

where we can see that the pot setting affects the damping ratio and that's what changes the effect for a given stage.
Note: U is the upper pot section resistance, V is the lower pot section resistance.
We can also see that the gain G=R6/R5 has a similar effect except it controls both cut and boost.
Z,Y,K all constants that depend on the circuit component values.

Oh nice never thought about using an LC filter stage instead before, one thing is that G value doesn't seem to be affecting the circuit in the way it should in terms of maximum boost or cut in the simulations. It doesn't seem to follow the G = 1 + A at max boost and G = 1/(A+1) at maximum cut convention

 

Oh okay I think I understood the syntax from having a look at it, Its basically (from what I gather) an if-then-else
So it plots for one band boosted (with the others at no boost/cut), then the second band boosted (with the others at no boost/cut) etc until the last band is boosted.

That's what the then 0.5 else 0.8 of the if statement does. One thing i've noticed, if i set m to be 0.5 so that it plots for all pots at 0% boost/cut I get a flat response @ -6dB...and in the higher frequencies I get some ripple. Shouldn't it be 0dB for all pots centered? If it flat lines at -6dB I believe the audio signal will be attenuated at -6dB across all bands (well most) until it hits the ripple in the higher frequencies, so some higher frequencies would be boosted with reference to the -6dB flat line.

Setting m to 1 I see that the maximum boost is about 7dB if I remember correctly, and setting m to 0 the maximum cut was about -20dB (I think it should be +/- 12 dB based on the 39k resistor for R6)


Oh nice never thought about using an LC filter stage instead before, one thing is that G value doesn't seem to be affecting the circuit in the way it should in terms of maximum boost or cut in the simulations. It doesn't seem to follow the G = 1 + A at max boost and G = 1/(A+1) at maximum cut convention

Hi again,

Did you replace L with the appropriate values?
I did not actually check it yet but assumed it was working the same.
The thing about the LC section is that it has the same transfer function as the original op amp BP filter, so if it doesnt work then maybe i did something wrong in the formulation.
The form of the much simpler transfer function should be correct though because that just shows the basic function of this kind of circuit and points out that the gain is in both the numerator and denominator, and gets controlled by the pot setting in both places.

I hope you made out better with the LT spice sim now.

[LATER]
It looks like G is actually supposed to be -G in that previous formulation, but the simpler form i think is more informative as it shows how the gain is cut out of the denominator of the equation if R22=0 and how it is cut out of the numerator if R23=0.
And it also shows how it changes the damping ratio.

 

Hi again,

Did you replace L with the appropriate values?
I did not actually check it yet but assumed it was working the same.
The thing about the LC section is that it has the same transfer function as the original op amp BP filter, so if it doesnt work then maybe i did something wrong in the formulation.
The form of the much simpler transfer function should be correct though because that just shows the basic function of this kind of circuit and points out that the gain is in both the numerator and denominator, and gets controlled by the pot setting in both places.

I hope you made out better with the LT spice sim now.

[LATER]
It looks like G is actually supposed to be -G in that previous formulation, but the simpler form i think is more informative as it shows how the gain is cut out of the denominator of the equation if R22=0 and how it is cut out of the numerator if R23=0.
And it also shows how it changes the damping ratio.

I haven't tried it on Spice as yet.
But the tf I got with the RC section has the gain in the numerator and denominator and it's scaled by the pots...looks to be along the same line as yours. I'll substitute in H(s) as the tf of a filter stage (i'll try using maple this time to simplify everything instead of doing it on paper) and i'll post up my results.

It's still confusing though since the theory says one thing regarding the gain, then when simulating it it's a different story in terms of the max boost/cut.

 

This is what I'm getting for the tf (Ignoring the RC section)...not sure if yours was similar to that.
If I had to include the RC section the resulting tf would be the second image.
I replaced R1 in the first stage with Rf1 since the filters use R1 already.

 

This is what I'm getting for the tf (Ignoring the RC section)...not sure if yours was similar to that.
If I had to include the RC section the resulting tf would be the second image.
I replaced R1 in the first stage with Rf1 since the filters use R1 already.

Do you want verification of your result?

Just to be sure there's no confusion, please post the schematic of just what circuit you're calculating the TF. Clearly mark the input and output.

 

I haven't tried it on Spice as yet.
But the tf I got with the RC section has the gain in the numerator and denominator and it's scaled by the pots...looks to be along the same line as yours. I'll substitute in H(s) as the tf of a filter stage (i'll try using maple this time to simplify everything instead of doing it on paper) and i'll post up my results.

It's still confusing though since the theory says one thing regarding the gain, then when simulating it it's a different story in terms of the max boost/cut.

Hello again,

I think you should show where your values came from, like RA and RB, as well as what H(s) is. Is that again the BP transfer function?

What 'theory' are you referring to? The theory i see from pure circuit analysis with ideal components says that the mid band gain peak is related to A+1 not A, so not sure if you are referring to that again or not. But here is another rendering of the circuit in LC format and it becomes clear that A+1 is the controlling gain not A alone.
I checked it this time, but you are free to check it yourself too.
Also, did your equations come out to show A+1 or just A?

Here is the reduced schematic. Note that once we replace the filter with the LC section we can reduce the gain stage to just "G" which will be 4 for your circuit, but in the equation Gp=G+1 so it should be clear that Gp controls the mid band gain not just G (or A in your equation) alone.
Also, using another simulator i see a midband boost gain of 5 when i set G=4 in that circuit and Gp=5 in this new transfer function.

This is the transfer function for the entire circuit with one BP stage.
H(s)=
(s^2*C1*L1*R1*R22*R23+R1*R22*R23+s*L1*R22*R23+s^2*C1*L1*R1*
R23+R1*R23+Gp*s*L1*R23+s^2*C1*L1*R1*R22+R1*R22+s*L1*R22)/(s^2*C1*L1*
R1*R22*R23+R1*R22*R23+s*L1*R22*R23+s^2*C1*L1*R1*R23+R1*R23+s*L1*R23+
s^2*C1*L1*R1*R22+R1*R22+Gp*s*L1*R22)

and again R1 is the original R1, and R22 and R23 are scaled from the previous values and expressed in 10's of Kohms (so R23=10-R22 because the whole pot value becomes 10 ohms now).

Also as before:
L1=(2*C1*R1^2*R2)/(R2+R1)
and
R1, R2, C1 are the original values used in the op amp band pass filter section.
R4=1 ohm as are other previously 10k ohm resistors although the equation was normalized for that so it does not have to appear in the transfer function anymore.

The key point here though is that Gp is equal to G+1 or if you use A for the gain of that R5,R6 gain stage (where A is the positive value of the gain not the negative value) then Gp=A+1.

Here is the schematic i meant to upload earlier...

 

Here is the reduced schematic. Note that once we replace the filter with the LC section we can reduce the gain stage to just "G" which will be 4 for your circuit, but in the equation Gp=G+1 so it should be clear that Gp controls the mid band gain not just G (or A in your equation) alone.
Also, using another simulator i see a midband boost gain of 5 when i set G=4 in that circuit and Gp=5 in this new transfer function.

Where is the reduced schematic? In particular, would you post a schematic showing the replacement of the opamp filter with L and C?

 

Do you want verification of your result?

Just to be sure there's no confusion, please post the schematic of just what circuit you're calculating the TF. Clearly mark the input and output.

Yes thanks, here's the circuit I used with all my working for the analysis:

H(s) is the transfer function of the Bandpass section.

The last screenshot's from Maple is how it reduced when I substituted H(s)

Hello again,

I think you should show where your values came from, like RA and RB, as well as what H(s) is. Is that again the BP transfer function?

What 'theory' are you referring to? The theory i see from pure circuit analysis with ideal components says that the mid band gain peak is related to A+1 not A, so not sure if you are referring to that again or not. But here is another rendering of the circuit in LC format and it becomes clear that A+1 is the controlling gain not A alone.
I checked it this time, but you are free to check it yourself too.
Also, did your equations come out to show A+1 or just A?

Here is the reduced schematic. Note that once we replace the filter with the LC section we can reduce the gain stage to just "G" which will be 4 for your circuit, but in the equation Gp=G+1 so it should be clear that Gp controls the mid band gain not just G (or A in your equation) alone.
Also, using another simulator i see a midband boost gain of 5 when i set G=4 in that circuit and Gp=5 in this new transfer function.

This is the transfer function for the entire circuit with one BP stage.
H(s)=
(s^2*C1*L1*R1*R22*R23+R1*R22*R23+s*L1*R22*R23+s^2*C1*L1*R1*
R23+R1*R23+Gp*s*L1*R23+s^2*C1*L1*R1*R22+R1*R22+s*L1*R22)/(s^2*C1*L1*
R1*R22*R23+R1*R22*R23+s*L1*R22*R23+s^2*C1*L1*R1*R23+R1*R23+s*L1*R23+
s^2*C1*L1*R1*R22+R1*R22+Gp*s*L1*R22)

and again R1 is the original R1, and R22 and R23 are scaled from the previous values and expressed in 10's of Kohms (so R23=10-R22 because the whole pot value becomes 10 ohms now).

Also as before:
L1=(2*C1*R1^2*R2)/(R2+R1)
and
R1, R2, C1 are the original values used in the op amp band pass filter section.
R4=1 ohm as are other previously 10k ohm resistors although the equation was normalized for that so it does not have to appear in the transfer function anymore.

The key point here though is that Gp is equal to G+1 or if you use A for the gain of that R5,R6 gain stage (where A is the positive value of the gain not the negative value) then Gp=A+1.

Mine didn't reduce to A+1 when I found the gain...so I guess mine is different to yours

I'll work it over but for now that's what I got

 

Yes thanks, here's the circuit I used with all my working for the analysis:

H(s) is the transfer function of the Bandpass section.

The last screenshot's from Maple is how it reduced when I substituted H(s)



Mine didn't reduce to A+1 when I found the gain...so I guess mine is different to yours

I'll work it over but for now that's what I got

Hi,

Remember that i had obtained a center freq gain of 5 in full boost mode and the simulation showed that too.

So what did yours reduce to then?
Did you remember to remove the two 10 ohm resistors to get it into the simpler form?

What is RA and RB?

 

Where is the reduced schematic? In particular, would you post a schematic showing the replacement of the opamp filter with L and C?

Hi,

Yeah i cant believe i forgot to post that
What happened was i was doing some other things too and after i said 'here is the schematic' i went on to talk about some other things then forgot to upload it. Anyway, here it is. It makes the analysis much simpler.
We could probably reduce more too.

G is a gain block, which could be an op amp with non inverting gain, and the gain is 4.
In the equation, Gp is G+1.
Since G is a gain block, it has infinite input impedance and zero output impedance and 0 to infinite frequency response. Therefore it does not load the LC section at all, and nothing loads the output.

 

Hi,

Remember that i had obtained a center freq gain of 5 in full boost mode and the simulation showed that too.

So what did yours reduce to then?
Did you remember to remove the two 10 ohm resistors to get it into the simpler form?

What is RA and RB?

Oh I forgot to remove those resistors, RA and RB is how i modelled the pot, so RB = R-RA for me where R is the size of the pot.

What are you using to reduce these equations btw?
I'm trying to do it by hand and that's realllly tedious and I think that's where i'm making the mistakes.

This is what I'm getting for the tf (Ignoring the RC section)...not sure if yours was similar to that.
If I had to include the RC section the resulting tf would be the second image.
I replaced R1 in the first stage with Rf1 since the filters use R1 already.

Mine reduced down to this(attached images from that post) for the overall tf when I substituted in H(s) in Maple.

If I find the magnitude of that whole thing as

G = Magnitude(Numerator)/Magnitude(Denominator)
= sqrt( (Re(Numerator))^2 + (Im(Numerator))^2) / (Re(Denominator))^2 + (Im(Denominator))^2)

It just worked out to be a nasty square-root equation with no way of it to come down to A or A+1 in full boost or cut mode

 

Oh I forgot to remove those resistors, RA and RB is how i modelled the pot, so RB = R-RA for me where R is the size of the pot.

What are you using to reduce these equations btw?
I'm trying to do it by hand and that's realllly tedious and I think that's where i'm making the mistakes.



Mine reduced down to this(attached images from that post) for the overall tf when I substituted in H(s) in Maple.

If I find the magnitude of that whole thing as

G = Magnitude(Numerator)/Magnitude(Denominator)
= sqrt( (Re(Numerator))^2 + (Im(Numerator))^2) / (Re(Denominator))^2 + (Im(Denominator))^2)

It just worked out to be a nasty square-root equation with no way of it to come down to A or A+1 in full boost or cut mode

Hello again,

Oh yes depending on how you approach the analysis you could end up with a very hard to reduce equation. But to reduce effectively you MUST set R10 and R11 to zero. Did you do that?

For now i am using Maxima, which is a symbolic algebra program that does symbolic math. You really need something like that. Is that what you are using 'Maple' for? I have not used that.

I was able to prove the A+1 idea in a different way, by transforming the circuit into a purely boost mode, then see if the gain came out to A+1. The resulting equation for the proof came down to this as approximated with floats:
0.9999999*(G+1)^2+1.18499e-7=(G+1)^2

and if you examine this you will see that 1.18499e-7 is very small compared to the other quantities, so an approximation would be:
0.9999999*(G+1)^2=(G+1)^2

and since 0.9999999 is so close to 1 an approximation would be:
(G+1)^2=(G+1)^2

which proves that the gain is G+1 because both sides are equal.
Note the reason for the approximations is because floating point causes small inaccuracies. If this was done with pure integer math it would come out exact.

So we have as proofs:
1. My personal analysis.
2. Two different simulators both show a gain of 5 with A=4.
3. This new proof which because of the equality shows that the original analysis that produced the left hand side was exactly equal to G+1 and that is the gain of course.

See attachment for the origin of the left hand side of that third proof.
Isnt it amazing that gain expression in the attachment equals G+1 at w=w0.
The approximate expression below comes after subst some values and using floating point math and equating to G+1, where squaring was needed to get to a simpler form.

 

Yes thanks, here's the circuit I used with all my working for the analysis:

H(s) is the transfer function of the Bandpass section.

The last screenshot's from Maple is how it reduced when I substituted H(s)

In post #49, what I'm asking you to do is to post a schematic that shows only the components you're using to get the tf shown in image file Mapletf.png from post #48. The expressions in that image file show a number of items that are not defined in the schematic image file Scan0062.jpg from post #52. For example, where is A,RA, RB, C? If A is the gain of the opamps, say so, or put an "A" inside the opamp symbol. There is no capacitor labeled just C; the capacitors I see are C1 at the far left near Vin, and capacitors labeled C1 and C2 in the filters. Your schematic shows two different capacitors, C1 and C2, in the MFB filter, but elsewhere in the thread you show that C1 and C2 are the same value; when C1 and C2 are the same, just label them both just C in the MFB filter part of the schematic.

When you say in post #48 that you are ignoring the RC section, what RC section are you ignoring? Is it opamp U1A and the associated C1, R1, R2? If that is so, then show a schematic with those components wiped out, and Vin shown at the left end of R3.

For the tf in image file Mapletf.png, the expression in that file shows what appears to be the tf of the MFB filter plus a fraction involving A, RA, RB and R11. As I said above, RA and RB are not shown in the schematic; how would somebody looking at the schematic know what they are?

The idea is to have a schematic that includes everything used to produce a tf, and eliminates anything not used, so the person reading the post doesn't have to go back over the entire thread. It is becoming confusing to follow all the side paths this thread has gone down.

Also, I think you have an error in your expression for the H(s) of a MFB filter. For this filter:



You have this as the transfer function:



But I don't this is correct; I think it should be:

 

Here's an example of what I mean for a shematic:

 

Oh sorry about that, didn't know that's what you had meant.
I edited it to add in some more labels so it would be consistent with what I wrote out.

And yes your MFB is correct, I derived that MFB eqn from another circuit which had R3 and R2 basically in the other positions, I didn't even notice it was oriented differently there in this circuit.

Hello again,

Oh yes depending on how you approach the analysis you could end up with a very hard to reduce equation. But to reduce effectively you MUST set R10 and R11 to zero. Did you do that?

For now i am using Maxima, which is a symbolic algebra program that does symbolic math. You really need something like that. Is that what you are using 'Maple' for? I have not used that.

I was able to prove the A+1 idea in a different way, by transforming the circuit into a purely boost mode, then see if the gain came out to A+1. The resulting equation for the proof came down to this as approximated with floats:
0.9999999*(G+1)^2+1.18499e-7=(G+1)^2

and if you examine this you will see that 1.18499e-7 is very small compared to the other quantities, so an approximation would be:
0.9999999*(G+1)^2=(G+1)^2

and since 0.9999999 is so close to 1 an approximation would be:
(G+1)^2=(G+1)^2

which proves that the gain is G+1 because both sides are equal.
Note the reason for the approximations is because floating point causes small inaccuracies. If this was done with pure integer math it would come out exact.

So we have as proofs:
1. My personal analysis.
2. Two different simulators both show a gain of 5 with A=4.
3. This new proof which because of the equality shows that the original analysis that produced the left hand side was exactly equal to G+1 and that is the gain of course.

See attachment for the origin of the left hand side of that third proof.
Isnt it amazing that gain expression in the attachment equals G+1 at w=w0.
The approximate expression below comes after subst some values and using floating point math and equating to G+1, where squaring was needed to get to a simpler form.

View attachment 162846

Ohh I see, I analyzed it with the pot in any arbitrary position. My goal was to do it for any position so that I can substitute Ra and Rb accordingly for any boost/cut. And well of course Ra= 0 for Full boost or Rb=0 for Full cut.

 

Hello again,

Here is a symbolic proof of the G+1 idea:
(4*(G+1)^2*R1^2*R2)/(R2+R1)^2=(4*(G+1)^2*R1^2*R2)/(R2+R1)^2

and again that comes from the full boost mode only simplification. Note that we really only have to prove the full boost mode, the full cut more, the no boost or cut mode, and that the gain is monotonically increasing or decreasing. Showing the boost mode is G+1 says a lot by itself though.

I get for the transfer function:
Hs1=-(s*C2*R2*R3)/(s^2*C1*C2*R1*R2*R3+s*C2*R1*R2+s*C1*R1*R2+R2+R1)

and with C2=C1 we get:
Hs2=-(s*C1*R2*R3)/(s^2*C1^2*R1*R2*R3+2*s*C1*R1*R2+R2+R1)

and because we always set R3=2*R1 we end up with:
Hs3=-(2*s*C1*R1*R2)/(2*s^2*C1^2*R1^2*R2+2*s*C1*R1*R2+R2+R1)