Op-amp stability - inverting attenuator and Sallen-Key LPF

Thread Starter

ttshaw1

Joined Jan 24, 2016
14
I want to implement this upload_2019-5-11_11-22-9.png
analog front end with the op-amps in the PIC32MK1024GPD064-I/PT. I might buffer the inputs before R5 and R6 too, but for now let's assume that the source has just a 50k input resistance to the inverting input of U1. The datasheet gives the minimum stable closed-loop gain, OA4 and OA16, as 8. Open loop gain is 90dB and GBWP is 10MHz. Both op-amp circuits have a nominal closed-loop gain of 1, so the first op-amp pole should be around 10MHz as well.

I added C5 to form a 60kHz low-pass with R7 in the feedback loop, so U1's feedback network's gain should drop to -40dB by 10MHz. The R9-C3 lowpass in U2's feedback loop will bring the feedback network gain there close to -60dB by 10MHz. Each of these filters will give 90 degrees of phase shift well before 10MHz. So my question is, is there any chance the op-amp's poles will drop the open-loop gain 50dB by the time they add 90 degrees of phase shift? If so, this should result in a loop gain of less than 1 at 180 degrees phase shift, which will make each circuit stable. But it seems like filters will typically show a phase shift much more quickly than they'll attenuate to that degree.

Is there anything I'm missing in this analysis? How is dominant-pole frequency compensation of this sort viable outside of extremely low signal frequencies or extremely fast op-amps?
 

MrAl

Joined Jun 17, 2014
7,762
Hi,

Connecting C5 in that manner is not the right way to make a low pass filter. You dont mess with the inverting terminal in that way you either make a low pass filter the right way or you dont make any low pass filter.
What you have done in effect is create both a low pass filter and another high pass filter at the same time. The high pass is because the cap is also present in the feedback loop and because it looks like a low pass as viewed by the output of the op amp, it is actually a high pass to the input of the circuit (inverse function theory).

We could look at the analysis of this idea but it would just be easier to do it the right way. Perhaps just a resistor between C5 and R7 connections so C5 does not connect directly to the inverting input.
 

OBW0549

Joined Mar 2, 2015
3,514
Is there anything I'm missing in this analysis?
Yes. Read Section 4, Capacitance at the Inverting Input, of the attached application note, with special attention to section 4.1.1. The way you have it, you've not created a low-pass filter but rather an oscillator (or, at best, a marginally-stable circuit on the brink of oscillation).
 

Attachments

MrAl

Joined Jun 17, 2014
7,762
Yes. Read Section 4, Capacitance at the Inverting Input, of the attached application note, with special attention to section 4.1.1. The way you have it, you've not created a low-pass filter but rather an oscillator (or, at best, a marginally-stable circuit on the brink of oscillation).
Hi,

Yes that is interesting too.

I did the analysis a couple times and always come up with an output with step input in the time domain of the form:

Vout/Vin=e^(-a*t)*(A*sin(w*t)+B*cos(w*t))-K

with B>>A so approximately:
Vout/Vin=e^(-a*t)*(B*cos(w*t))-K

so it is a damped sinusoid, but that could mean many oscillations before it comes to rest at any one DC level.
This is far from a low pass filter with that high pass filter effect kicking in to combine with the low pass which creates some sort of nasty band pass.
 

Thread Starter

ttshaw1

Joined Jan 24, 2016
14
Alright, so no to the capacitor on the inverting input. I could put in a low-pass filter elsewhere. But is dominant pole compensation even viable for this circuit? The gain-bandwidth product and open loop gain give a first pole at about 300Hz for open loop. Someone else told me that adding in negative feedback doesn't move that pole, which would require me to add a dominant pole with a frequency much lower than 300Hz. Is he right about the pole not moving? I have a textbook which seems to imply that the pole does indeed move to the knee frequency of the closed-loop gain, which would in this case be around 10MHz.

Then why do you have them operating with a closed-loop gain of 1? :confused:
Mainly for fun. I could attenuate the input signal to the inverting op-amp then increase the gain and use a unity-gain-stable buffer included in the MCU for the Sallen-Key, but then I wouldn't learn as much.
 

MrAl

Joined Jun 17, 2014
7,762
Hi,

Yeah well as crutschow pointed out in post #4 you are using the op amps outside of their range.
Did you look at that yet? I think that would be the first step.

Also, any ideas you have can be tested with a little circuit analysis. We can do that here too.

Your last stage LP filter has a cutoff frequency of 19871.02 Hz for example.
 

Thread Starter

ttshaw1

Joined Jan 24, 2016
14
Yeah, I don't have any issues designing and simulating a circuit that keeps the amplifiers above a gain of 8. But that's not what I want to do. I want to do dominant pole compensation, or at least learn if it's possible. This is a hobby project, so it's almost entirely a learning exercise. And this is a good opportunity for me to learn dominant pole compensation. Compensating by increasing the gain isn't a satisfactory solution only because I won't learn enough from it; I'll do it only if dominant pole compensation isn't feasible. Which leads me back to the questions from my last post.
 

MrAl

Joined Jun 17, 2014
7,762
Yeah, I don't have any issues designing and simulating a circuit that keeps the amplifiers above a gain of 8. But that's not what I want to do. I want to do dominant pole compensation, or at least learn if it's possible. This is a hobby project, so it's almost entirely a learning exercise. And this is a good opportunity for me to learn dominant pole compensation. Compensating by increasing the gain isn't a satisfactory solution only because I won't learn enough from it; I'll do it only if dominant pole compensation isn't feasible. Which leads me back to the questions from my last post.
Hi,

Well you could use a lag network on the output of one of the op amps and see how that works for you.
You can read the entry on Wikipedia too on that topic.
Place the new pole just under the highest frequency pole.
 

danadak

Joined Mar 10, 2018
4,057
There are calculations here using pole shrinkage equation, eg, compute
Fc for 1 pole, same for 2 pole, and apply the equation for final cutoff.

https://www.brainkart.com/article/Frequency-Response-of-Multistage-Amplifiers_13275/

Note in general these are calculations for N identical stages. In order to get your
specific design best way is to use LaPlace and do your own calculation. Signal
flow graph fastest way to get to T(s) by the way, otherwise calculation pretty
tedious.


Regards, Dana.
 

MrAl

Joined Jun 17, 2014
7,762
There are calculations here using pole shrinkage equation, eg, compute
Fc for 1 pole, same for 2 pole, and apply the equation for final cutoff.

https://www.brainkart.com/article/Frequency-Response-of-Multistage-Amplifiers_13275/

Note in general these are calculations for N identical stages. In order to get your
specific design best way is to use LaPlace and do your own calculation. Signal
flow graph fastest way to get to T(s) by the way, otherwise calculation pretty
tedious.


Regards, Dana.

Hi,

Oh i wasnt trying to imply that i could not calculate the cutoff frequency myself, just that the website itself did not have that calculation anywhere. It had the 'Bode' plot but note how hard it would be to read the cutoff frequency from that graph.
What was strange i though was that they calculated everything else directly but that :)

Yeah the transfer function isnt that hard to get using Laplace Transforms, and the cutoff isnt that hard to get after that either just a little more work that's all.

Could it be that the web author thought it was too hard to calculate?
 

MrAl

Joined Jun 17, 2014
7,762
Hello again,

I think this is the only solution to the cut off frequency:

With resistors R3 R1 R2 in that order in the schematic and
C1 non inverting input to ground,
C2 feedback cap,
C3 first cap R3 to ground.

Two examples follow this rather huge formula.

w0^2=((sqrt(-(((C1*C3^8+(4*C1*C2+4*C1^2)*C3^7+(6*C1*C2^2+16*C1^2*C2+6*C1^3)*C3^6+(4*C1*C2^3+24*C1^2*C2^2+24*C1^3*C2+4*C1^4)*C3^5+(C1*C2^4+16*C1^2*C2^3+36*C1^3*C2^2+16*C1^4*C2+C1^5)*C3^4+(4*C1^2*C2^4+24*C1^3*C2^3+24*C1^4*C2^2+4*C1^5*C2)*C3^3+(6*C1^3*C2^4+16*C1^4*C2^3+6*C1^5*C2^2)*C3^2+(4*C1^4*C2^4+4*C1^5*C2^3)*C3+C1^5*C2^4)*R2^4+((4*C1-4*C2)*C3^8+(-8*C2^2-8*C1*C2+16*C1^2)*C3^7+(-4*C2^3-28*C1*C2^2+8*C1^2*C2+24*C1^3)*C3^6+(-16*C1*C2^3-32*C1^2*C2^2+32*C1^3*C2+16*C1^4)*C3^5+(-24*C1^2*C2^3-8*C1^3*C2^2+28*C1^4*C2+4*C1^5)*C3^4+(-16*C1^3*C2^3+8*C1^4*C2^2+8*C1^5*C2)*C3^3+(4*C1^5*C2^2-4*C1^4*C2^3)*C3^2)*R1*R2^3+((6*C1-8*C2)*C3^8+(-8*C2^2-28*C1*C2+24*C1^2)*C3^7+(-34*C1*C2^2-32*C1^2*C2+36*C1^3)*C3^6+(-56*C1^2*C2^2-8*C1^3*C2+24*C1^4)*C3^5+(-44*C1^3*C2^2+8*C1^4*C2+6*C1^5)*C3^4+(4*C1^5*C2-16*C1^4*C2^2)*C3^3-2*C1^5*C2^2*C3^2)*R1^2*R2^2+((4*C1-4*C2)*C3^8+(16*C1^2-16*C1*C2)*C3^7+(24*C1^3-24*C1^2*C2)*C3^6+(16*C1^4-16*C1^3*C2)*C3^5+(4*C1^5-4*C1^4*C2)*C3^4)*R1^3*R2+(C1*C3^8+4*C1^2*C3^7+6*C1^3*C3^6+4*C1^4*C3^5+C1^5*C3^4)*R1^4)*R3^8+(((4*C1^3-4*C1^2*C2)*C3^6+(-16*C1^2*C2^2+8*C1^3*C2+8*C1^4)*C3^5+(-24*C1^2*C2^3-8*C1^3*C2^2+28*C1^4*C2+4*C1^5)*C3^4+(-16*C1^2*C2^4-32*C1^3*C2^3+32*C1^4*C2^2+16*C1^5*C2)*C3^3+(-4*C1^2*C2^5-28*C1^3*C2^4+8*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(-8*C1^3*C2^5-8*C1^4*C2^4+16*C1^5*C2^3)*C3-4*C1^4*C2^5+4*C1^5*C2^4)*R2^5+((20*C1*C2^2-32*C1^2*C2+20*C1^3)*C3^6+(40*C1*C2^3-16*C1^2*C2^2-16*C1^3*C2+40*C1^4)*C3^5+(20*C1*C2^4+64*C1^2*C2^3-48*C1^3*C2^2+64*C1^4*C2+20*C1^5)*C3^4+(48*C1^2*C2^4+32*C1^3*C2^3+32*C1^4*C2^2+48*C1^5*C2)*C3^3+(44*C1^3*C2^4+32*C1^4*C2^3+44*C1^5*C2^2)*C3^2+(24*C1^4*C2^4+24*C1^5*C2^3)*C3+8*C1^5*C2^4)*R1*R2^4+((32*C1*C2^2-72*C1^2*C2+40*C1^3)*C3^6+(32*C1*C2^3-16*C1^2*C2^2-96*C1^3*C2+80*C1^4)*C3^5+(56*C1^2*C2^3-120*C1^3*C2^2+24*C1^4*C2+40*C1^5)*C3^4+(16*C1^3*C2^3-64*C1^4*C2^2+48*C1^5*C2)*C3^3+(8*C1^5*C2^2-8*C1^4*C2^3)*C3^2)*R1^2*R2^3+((12*C1*C2^2-64*C1^2*C2+40*C1^3)*C3^6+(-16*C1^2*C2^2-112*C1^3*C2+80*C1^4)*C3^5+(-80*C1^3*C2^2-32*C1^4*C2+40*C1^5)*C3^4+(16*C1^5*C2-64*C1^4*C2^2)*C3^3-12*C1^5*C2^2*C3^2)*R1^3*R2^2+((20*C1^3-20*C1^2*C2)*C3^6+(40*C1^4-40*C1^3*C2)*C3^5+(20*C1^5-20*C1^4*C2)*C3^4)*R1^4*R2+(4*C1^3*C3^6+8*C1^4*C3^5+4*C1^5*C3^4)*R1^5)*R3^7+((6*C1^3*C3^6+(32*C1^3*C2+4*C1^4)*C3^5+(76*C1^3*C2^2+8*C1^4*C2+6*C1^5)*C3^4+(104*C1^3*C2^3-8*C1^4*C2^2+24*C1^5*C2)*C3^3+(86*C1^3*C2^4-32*C1^4*C2^3+36*C1^5*C2^2)*C3^2+(40*C1^3*C2^5-28*C1^4*C2^4+24*C1^5*C2^3)*C3+8*C1^3*C2^6-8*C1^4*C2^5+6*C1^5*C2^4)*R2^6+((36*C1^3-36*C1^2*C2)*C3^6+(-128*C1^2*C2^2+104*C1^3*C2+24*C1^4)*C3^5+(-208*C1^2*C2^3+176*C1^3*C2^2-4*C1^4*C2+36*C1^5)*C3^4+(-176*C1^2*C2^4+144*C1^3*C2^3-64*C1^4*C2^2+96*C1^5*C2)*C3^3+(-60*C1^2*C2^5-4*C1^3*C2^4-48*C1^4*C2^3+112*C1^5*C2^2)*C3^2+(-40*C1^3*C2^5-40*C1^4*C2^4+80*C1^5*C2^3)*C3-28*C1^4*C2^5+28*C1^5*C2^4)*R1*R2^5+((44*C1*C2^2-144*C1^2*C2+90*C1^3)*C3^6+(96*C1*C2^3-284*C1^2*C2^2+96*C1^3*C2+60*C1^4)*C3^5+(52*C1*C2^4-152*C1^2*C2^3+106*C1^3*C2^2-96*C1^4*C2+90*C1^5)*C3^4+(-12*C1^2*C2^4+188*C1^3*C2^3-200*C1^4*C2^2+144*C1^5*C2)*C3^3+(88*C1^3*C2^4+16*C1^4*C2^3+86*C1^5*C2^2)*C3^2+(60*C1^4*C2^4+60*C1^5*C2^3)*C3+28*C1^5*C2^4)*R1^2*R2^4+((8*C2^3+88*C1*C2^2-216*C1^2*C2+120*C1^3)*C3^6+(120*C1*C2^3-184*C1^2*C2^2-16*C1^3*C2+80*C1^4)*C3^5+(76*C1^2*C2^3-12*C1^3*C2^2-184*C1^4*C2+120*C1^5)*C3^4+(144*C1^3*C2^3-240*C1^4*C2^2+96*C1^5*C2)*C3^3+(20*C1^4*C2^3-20*C1^5*C2^2)*C3^2)*R1^3*R2^3+((44*C1*C2^2-144*C1^2*C2+90*C1^3)*C3^6+(-28*C1^2*C2^2-64*C1^3*C2+60*C1^4)*C3^5+(-18*C1^3*C2^2-136*C1^4*C2+90*C1^5)*C3^4+(24*C1^5*C2-96*C1^4*C2^2)*C3^3-30*C1^5*C2^2*C3^2)*R1^4*R2^2+((36*C1^3-36*C1^2*C2)*C3^6+(24*C1^4-24*C1^3*C2)*C3^5+(36*C1^5-36*C1^4*C2)*C3^4)*R1^5*R2+(6*C1^3*C3^6+4*C1^4*C3^5+6*C1^5*C3^4)*R1^6)*R3^6+(((4*C1^5-4*C1^4*C2)*C3^4+(16*C1^5*C2-16*C1^4*C2^2)*C3^3+(24*C1^5*C2^2-24*C1^4*C2^3)*C3^2+(16*C1^5*C2^3-16*C1^4*C2^4)*C3-4*C1^4*C2^5+4*C1^5*C2^4)*R2^7+((56*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(176*C1^3*C2^3-112*C1^4*C2^2+80*C1^5*C2)*C3^3+(232*C1^3*C2^4-128*C1^4*C2^3+112*C1^5*C2^2)*C3^2+(160*C1^3*C2^5-112*C1^4*C2^4+96*C1^5*C2^3)*C3+48*C1^3*C2^6-48*C1^4*C2^5+36*C1^5*C2^4)*R1*R2^6+((-100*C1^2*C2^3+196*C1^3*C2^2-180*C1^4*C2+84*C1^5)*C3^4+(-232*C1^2*C2^4+376*C1^3*C2^3-304*C1^4*C2^2+160*C1^5*C2)*C3^3+(-132*C1^2*C2^5+100*C1^3*C2^4-120*C1^4*C2^3+152*C1^5*C2^2)*C3^2+(-80*C1^3*C2^5-80*C1^4*C2^4+160*C1^5*C2^3)*C3-84*C1^4*C2^5+84*C1^5*C2^4)*R1^2*R2^5+((-24*C1*C2^4-80*C1^2*C2^3+252*C1^3*C2^2-320*C1^4*C2+140*C1^5)*C3^4+(-152*C1^2*C2^4+376*C1^3*C2^3-400*C1^4*C2^2+160*C1^5*C2)*C3^3+(32*C1^3*C2^4+64*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(80*C1^4*C2^4+80*C1^5*C2^3)*C3+56*C1^5*C2^4)*R1^3*R2^4+((20*C1^2*C2^3+140*C1^3*C2^2-300*C1^4*C2+140*C1^5)*C3^4+(176*C1^3*C2^3-256*C1^4*C2^2+80*C1^5*C2)*C3^3+(80*C1^4*C2^3-80*C1^5*C2^2)*C3^2)*R1^4*R2^3+((28*C1^3*C2^2-144*C1^4*C2+84*C1^5)*C3^4+(16*C1^5*C2-64*C1^4*C2^2)*C3^3-40*C1^5*C2^2*C3^2)*R1^5*R2^2+(28*C1^5-28*C1^4*C2)*C3^4*R1^6*R2+4*C1^5*C3^4*R1^7)*R3^5+((C1^5*C3^4+4*C1^5*C2*C3^3+6*C1^5*C2^2*C3^2+4*C1^5*C2^3*C3+C1^5*C2^4)*R2^8+((8*C1^5-8*C1^4*C2)*C3^4+(24*C1^5*C2-24*C1^4*C2^2)*C3^3+(44*C1^5*C2^2-44*C1^4*C2^3)*C3^2+(48*C1^5*C2^3-48*C1^4*C2^4)*C3-20*C1^4*C2^5+20*C1^5*C2^4)*R1*R2^7+((10*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(68*C1^3*C2^3-112*C1^4*C2^2+60*C1^5*C2)*C3^3+(178*C1^3*C2^4-112*C1^4*C2^3+86*C1^5*C2^2)*C3^2+(240*C1^3*C2^5-168*C1^4*C2^4+144*C1^5*C2^3)*C3+120*C1^3*C2^6-120*C1^4*C2^5+90*C1^5*C2^4)*R1^2*R2^6+((24*C1^2*C2^3+40*C1^3*C2^2-120*C1^4*C2+56*C1^5)*C3^4+(-72*C1^2*C2^4+200*C1^3*C2^3-208*C1^4*C2^2+80*C1^5*C2)*C3^3+(-76*C1^2*C2^5+44*C1^3*C2^4+8*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(-80*C1^3*C2^5-80*C1^4*C2^4+160*C1^5*C2^3)*C3-140*C1^4*C2^5+140*C1^5*C2^4)*R1^3*R2^5+((-84*C1*C2^4+48*C1^2*C2^3+60*C1^3*C2^2-160*C1^4*C2+70*C1^5)*C3^4+(-96*C1^2*C2^4+196*C1^3*C2^3-192*C1^4*C2^2+60*C1^5*C2)*C3^3+(-82*C1^3*C2^4+176*C1^4*C2^3-94*C1^5*C2^2)*C3^2+(60*C1^4*C2^4+60*C1^5*C2^3)*C3+70*C1^5*C2^4)*R1^4*R2^4+((24*C1^2*C2^3+40*C1^3*C2^2-120*C1^4*C2+56*C1^5)*C3^4+(64*C1^3*C2^3-88*C1^4*C2^2+24*C1^5*C2)*C3^3+(100*C1^4*C2^3-100*C1^5*C2^2)*C3^2)*R1^5*R2^3+((10*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(4*C1^5*C2-16*C1^4*C2^2)*C3^3-30*C1^5*C2^2*C3^2)*R1^6*R2^2+(8*C1^5-8*C1^4*C2)*C3^4*R1^7*R2+C1^5*C3^4*R1^8)*R3^4+((4*C1^5*C2^2*C3^2+8*C1^5*C2^3*C3+4*C1^5*C2^4)*R1*R2^8+((8*C1^5*C2^2-8*C1^4*C2^3)*C3^2+(48*C1^5*C2^3-48*C1^4*C2^4)*C3-40*C1^4*C2^5+40*C1^5*C2^4)*R1^2*R2^7+((4*C1^3*C2^4+32*C1^4*C2^3-20*C1^5*C2^2)*C3^2+(160*C1^3*C2^5-112*C1^4*C2^4+96*C1^5*C2^3)*C3+160*C1^3*C2^6-160*C1^4*C2^5+120*C1^5*C2^4)*R1^3*R2^6+((24*C1^2*C2^5-88*C1^3*C2^4+144*C1^4*C2^3-80*C1^5*C2^2)*C3^2+(-40*C1^3*C2^5-40*C1^4*C2^4+80*C1^5*C2^3)*C3-140*C1^4*C2^5+140*C1^5*C2^4)*R1^4*R2^5+((-92*C1^3*C2^4+160*C1^4*C2^3-100*C1^5*C2^2)*C3^2+(24*C1^4*C2^4+24*C1^5*C2^3)*C3+56*C1^5*C2^4)*R1^5*R2^4+(56*C1^4*C2^3-56*C1^5*C2^2)*C3^2*R1^6*R2^3-12*C1^5*C2^2*C3^2*R1^7*R2^2)*R3^3+((-2*C1^5*C2^2*C3^2+4*C1^5*C2^3*C3+6*C1^5*C2^4)*R1^2*R2^8+((12*C1^4*C2^3-12*C1^5*C2^2)*C3^2+(16*C1^5*C2^3-16*C1^4*C2^4)*C3-40*C1^4*C2^5+40*C1^5*C2^4)*R1^3*R2^7+((-28*C1^3*C2^4+48*C1^4*C2^3-30*C1^5*C2^2)*C3^2+(40*C1^3*C2^5-28*C1^4*C2^4+24*C1^5*C2^3)*C3+120*C1^3*C2^6-120*C1^4*C2^5+90*C1^5*C2^4)*R1^4*R2^6+((24*C1^2*C2^5-56*C1^3*C2^4+72*C1^4*C2^3-40*C1^5*C2^2)*C3^2+(-8*C1^3*C2^5-8*C1^4*C2^4+16*C1^5*C2^3)*C3-84*C1^4*C2^5+84*C1^5*C2^4)*R1^5*R2^5+((-28*C1^3*C2^4+48*C1^4*C2^3-30*C1^5*C2^2)*C3^2+(4*C1^4*C2^4+4*C1^5*C2^3)*C3+28*C1^5*C2^4)*R1^6*R2^4+(12*C1^4*C2^3-12*C1^5*C2^2)*C3^2*R1^7*R2^3-2*C1^5*C2^2*C3^2*R1^8*R2^2)*R3^2+(4*C1^5*C2^4*R1^3*R2^8+(20*C1^5*C2^4-20*C1^4*C2^5)*R1^4*R2^7+(48*C1^3*C2^6-48*C1^4*C2^5+36*C1^5*C2^4)*R1^5*R2^6+(28*C1^5*C2^4-28*C1^4*C2^5)*R1^6*R2^5+8*C1^5*C2^4*R1^7*R2^4)*R3+C1^5*C2^4*R1^4*R2^8+(4*C1^5*C2^4-4*C1^4*C2^5)*R1^5*R2^7+(8*C1^3*C2^6-8*C1^4*C2^5+6*C1^5*C2^4)*R1^6*R2^6+(4*C1^5*C2^4-4*C1^4*C2^5)*R1^7*R2^5+C1^5*C2^4*R1^8*R2^4)/(C1)))/(2*3^(3/2)*C1^2*C2^4*C3^4*R1^4*R2^4*R3^4)-(C1^3*(C2^2*C3^4*(R1*(72*R2^5*R3^6+12*R2^6*R3^5)+R1^2*(51*R2^4*R3^6+30*R2^5*R3^5-3*R2^6*R3^4)+R1^3*(6*R2^3*R3^6+18*R2^4*R3^5-12*R2^5*R3^4)+R1^4*(-3*R2^2*R3^6-6*R2^3*R3^5-18*R2^4*R3^4)+30*R2^6*R3^6+R1^5*(-6*R2^2*R3^5-12*R2^3*R3^4)-3*R1^6*R2^2*R3^4)+C2^3*C3^3*(R1*(48*R2^5*R3^6+48*R2^6*R3^5)+R1^2*(6*R2^4*R3^6+60*R2^5*R3^5+6*R2^6*R3^4)+40*R2^6*R3^6+R1^3*(12*R2^4*R3^5+12*R2^5*R3^4)+6*R1^4*R2^4*R3^4)+C2^4*C3^2*(R1*(12*R2^5*R3^6+72*R2^6*R3^5)+R1^2*(-3*R2^4*R3^6+30*R2^5*R3^5+51*R2^6*R3^4)+30*R2^6*R3^6+R1^3*(-12*R2^4*R3^5+18*R2^5*R3^4+6*R2^6*R3^3)+R1^4*(-18*R2^4*R3^4-6*R2^5*R3^3-3*R2^6*R3^2)+R1^5*(-12*R2^4*R3^3-6*R2^5*R3^2)-3*R1^6*R2^4*R3^2)+C2*C3^5*(12*R2^6*R3^6+48*R1*R2^5*R3^6+72*R1^2*R2^4*R3^6+48*R1^3*R2^3*R3^6+12*R1^4*R2^2*R3^6)+C2^5*C3*(12*R2^6*R3^6+48*R1*R2^6*R3^5+72*R1^2*R2^6*R3^4+48*R1^3*R2^6*R3^3+12*R1^4*R2^6*R3^2)+C3^6*(2*R2^6*R3^6+12*R1*R2^5*R3^6+30*R1^2*R2^4*R3^6+40*R1^3*R2^3*R3^6+30*R1^4*R2^2*R3^6+12*R1^5*R2*R3^6+2*R1^6*R3^6)+C2^6*(2*R2^6*R3^6+12*R1*R2^6*R3^5+30*R1^2*R2^6*R3^4+40*R1^3*R2^6*R3^3+30*R1^4*R2^6*R3^2+12*R1^5*R2^6*R3+2*R1^6*R2^6))+C1^2*(C2^4*C3^3*(R1^2*(-18*R2^4*R3^6-60*R2^5*R3^5)-48*R1*R2^5*R3^6+R1^3*(-36*R2^4*R3^5-12*R2^5*R3^4)-18*R1^4*R2^4*R3^4)+C2^3*C3^4*(R1^2*(-84*R2^4*R3^6-30*R2^5*R3^5)+R1^3*(-6*R2^3*R3^6-24*R2^4*R3^5+12*R2^5*R3^4)-72*R1*R2^5*R3^6+R1^4*(6*R2^3*R3^5+24*R2^4*R3^4)+12*R1^5*R2^3*R3^4)+C2*C3^6*(-12*R1*R2^5*R3^6-48*R1^2*R2^4*R3^6-72*R1^3*R2^3*R3^6-48*R1^4*R2^2*R3^6-12*R1^5*R2*R3^6)+C2^5*C3^2*(-12*R1*R2^5*R3^6-30*R1^2*R2^5*R3^5-18*R1^3*R2^5*R3^4+6*R1^4*R2^5*R3^3+6*R1^5*R2^5*R3^2)+C2^2*C3^5*(-48*R1*R2^5*R3^6-114*R1^2*R2^4*R3^6-84*R1^3*R2^3*R3^6-18*R1^4*R2^2*R3^6))+C1*(C2^3*C3^5*(30*R1^2*R2^4*R3^6+36*R1^3*R2^3*R3^6)+C2^2*C3^6*(15*R1^2*R2^4*R3^6+30*R1^3*R2^3*R3^6+15*R1^4*R2^2*R3^6)+C2^4*C3^4*(15*R1^2*R2^4*R3^6-6*R1^3*R2^4*R3^5-48*R1^4*R2^4*R3^4))+2*C2^3*C3^6*R1^3*R2^3*R3^6)/(54*C1^3*C2^6*C3^6*R1^6*R2^6*R3^6))^(1/3)+(C1^2*(C2^2*C3^2*(R1*(4*R2^3*R3^4+4*R2^4*R3^3)+R1^2*(-R2^2*R3^4+2*R2^3*R3^3-R2^4*R3^2)+6*R2^4*R3^4+R1^3*(-2*R2^2*R3^3-2*R2^3*R3^2)-R1^4*R2^2*R3^2)+C2*C3^3*(4*R2^4*R3^4+8*R1*R2^3*R3^4+4*R1^2*R2^2*R3^4)+C2^3*C3*(4*R2^4*R3^4+8*R1*R2^4*R3^3+4*R1^2*R2^4*R3^2)+C3^4*(R2^4*R3^4+4*R1*R2^3*R3^4+6*R1^2*R2^2*R3^4+4*R1^3*R2*R3^4+R1^4*R3^4)+C2^4*(R2^4*R3^4+4*R1*R2^4*R3^3+6*R1^2*R2^4*R3^2+4*R1^3*R2^4*R3+R1^4*R2^4))+C1*(C2*C3^4*(-4*R1*R2^3*R3^4-8*R1^2*R2^2*R3^4-4*R1^3*R2*R3^4)+C2^3*C3^2*(-4*R1*R2^3*R3^4-2*R1^2*R2^3*R3^3+2*R1^3*R2^3*R3^2)+C2^2*C3^3*(-8*R1*R2^3*R3^4-6*R1^2*R2^2*R3^4))+C2^2*C3^4*R1^2*R2^2*R3^4)/(9*C1^2*C2^4*C3^4*R1^4*R2^4*R3^4*((sqrt(-(((C1*C3^8+(4*C1*C2+4*C1^2)*C3^7+(6*C1*C2^2+16*C1^2*C2+6*C1^3)*C3^6+(4*C1*C2^3+24*C1^2*C2^2+24*C1^3*C2+4*C1^4)*C3^5+(C1*C2^4+16*C1^2*C2^3+36*C1^3*C2^2+16*C1^4*C2+C1^5)*C3^4+(4*C1^2*C2^4+24*C1^3*C2^3+24*C1^4*C2^2+4*C1^5*C2)*C3^3+(6*C1^3*C2^4+16*C1^4*C2^3+6*C1^5*C2^2)*C3^2+(4*C1^4*C2^4+4*C1^5*C2^3)*C3+C1^5*C2^4)*R2^4+((4*C1-4*C2)*C3^8+(-8*C2^2-8*C1*C2+16*C1^2)*C3^7+(-4*C2^3-28*C1*C2^2+8*C1^2*C2+24*C1^3)*C3^6+(-16*C1*C2^3-32*C1^2*C2^2+32*C1^3*C2+16*C1^4)*C3^5+(-24*C1^2*C2^3-8*C1^3*C2^2+28*C1^4*C2+4*C1^5)*C3^4+(-16*C1^3*C2^3+8*C1^4*C2^2+8*C1^5*C2)*C3^3+(4*C1^5*C2^2-4*C1^4*C2^3)*C3^2)*R1*R2^3+((6*C1-8*C2)*C3^8+(-8*C2^2-28*C1*C2+24*C1^2)*C3^7+(-34*C1*C2^2-32*C1^2*C2+36*C1^3)*C3^6+(-56*C1^2*C2^2-8*C1^3*C2+24*C1^4)*C3^5+(-44*C1^3*C2^2+8*C1^4*C2+6*C1^5)*C3^4+(4*C1^5*C2-16*C1^4*C2^2)*C3^3-2*C1^5*C2^2*C3^2)*R1^2*R2^2+((4*C1-4*C2)*C3^8+(16*C1^2-16*C1*C2)*C3^7+(24*C1^3-24*C1^2*C2)*C3^6+(16*C1^4-16*C1^3*C2)*C3^5+(4*C1^5-4*C1^4*C2)*C3^4)*R1^3*R2+(C1*C3^8+4*C1^2*C3^7+6*C1^3*C3^6+4*C1^4*C3^5+C1^5*C3^4)*R1^4)*R3^8+(((4*C1^3-4*C1^2*C2)*C3^6+(-16*C1^2*C2^2+8*C1^3*C2+8*C1^4)*C3^5+(-24*C1^2*C2^3-8*C1^3*C2^2+28*C1^4*C2+4*C1^5)*C3^4+(-16*C1^2*C2^4-32*C1^3*C2^3+32*C1^4*C2^2+16*C1^5*C2)*C3^3+(-4*C1^2*C2^5-28*C1^3*C2^4+8*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(-8*C1^3*C2^5-8*C1^4*C2^4+16*C1^5*C2^3)*C3-4*C1^4*C2^5+4*C1^5*C2^4)*R2^5+((20*C1*C2^2-32*C1^2*C2+20*C1^3)*C3^6+(40*C1*C2^3-16*C1^2*C2^2-16*C1^3*C2+40*C1^4)*C3^5+(20*C1*C2^4+64*C1^2*C2^3-48*C1^3*C2^2+64*C1^4*C2+20*C1^5)*C3^4+(48*C1^2*C2^4+32*C1^3*C2^3+32*C1^4*C2^2+48*C1^5*C2)*C3^3+(44*C1^3*C2^4+32*C1^4*C2^3+44*C1^5*C2^2)*C3^2+(24*C1^4*C2^4+24*C1^5*C2^3)*C3+8*C1^5*C2^4)*R1*R2^4+((32*C1*C2^2-72*C1^2*C2+40*C1^3)*C3^6+(32*C1*C2^3-16*C1^2*C2^2-96*C1^3*C2+80*C1^4)*C3^5+(56*C1^2*C2^3-120*C1^3*C2^2+24*C1^4*C2+40*C1^5)*C3^4+(16*C1^3*C2^3-64*C1^4*C2^2+48*C1^5*C2)*C3^3+(8*C1^5*C2^2-8*C1^4*C2^3)*C3^2)*R1^2*R2^3+((12*C1*C2^2-64*C1^2*C2+40*C1^3)*C3^6+(-16*C1^2*C2^2-112*C1^3*C2+80*C1^4)*C3^5+(-80*C1^3*C2^2-32*C1^4*C2+40*C1^5)*C3^4+(16*C1^5*C2-64*C1^4*C2^2)*C3^3-12*C1^5*C2^2*C3^2)*R1^3*R2^2+((20*C1^3-20*C1^2*C2)*C3^6+(40*C1^4-40*C1^3*C2)*C3^5+(20*C1^5-20*C1^4*C2)*C3^4)*R1^4*R2+(4*C1^3*C3^6+8*C1^4*C3^5+4*C1^5*C3^4)*R1^5)*R3^7+((6*C1^3*C3^6+(32*C1^3*C2+4*C1^4)*C3^5+(76*C1^3*C2^2+8*C1^4*C2+6*C1^5)*C3^4+(104*C1^3*C2^3-8*C1^4*C2^2+24*C1^5*C2)*C3^3+(86*C1^3*C2^4-32*C1^4*C2^3+36*C1^5*C2^2)*C3^2+(40*C1^3*C2^5-28*C1^4*C2^4+24*C1^5*C2^3)*C3+8*C1^3*C2^6-8*C1^4*C2^5+6*C1^5*C2^4)*R2^6+((36*C1^3-36*C1^2*C2)*C3^6+(-128*C1^2*C2^2+104*C1^3*C2+24*C1^4)*C3^5+(-208*C1^2*C2^3+176*C1^3*C2^2-4*C1^4*C2+36*C1^5)*C3^4+(-176*C1^2*C2^4+144*C1^3*C2^3-64*C1^4*C2^2+96*C1^5*C2)*C3^3+(-60*C1^2*C2^5-4*C1^3*C2^4-48*C1^4*C2^3+112*C1^5*C2^2)*C3^2+(-40*C1^3*C2^5-40*C1^4*C2^4+80*C1^5*C2^3)*C3-28*C1^4*C2^5+28*C1^5*C2^4)*R1*R2^5+((44*C1*C2^2-144*C1^2*C2+90*C1^3)*C3^6+(96*C1*C2^3-284*C1^2*C2^2+96*C1^3*C2+60*C1^4)*C3^5+(52*C1*C2^4-152*C1^2*C2^3+106*C1^3*C2^2-96*C1^4*C2+90*C1^5)*C3^4+(-12*C1^2*C2^4+188*C1^3*C2^3-200*C1^4*C2^2+144*C1^5*C2)*C3^3+(88*C1^3*C2^4+16*C1^4*C2^3+86*C1^5*C2^2)*C3^2+(60*C1^4*C2^4+60*C1^5*C2^3)*C3+28*C1^5*C2^4)*R1^2*R2^4+((8*C2^3+88*C1*C2^2-216*C1^2*C2+120*C1^3)*C3^6+(120*C1*C2^3-184*C1^2*C2^2-16*C1^3*C2+80*C1^4)*C3^5+(76*C1^2*C2^3-12*C1^3*C2^2-184*C1^4*C2+120*C1^5)*C3^4+(144*C1^3*C2^3-240*C1^4*C2^2+96*C1^5*C2)*C3^3+(20*C1^4*C2^3-20*C1^5*C2^2)*C3^2)*R1^3*R2^3+((44*C1*C2^2-144*C1^2*C2+90*C1^3)*C3^6+(-28*C1^2*C2^2-64*C1^3*C2+60*C1^4)*C3^5+(-18*C1^3*C2^2-136*C1^4*C2+90*C1^5)*C3^4+(24*C1^5*C2-96*C1^4*C2^2)*C3^3-30*C1^5*C2^2*C3^2)*R1^4*R2^2+((36*C1^3-36*C1^2*C2)*C3^6+(24*C1^4-24*C1^3*C2)*C3^5+(36*C1^5-36*C1^4*C2)*C3^4)*R1^5*R2+(6*C1^3*C3^6+4*C1^4*C3^5+6*C1^5*C3^4)*R1^6)*R3^6+(((4*C1^5-4*C1^4*C2)*C3^4+(16*C1^5*C2-16*C1^4*C2^2)*C3^3+(24*C1^5*C2^2-24*C1^4*C2^3)*C3^2+(16*C1^5*C2^3-16*C1^4*C2^4)*C3-4*C1^4*C2^5+4*C1^5*C2^4)*R2^7+((56*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(176*C1^3*C2^3-112*C1^4*C2^2+80*C1^5*C2)*C3^3+(232*C1^3*C2^4-128*C1^4*C2^3+112*C1^5*C2^2)*C3^2+(160*C1^3*C2^5-112*C1^4*C2^4+96*C1^5*C2^3)*C3+48*C1^3*C2^6-48*C1^4*C2^5+36*C1^5*C2^4)*R1*R2^6+((-100*C1^2*C2^3+196*C1^3*C2^2-180*C1^4*C2+84*C1^5)*C3^4+(-232*C1^2*C2^4+376*C1^3*C2^3-304*C1^4*C2^2+160*C1^5*C2)*C3^3+(-132*C1^2*C2^5+100*C1^3*C2^4-120*C1^4*C2^3+152*C1^5*C2^2)*C3^2+(-80*C1^3*C2^5-80*C1^4*C2^4+160*C1^5*C2^3)*C3-84*C1^4*C2^5+84*C1^5*C2^4)*R1^2*R2^5+((-24*C1*C2^4-80*C1^2*C2^3+252*C1^3*C2^2-320*C1^4*C2+140*C1^5)*C3^4+(-152*C1^2*C2^4+376*C1^3*C2^3-400*C1^4*C2^2+160*C1^5*C2)*C3^3+(32*C1^3*C2^4+64*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(80*C1^4*C2^4+80*C1^5*C2^3)*C3+56*C1^5*C2^4)*R1^3*R2^4+((20*C1^2*C2^3+140*C1^3*C2^2-300*C1^4*C2+140*C1^5)*C3^4+(176*C1^3*C2^3-256*C1^4*C2^2+80*C1^5*C2)*C3^3+(80*C1^4*C2^3-80*C1^5*C2^2)*C3^2)*R1^4*R2^3+((28*C1^3*C2^2-144*C1^4*C2+84*C1^5)*C3^4+(16*C1^5*C2-64*C1^4*C2^2)*C3^3-40*C1^5*C2^2*C3^2)*R1^5*R2^2+(28*C1^5-28*C1^4*C2)*C3^4*R1^6*R2+4*C1^5*C3^4*R1^7)*R3^5+((C1^5*C3^4+4*C1^5*C2*C3^3+6*C1^5*C2^2*C3^2+4*C1^5*C2^3*C3+C1^5*C2^4)*R2^8+((8*C1^5-8*C1^4*C2)*C3^4+(24*C1^5*C2-24*C1^4*C2^2)*C3^3+(44*C1^5*C2^2-44*C1^4*C2^3)*C3^2+(48*C1^5*C2^3-48*C1^4*C2^4)*C3-20*C1^4*C2^5+20*C1^5*C2^4)*R1*R2^7+((10*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(68*C1^3*C2^3-112*C1^4*C2^2+60*C1^5*C2)*C3^3+(178*C1^3*C2^4-112*C1^4*C2^3+86*C1^5*C2^2)*C3^2+(240*C1^3*C2^5-168*C1^4*C2^4+144*C1^5*C2^3)*C3+120*C1^3*C2^6-120*C1^4*C2^5+90*C1^5*C2^4)*R1^2*R2^6+((24*C1^2*C2^3+40*C1^3*C2^2-120*C1^4*C2+56*C1^5)*C3^4+(-72*C1^2*C2^4+200*C1^3*C2^3-208*C1^4*C2^2+80*C1^5*C2)*C3^3+(-76*C1^2*C2^5+44*C1^3*C2^4+8*C1^4*C2^3+24*C1^5*C2^2)*C3^2+(-80*C1^3*C2^5-80*C1^4*C2^4+160*C1^5*C2^3)*C3-140*C1^4*C2^5+140*C1^5*C2^4)*R1^3*R2^5+((-84*C1*C2^4+48*C1^2*C2^3+60*C1^3*C2^2-160*C1^4*C2+70*C1^5)*C3^4+(-96*C1^2*C2^4+196*C1^3*C2^3-192*C1^4*C2^2+60*C1^5*C2)*C3^3+(-82*C1^3*C2^4+176*C1^4*C2^3-94*C1^5*C2^2)*C3^2+(60*C1^4*C2^4+60*C1^5*C2^3)*C3+70*C1^5*C2^4)*R1^4*R2^4+((24*C1^2*C2^3+40*C1^3*C2^2-120*C1^4*C2+56*C1^5)*C3^4+(64*C1^3*C2^3-88*C1^4*C2^2+24*C1^5*C2)*C3^3+(100*C1^4*C2^3-100*C1^5*C2^2)*C3^2)*R1^5*R2^3+((10*C1^3*C2^2-48*C1^4*C2+28*C1^5)*C3^4+(4*C1^5*C2-16*C1^4*C2^2)*C3^3-30*C1^5*C2^2*C3^2)*R1^6*R2^2+(8*C1^5-8*C1^4*C2)*C3^4*R1^7*R2+C1^5*C3^4*R1^8)*R3^4+((4*C1^5*C2^2*C3^2+8*C1^5*C2^3*C3+4*C1^5*C2^4)*R1*R2^8+((8*C1^5*C2^2-8*C1^4*C2^3)*C3^2+(48*C1^5*C2^3-48*C1^4*C2^4)*C3-40*C1^4*C2^5+40*C1^5*C2^4)*R1^2*R2^7+((4*C1^3*C2^4+32*C1^4*C2^3-20*C1^5*C2^2)*C3^2+(160*C1^3*C2^5-112*C1^4*C2^4+96*C1^5*C2^3)*C3+160*C1^3*C2^6-160*C1^4*C2^5+120*C1^5*C2^4)*R1^3*R2^6+((24*C1^2*C2^5-88*C1^3*C2^4+144*C1^4*C2^3-80*C1^5*C2^2)*C3^2+(-40*C1^3*C2^5-40*C1^4*C2^4+80*C1^5*C2^3)*C3-140*C1^4*C2^5+140*C1^5*C2^4)*R1^4*R2^5+((-92*C1^3*C2^4+160*C1^4*C2^3-100*C1^5*C2^2)*C3^2+(24*C1^4*C2^4+24*C1^5*C2^3)*C3+56*C1^5*C2^4)*R1^5*R2^4+(56*C1^4*C2^3-56*C1^5*C2^2)*C3^2*R1^6*R2^3-12*C1^5*C2^2*C3^2*R1^7*R2^2)*R3^3+((-2*C1^5*C2^2*C3^2+4*C1^5*C2^3*C3+6*C1^5*C2^4)*R1^2*R2^8+((12*C1^4*C2^3-12*C1^5*C2^2)*C3^2+(16*C1^5*C2^3-16*C1^4*C2^4)*C3-40*C1^4*C2^5+40*C1^5*C2^4)*R1^3*R2^7+((-28*C1^3*C2^4+48*C1^4*C2^3-30*C1^5*C2^2)*C3^2+(40*C1^3*C2^5-28*C1^4*C2^4+24*C1^5*C2^3)*C3+120*C1^3*C2^6-120*C1^4*C2^5+90*C1^5*C2^4)*R1^4*R2^6+((24*C1^2*C2^5-56*C1^3*C2^4+72*C1^4*C2^3-40*C1^5*C2^2)*C3^2+(-8*C1^3*C2^5-8*C1^4*C2^4+16*C1^5*C2^3)*C3-84*C1^4*C2^5+84*C1^5*C2^4)*R1^5*R2^5+((-28*C1^3*C2^4+48*C1^4*C2^3-30*C1^5*C2^2)*C3^2+(4*C1^4*C2^4+4*C1^5*C2^3)*C3+28*C1^5*C2^4)*R1^6*R2^4+(12*C1^4*C2^3-12*C1^5*C2^2)*C3^2*R1^7*R2^3-2*C1^5*C2^2*C3^2*R1^8*R2^2)*R3^2+(4*C1^5*C2^4*R1^3*R2^8+(20*C1^5*C2^4-20*C1^4*C2^5)*R1^4*R2^7+(48*C1^3*C2^6-48*C1^4*C2^5+36*C1^5*C2^4)*R1^5*R2^6+(28*C1^5*C2^4-28*C1^4*C2^5)*R1^6*R2^5+8*C1^5*C2^4*R1^7*R2^4)*R3+C1^5*C2^4*R1^4*R2^8+(4*C1^5*C2^4-4*C1^4*C2^5)*R1^5*R2^7+(8*C1^3*C2^6-8*C1^4*C2^5+6*C1^5*C2^4)*R1^6*R2^6+(4*C1^5*C2^4-4*C1^4*C2^5)*R1^7*R2^5+C1^5*C2^4*R1^8*R2^4)/(C1)))/(2*3^(3/2)*C1^2*C2^4*C3^4*R1^4*R2^4*R3^4)-(C1^3*(C2^2*C3^4*(R1*(72*R2^5*R3^6+12*R2^6*R3^5)+R1^2*(51*R2^4*R3^6+30*R2^5*R3^5-3*R2^6*R3^4)+R1^3*(6*R2^3*R3^6+18*R2^4*R3^5-12*R2^5*R3^4)+R1^4*(-3*R2^2*R3^6-6*R2^3*R3^5-18*R2^4*R3^4)+30*R2^6*R3^6+R1^5*(-6*R2^2*R3^5-12*R2^3*R3^4)-3*R1^6*R2^2*R3^4)+C2^3*C3^3*(R1*(48*R2^5*R3^6+48*R2^6*R3^5)+R1^2*(6*R2^4*R3^6+60*R2^5*R3^5+6*R2^6*R3^4)+40*R2^6*R3^6+R1^3*(12*R2^4*R3^5+12*R2^5*R3^4)+6*R1^4*R2^4*R3^4)+C2^4*C3^2*(R1*(12*R2^5*R3^6+72*R2^6*R3^5)+R1^2*(-3*R2^4*R3^6+30*R2^5*R3^5+51*R2^6*R3^4)+30*R2^6*R3^6+R1^3*(-12*R2^4*R3^5+18*R2^5*R3^4+6*R2^6*R3^3)+R1^4*(-18*R2^4*R3^4-6*R2^5*R3^3-3*R2^6*R3^2)+R1^5*(-12*R2^4*R3^3-6*R2^5*R3^2)-3*R1^6*R2^4*R3^2)+C2*C3^5*(12*R2^6*R3^6+48*R1*R2^5*R3^6+72*R1^2*R2^4*R3^6+48*R1^3*R2^3*R3^6+12*R1^4*R2^2*R3^6)+C2^5*C3*(12*R2^6*R3^6+48*R1*R2^6*R3^5+72*R1^2*R2^6*R3^4+48*R1^3*R2^6*R3^3+12*R1^4*R2^6*R3^2)+C3^6*(2*R2^6*R3^6+12*R1*R2^5*R3^6+30*R1^2*R2^4*R3^6+40*R1^3*R2^3*R3^6+30*R1^4*R2^2*R3^6+12*R1^5*R2*R3^6+2*R1^6*R3^6)+C2^6*(2*R2^6*R3^6+12*R1*R2^6*R3^5+30*R1^2*R2^6*R3^4+40*R1^3*R2^6*R3^3+30*R1^4*R2^6*R3^2+12*R1^5*R2^6*R3+2*R1^6*R2^6))+C1^2*(C2^4*C3^3*(R1^2*(-18*R2^4*R3^6-60*R2^5*R3^5)-48*R1*R2^5*R3^6+R1^3*(-36*R2^4*R3^5-12*R2^5*R3^4)-18*R1^4*R2^4*R3^4)+C2^3*C3^4*(R1^2*(-84*R2^4*R3^6-30*R2^5*R3^5)+R1^3*(-6*R2^3*R3^6-24*R2^4*R3^5+12*R2^5*R3^4)-72*R1*R2^5*R3^6+R1^4*(6*R2^3*R3^5+24*R2^4*R3^4)+12*R1^5*R2^3*R3^4)+C2*C3^6*(-12*R1*R2^5*R3^6-48*R1^2*R2^4*R3^6-72*R1^3*R2^3*R3^6-48*R1^4*R2^2*R3^6-12*R1^5*R2*R3^6)+C2^5*C3^2*(-12*R1*R2^5*R3^6-30*R1^2*R2^5*R3^5-18*R1^3*R2^5*R3^4+6*R1^4*R2^5*R3^3+6*R1^5*R2^5*R3^2)+C2^2*C3^5*(-48*R1*R2^5*R3^6-114*R1^2*R2^4*R3^6-84*R1^3*R2^3*R3^6-18*R1^4*R2^2*R3^6))+C1*(C2^3*C3^5*(30*R1^2*R2^4*R3^6+36*R1^3*R2^3*R3^6)+C2^2*C3^6*(15*R1^2*R2^4*R3^6+30*R1^3*R2^3*R3^6+15*R1^4*R2^2*R3^6)+C2^4*C3^4*(15*R1^2*R2^4*R3^6-6*R1^3*R2^4*R3^5-48*R1^4*R2^4*R3^4))+2*C2^3*C3^6*R1^3*R2^3*R3^6)/(54*C1^3*C2^6*C3^6*R1^6*R2^6*R3^6))^(1/3))-(C1*(C3^2*(R2^2*R3^2+2*R1*R2*R3^2+R1^2*R3^2)+C2^2*(R2^2*R3^2+2*R1*R2^2*R3+R1^2*R2^2)+2*C2*C3*R2^2*R3^2)-2*C2*C3^2*R1*R2*R3^2)/(3*C1*C2^2*C3^2*R1^2*R2^2*R3^2)

That's kind of complicated though.
If we make some components equal it gets simpler.

So with R2=R1 aned R3=R1, and C2=C1 and C3=C1 and R1=1k and C1=0.1uf we get:
f0=533.1765 Hz.

and if instead C1=0.01uf we get:
f0=5331.765 Hz

which is 10 times higher than with C1=0.1uf.

We could check this with a simulator of course.
 
Last edited:

danadak

Joined Mar 10, 2018
4,057
That can't be the solution formula.

Clearly the the T(s) denominator is 3'rd order, numerator cant be more than 3'rd order.

I am assuming you considered ideal OpAmp.

The easy way to solves this is label each element z1, z2,.....z6, solve, then plug in
Z element LaPlace equivalent and simplify.

There is a paper on 3 order salen key LPF online, but looks like you have to
pay for it.

http://sim.okawa-denshi.jp/en/detatukeisan.htm

https://en.wikipedia.org/wiki/Symbolic_circuit_analysis



Regards, Dana.
 
Last edited:

MrAl

Joined Jun 17, 2014
7,762
That can't be the solution formula.

Clearly the the T(s) denominator is 3'rd order, numerator cant be more than 3'rd order.

I am assuming you considered ideal OpAmp.

The easy way to solves this is label each element z1, z2,.....z6, solve, then plug in
Z element LaPlace equivalent and simplify.

There is a paper on 3 order salen key LPF online, but looks like you have to
pay for it.

http://sim.okawa-denshi.jp/en/detatukeisan.htm

https://en.wikipedia.org/wiki/Symbolic_circuit_analysis



Regards, Dana.

Hi,

He he, intuition does not serve us very well sometimes.
For example, did you consider that there is a square root, and the solution is for w0^2 not w0 so we have to take the square root of that again anyway. So if we had a power of say x^12 in there when we take the square root we get some power like x^6 then take the square root again and get some power like x^3.
The highest power i see in the numerator of the factored version is 8 and also in the denominator. Taking the square root twice, we get a power of 2.

There may be a simplification to that too i didnt look for, but the bottom line is that it has been checked with another formula that uses the f1,f2,f3 method of cascading filter sections. Both methods come out with the same cutoff frequency. A simulation should be easy though.

Also i just want to mention that when i said that it was the 'only' solution what i meant was that it was probably the only solution out of the three possible solutions for w0^2 and that is because the other two will probably lead to an imaginary value for w0^2 (or just w0) and it must be real.
So it's not that there are no other ways to find w0^2, it's just that when we solve it with a brute force amplitude type solution we get three solutions only one of them is real or at least not complex just completely real.

If all R's are R and all C's are C, we get this:
w0=0.33500470260371/(R*C)

so that makes it easier to test.
The initial rendition is like this:
w0=sqrt((9*sqrt(2917)*i-421*sqrt(3))^(2/3)-2^(10/3)*3^(1/6)*(9*sqrt(2917)*i-421*sqrt(3))^(1/3)+5*2^(11/3)*3^(1/3))/(2^(1/6)*3^(7/12)*(9*sqrt(2917)*i-421*sqrt(3))^(1/6)*R*C)

and the imaginary operators 'i' make it look a little strange, but that resolves into a real number as above.
And all this comes directly from that huge formula above.

There is also a trick we can use if we make one of the resistors (i think it is R2 in the original schematic) maybe 10 or more times the first resistor (i think is R1 in the original and that makes the first stage somewhat isolated for current from the actual 2nd order Sallen Key filter. This is really just a passive RC combined with a 2nd order Sallen Key anyway.
 
Last edited:

danadak

Joined Mar 10, 2018
4,057
As an aside its possible to use Microsoft word, or other equation writer, and
post as an image so that its more readable.

Just a thought.

I follow your development now. Has a pretty high ugh factor :).....


Regards, Dana.
 

MrAl

Joined Jun 17, 2014
7,762
As an aside its possible to use Microsoft word, or other equation writer, and
post as an image so that its more readable.

Just a thought.

I follow your development now. Has a pretty high ugh factor :).....


Regards, Dana.

Hi,

Caveman awake ... ugh ugh, ha ha. Reminds me of an arcade game from back in the 1980's we used to play down in Keansburg NJ.

Yeah i post sometimes in pure text because that allows someone to copy it as text and then paste it into their own math program and then use it just like a formula. So you can copy and past that, then enter the values for the components, then simplify a little and then you get the cutoff frequency. That is the combined cutoff so that's the cutoff of the whole circuit. I dont think it is easy to get that very readable in any display form, unfortunately.

There may be a simpler formula but i havent looked too hard yet.
I'll also check with a simulator later today.
 
Top