This question is in relation to this question asked before (Settling time of sixth order denominator transfer function), but formulated in greater details.

So I have a transfer function which looks as follows: $$ \frac{V_o(s)}{V_i(s)}=\frac {\text{numerator}(s)}{a_1 s^6 + a_2 s^5 + a_3 s^4 + a_4 s^3 + a_5 s^2 +a_6 s + a_7}. $$ Each coefficient contains number of variables, but there are only two variables i have the freedom to play with, the rest are fixed. For example: if \$ a6= C1\cdot Rx+C1\cdot R1+C2\cdot R2+C3\cdot Ry + C3\cdot (R1+R2)\$, then I have the freedom to play with \$ R1 \$ and \$R2 \$ and other variables are constant. The rest of the co-efficients i.e \$a1,a2,a3,a4,a5\$ are an algebraic addition and subtraction of \$ R1 \$ and \$R2 \$ and more variable i.e \$a1,a2...a6=f(R1,R2,Rx,Cx,Lx...) \$. Also \$ R1 \$ and \$R2 \$ do not appear in the numerator and hence numerator is ignored. Unfortunately it is not possible to post the circuit due to legal reasons.

The goal is to find how these two variables affect settling time.

Now I need to elaborate my question, which I found lacking in the previous post as pointed out by others. What I would like to know is which direction should i proceed from the present values of \$ R1 \$ and \$R2 \$ such that settling time is minimized. I know there is no analytical solution to the sixth order denominator, but I am not trying to find a solution, but a relation between these two variables and settling time. Numerical analysis will give me two values of \$ R1 \$ and \$R2 \$ for which minimum settling time can be obtained. But is it possible by numerical analysis to graph the approximate relation between \$ R1 \$, \$R2 \$ and settling time?

What is the purpose or use case to find the relation?
This is a complicated circuit, and based on the environment this circuit is employed in, the values of other variables, apart from \$ R1 \$ and \$R2 \$ can change slightly, and hence \$ R1 \$ and \$R2 \$ need to be tweaked in order to either minimize the settling time or to bring it back to original value, if the settling time has increased. I will not always have access to a computer, let alone numerical analysis tools. Also in controlled environment there is a need to understand the effects of the two variables on settling time.

From the post earlier, I have come to following inferences.

  1. The denominator is of the sixth order hence there are six poles. If the denominator can be factorized, assuming there are complex conjugate poles \$A+Bj\$ and \$A-Bj\$ to seperate out \$s^2 + 2As + A^2 + B^2\$ from the denominator, then there are four more poles remaining, which when expanded lead to 4th order polynomial, i.e \$den=(s+A+Bj)\cdot (s+A-Bj) \cdot(s+p3) \cdot(s+p4) \cdot(s+p5) \cdot(s+p6)\$. Now if A is very close to zero then it wont affect \$ s+p6\$ when multiplied, hence the effect of sixth order term can be isolated. The problem however is the fact that the complex poles only exist for a specific set of values of \$ R1 \$ and \$R2 \$. The moment i change the values of \$ R1 \$ and \$R2 \$, the complex roots might not exist close to zero and hence this analysis wont be valid. Although a very nice idea, courtesy @user287001. It could be helpful to to others, and I hope i understood user287001's answer correctly.

  2. Another way is to take the derivative of coefficients to settling time. This will tell me how much does the settling time change by change of one coefficient, keeping other coefficients unchanged. That is to take \$ \frac{d settling time}{d a1},\frac{d settling time}{d a2}...\$. individually. In this way one can find which co-efficient affects the most on settling time. For example if \$ a5 \$ affects the most then one can concentrate only on \$ a5 \$ leading to an approximate graphical solution which can be converted to an analytical solution by curve fitting. I can change the absolute values of each coefficient after substituting the numerical value of \$ R1 \$ and \$ R2 \$. If i find out that \$ a3 \$ affects the settling time the most, then changing values of \$ R1 \$ and \$ R2 \$ in \$ a3\$ will also change the other coefficients (\$a1,a2,a4,a5,a6\$, since they are a function of \$ R1 \$ and \$ R2 \$) which could make the system unstable. This is chicken and egg problem. Had \$ R1 \$ and \$ R2 \$ appeared in only three or less coefficients i think this solution could have been feasible.

  3. At all times system stability needs to be guaranteed. Hence finding the bounds of each coefficient keeping other coefficients unchanged will atleast put some constrains on \$ R1 \$ and \$ R2 \$. But I think it will lead to the same problem as in the second point.

I have run out of ideas, but I am certain that this is not a new problem, and there must be people out there dealing with higher order systems to find approximate analysis. Like I mentioned before, I am not after the solution of the denominator polynomial, but to use numerical analysis to get to an approximate relation between \$ R1 \$ and \$ R2 \$ and settling time. Any thoughts comments will be more than welcome.

  • \$\begingroup\$ When you say this: I will not always have access to a computer, let alone numerical analysis tools. - are you actually saying that at those times you still wish to be able to make an assessment despite not having tools available? \$\endgroup\$
    – Andy aka
    Mar 29, 2021 at 10:40
  • \$\begingroup\$ @Andyaka yes, I might have to work in the field. \$\endgroup\$
    – RAN
    Mar 29, 2021 at 10:41
  • \$\begingroup\$ Have you simulated the problem yet? In a nutshell, what are you trying to achieve? \$\endgroup\$
    – Andy aka
    Mar 29, 2021 at 10:49
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    \$\begingroup\$ Just like the other question, this, too, sounds like an XY problem: given your statement in the beginning about the a6 term, it sounds as if you have a transfer function based on several second order circuits implemented with opamps. If that's the case, each stage has an analytical Q, and all of them will contribute to the general transfer function -- which you shouldn't try to analyze as a whole in the first place. Certainly not without proper tools, unless it's a well known t.f. (e.g. Chebyshev, Pascal, etc). And if it's a compensator, you should already know the roots. \$\endgroup\$ Mar 29, 2021 at 11:35
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    \$\begingroup\$ Why don't you divide through by a1, which will leave only(!) six unknown parameters? \$\endgroup\$
    – Chu
    Mar 29, 2021 at 16:22

1 Answer 1


I'll reply given your information that you added in the comments:

it is a matching network with only resistors caps and inductors, hence it is difficult to isolate them into second order systems. No opamps in the circuit, hence no high impedance nodes

What you want is impossible without a numerical solver, or a symbolic one that can handle that polynomial. This is because a completely passive network cannot be isolated in sections due to the loading effect. Therefore the transfer function for such a circuit must be determined a priori and, once placed in the circuit, it should not be altered. Any modifications will affect the entire transfer function, because almost all the coefficients will be affected and, thus, all the roots will be displaced.

Simple 3rd order case:


The transfer function is seen in the Laplace source, above (ignore the opening '+' at the beginning of the lines). The two plots show the differences between the original transfer function and the one with only a modified Rx (0.4, instead of 0.3). The response is shown with the group delay, rather than the phase, and the impulse response, to show the time domain differences. And here are the roots, shown in full precision to show that even the single, real root is affected (a bit), despite the plot below:




Small differences but you can see effects overall. Therefore the conclusion remains: an exact solution is not possible without numerical solvers, and changing any component will affect the entire tranfer function.

If you can live with these differences then I have to wonder why bother asking so many questions about 6th order polynomials when you can simply "ogle" and change values as needed. This is why your whole question seems strange: you say you need to perform changes "on the field" and without helpers, but you are looking for deep analysis of 6th order polynomials; you say it's an impedance matching network, yet you use the , , tags. What sort of field intervention requires one changing a passive network for impedance adaptation, used in a control system with stability concerns, and without proper tools but which needs a polynomial evaluation in order to reduce the step response?

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    \$\begingroup\$ This discussion between you and Op is reminding me of this link dealing with a 12-pole filter. Read it and enjoy. Oh, and +1 of course. \$\endgroup\$
    – jonk
    Mar 29, 2021 at 19:06
  • \$\begingroup\$ Oh and why isn't the OP dividing through by \$a_7\$ to get rid of at least one constant. For \$b_2\,s^2+b_1\,s+b_0\$, one can factor this into \$b_0\left[\left(\frac{s}{\sqrt{\frac{b_0}{b_2}}}\right)^2+\frac{s}{\sqrt{\frac{b_0}{b_2}}}\cdot\frac{b_1}{\sqrt{b_0\,b_2}}+1\right]=b_0\left[\left(\frac{s}{\omega_{_0}}\right)^2+d\,\left(\frac{s}{\omega_{_0}}\right)+1\right]\$, where \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$d=\frac{b_1}{\sqrt{b_0\,b_2}}\$. I'd imagine that with some work on the OP's part, some motion in the right direction could be had in breaking this down, more. \$\endgroup\$
    – jonk
    Mar 29, 2021 at 19:15
  • \$\begingroup\$ (Though I've no idea what's the right direction for the OP, just yet.) \$\endgroup\$
    – jonk
    Mar 29, 2021 at 19:15
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    \$\begingroup\$ @jonk That's a very useful link (thank you, for both), and goes along my lines at the end: if OP can do with "approximate" tuning, then a bit of hammer time will probably get you close enough to your needs. Sure, start from the top and, bit by bit, try to tame the overall response. For the link, it won't only affect one zero, but the adjacent ones, too (a bit, or a bit more), but it will work. But then there's no need for any of the high order polynomial analysis for the specific task of minimizing the step response. Because if it's part of a control loop, the roots should already be known. \$\endgroup\$ Mar 29, 2021 at 21:59
  • \$\begingroup\$ @aconcernedcitizen thank you for the elaborate answer. This is what i did, i simply swept both the variables in numerical analysis, and found the settling time. Then plotting a 3d graph and curve fitting it to a 5th order polynomial (a sort of analytical) led me to the conclusion than one of the variables R1, does not affect the settling time "so much", reducing the equation even further. Now I have an equation in one variable, hopefully that will make my life simpler. I am trying to see if i can reduce it further. \$\endgroup\$
    – RAN
    Mar 31, 2021 at 9:12

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