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I have a system whose transfer function is as follows

$$ \frac{V_o(s)}{V_i(s)}=\frac {\text{numerator}(s)}{a_1 s^6 + a_2 s^5 + a_3 s^4 + a_4 s^3 + a_5 s^2 +a_6 s + a_7}. $$

I am interested in the settling time exhibited by the system, hence the numerator is not in focus (hope I am correct there). I do know that the sixth order polynomial has no analytical solution. Hence I used Matlab to find the numerical solution and settling time. I would like to know which coefficients of the denominator has the most influence on settling time. Any ideas on how to analyze that?

On the other hand, for a given set of values for coefficients, if I am able to figure it out the dependence of settling time on coefficients, for example \$ \text{Settling time} = f(\frac{a_1}{a_2})\$ and then if I change the values of \$a_1\$ and/or \$a_2\$, to lower the settling time then will that work? I reckon that the moment I change the coefficents then the settling time will now be dependent on other coefficients, such as, for instance \$\text{Settling time} = f(\frac{a_3}{a_4})\$, am I correct? The aim is to reduce settling time by choosing right values of two variables which appear only in the denominator of the transfer function

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    \$\begingroup\$ Assuming out/in for your function, you'd need to first define "settling time." There are usually two factors here, the phase delay and the transient decay time. You may mean a settling time defined as the delay needed for a transient wave to settle to within some specified % of steady state. But definitions matter. In any case, the numerator also matters since bandpass filters take more time to settle than low- or high- pass. Also, the difference between the applied frequency and the filter cut-off matters because the closer they are the longer the settling time. \$\endgroup\$
    – jonk
    Commented Mar 18, 2021 at 18:15
  • \$\begingroup\$ Yes, by settling team I mean the time required for the transient wave to settle within required %. Also the reason why I have omitted the numerator is because the variable that i have the freedom to play with appears only in the denominator. \$\endgroup\$
    – RAN
    Commented Mar 18, 2021 at 18:34
  • \$\begingroup\$ Is your expression already multiplied by \$V_i\left(s\right)\$ so that we have \$V_o\left(s\right)=V_i\left(s\right)H\left(s\right)\$? I was kind of thinking not, earlier. But now I'm not so sure. In other words, does it already include the \$\frac{\omega}{s^2+\omega^2}\$ for the sine wave? \$\endgroup\$
    – jonk
    Commented Mar 18, 2021 at 19:05
  • \$\begingroup\$ One more hint to suggest. Suppose you can factor your denominator into 2nd and 1st order sections. Then each 2nd order section will have a distinct Q. Assuming you can find one that does dominate, then the one with the highest Q will dominate the settling time and you can "mostly ignore" the others. \$\endgroup\$
    – jonk
    Commented Mar 19, 2021 at 5:45
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    \$\begingroup\$ Hi, no it is of the form \$V_o(s)/V_i(s)\$ assuming \$V_i(s)=1/s$. In Matlab, using 'damp' function, i can find all the poles of the system, but numerically. There are six poles, hence I can get it into the form of three second order equations. However the problem remains, it is numerical analysis, hence I will never know which of the coefficients really dominate the system. \$\endgroup\$
    – RAN
    Commented Mar 19, 2021 at 6:43

3 Answers 3

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You cannot find the rule you hope. Assuming the system is stable a long settling time is caused by a pole or conjugate pole pair which has near zero negative real part. You may separate from the denominator factor \$s^2 + 2As + A^2 + B^2\$ if you have complex pole pair \$A+Bj\$ and \$A-Bj\$. Let A be radically closer to zero than the real parts of other poles. When this is multiplied with the other (4th order) part of the denominator, the critical real part \$A\$ affects all but the sixth order term.

ADD after more info was inserted to the question:

So, you are able to use advanced software like Matlab and you already have made simulation runs. Go on and find a good combination of the 2 variables which you can set. I guess Matlab can be programmed to make the search for you. Unfortunately I'm not a programmer nor have such advanced software. Sorry.

BTW. Often people must take into the account also other things than the settling time. The general method is to build a cost function which is tried to be minimized.

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  • \$\begingroup\$ That is a good idea. Another point that I noted is i can approximate this transfer function to that of 5th order, because the coefficient of s^6 term is negligible. Now 5th order polynomial has analytical solution under certain conditions. \$\endgroup\$
    – RAN
    Commented Mar 18, 2021 at 18:37
  • \$\begingroup\$ do you have any literature on settling time analysis of higher order (greater than 4), I am not sure whether I completely understand your solution \$\endgroup\$
    – RAN
    Commented Mar 22, 2021 at 9:06
  • \$\begingroup\$ No, I haven't. My answer was an excuse actually written for myself why I should believe that searching the formula you want is useless i.e. nothing will be found. It's not a rigorous proof. Mathematician Niels Abel proved fully undeniably that no expression made of arithmetic operations, powers and roots presents the zeros of the general 5th or higher order polynomial. That means: You have no formula which connects the factors a1...a7 to the poles. That connection is a must to be able to divide the system to cascaded lower order subsystems which can be handled analytically. \$\endgroup\$
    – user136077
    Commented Mar 22, 2021 at 11:37
  • \$\begingroup\$ @RAN (continued) We have here used to see that often people hide their actual problem and ask something that they imagine to be useful in solving the hidden problem. We call them XY-question cases. Do you have an actual hidden problem? Someone can know how to help you with it. \$\endgroup\$
    – user136077
    Commented Mar 22, 2021 at 11:43
  • \$\begingroup\$ The problem is as I have mentioned. I have a system which takes too long to settle, and there are only two variables i can tweak which appear in the denominator of the transfer function, in the expression of almost all coefficients. I have yet to toy with your idea, if I am able to cascade the denominator to lower order, but I believe that is possible. I still do not understand, from the recent posts on finding the analytical equation for 5th order system by reducing it to Bring-Jerrard form is not applicable here \$\endgroup\$
    – RAN
    Commented Mar 23, 2021 at 10:18
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I'll add a few bits besides user287001's answer (who has the answer you should be selecting) and jonk's comments. The settling time is generally accepted to be the time it takes for the response to settle within 1% of the final value. As per the answer above, the closer the real parts of the roots are to the \$j\omega\$ axis, the longer the settling time, and the lower frequency these poles have, the more they influence the output.

Consider three sets of roots, each with three complex conjugate pairs: r1 has a pole closer to the origin by a factor of around 10, compared to the others, r2 has twice that value of the real part for the first pair, and r3 has the realpart of the 3rd pair close to the \$j\omega\$ axis. The three resulting polynomials will be their denominators. For convenience, the transfer functions are for an all-pole lowpass.

r1=[-1-i, -1+i, -10-5i, -10+5i, -20-30i, -20+30i];
r2=[-1-r1(1:2) r1(3:end)];
r3=[r1(1:4) 19+r1(5:6)];
p1=poly(r1),
p2=poly(r2),
p3=poly(r3),
p1 =

        1       62     2347    35570   228950   387000   325000

p2 =

        1       64     2470    40200   297625   805000   812500

p3 =

        1       24     1112    20446   151297   261790   225250

Their values are all quite similar, with the minor exceptions that p2 has 8.125e5 instead of 3.25e5, and p3 has lower \$s^5\$ term (and onward), but similar for \$s^0\$. If you were to look at any of these, without any other comparison, would you be able to tell which one had the fastest settling time? Looking at all three, you might guess that the \$s^0\$ term for all will influence the most their corner frequency and, while true, it wouldn't be quite so true numerically:

freq

Despite the large difference between p1 and p2, the difference is less than 20%, and that's because the last term has, in fact, the power of 6. OTOH, that doesn't tell you anything about the damping factor(s) of the forming 2nd order sections, those would influence the other terms except \$s^6\$, and it's these ones that affect the settling time:

time

Out of the three, p2 influences the most the settling time because the magnitude of the pole is 1.58 times greater and still closer to the origin than the others, while p2 and p3, despite an almost twice as large difference in magnitude for the last pole, their responses are similar because they are also 20 times farther from the unit circle.

So, the conclusion to take is that there is no analytical solution, and even "informed guessing" can't help much. Also, what you said in the comment is not applicable, because if you can simply discard the greater term, what's stopping you from discarding the next largest, and then on and on? And a 5th order has no analytical roots, a 4th does, but have you seen the horrors? This is one root for the generic \$x^4+ax^3+bx^2+cx+d\$:

4th

I dare you to find a numerically-friendly way to calculate this (i mean in terms of numerical accuracy). You, yourself, are talking about root finding, which are complex operations -- doesn't this make you ponder about the possibility of finding simplicity in such complex operations?

Still, if the settling time is not your concern but the step response is (the time for the response to reach the 50% value), then you could use free term and the larger the value, the faster the response. As for the rising time (10% to 90%), that's also dependent on the damping(s), so that's also off.

All in all, I'd say that your quest for finding that one magical formula to determine the settling time for higher order systems is not worth it, because higher order systems are inherently complex and, thus, impossible to describe in a few words. Except, maybe, "they're complex".

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  • \$\begingroup\$ " Also, what you said in the comment is not applicable, because if you can simply discard the greater term, what's stopping you from discarding the next largest, and then on and on?" --> That is because the coefficient a1 is almost equal to zero, if i plot the step response with and without the sixth degree, the responses are almost identical, which i can live with, that is not true with a2, and hence i have to consider the s^5 \$\endgroup\$
    – RAN
    Commented Mar 19, 2021 at 9:43
  • \$\begingroup\$ "And a 5th order has no analytical roots". That is not entirely true. Please check this post math.stackexchange.com/questions/1555743/… \$\endgroup\$
    – RAN
    Commented Mar 19, 2021 at 9:44
  • \$\begingroup\$ I am inclined to agree to your conclusion. The exercise to find analytical roots, leading to a n analytical solution for settling time might prove to be cumbersome and fruitless. However it is a need to find the effect of two variables on setting time and to that end, I need to try at the very least. \$\endgroup\$
    – RAN
    Commented Mar 19, 2021 at 9:47
  • \$\begingroup\$ @RAN "the coefficient a1 is almost equal to zero" I see no value, but if you mean that 1 is almost zero compared to the others, that's not a very healthy attitude, because what I said still applies: you can reduce to a2, then normalize to a2, then reduce a2 to a3, then normalize to a3, etc. You'll find that what both answers/comments said about the dominant pole holds. And if you, somehow, manage to reduce it to a 5th order, then good luck with the analytical roots, because all imply having reduced forms of the polynomial, not generic cases such as you present. Therefore, what I said (cont) \$\endgroup\$ Commented Mar 19, 2021 at 9:53
  • \$\begingroup\$ (cont) holds true: you need to be able to compare polynomials to see which one has a greater \$s^0\$ term, which ones have lesser \$s^{5,4,...}\$ terms, etc. So you need at least two polynomials, otherwise you're back at trying to solve for one root or another, to determine, or at least, try to guess the lowest root and its damping, which brings you back to quare 1: Matlab/Octave/etc. (therefore not much analytical in is) \$\endgroup\$ Commented Mar 19, 2021 at 9:55
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But firstly, you have to determine if the system is stable or not or else it will never settle. Use RH criteria to determine the system stability. This will give you a lot of constraints on the value of coefficients in denominator. That would be the first step of analyzing.

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  • \$\begingroup\$ Hi, make sense, that will already narrow down to the set of values i can use, \$\endgroup\$
    – RAN
    Commented Mar 24, 2021 at 21:03

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