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The question is simple and I rather need a reference point.

How the parameters of transients are estimated (as in the picture) from an arbitrary linear transfer function (formula is given).

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dtn
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    Are you able to find the step response using the inverse Laplace transform of $\frac{G(s)}{s}$? – AJN Aug 15 '21 at 04:42
  • @AJN Yes. This is easily done using the inverse Laplace transform. I understand where you are leading. I want to note that sometimes the inverse Laplace transform turns out to be very cumbersome. Can we use zeros, poles, hodographs, bode plot? – dtn Aug 15 '21 at 04:45
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    where's the problem? If you have knowledge of all p_m and z_n, computing the step response numerically is straightforward. If you want to do it analytically, first consider the case without zeros. Then you can look for a dominant pole or pole-pair, and if there is one, approximate the result using properties of a 2nd order system using the dominant pole-pair. Zeros also cause overshoot, so in the general case, that complicates it a bit. – Pete W Aug 15 '21 at 19:44
  • @PeteW Dominant pole approximation? I.e. reduction sto second order system? – dtn Aug 16 '21 at 02:38
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    Yes. Dominant pole-pair, really. Like Alfredo's answer. But if there is a zero of similar or lower-frequency magnitude, then that will also be visible in the step response. (imagine the fourier transform of a step with a particular frequency removed... that's what a zero can do) – Pete W Aug 16 '21 at 02:49
  • @PeteW Or it may be that all the roots are in the negative half-plane and at the same time are equidistant from the center, i.e. are on the circle line? What in that case? – dtn Aug 16 '21 at 03:10
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    Then I don't think the 2 pole approximation would be good. But it might be a well known special case like a butterworth filter etc and you could maybe look that up – Pete W Aug 16 '21 at 03:40
  • @PeteW Nevermind. I considered one special case, which is not yet relevant to the question, but may "emerge". In general, as Alfredo showed, there are formulas for a second-order object. Are there such for the third or fourth? Yesterday I tried to derive these formulas for the 5th order myself, but nothing came of it. The computer simply could not calculate them, apparently due to the cumbersomeness of the expressions. – dtn Aug 16 '21 at 03:43

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Those are parameters on time domain, so calculate them in spectrum domain (laplace or fourier) is almost impossible.

So, you should apply inverse Laplace transform to get the solution in time domain, then using calculus and algebra (like first derivative zero).

The formula for each parameter is generalized so you could find it in a table. but if you want to know how calculate them, using the time domain plus evaluate the function on their derivatives is the way.

  • peak time: time at first derivative equal to zero.
  • Overshoot: function evaluated at 'peak time'.
  • $e_{ss}$: (this is easy in frequency domain) i think is setpoint minus the static gain
  • Rise time: time at first f(x)=1
  • $T_r$: time at f(x)=0.9 minus time at f(x)=0.1
  • Setting time: evaluate the exponential decay when the envelope is 95% (delta 5%)

*The formulas for those parameters are calculated starting with the standard formulation of transfer function:

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reference 1

NMech
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Alfredo Maussa
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  • Thank you for the answer. I agree with you. – dtn Aug 15 '21 at 12:14
  • However, maybe there are some formulas connecting the location of the poles and zeros to these parameters? – dtn Aug 15 '21 at 12:15
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    Information is not lost, so you can always take advantage 1by1 relation to map the variable of interest. In order to put the parameters in term of the poles locations you could use the 'quadratic formula'. – Alfredo Maussa Aug 15 '21 at 12:31
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    time domain specifications: [formulas](https://slidetodoc.com/presentation_image/45761dee4d3c5da300ae64b3dbaa71d8/image-17.jpg). write the poles in term of zeta and omega should associate the parameters with the poles. – Alfredo Maussa Aug 15 '21 at 12:35
  • Yes, it looks like what I'm looking for, but this is for a second order object. And how will the formulas change if the transfer function of the object has a higher order? – dtn Aug 15 '21 at 12:58
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    some months ago i made the same question, i found that any number of poles could be written as multiple second order behavior (look for 'partial fractions'). The main problem is to factorise the polynomial to get the poles, since there are no high order general formula for roots. – Alfredo Maussa Aug 15 '21 at 13:02
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    btw, the shape of the response wouldn't be clear to define parameters at high order system... is preferible to approximate to second order taking the two closest poles to zero. But with the poles you should be able to write the function as a linear combination of exponential decays (sometime phasor/ complex values), then find similar requirements like "time at first derivative zero/ backward motion begins". Note that clear the argument of multiple exponential decays is not easy. – Alfredo Maussa Aug 15 '21 at 13:12
  • Alfredo, very interesting. And the following can somehow help in solving the problem? Let's say the structure of an object is chosen depending on the required order $n$. Can such a structure of the control object simplify the task? If $n$ is even, then $\prod _{k=1}^{\frac{n}{2}} \frac{1}{s \left(a_k-i b_k+s\right) \left(a_k+i b_k+s\right)}$. If $n$ is odd, then $\frac{\prod _{k=2}^{\frac{1-n}{2}+n} \frac{1}{\left(a_k-i b_k+s\right) \left(a_k+i b_k+s\right)}}{s \left(a_1+s\right)}$. Parameters $a_k$ and $b_k$ - known real and imag parts of roots. – dtn Aug 15 '21 at 13:45
  • Alfredo, please tell about the product of several second-order objects as an approximation. – dtn Aug 16 '21 at 03:40