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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).

enter image description here

enter image description here

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    $\begingroup$ Are you able to find the step response using the inverse Laplace transform of $\frac{G(s)}{s}$? $\endgroup$
    – AJN
    Aug 15 at 4:42
  • $\begingroup$ @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? $\endgroup$
    – dtn
    Aug 15 at 4:45
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    $\begingroup$ 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. $\endgroup$
    – Pete W
    Aug 15 at 19:44
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    $\begingroup$ 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) $\endgroup$
    – Pete W
    Aug 16 at 2:49
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    $\begingroup$ 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 $\endgroup$
    – Pete W
    Aug 16 at 3:40
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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:

enter image description here

reference 1

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  • $\begingroup$ Thank you for the answer. I agree with you. $\endgroup$
    – dtn
    Aug 15 at 12:14
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    $\begingroup$ 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'. $\endgroup$ Aug 15 at 12:31
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    $\begingroup$ time domain specifications: formulas. write the poles in term of zeta and omega should associate the parameters with the poles. $\endgroup$ Aug 15 at 12:35
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    $\begingroup$ 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. $\endgroup$ Aug 15 at 13:02
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    $\begingroup$ 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. $\endgroup$ Aug 15 at 13:12

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