Time constant of a transfer function

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FactChecker said:
In situations where the system can become unstable, maintaining large enough phase and gain margins is one of the top level requirements. The margins allow for some uncertainty in the modeling and the equipment condition (age, damage, dirt, etc.). In those cases, getting fast response often makes the designs come close to the margin limits. So they become important.

Yes - full agreement.
The classical procedure for designing/tuning a control system is to find a controller that satisfies the requirements in the time domain (response time, overshoot, etc) and - at the same time - in the frequency domain (stability properties, safety margins)
 
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I didn't understand. Can you please confirm this:
given transfer function.
Find output for step response using laplace, multiply, then inverse laplace.
Differentiate y(t) to find slope. This gives y'(t)
Then what?
How to find time constant from here?
 
Calculating the time constant is possible, but it is rather involved.:

* Find the step response, which should be named g(t)
* Calculate the first derivative g´(t)
* Find the value for g´(t) for very large times (you need the asymptotic line)
* Now you have a slope of a straight line. For finding the equation of this line you need one point on the line.
* Select one point on the g(t) curve for very large times.
* Applying math basics for calculating the equation for the wanted asymptotic line. .
* Find the time t=T where this line crosses the horizontal time axis.
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This is exactly what you can do (much quicker) graphically (crossing of the asymptotic line).
 
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I have corrected an error in my above answer; I have forgotten the step for calculating the first derivative g´(t).
 
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