Example H4.8
Matlab code for example 4.8 from the book "Regeltechniek voor het HBO"
Experimental determination of the transferfunction from the step response
- Date : 31/03/2022
- Revision : 1.0
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Table of Contents
Determine three parameters from the step response
Suppose in figure 4.25 the step response with step size 4 has a result as shown in figure 4.26. Because of the overshoot we clearly see that this response is part of a (at least dominant) second order process with damping factor 
. The horizontal start of the response in 
underlines the second order behaviour. Figure 4.26 immediately shows the relevant values of the static amplification 
, overshoot D and period time 
of the oscillation:
. The horizontal start of the response in underlines the second order behaviour. Figure 4.26 immediately shows the relevant values of the static amplification , overshoot D and period time of the oscillation:
Calculate the parameters 
From this we can perform all calculations in a certain sequence (see formula sheet in appendix D):
From 
: 
rad/s.
: rad/s.From 
=0.4:
=0.4: 
; as λ is negative → 
.Calculate the parameters β and 
From 
→ 
→ 
(a negative β is physically not possible).(maybe the formula 
had been preferrable in this case; now to late!)
had been preferrable in this case; now to late!)From 
: 
rad/s.
: rad/s.From 
: 
: In summary
We derrived two sets of parameters:
(
) = (6.72, -0.46, 1.57)
) = (6.72, -0.46, 1.57) (
) = (2.5 0.29, 1.64)
) = (2.5 0.29, 1.64)In transfer function (see 4.11):
You can check again the solution by simulating the step response in MatLab (step size 4 of the input!) and compare the result with figure 4.26. You can do it yourself now, or see the previous examples of this chapter.
In this example we had to do a lot of calculations. To a certain extent that is secondary. What matters is understanding the theory, especially the physical and /or mathematical meaning of the parameters and .
and .