Example H4.9
MATLAB code for example 4.9 from the book "Regeltechniek voor het HBO"
Detecting a transfer function through the frequency domain
- Date : 02/03/2022
- Revision : 1.0
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Table of Contents
The assignment
In figure 4.28 you see a so called servo controller to control the speed 
of a motor connected to a load. How this works is not of our interest at the moment. The Bode plot that you can see in figure 4.29 is that of the transfer function
of a motor connected to a load. How this works is not of our interest at the moment. The Bode plot that you can see in figure 4.29 is that of the transfer function
of the open system (so without feedback). The assignment is: determine 
.
.Analysis of the bode diagram
Because there is no peaking in the amplitude diagram we conclude that the poles are real (not complex) and the system can be described by one or more time constants (we could also say: damping 
, although we do not know until now if it is a pure second order process. Only then β is connected with such a process). These time constants τ we can associate in the amplitude diagram with the frequency 
or 
of the frequency breakpoints : 
(see section 4.8.2 and figure 4.20).
, although we do not know until now if it is a pure second order process. Only then β is connected with such a process). These time constants τ we can associate in the amplitude diagram with the frequency or of the frequency breakpoints : (see section 4.8.2 and figure 4.20). Finding the breakpoints
To find these τ's we draw in the amplitude diagram asymptotes of 0, -20, -40 etc. dB/decade (which equals 0, -6. -12 etc. dB/octave). Then we slide the 0 dB asymptote from above against the curve and the other asymptotes from the right against the curve. The frequency breakpoints are there where the asymptotes intersect. In figure 4.30 this has been realised and we see that a horizontal 0 dB asymptote is not present; it cannot be pushed against the amplitude curve. That is not the case with the -20, -40 etc. dB/decade asymptotes; they can be pushed against the curve. In combination with a look at the phase diagram (at low frequencies a constant phase shift of -90 degrees) shows, that the -20 dB/decade asymptote must be the result of an integrating factor 
in the system.
in the system. The intersection of the -20 and -40 dB/decade asymptotes yields the folowing two break points (unfortunately difficult to see, but already indicated in figure 4.30):
and 
rad/s. In the transfer function this means the factor:
=
.In total until now: 
.
.Finding the gain
We still have to look for 
. For 
we see that 
in our result until now equals 
. In figure 4.29 it equals about 10 to 20 dB; we decide it is 14 dB. This must have been caused by . So for 
we can conclude: 
. And thus 
, so 
.
. For we see that in our result until now equals . In figure 4.29 it equals about 10 to 20 dB; we decide it is 14 dB. This must have been caused by . So for we can conclude: . And thus , so .In total: 
.
.Analyze possible time delay
We still are not ready, because maybe there is a time delay. In that case the phase characteristic would keep increasing, and it does not. In figure 4.29 we see it stabilizes for big ω's at -270 degrees. This matches with the transfer function we detected until now. So we are done!
Conclusion
In s-notation the transfer function becomes (simply replace 
with s): 
with s): You could check the result by simulating the Bode plots in MATLAB