Stabilizing the inverse transform of a system

Question

Currently I try to understand an unknown mechatronic system. I used white noise as an input signal and measured the output of the system. Then I estimated a 2nd order transfer function (Output/Input) in the frequency domain using Matlab:

sys=Out/In
h = idfrd(sys, freq*2*pi, Ts);
np = 2;
nz = 2;
Gs = tfest(h, np, nz)
opts = c2dOptions('Method', 'tustin', 'PrewarpFrequency', 2*pi*120);
Gz = c2d(Gs, 1/4000, opts);

$$ Gs=\frac{3.765\cdot s^2 - 528.7 s - 5.113 \cdot 10^4}{s^2+183 \cdot s + 4.117 \cdot 10^5} $$

A bode Plot of the derived 2nd order system Gs (blue) and the measured data (orange) shows good alignment in amplitude and phase above 60Hz:

Further a pole-zero-plot of Gz shows that all poles are stable (inside unit circle of z-plane), but that there is one zero outside of unit circle of the z-plane:

Now I want to counteract the behaviour of the mechatronic system in a way that I achieve a unity output. So by modifying the input to the system I want to basically get rid of the system characteristics and achieve a pure unit transfer behaviour.

IGz=1/Gz; % so that IGz * Gz = 1

Mathematically I need to multiplicate the input signal with the inverse transform of the system. Here a bode plot of the system Gs (blue) and its inverse transform IGz (orange) basically shows what I want to do:

But because of the zero outside the unit circle the inverse transfer function now has a pole outside the unit circle and thus is unstable. Therefor I wonder how can I counteract the behaviour of the system while not getting an unstable system?

So far I have tried to replace the unstable pole by reflecting the pole inside the unit circle like it is advised here: $(z-p)$ by $(z-1/p)$ (as the unstable pole is purely real I don't need to do the complex conjungate). This keeps the amplitude response, but changes the phase response (blue: unstable IGz, orange: stabilized IGz):

However I would like to keep the phase response as closely to my inverse transfer function as well. But as I don't care so much about what happens below 60 Hz I wonder if there is an approximation of my inverse system which is good above ~60Hz in amplitude and in phase, but can deviate below ~60Hz and is stable. I wonder if there is an established way how one would derive such a system? Here is what I have tried so far:

pIG=pole(IGz);
zIG=zero(IGz);

z=tf('z',1/4000);
Test=((z-zIG(1))*(z-zIG(2)))/((z-1/pIG(1))*(z-pIG(2)));
Fac=evalfr(IGz,100)/evalfr(Test,100);
Test1=Fac*Test;
bode(IGz,Test1)

The result looks good at higher frequencies, however the phase could be better below 100Hz or so (blue: unstable IGz, orange: stabilised IGz with phase kept roughly the same above 100Hz):

Edit

I just looked at the impulse responses. The system $Gs$ of course is oscillationg, but decaying:

However the impulse response of the inverse system is instable of course:

The Impulse Response of the 2nd stabilized system is still overshooting with high amplitude. I wonder if this works, as I of course am limited in output with my controller:

With my Input u to the controller and the output y I want to do the following: u -> IGz (controller) -> y -> Gz (mechatronic system) -> y2 During all this the output y of the controller can't get too big. And it should be: y2/u=1

How can I best come up with a IGz that achieves this.

Answer 1

Like you mentionned, you cannot cancel a right-half-plane zero (or a zero outside the unit circle) by placing a pole on it. A unstable pole in your compensator will make the command of your controller unbounded (i.e. it will reach infinity).

There are no ways to cancel a right-half-plane zero. It's sometimes possible to "remove" a right-half-plane zero by chansing a sensor position.

From Murray's book http://www.cds.caltech.edu/~murray/books/AM08/pdf/fbs-limits_18Aug2019.pdf

The existence of the right half-plane zero can be removed if we choose to measure the location of the vehicle by the position of the center of the rear wheels instead of the center of mass.

Edit : A right-half plane zero can also be caused by a time delay. Delays to computation can be reduced, but some other delays cannot be avoided.

Edit 2 : I will try to answer your other questions later.