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Laplace integrator at input of system - expected issue?

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SouthPark

Electrical
Jun 28, 2018
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Hi all.

From automatic control theory involving Laplace blocks, I can see that the two systems (drawn in the figure) are 'equivalent'.
However, the lower figure with the pure integrator at the input becomes a practical problem, or something that is not practical or workable, right? My thinking is that the 1/s block is a pure integrator, so if the input signal is a constant value, then the output of the integrator just keeps getting larger and larger indefinitely, right? So if this were to be done with a real-world system, then this wouldn't work, right? Thanks all!

integrator1_jmx0fz.jpg
 
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The problem with Laplace transforms is that they ignore the limitations of electrical hardware.
The first diagram could be a model of a motor in position mode.
The second diagram would work as long as the output of the first integrator wasn't allow to integrate up to the limits of the power supply or some other limiting factor. For instance a sine wave could be the input that has an offset of 0. Both should work the same way as long as the sine wave had an offset of 0 and the amplitude was not big enough to cause the power supply voltage limitations to be reached.

So the two are equivalent but with different input restrictions.

Peter Nachtwey
Delta Computer Systems
 
Integrators typically need to have both an initial condition and max/min saturation conditions applied when used for modeling and simulation or control purposes.

xnuke
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PN and XN ...... thanks for your time and kind help!! Totally appreciated. This really helped a lot, as I was thinking about equivalent diagrams..... and noticed some possible issue with applying the lower diagram.
The note about hardware limitations and integrator at the input was really good.
Thanks again.
 
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