Motor Position Control (State Space Matrix Question)

@killingfloor-wNO4xj • Oct 13, 2024

Oct 13, 2024

1.1K

I have an exam on Monday and haven't been able to get in touch with my lecturer for the topic so I'm in need of a little help understanding plugging the numbers in to a state space matrix.

It's a motor position problem taking into account disturbances.

Standard form is x(dot) = Ax + Bu + Bw

Where Bw is the disturbance.

I've taken a screenshot of the slides I have to show the problem but my problem here is plugging the numbers into a matrix for A. I think I get the 1 in the top right but I'm not quite sure why it's -2 instead of 2. Could anyone explain to me why?

Thank you in advance.

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KillingFloor

@killingfloor-wNO4xj • May 8, 2009

Nevermind. I think I've cracked it.

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Kaustubh Katdare

@thebigk • May 8, 2009

@KillingFloor : How about sharing it with us? It'll help others.

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KillingFloor

@killingfloor-wNO4xj • May 8, 2009
Ok. Sorry, I would have posted it but I think I've just got the grasp of it. I'll give it a go though...

The 0s on the left of matrix A can both be agreed to be 0 because there are no y's (i.e. x1).

Now bare with me on the rest but x1(dot) can be agreed to be the same as x2 from the state variables defined and will therefore be 1. As for the final bottom right corner of the matrix, I found:
```
_    _       _                              _
l x1' l    l    0        1       0     etc. l
l x2' l =  l    0        0       1     etc. l
l etc.l    l                                l
l_xn'_l    l -a0/an   -a1/an  -a2/an   etc. l
            _                              _
```
Therefore, the bottom left corner is -2.

You then work out the determinant of (sI-A) + BK, which you arrange into the standard form of the characterisitic polynomial and then equate the coefficients.

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