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‫Welcome back.

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‫So now let's go back to our feedback legalization controller.

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‫This was it schematics.

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‫And you can see that you have three control laws here.

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‫One was here for the X dimension then.

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‫One was for the Y dimension and one was for the Z dimension.

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‫And just to remind you that in the vector matrix form, it was like this, so you had K one here and

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‫then K two here, regardless of the dimension, and then you would multiplied by an error and error

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‫dot, and then you received this scalar quantity for each dimension.

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‫And so what we had also earlier established was that our you x this one here, USIP X, it turned out

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‫to be our error double dot for the X dimension.

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‫Then this use of why was our error double that for the why dimension and the use of BCU was our error

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‫double dot for the Z dimension, that means that you have three homogeneous LTI differential equations

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‫that are second order because your use of X is the error double derivative with respect to time use

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‫of Y the same thing and use of Z, the same thing.

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‫For the X, Y and Z dimension, respectively, so this differential equation here is for the X dimension,

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‫this differential equation is for the Y dimension.

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‫And this differential equation is for the Z dimension and so for each dimension, for each differential

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‫equation, the polls that I chose are minus one and minus two because they showed very good tracking.

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‫Therefore, Lambda one for the X dimension equals minus one and lambda two for the X dimension equals

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‫minus two.

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‫The same thing for the Y dimension.

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‫And the same thing for the Z dimension.

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‫And so if you put these polls in the equations that we had derived before, then you will get your key

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‫values.

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‫Your key one for the X dimension will be minus two.

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‫And then your key to for the X dimension will be minus three.

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‫The same thing for the Y dimension and the same thing for the Z dimension.

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‫So we chose our numbers to be minus one and minus two because they would give us negative exponents.

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‫And so that means that they would drive our errors down to zero like this and we would avoid isolation

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‫because we don't have any imaginary poles here.

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‫And based on our choice for Poles, we computed our key values to be minus two and minus three for all

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‫our three dimensions.

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‫And now let me show you the results with the simulation.