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Hello I have a rather specific question.

I am trying to follow a derivation and can't work out the last step. Given: $$s\,X(s) = (A-B\,K\,C) X(s) + B\,K\,V(s)$$ and $$Y(s) = CX(s)$$

Multiplying $Y(s) = CX(s)$ through by s, then substituting in $sX(s)$....

Here is where I am at: $$ sY(s) = C{((A-BKC)X(s) + BKV(s))} $$

A is an n x n matrix, K is 1 x n, X(s) is n x 1.

However, the final form is given as this: $$ Y(s) = C(sI-A+BKC)^{-1}BK V(s) $$

Hoping someone can help me with the intermediate steps.

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    Is $V$ also an $n\times1$ matrix? – 2017-02-23
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    Yes I think so, or maybe a 1x1... – 2017-02-23
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    I think maybe Y and V are 1x1 – 2017-02-23
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    Could it be that the first equation actually should be: $s\,X(s) = (A-B\,K\,C) X(s) + B\,K\,V(s)$ and $Y(s)=C\,X(s)$? – 2017-02-23
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    Thanks for your comments. Yes, you are absolutely right. I have perhaps tried to jump a step here. Yes, you are absolutely correct. – 2017-02-23
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    I was thinking to multiply $$Y(s) = CX(s)$$ by s and then substitute for $$sX(s)$$ – 2017-02-23

1 Answers 1

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Starting from the equation expressed in $X(s)$ and $V(s)$ and solving it for $X(s)$ yields,

$$ s\, X(s) - \left(A - B\, K\, C\right) X(s) = \left(s\, I - A + B\, K\, C\right) X(s) = B\, K\, V(s), $$

$$ X(s) = \left(s\, I - A + B\, K\, C\right)^{-1} B\, K\, V(s). $$

Now using the definition of $Y(s)$ yields,

$$ Y(s) = C\, X(s) = C \left(s\, I - A + B\, K\, C\right)^{-1} B\, K\, V(s). $$