# What happens at step 3 of this Feedback Control Loop proof [closed]

So input $$= R(s)$$ and output $$= C(s)$$ and forward transfer $$= G(s)$$ and feedback transfer $$= G_2(s)$$

deriving feedback now $$= G_2(s)C(s)$$ forward signal $$= R(s) - \text{feedback}$$$$=R(s) - G_2(s)C(s)$$ output $$C(s) = G(s)(R(s) - G_2(s)C(s))$$

Now I don't understand this step The text says next is $$C(s) = \frac{G(s)}{1 + G_2(s)G(s)}R(s)$$

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## closed as not a real question by Rasmus, Michael Greinecker♦, Old John, William, SashaAug 31 '12 at 14:42

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This is straightforward algebraic manipulation, which would be easier to follow if you just suppressed all those $\cdot(s)$ things. Anyway, I'll keep them for now.

$$C(s) = G(s)\left(R(s)-G_2(s)C(s)\right)$$

can be re-written as

$$C(s) = G(s)R(s) - G(s)G_2(s)C(s)$$

(this is the distributive law), and moving the last term to the right yields

$$C(s)+C(s)G(s)G_2(s) = G(s)R(s)$$

or

$$C(s)(1+G(s)G_2(s)) = G(s)R(s)$$

and dividing through by $1+G(s)G_2(s)$ gives

$$C(s) = \frac{G(s)R(s)}{1+G_2(s)G(s)}$$

which is what you wanted.

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I can't believe how I overlooked that... Thanks. You are probably aware, the $(s)$ is because it is a Laplace domain equation for a feedback controller. In my case, I was after the form just $\frac{G(s)}{1 + G(s)}$ as my feedback was unity. Thanks. –  Supernovah Nov 8 '11 at 8:24
@Supernovah : I guess what Alon meant is that usually we don't write the (s) for the signals but only keep it for the systems. Moreover, to make a visual distinction people only capitalize the system such as $y=G(s)u$. –  user13838 Nov 8 '11 at 16:06