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

1 Answers 1

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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.

Your penultimate equation

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

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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    @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 $y=G(s)u$.2011-11-08