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Why this relation is not right?

C(s) = R(s) * [G(s).G(p)+D(s)] / [1+G(s).G(p)+D(s)]

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  • $\begingroup$ What would happen to the output if $R(s)=0$ and $D(s)\neq0$? Your solution would always return zero is that case. $\endgroup$ – fibonatic Jan 1 '19 at 20:37
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First step is the summing junction:

$$ R(s) - C(s) \\ $$

Then the controller:

$$ G_c(s)\left(R(s) - C(s)\right) \\ $$

Then the plant:

$$ G_p(s)\left(G_c(s)\left(R(s) - C(s)\right)\right) \\ $$

Then the addition of the disturbance input:

$$ D(s) + G_p(s)\left(G_c(s)\left(R(s) - C(s)\right)\right) \\ $$

Finally, all of the last step above is equal to the output, $C(s)$:

$$ C(s) = D(s) + G_p(s)\left(G_c(s)\left(R(s) - C(s)\right)\right) \\ $$

Multiply out the business on the right:

$$ C(s) = D(s) + G_p(s)\left(G_c(s)\left(R(s) - C(s)\right)\right) \\ C(s) = D(s) + G_p(s)G_c(s)R(s) - G_p(s)G_c(s)C(s) \\ $$

Then rearrange to get the $C(s)$ terms all on one side:

$$ C(s) + G_p(s)G_c(s)C(s) = D(s) + G_p(s)G_c(s)R(s) \\ $$

Factor out the $C(s)$ term and divide to get the answer:

$$ C(s)\left(1 + G_p(s)G_c(s)\right) = D(s) + G_p(s)G_c(s)R(s) \\ C(s) = R(s)\left(\frac{G_p(s)G_c(s)}{1 + G_p(s)G_c(s)}\right) + \frac{D(s)}{1 + G_p(s)G_c(s)} \\ $$

So it looks like your answer isn't correct because you have what distributes to be an $R(s)D(s)$ term, when the disturbance only sums with the result of the plant (and doesn't multiply), and you also have the disturbance term in the denominator, which is also incorrect.

There are all kinds of "thumbrules" on how you do block diagram reduction, but I always just multiply it out. It only takes a little bit longer, but I personally always get something wrong when I try the (N/(1+N)) method or something like that.

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  • $\begingroup$ Mathematics as an abstraction method for plant-prototyping has limits. It works only for linear-systems and for toy problems. A real plant can't be realized with equations in the form y=f(x). The problem is, that after transfer the problem into simple equations, most properties of the problem gets lost. Even a simple pid controller needs a different kind of prototype than given here. $\endgroup$ – Manuel Rodriguez Jan 1 '19 at 9:18
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    $\begingroup$ @ManuelRodriguez - I have no idea what you're trying to say. The question is about block diagram equivalency. This is the answer. Please post your own answer that conforms with whatever framework you're trying to explain; I'm curious to see what you're talking about. $\endgroup$ – Chuck Jan 1 '19 at 11:15

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