# Closed loop plant-control system

I have the next system, and I want to find the transfer function from d to y.

So I've got the next equations

v = Ce = C(r-y)

e = r - y

u = d + v

y = Pu = Pd + PCr - PCy

Now I know that:

$e/r = \frac{1}{1+PC}$

So eventaully if I am not mistaken I arrive at:

$$y/d = \frac{P}{1+PC}+\frac{PC}{1+PC} \frac{r}{d}$$

How do I eliminate the dependence on $r/d$? i.e, I want y/d to be a function of P and C.

Edit: actually, I arrive at:

$$y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r$$

So the transfer function from d to y should be: $P/(1+PC)$, correct?

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If you want to find the transfer function of y/d you have to ignore r, ie, set it to 0. It has no place in the y/d transfer function. – AngryEE Dec 17 '12 at 19:07

$$y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r$$ r is desired input and d is disturbance so we have to reduce effect of d if gain of C picked big so we have: $$if \space C \uparrow \hspace{8 mm} \frac{P}{1+PC} \downarrow \hspace{8mm} and \hspace{8mm} \frac{PC}{1+PC} \simeq 1$$ so $$y \rightarrow r$$ warning : be careful about stability.if C picked very big stability of system would be at risk