Closed loop plant-control system

Question

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: \$\dfrac{P}{1+PC}\$, correct?

Answer 1

Your solution is correct but an easier method (I think) is to use superposition, first we will suppress r ( i.e. we will ignore r ), then get the transfer function, then suppress d then get the other transfer function, then sum the two up to get the final transfer function.

When we suppress r will then get

$$ \frac{Y}{d} = \frac{P}{1 + PC} $$

and when we suppress d we will get $$ \frac{V}{r} = \frac{C}{1 + PC} $$

but Y = VP meaning

$$ \frac{Y}{r} = \frac{PC}{1 + PC} $$

We the sum up the two values of Y to get the final response of

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

There is no transfer function from d to y because y depends on both d and r, you can't evaluate the value y without knowing both d and r values.

Answer 2

$$ y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r $$ \$r\$ is desired input and \$d\$ is disturbance, so we have to reduce the effect of \$d\$.

If we pick a large gain for C, 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, $$ y \rightarrow r $$

Be careful about stability: if C is too big, system stability is at risk.

Answer 3

From what I see is that -

$$Y/d = \frac{P}{1+PC} + \frac{PCr}{1+PC}d$$

$$y/d = \frac{PC}{1+PC}(\frac{1}{C} + \frac{r}{d})$$

Hence if I choose my \$C << d/r\$, then my \$1/C + r/d\$ will be almost equal to \$1/C\$. Hence you will have -

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