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

• 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
• Is this a question about solving a maths problem or are you interested in understanding how to minimize the effect of a disturbance (for example) whilst still allowing the control loop to react to the reference? – akellyirl Feb 28 '14 at 15:27
• It was a math question. – MathematicalPhysicist Aug 1 '14 at 16:19

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.

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

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 -

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