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

enter image description here

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?

  • 2
    \$\begingroup\$ 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. \$\endgroup\$
    – AngryEE
    Dec 17, 2012 at 19:07
  • 1
    \$\begingroup\$ 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? \$\endgroup\$
    – akellyirl
    Feb 28, 2014 at 15:27
  • \$\begingroup\$ It was a math question. \$\endgroup\$ Aug 1, 2014 at 16:19

3 Answers 3


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 -

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.