Suppose I have a system with feedback. Consider the input \$d\$ that represents disturbances to the plant; also, consider the input \$V\$ that represents sensor noise on the feedback loop. I can calculate the closed-loop transfer function \$\dfrac{Y(s)}{d(s)}\$ and \$\dfrac{Y(s)}{V(s)}\$. Then, my question is: does it even make sense to talk about the transfer function \$\dfrac{D(s)}{V(s)}\$, if so, could it be calculated as \$\dfrac{d(s)}{V(s)}=\dfrac{d(s)}{Y(s)}\dfrac{Y(s)}{V(s)}\$?


  • \$\begingroup\$ For two different inputs (e.g. reference and disturbance) we have, of course, two different transfer functions. It makes no sense to combine both. \$\endgroup\$
    – LvW
    Commented Jan 28, 2021 at 8:25

1 Answer 1


Unless the sensor noise and the disturbances are related somehow, there is no "transfer function" that exists between them. So no, it probably doesn't make sense.

There may be some sensible reason to calculate \$D(s)/V(s)\$, but I wouldn't call it a "transfer function", under pretty much any circumstances.

  • \$\begingroup\$ So, assuming there is a sensible reason to calculate such quantity, would my procedure be correct? \$\endgroup\$
    – Schach21
    Commented Jan 28, 2021 at 0:23
  • 1
    \$\begingroup\$ That's kind of a null question -- if a relationship exists between two otherwise independent inputs to a system, it's because something outside that system is affecting them. So the system's behavior doesn't really enter into things. \$\endgroup\$
    – TimWescott
    Commented Jan 28, 2021 at 0:28

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.