simulate this circuit – Schematic created using CircuitLab

I know the difference between those two functions , but i want to see the relation between them , i think that the error gain function , is given by the transfer function , and the best transfer function which is

  1. stable.
  2. has an error gain function equal to 1.

if this is true , how can i derive the error gain function , from transfer function ?

  • \$\begingroup\$ The stable output condition for an input of 1.0 is 0.5. Does that help at all? Regarding the rest of your question, I'm not entirely sure what you mean. \$\endgroup\$
    – Andy aka
    Apr 10, 2014 at 18:50
  • \$\begingroup\$ @Andyaka since E(s) = Y(s) - X(s)G(S) , where X : is the desired output(input to system) , and G is the error gain function , then X(s)G(S) = X(s) . which is equal to Y(s) , then E(s) = 0. \$\endgroup\$
    – hbak
    Apr 10, 2014 at 19:02

1 Answer 1


Let me label your circuit nodes for better understanding of your question


simulate this circuit – Schematic created using CircuitLab

To answer your question I must first clarify some definitions:

  • Error gain function: a gain k that multiplies the error e = r-y
  • Transfer function: the behavior of the closed-loop system y/r


$$ y=k\cdot e $$ $$ e=r-y $$ $$ y=k\cdot (r-y) $$

So, the error gain function follows as

$$ k=\frac{y}{r-y} $$

So, you don't even need the transfer function to establish the value of k. You just need y and r.

Anyway the transfer function is

$$ TF=\frac{y}{r}=\frac{k}{k+1} $$

So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as

$$ k=\frac{TF}{1-TF} $$

Note also that, with negative feedback, in stable system conditions, and with finite k:

$$ 0\leq TF<1 $$

  • \$\begingroup\$ the transfer function that i meant , that the function that you entered on it an input and give you an output , but the transfer function that you define , is the function that give it desired output r , and give you an output y , in another woed , you define the system above as transfer function , and i want the transfer function that described by this error gain function .... \$\endgroup\$
    – hbak
    Apr 19, 2014 at 9:33
  • \$\begingroup\$ Can you be clearer? r is an input, not an output. Anyway the transfer function described by the error gain function (assuming you agree that this error gain function is just k, otherwise I don't understand your question) is what I defined as TF. \$\endgroup\$
    – raggot
    Apr 28, 2014 at 13:11

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