Thanks in advance for having a look.

I finally have some time on my hands to go back through my controls systems text and I want to understand how the general expression is derived; rather than just use the equation without understanding where it came from.

I'm hoping someone can help me find out why I have one extra term [1/G(S)] in the general equation which defines the steady state error of a non-unity gain feedback control system with an external disturbance.

I've searched the Internet and have found and read numerous papers about the topic (mostly from university lectures) but, none show the development of the general equation. The papers usually show the general equation without any development, only how to use the formula, or the paper covers only the unity gain case.

I've uploaded images showing the steps I've taken during the derivation of the general equation along with the final result.

My work matches the equation in my control system's text except for the 1/G(S) term.

Page One of dev work Page Two of dev work Page Three of dev work Textbook Equation

Thank you.



3 Answers 3


The answer to this solution is to first find the transfer function for the system and then to use the general error equation R(s) - C(s) which is equal to C(s).

The math is provided below...

\$C(s) = G2(s){[R(s) - H(s)C(s)]G1(s) + D(s)}\$
\$C(s) + C(s)H(s)G1(s)G2(s) = R(s)G1(s)G2(s) + D(s)G2(s)\$
\$C(s)[1 + H(s)G1(s)G2(s)] = R(s)G1(s)G2(s) + D(s)G2(s)\$
\$C(s) = \frac{R(s)G1(s)G2(s) + D(s)G2(s)}{1 + H(s)G1(s)G2(s)}\$

\$E(s) = R(s) - C(s)\$
\$E(s) = R(s) - \frac{R(s)G1(s)G2(s) + D(s)G2(s)}{1 + H(s)G1(s)G2(s)}\$
\$E(s) = \frac{R(s)[1 - G1(s)G2(s)] - D(s)G2(s)}{1 + H(s)G1(s)G2(s)}\$
\$E(s) = \frac{R(s)(1 - G1(s)G2(s))}{1 + H(s)G1(s)G2(s)}- \frac{D(s)G2(s)}{1 + H(s)G1(s)G2(s)} \$


Starting from the system equations (upper caps for transfer functions, lower caps for signals):

$$ \begin{cases} e = r - Hc\\ c = G_2(d+G_1e) \end{cases} $$ $$ e = r-HG_2d-HG_1G_2e $$ $$ e = \frac{r-HG_2d}{1+HG_1G_2} $$ $$ e = r\left(\frac{1}{1+G_1G_2H}\right)-d\left(\frac{G_2H}{1+G_1G_2H}\right) $$

In comparison, your method is too convoluted, but I followed the development and believe the issue lies in the very beginning, moving \$d\$ to the left of \$G_1\$. This ends up changing the error signal to $$e = r+\frac{1}{G_1}d-Hc$$, which is incorrect.

  • \$\begingroup\$ Hi Vicente, Thank you kindly for having a look and spending the time working through not only my work but, also solving it a different way. Your work, along with earlier work I did, is making me wonder if my textbook has an error. I tried to solve it the same way you did in some earlier work and ended up with the same equation you did. There we both have H(S) in the second term and no -1 factor in the first term. My text only has G2(S) in the numerator of the second term with no H(S) and has a -1 in the first term. I'd like to upload the image for you. Don't see a way to \$\endgroup\$
    – graybeard
    Apr 24, 2019 at 22:41
  • \$\begingroup\$ What is the title and author of the book? I'll have a look on it. \$\endgroup\$ Apr 25, 2019 at 2:13
  • \$\begingroup\$ Hi Vincente, It is "Control Systems Engineering" by Norman Nise; 2nd Edition. I learned that I needed to "edit" my question in order to post another image. So, I was able to post the pertinent page of the text. Thank you, Ed \$\endgroup\$
    – graybeard
    Apr 25, 2019 at 4:02

I don't know if you were able to get the answer or not but here is the hint, (I am trying not to solve anyone's HW problem by answering this).

There are two sources of errors in this system:

  1. Due to a mismatch in the input and the response
  2. Due to the disturbance.

Apply the basic error formula considering inputs separately, read the topic from the same book "Steady-state error in terms of T(s)". There are two inputs and there are two errors. Apply the superposition principle and add the errors to get the desired expression.


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