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In the textbooks and reference material which I have been using during my course on control systems, a common definition of steady-state error is as follows: $$E(s)=R(s)-C(s)$$ where E(s) is the error (and also the signal carried forward directly from the summing node), R(s) input and C(s) output.

This definition has the slightly unsettling effect of yielding a negative error in the case that the output is above the reference signal, and vice versa. Wouldn't it be more logical to flip the RHS expression?

Is there any particular reasoning behind selecting this convention?

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where E(s) is the error (and also the signal carried forward directly from the summing node), R(s) input and C(s) output.

The error is "demand" minus "output" and the output and the demand are desired to be equal hence, the "thing that does the math" is a subtractor: -

enter image description here

Picture from here.

This definition has the slightly unsettling effect of yielding a negative error in the case that the output is above the reference signal, and vice versa.

If the controller, feedback network and plant (as shown above) are non-inverting, then "demand" minus "output" is absolutely correct in that the error produced drives the system towards closer accuracy.

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  • \$\begingroup\$ I used "summing node" as a catch-all term for "the thing that does the math", without regard for the sign of its inputs, as ultimately subtraction is summation with negatives, so I don't think it was here that my understanding fell short. \$\endgroup\$ – Benjamin Crawford Ctrl-Alt-Tut May 3 at 19:27
  • \$\begingroup\$ Ah, viewing the error as the "driving force" as opposed to a simple error (in the English sense) gives some insight. It would therefore drive the system's output down for an output which is too high. Thanks. \$\endgroup\$ – Benjamin Crawford Ctrl-Alt-Tut May 3 at 19:33
  • \$\begingroup\$ Most folk regard the controller and plant as having a default "sign" as being positive. Slightly less than most folk might regard the feedback network as having a positive "sign" also and so, if you flipped the subtractor by adding a stage of error inversion (for instance) then you would have positive feedback and an almost instantaneously unstable system. The sign is important. \$\endgroup\$ – Andy aka May 3 at 19:33
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To me, the convention here makes much more sense than what you propose, but I think that's really just personal opinion!

However, this definition simply makes the necessary correction have the same sign as the error, and that, I feel, has a beneficial effect on the simplicity of the system block diagram. Again, pure opinion – I don't think there's a "factual" reason; after all, every proportionality can have a negative sign, so there's nothing inherently forcing any specific signs.

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  • \$\begingroup\$ I know I wasn't alone in class for confusing the sign of the error, but I can see the benefit. For me, an error is always something to be removed (i.e. subtraction) from the output. \$\endgroup\$ – Benjamin Crawford Ctrl-Alt-Tut May 3 at 19:23

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