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I read this line in a Control Systems book and I'm wondering if someone could provide an intuitive explanation of how the steady-state error is related to the DC gain of closed-loop transfer function?

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    \$\begingroup\$ Did you notice how you changed their wording, low-frequency magnitude responses, to DC gain in your final question? Do you think they meant what they said? Or what you said? \$\endgroup\$
    – jonk
    Commented Jul 14, 2021 at 23:27
  • \$\begingroup\$ Yes, I did notice that. I thought they meant the DC gain here since it's talking about steady-state error? \$\endgroup\$ Commented Jul 16, 2021 at 17:54
  • \$\begingroup\$ Well, I guess the DC gain is related to low-frequency magnitude responses.. So there's that. But it takes infinite time, too. I liked their wording better as it helps avoid certain mistakes that can occur by an over-simplified way of thinking. \$\endgroup\$
    – jonk
    Commented Jul 16, 2021 at 19:03

2 Answers 2

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  • The drive to the plant is how hard you push it.
  • The error between the plant output and the set point is what drives the controller.
  • The bigger the controller DC gain is, the harder it pushes on the plant for a given error, once everything has settled out.

If the controller pushes harder on the plant for a given error, then to get the plant to go somewhere, it takes less error to drive the controller to that point. It's the difference between someone who looks at a crooked painting and says "yeah, whatever" and someone who has to adjust it until it's perfect -- the first person will leave it crooked (low DC gain); the second won't rest until it's perfect (high DC gain).

So -- high DC gain means more "push" for less error. The plant needs a certain amount of push to stay still. So more push for less error means, in the end, less error.

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  • \$\begingroup\$ Oh I see. So it's kind of like the accuracy of a regulation loop? How close will the controller bring the output to the target value before it is satisfied? For example, for a buck converter, if I have a target reference output level of 5V and my actual output is 4.95V, if I increase the DC gain, I should be able to achieve a value closer to 5V (e.g 4.97V) \$\endgroup\$ Commented Jul 14, 2021 at 22:42
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    \$\begingroup\$ As more simple example, look at a simple inverter amplifier with quasi ideal opamp (use DC Gain only). Calculate the error at output when Gain vary from -10, -100, -1000 ... to -infinity by step. \$\endgroup\$
    – Antonio51
    Commented Jul 15, 2021 at 7:32
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The language used in control systems is not always clear. Here are some interpretations of the phrases you asked about (I'd be interested to hear any comments no this):

Gain: Ratio of output to input, usually the symbol K. Computed as a result of applying a unit step input and using the final-value theorem to compute the final value. Since the input is one, the output we get is the ratio output/input = output. Sometimes you hear people refer to the gain of a transfer function but it is implied that a unit step is applied to it. If we didn't the gain would be zero assuming the system is stable. eg the gain of a first-order transfer function is 0 because if:

$$ H(s) = \frac{K}{\tau s + 1} $$

If we used the final-value theorem on this (without a unit step input) we get:

Recall first the final-value theorem:

$$ h(\infty) = \lim_{s \rightarrow 0} s H(s) $$

Therefore

$$ h(\infty) = s \frac{K}{\tau s + 1} = \frac{K}{\tau + 1/s} = 0 $$

A transfer function without an explicit input is equivalent to a system with a unit impulse as the input. If the system is stable and as the impulse is transitory and the system will simply decay back to it original state, hence [h(\infty)= 0]

People will also use the following shortcut to compute the gain:

$$ K = H(0) $$

You'll see that a lot and it can be proved that this statement is true. But operationally there is an implied unit step input.

Steady-State (also referred to as the steady-state gain): This is really synonymous to the gain but is more general in that one could find the steady state for any input, assuming a steady state exists. I would therefore be inclined to use gain for unit step input and steady state for any input, though others might disagree.

Steady-State error: This is error between the output and input in a negative feedback system. It is computed by obtaining E(s) and using the final-value theorem to compute it's time domain steady-state value. There is no other input applied to E(s), you apply the final-value to theorem directly to E(s) (multiplying by s of course as part of the theorem).

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