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(Note: I'm currently learning about this, but I'm having a hard time understanding why this system I am modelling is giving unexpected results when finding the steady state error)

I have a system with a transfer function $G(s)=\frac{48}{s^3 + 7s^2 + 6s}$ (in the Laplace domain). This system has unity negative feedback, meaning the overall transfer function is $\frac{O(s)}{I(s)}=\frac{48}{s^3 + 7s^2 + 6s + 48}$. This transfer function is unstable because the system has poles in the right half of the s-plane. For a step input, the output increases without bound to infinity.

My issue is this equation for the steady state error: $$e(\infty)=\lim_{s\to0}\frac{sR(s)}{1+G(s)}$$

Plugging the values $R(s) = \frac{1}{s}$ (for a step input) and $G(s)=\frac{48}{s^3 + 7s^2 + 6s}$ gives: $$\begin{align*}e(\infty) &= \lim_{s\to0}\frac{s\frac{1}{s}}{1+\frac{48}{s^3 + 7s^2 + 6s}}\\ &= \lim_{s\to0}\frac{1}{1+\frac{48}{s^3 + 7s^2 + 6s}}\\ &= \lim_{s\to0}\frac{s^3 + 7s^2 + 6s}{s^3 + 7s^2 + 6s+48}\\ &= \frac{0^3+7\cdot0^2+6\cdot0}{0^3+7\cdot0^2+6\cdot0+48}\\ &= \frac{0}{48}\\ &= 0 \end{align*}$$

This means the steady state error is 0, and since the input was a unit step then (to my understanding) this means the system has a steady state at an amplitude of 1. However, as stated earlier, the output never actually settles, so surely the steady state error should be infinity. Where have I gone wrong?

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1 Answer 1

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The finite value theorem (FVT) of the Laplace transform does only apply if the considered transfer function $F(s)$ is stable. In your case, you have poles with positive real part, so you can't deduce the steady state error in this case.

As stated on Wikipedia: "If $f$ is bounded on $(0,\infty)$ and

$$ \lim_{t \rightarrow \infty} f(t) $$

has a finite value then

$$ \lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} s F(s) $$

where $F(s)$ is the (unilateral) Laplace transform of $f(t)$." A link to a proof of this statement is also given there.

In your case, the limit has obviously no finite value and $f(t)$ is also not bounded, as the system is unstable.

Actually, Wikipedia has an example very similar to your case: Example where FVT does not hold.

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  • $\begingroup$ I thought that could've been the case. However, I thought I was wrong because $e(\infty)$ only has a $G(s)$ term, not an $sG(s)$ term. Does that mean for an unbounded system, the steady state error using the equation for $e(\infty)$ is incorrect? Is there any way to tell beforehand that a system is going to be unbounded without needing to find the poles (because it can be hard to find the poles for a third order system)? $\endgroup$
    – JolonB
    Commented Apr 11, 2019 at 20:27
  • $\begingroup$ Yes, it means the equation for $e(\infty)$ is incorrect for an unbounded system. You don't need to find the poles to check if the system is bounded. You can instead use for example the Routh-test for continous systems or the Jury-test for discrete systems. $\endgroup$
    – SampleTime
    Commented Apr 11, 2019 at 20:40

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