(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?


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.

  • $\begingroup$ 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 Apr 11 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 Apr 11 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.