# Why is the steady state error in this system incorrect?

(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.

• 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)? Commented Apr 11, 2019 at 20:27
• 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. Commented Apr 11, 2019 at 20:40