You took the transform of something like $\frac{1}{9}\sin(9t)$, got $F(s) = \frac{1}{s^2+81}$, multiplied by s, set s = 0, and read off 0 — so you wrote down that the steady state is zero. The recurring confusion is exactly this: the Final Value Theorem returned a clean number, so it felt like a valid answer. But a pure sine oscillates between $-\frac{1}{9}$ and $+\frac{1}{9}$ forever; it has no steady state at all. The 0 is fiction.
The theorem only holds when every pole of sF(s) sits strictly in the open left half-plane. A pure sine has poles on the imaginary axis, so there is no final value to find. The formula still hands you a number, but that number is meaningless. Check the poles before you trust the limit.
That's the whole answer. The rest of this is why it's so easy to get burned, and how to spot the trap in two seconds.
The shortcut that almost always works
The Final Value Theorem says:
$$\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$$
The appeal is obvious. You don't have to do partial fractions, you don't have to invert the transform, you don't have to know what f(t) even looks like. Multiply F(s) by s, plug in s = 0, read off the steady state. Two lines.
And on a normal homework problem, it works. A first-order RC circuit, a critically damped second-order system, anything that settles — the theorem nails the answer every time. So you build a habit: see "find the steady-state value," multiply by s, set s = 0, done.
The habit is the problem. The theorem produces a number for almost any F(s) you feed it, including ones where no final value exists. It rarely throws an error. A clean number feels like a valid answer, and the one precondition that would have stopped you is usually a single sentence buried in the lecture notes.
What the precondition actually requires
Here it is in full: the theorem is valid only if every pole of sF(s) lies strictly in the open left half-plane (real part < 0).
Two failure modes break it:
- A pole on the imaginary axis (real part = 0) means sustained oscillation. The signal never settles, so $\lim_{t\to\infty} f(t)$ does not exist.
- A pole in the right half-plane (real part > 0) means the signal grows without bound. Again, no final value.
In both cases the $s \to 0$ limit still evaluates to something, but it has no connection to the actual behavior of f(t). A finite number from the formula is necessary but not sufficient. You have to factor the denominator of sF(s) and look at where the roots land first.
The worked example: same algebra, two different truths
Take the transform of a pure sine. With $f(t) = \frac{1}{9}\sin(9t)$, the Laplace transform is:
$$F(s) = \frac{1}{s^2 + 81}$$
Apply the theorem naively:
$$\lim_{s \to 0} sF(s) = \lim_{s \to 0} \frac{s}{s^2 + 81} = \frac{0}{81} = 0$$
The formula says f settles to 0. But $\frac{1}{9}\sin(9t)$ oscillates between $-\frac{1}{9}$ and $+\frac{1}{9}$ forever. It is never near 0 except as it passes through. The "steady state" of 0 is fiction.
Why did it lie? Factor the denominator: $s^2 + 81 = (s - 9j)(s + 9j)$. The poles are at $s = \pm 9j$ — sitting right on the imaginary axis, real part exactly 0. Multiplying by s doesn't move them off the axis (s = 0 isn't a pole here). The precondition fails, so the output is garbage.
Now a case that is valid, with deliberately similar-looking algebra:
$$F(s) = \frac{1}{s(s+5)}$$
Apply the theorem:
$$sF(s) = \frac{1}{s+5}, \qquad \lim_{s \to 0} \frac{1}{s+5} = \frac{1}{5} = 0.2$$
Check the precondition. The poles of $sF(s) = \frac{1}{s+5}$: just one, at $s = -5$. Real part $-5 < 0$, strictly in the left half-plane. Valid.
And it's correct. Inverting the transform gives $f(t) = 0.2\left(1 - e^{-5t}\right)$, which rises from 0 and approaches 0.2 as $t \to \infty$. The theorem matched reality.
Put them side by side:
| $F(s) = \dfrac{1}{s^2+81}$ | $F(s) = \dfrac{1}{s(s+5)}$ | |
|---|---|---|
| Multiply by s, set s=0 | gives 0 | gives 0.2 |
| Poles of sF(s) | $\pm 9j$ (imaginary axis) | $-5$ (left half-plane) |
| Precondition | fails | holds |
| Real f(t) | $\frac{1}{9}\sin(9t)$, never settles | $0.2(1-e^{-5t}) \to 0.2$ |
| The "answer" | a lie | true |
The algebra was equally trivial in both columns. The only thing separating a correct answer from a confident wrong one was the pole check — the step the shortcut tempts you to skip.
Why oscillating circuits trigger this constantly
This shows up most in s-domain circuit analysis, which is why the question keeps appearing on circuit and physics forums. An LC tank, an undamped resonant loop, any system with purely imaginary poles produces a transform whose denominator has an $s^2 + \omega^2$ factor. That factor is the imaginary-axis pole pair at $\pm j\omega$ — the same conjugate pairing that gives rise to the negative frequencies in a Fourier transform. The moment you see $s^2 + \omega^2$ with no real term in the denominator, the Final Value Theorem is off the table — the system rings forever and "steady state" isn't defined. (The Wikipedia entry for the theorem lists this precondition explicitly; it's easy to read past.)
The same goes for a marginally stable system with a pole pair exactly on the axis, or an unstable one with a right-half-plane pole. The formula will still return a finite number for many of these. Don't take the bait.
The two-second check before you apply it
- Form sF(s).
- Factor its denominator and find the poles.
- If every pole has real part strictly less than 0, apply the theorem and trust the number.
- If any pole has real part $\geq 0$ — on the imaginary axis or to the right — stop. The final value doesn't exist, and whatever the formula returns is meaningless.
The clean number is the trap. The pole locations are the truth. It's the same lesson as reading a dB figure without checking whether it's a power or a voltage ratio: the arithmetic is easy, and the one precondition you skipped is the whole answer.
Check yourself
For which of these is the Final Value Theorem valid, so that $\lim_{s\to 0} sF(s)$ actually equals $\lim_{t\to\infty} f(t)$?
A) $F(s) = \dfrac{10}{s^2 + 16}$ B) $F(s) = \dfrac{3}{s(s+2)}$ C) $F(s) = \dfrac{1}{s - 4}$ D) $F(s) = \dfrac{s+1}{s^2 + 9}$
Correct answer: B.
For B, $sF(s) = \frac{3}{s+2}$ has a single pole at $s = -2$, strictly in the left half-plane. The theorem applies and gives $\frac{3}{2} = 1.5$, matching $f(t) = 1.5(1 - e^{-2t}) \to 1.5$.
A has poles at $\pm 4j$ on the imaginary axis — that's a $\sin(4t)$-type signal that oscillates forever, so no final value exists. D has poles at $\pm 3j$, same problem. C has a pole at $s = +4$ in the right half-plane, so f(t) grows like $e^{4t}$ and blows up. In A, C, and D the formula still hands you a number ($\frac{10}{16}\cdot 0 = 0$, and so on), but every one of them is a lie.
Close the gap
The Final Value Theorem isn't broken — it's conditional, and the condition is the part that's easy to drop once the algebra starts feeling automatic. Catching where a clean answer is actually invalid is exactly the kind of gap that's hard to see on your own, because nothing flags it. Gradual Learning works through these case by case, watching for the step you skip and surfacing it before it becomes a habit.