Final value theorem

The final value theorem reads off the steady-state value of a signal directly from its Laplace transform, without inverting:

${\lim }_{t\to \infty }x(t)={\lim }_{s\to 0}sX(s),$

provided the time-domain limit on the left exists.

What “if the limit exists” means

The condition matters. If $x(t)$ is a pure sinusoid, ${\lim }_{t\to \infty }x(t)$ doesn’t exist (it just oscillates forever), but $sX(s)\to 0$ as $s\to 0$ anyway, so the FVT would say the limit is $0$. Wrong: the limit doesn’t exist.

Rule of thumb: the FVT is valid when all poles of $sX(s)$ are in the open left half-plane, with the possible exception of a single pole at $s=0$ that $sX(s)$ has cancelled. Sinusoidal poles on the imaginary axis (like $\pm j{\omega }_0$) make it invalid.

Worked example

For a control system with step input where $Y(s)=(s+4)/[s(s+1)(s+2)]$:

${\lim }_{s\to 0}sY(s)={\lim }_{s\to 0}\frac{s+4}{(s+1)(s+2)}=\frac{4}{2}=2.$

So the steady-state value is $y(\infty )=2$. Check by inverting (partial fractions, etc.): $y(t)=2-3e^{-t}+e^{-2t}$ for $t\ge 0$, and indeed ${\lim }_{t\to \infty }y(t)=2$. ✓

The transient terms $-3e^{-t}$ and $e^{-2t}$ decay; the constant $2$ remains.

When it’s useful

The FVT is the standard tool for finding the DC gain or steady-state response of an LTI system to a step input. For a unit-step input with transfer function $H(s)$:

$y(\infty )={\lim }_{s\to 0}sH(s)\frac{1}{s}={\lim }_{s\to 0}H(s)=H(0).$

So the DC gain $H(0)$ is the steady-state response to a unit step.

Beware of misuse

A common mistake is applying the FVT to systems with poles in the right half-plane (unstable) or on the imaginary axis (oscillatory). For these, ${\lim }_{t\to \infty }x(t)$ doesn’t exist, but $sX(s)$ may still have a finite limit at $s=0$, giving a wrong answer. Always check pole locations first: all poles of $sX(s)$ must be in the open left half-plane.

See also

The Initial value theorem gives $x(0^{+})={\lim }_{s\to \infty }sX(s)$. For a sinusoidal input, where the FVT doesn’t apply, use the frequency response $H(j\omega )$ directly.