# Final value theorem

In mathematical analysis, the final value theorem (FVT) is one of several similar theorems used to relate frequency domain expressions to the time domain behavior as time approaches infinity. A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression, as opposed to converting to a time domain expression and taking its limit.

Mathematically, if ${\displaystyle f}$ is bounded on ${\displaystyle (0,\infty )}$ and

${\displaystyle \lim _{t\to \infty }f(t)}$

has a finite limit, then

${\displaystyle \lim _{t\to \infty }f(t)=\lim _{s\,\searrow \,0}{sF(s)}}$

where ${\displaystyle F(s)}$ is the (unilateral) Laplace transform of ${\displaystyle f(t)}$.[1][2]

Likewise, in discrete time

${\displaystyle \lim _{k\to \infty }f[k]=\lim _{z\to 1}{(z-1)F(z)}}$

where ${\displaystyle F(z)}$ is the (unilateral) Z-transform of ${\displaystyle f[k]}$.[2]

## Proofs

The first proof below has the virtue of being totally elementary and self-contained. If one is willing to use not-quite-so-elementary convergence theorems one can give a one-line proof using the Dominated Convergence Theorem.

### Elementary proof

Suppose for convenience that ${\displaystyle |f(t)|\leq 1}$ on ${\displaystyle (0,\infty )}$, and let ${\displaystyle \alpha =\lim _{t\to \infty }f(t)}$. Let ${\displaystyle \epsilon >0}$, and choose ${\displaystyle A}$ so that ${\displaystyle |f(t)-\alpha |<\epsilon }$ for all ${\displaystyle t>A}$. Since ${\displaystyle s\int _{0}^{\infty }e^{-st}\,dt=1}$, for every ${\displaystyle s>0}$ we have

${\displaystyle sF(s)-\alpha =s\int _{0}^{\infty }(f(t)-\alpha )e^{-st}\,dt;}$

hence

${\displaystyle |sF(s)-\alpha |\leq s\int _{0}^{A}|f(t)-\alpha |e^{-st}\,dt+s\int _{A}^{\infty }|f(t)-\alpha |e^{-st}\,dt\leq 2s\int _{0}^{A}e^{-st}\,dt+\epsilon s\int _{A}^{\infty }e^{-st}\,dt=I+II.}$

Now for every ${\displaystyle s>0}$ we have

${\displaystyle II<\epsilon s\int _{0}^{\infty }e^{-st}\,dt=\epsilon }$.

On the other hand, since ${\displaystyle A<\infty }$ is fixed it is clear that ${\displaystyle \lim _{s\to 0}I=0}$, and so ${\displaystyle |sF(s)-\alpha |<2\epsilon }$ if ${\displaystyle s>0}$ is small enough.

### Proof using the dominated convergence theorem

Let ${\displaystyle \alpha =\lim _{t\to \infty }f(t)}$. A change of variable in the integral defining ${\displaystyle F(s)}$ shows that

${\displaystyle sF(s)=\int _{0}^{\infty }f\left({\frac {t}{s}}\right)e^{-t}\,dt.}$

Since ${\displaystyle f}$ is bounded, there exists ${\displaystyle M>0}$ such that ${\displaystyle \left|f\left({\frac {t}{s}}\right)e^{-t}\right|}$ is dominated by the integrable function ${\displaystyle g(t):=Me^{-t}.}$

Thus, by the Dominated Convergence Theorem,

${\displaystyle \lim _{s\,\searrow \,0}sF(s)=\int _{0}^{\infty }\alpha e^{-t}\,dt=\alpha .}$

## Examples

### Example where FVT holds

For example, for a system described by transfer function

${\displaystyle H(s)={\frac {6}{s+2}},}$

and so the impulse response converges to

${\displaystyle \lim _{t\to \infty }h(t)=\lim _{s\,\searrow \,0}{\frac {6s}{s+2}}=0.}$

That is, the system returns to zero after being disturbed by a short impulse. However, the Laplace transform of the unit step response is

${\displaystyle G(s)={\frac {1}{s}}{\frac {6}{s+2}}}$

and so the step response converges to

${\displaystyle \lim _{t\to \infty }g(t)=\lim _{s\,\searrow \,0}{\frac {s}{s}}{\frac {6}{s+2}}={\frac {6}{2}}=3}$

and so a zero-state system will follow an exponential rise to a final value of 3.

### Example where FVT does not hold

For a system described by the transfer function

${\displaystyle H(s)={\frac {9}{s^{2}+9}},}$

the final value theorem appears to predict the final value of the impulse response to be 0 and the final value of the step response to be 1. However, neither time-domain limit exists, and so the final value theorem predictions are not valid. In fact, both the impulse response and step response oscillate, and (in this special case) the final value theorem describes the average values around which the responses oscillate.

There are two checks performed in Control theory which confirm valid results for the Final Value Theorem:

1. All non-zero roots of the denominator of ${\displaystyle H(s)}$ must have negative real parts.
2. ${\displaystyle H(s)}$ must not have more than one pole at the origin.

Rule 1 was not satisfied in this example, in that the roots of the denominator are ${\displaystyle 0+j3}$ and ${\displaystyle 0-j3}$.