Final Value Theorem

Final Value Theorem

Let $p(t)$/ $g[k]$ be a signal with Laplace/Z-transform that is real, rational, and proper. Then:

(a.) If all poles of $P(s)$/$G[z]$ lie in ${\mathbb{C}}^{-}$/$\mathbb{D}$,

$$\underset{t\to \infty }{\lim }p(t)=0/\underset{k\to \infty }{\lim }g[k]=0$$

(b.) If all poles of $P(s)$/$G[z]$ lie in ${\mathbb{C}}^{-}$/$\mathbb{D}$ except for exactly one pole at $0$/$1$:

$$\underset{t\to \infty }{\lim }p(t)=\underset{s\to 0}{\lim }sP(s)/\underset{k\to \infty }{\lim }g[k]=\underset{z\to 1}{\lim }(z-1)G[z]$$

Proof.

Let $G[z]$ be real, rational, and proper. Then

$$\begin{array}{rlr}g[k]& =Z^{-1}(G[z])& [\text{def of impulse response}]\\ & =Z^{-1}\left(G[\infty ]+\sum _{i=1}^n\sum _{j=1}^{n_i}\frac{c_{i,j}}{(z-p_i{)}^j}\right)& [\text{partial frac. deocmposition}]\\ & =G[\infty ]Z^{-1}(1)+\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}{Z}^{-1}\left(\frac{1}{(z-p_i{)}^j}\right)& [\text{Lineary of }{Z}^{-1}]\\ & =G[\infty ]\delta [k]+\sum _{i=1}^n\sum _{i=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}& [\text{Known }{Z}^{-1}]\end{array}$$

where

$$\delta [k]=\{\begin{array}{ll}1,& k=1\\ 0& \text{otherwise}\end{array}$$

So, given that $G[z]$ is real, rational, and proper, we can write

$$g[k]=G[\infty ]\delta [k]+\sum _{i=1}^n\sum _{i=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}[\text{expression from class}]$$

(a.) All poles of $G[z]$ lie in $\mathbb{D}$:

$$\begin{array}{rl}\underset{k\to \infty }{\lim }g[k]& =\underset{k\to \infty }{\lim }\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}\underset{\text{polynomial growth}}{\underbrace{\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)}}\underset{\text{exponential decay}}{\underbrace{p_i^{k-j}}}[\delta [k]=0\forall k>0]\\ & =0[\text{exponential decay dominates polynomial growth},p_i\in \mathbb{D}\forall i]\end{array}$$

(b.) All poles of $G[z]$ lie in $\mathbb{D}$ except exactly one at $z=1$. Therefore, we can write

$$G[z]=G[\infty ]+\frac{c_n}{z-1}+\sum _{i=1}^{n-1}\sum _{j=1}^{n_i}\frac{c_{i,j}}{(z-p_i{)}^j}[\text{partial frac. decomposition}]$$

Then, we have

$$g[k]=G[\infty ]\delta [k]+c_n+\sum _{i=1}^{n-1}\sum _{j=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}[\text{expression for g[k] from class}]$$

which in turn gives

$$\underset{k\to \infty }{\lim }g[k]=c_n[\text{follows from part (a)}]$$

Then, we have

$$\underset{z\to 1}{\lim }(z-1)G(z)=\underset{z\to 1}{\lim }{c}_n+(z-1)\text{cancelto}0\left[G[\infty ]+\sum _{i=1}^{n-1}\sum _{j=1}^{n_i}\frac{c_{i,j}}{(z-p_i{)}^j}\right]=c_n=\underset{k\to \infty }{\lim }g[k]$$

They key reason this works is that all the poles are inside the open unit disk, except for the one at $z=1$.

FVT for Step Response

Let $G[z]/P(s)$ be a real, rational, proper, and stable transfer function. Let $u[k]=1[k]/u(t)=1(t)$ be the input to $G[z]/P(s)$, and let $y[k]/y(t)$ be its output. Then the step response is:

$$\begin{array}{rl}\underset{k\to \infty }{\lim }y[k]& =G[1]\\ \underset{t\to \infty }{\lim }y(t)& =G(1)\end{array}$$

ELEC 3200

The Final Value Theorem in control states that if all poles of $sY(s)$ are strictly stable or lie in the open left half-plane, i.e., have $\text{Re}(s)<0$, then

$$y(\infty )=\underset{s\to \infty }{\lim }sY(s)$$

The more general mathematical form of the theorem states that if ${\lim }_{t\to \infty }f(t)$ exists (it has a finite limit), then

$$\underset{t\to \infty }{\lim }f(t)=\underset{s\to 0}{\lim }sF(s)$$

where $F(s)$ is the one-sided Laplace transform of $f(t)$.