Final Value Theorem

The theorem states… Let $p(t)/g[k]$ be a signal with Laplace/z transform that is real, rational, and proper. Then

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

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

2. If all pose of $P(s)/G[z]$ like in ${\mathbb{C}}^{-}/\mathbb{D}$ excepts for exactly one pole at 0/1

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

, THE FINAL VALUE THEOREM IS WHEN $z$ GOES TO 1

Proof Setup

$$g[k]={\mathcal{Z}}^{-1}(G[z])\text{(Definition of impulse response)}$$ $$={\mathcal{Z}}^{-1}\left(G[\infty ]+\sum _{i=1}^n\sum _{j=1}^{n_i}\frac{c_{i,j}}{(z-p_i{)}^j}\right)$$ $$=G[\infty ]{\mathcal{Z}}^{-1}(1)+\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}{\mathcal{Z}}^{-1}\left(\frac{1}{(z-p_i{)}^j}\right)$$ $$=G[\infty ]\cdot \delta [k]+\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}\frac{n(n-1)\dots (n-k+1)}{k!}{p}_i^{k-j}$$

Proof of Part 1

$$\text{Given: All poles of }G[z]\text{ lie in }\mathbb{D}$$ $$\text{WTS: }\underset{k\to \infty }{\lim }g[k]=0$$ $$G[z]\text{ is real, rational, and proper}$$ $$⟹g[k]=G[\infty ]\delta [k]+\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}$$

where $\delta$ is 1 when $k=0$ and 0 otherwise. so that term is 0

$$⟹\underset{k\to \infty }{\lim }g[k]=\underset{k\to \infty }{\lim }\sum _{i=1}^n\sum _{j=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}=0$$ $$where\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n(n-1)\dots (n-k+1)}{k!}$$

Proof of Part 2

$$\text{Given: All poles of }G[z]\text{ lie in }\mathbb{D}\text{ except exactly one at }z=1$$ $$\text{WTS: }\underset{k\to \infty }{\lim }g[k]=\underset{z\to 1}{\lim }(z-1)G[z]$$

Let’s specifically single out the pole at 1 from the summation

$$⟹G[z]=G[\infty ]+\frac{c_n}{z-1}+\sum _{i=1}^{n-1}\sum _{j=1}^{n_i}\frac{c_{i,j}}{(z-p_i{)}^j}$$ $$⟹g[k]=G[\infty ]\delta [k]+c_n+\sum _{i=1}^{n-1}\sum _{j=1}^{n_i}{c}_{i,j}\left(\genfrac{}{}{0ex}{}{k-1}{j-1}\right)p_i^{k-j}$$ $$⟹\underset{k\to \infty }{\lim }g[k]=c_n$$ $$\underset{z\to 1}{\lim }(z-1)G[z]=c_n=\underset{k\to \infty }{\lim }g[k]$$ $$Y[z]=G[z]\cdot \frac{z}{z-1}$$ $$\underset{k\to \infty }{\lim }y[k]=\underset{z\to 1}{\lim }(z-1)Y[z]=\underset{z\to 1}{\lim }zG[z]=G[1]$$

Corollary

Let $G[z]/P(s)$ be a real, ration, proper, and stable transfer function. Let $u[k]/u(t)$ be a unit step input to such transfer functions. Then we will see that

$$\underset{k\to \infty }{\lim }y[k]=G[1]$$ $$\underset{t\to \infty }{\lim }y(t)=P(0)$$