# Initial/final value proofs

Proofs for Z-transform initial and final values used in signal processing, presented in the Z-Transforms article

## Proofs for Initial and Final Values Theorem

### Initial Value Theorem

The initial value theorem is similar to that in the Laplace transform. As $$z\to\infty$$, all terms except $$f[0]z^0$$ approach zero, leaving only $$f[0]$$

Let $$f[n]=0$$ for $$n\lt0$$

\begin{align} \lim_{z\to \infty}F(z) &=\lim_{z\to \infty}\sum_{n=-\infty}^{\infty}z^{-n}\,f[n]\nonumber\\ &=\sum_{n=-\infty}^{\infty}f[n]\,\lim_{z\to \infty}z^{-n}\nonumber\\ &=x[0]+\sum_{n=1}^{\infty}f[n]\,\cancelto{0}{\lim_{z\to \infty}z^{-n}}\nonumber\\ \end{align}

The initial value follows as

$$\shaded{ f[0]=\lim_{z\to\infty}F(z) }$$

### Final Value Theorem

Consider the difference between a shifted version of a function $$f[n+1]$$ and the function itself $$f[n]$$

$$f[n-1]-f[n] \label{eq:final0}$$

1) Apply the Z-transform and take the limit as $$z\to1$$ on both sides

\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} \lim_{z\to1}\left(f[n-1]-f[n]\right)\ztransform & \lim_{z\to1}\left(\sum_{n=0}^{\infty}z^{-n}\left(f[n+1]-f[n]\right)\right)= \nonumber \\ & \sum_{n=0}^{\infty}\left(f[n+1]-f[n]\right)\,\cancelto{1}{\lim_{z\to1}z^{-n}} \end{align}

write out the summation to find common terms

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \begin{split} \lim_{z\to1}\left( f[n+1]-f[n]\right)\ztransform \lim_{n\to\infty}&\left({\ccancel[red]{f[1]}}+{\ccancel[blue]{f[2]}}+{\ccancel[grey]{f[3]}}\ldots \\ +{\ccancel[orange]{f[n-1]}}+{\ccancel[teal]{f[n]}}+{f[n+1]}\\ -f[0]-{\ccancel[red]{f[1]}}-{\ccancel[blue]{f[2]}}-\ldots \\ -{\ccancel[grey]{f[n-2]}} – {\ccancel[orange]{f[n-1]}} – {\ccancel[teal]{f[n]}} \right)\\ = &-f[0] + \lim_{n\to\infty}f[n] \end{split} \label{eq:final1}$$

2) Apply the Z-transform to $$\eqref{eq:final0}$$ using the time advance property and take the limit for $$z\to1$$

\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \begin{align} \lim_{z\to1}f[n+1]-f[n]\ztransform &\lim_{z\to1}\left((zF(z)-zf[0])-F(z)\right)=\nonumber\\ &\lim_{z\to1}\left((z-1)F(z)-zf[0]\right)=\nonumber\\ &\lim_{z\to1}(z-1)F(z)-\cancelto{1}{\lim_{z\to1}z}f[0]=\nonumber\\ &-f[0]+\lim_{z\to1}(z-1)F(z) \label{eq:final2} \end{align}

Equating $$\eqref{eq:final1}$$ and $$\eqref{eq:final2}$$ we get

$$\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\lfzraised#1{\raise{10mu}{#1}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}} -{\ccancel[red]{f[0]}} + \lim_{n\to\infty}f[n] = -{\ccancel[red]{f[0]}}+\lim_{z\to1}(z-1)F(z)$$

The final value theorem, for when $$\lim_{n\to\infty}f[n]$$ exists, follows as

$$\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \shaded{\lim_{n\to\infty}f[n]=\lim_{z\to1}(z-1)F(z)}$$

Suggested next reading is Discrete transfer functions.