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

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