Proofs for Z-transform properties, presented in the Z-Transforms article.\(\)
Proofs for Properties
Linearity
Consider the time-domain function
$$
a\,f[n]+b\,g[n]
$$
This function transforms to the \(z\)-domain as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
a\,f[n]+b\,g[n]
\ztransform
&\sum_{n=0}^{\infty}\left(a\,f[n]+ b\,g[n] \right)
z^{-n}= \nonumber\\
&a\underbrace{\sum_{n=0}^{\infty} f[n]\,z^{-n}}_{F(z)} +
b\underbrace{\sum_{n=0}^{\infty} g[n]\,z^{-n}}_{G(z)}
\end{align}
$$
From which follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
a\,f[n]+b\,g[n] \ztransform a\,F(z) + b\,G(z)
}
\label{eq:linearity}
$$
Time Delay
Consider a sequence truncated at n=0, and delayed by \(a\) samples, where \(a\gt0\)
$$
f[n-a]\,\color{grey}{\gamma[n}-a\color{grey}{]}
\label{eq:delay0}
$$
Where the delayed step function \(\gamma[n-a]\) is defined as
$$
\gamma[n-a] =
\begin{cases}
0 & n\lt a \\
1 & n\geq a \\
\end{cases}
$$
The unilateral Z-transforms of \(\eqref{eq:delay0}\) is
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-a]\,\gamma[n-a]\,
\ztransform
\sum_{n=0}^\infty z^{-n}\ f[n-a]\,\gamma[n-a]
\label{eq:timedelay1}
$$
Apply \(\gamma[n-a]=0\) for all \(n\lt a\) by changing the start value of the summation
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-a]\,\gamma[n-a]\,
\ztransform
\sum_{\color{blue}{n=\mathbf{a}}}^{\color{blue}{\infty}} z^{-n}\ f[n-a]\label{eq:timedelay1b}
$$
Make the summation start at \(0\) by subtracting \(a\) on both sides of the summation start value: introduce \(m=n-a\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n-a]\,\gamma[n-a]\,
\ztransform
&\sum_{\color{blue}{m=0}}^\infty z^{-(\color{blue}{m+a})}\ f[\color{blue}{m}]\nonumber\\
\ztransform
&z^{-a}\underbrace{\sum_{m=0}^\infty z^{-m}\ f[m]}_{\color{blue}{=F(z)}}
\end{align}
$$
The unilateral Z-transform of the positive delay property follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
f[n-a]\,\color{grey}{\gamma[n}-a\color{grey}{]}\, \ztransform z^{-a}F(z)
}
\label{eq:timedelay}
$$
Time Delay #2
Consider the sequence from the previous Time Delay, where also \(f[-a]\ldots f[-1]\) are known. Once more, delayed by \(a\) samples, where \(a\gt0\)
$$
f[n-a]\,\color{grey}{\gamma[n]}
$$
The unilateral Z-transforms is
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-a]\,\gamma[n]\,
\ztransform
\sum_{n=0}^\infty z^{-n}\ f[n-a]\,\gamma[n]
$$
Apply the definition of the unit step function \(\eqref{eq:unitstep_def}\): \(\gamma[n]=0,\ \ \forall_{n\lt 0}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-a]\,\gamma[n]\,
\ztransform
\sum_{n=0}^\infty z^{-n}\ f[n-a]
$$
Substitute \(m=n-a\) to remove the \(-a\) offset from \(f[n-a]\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n-a]\,\gamma[n]\, \ztransform & \sum_{m=-a}^\infty z^{-(m+a)}\ f[m] \nonumber \\
\ztransform & z^{-a}\,\sum_{m=-a}^\infty z^{-m}\ f[m]
\end{align}
$$
To make the summation start at \(0\), subtract the first \(a\) terms, and then add these points back again.
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n-a]\,\gamma[n]\, \ztransform &\ z^{-a}\sum_{m=-a}^\infty z^{-m}\ f[m]\ \overbrace{\color{red}{-}z^{-a}\sum_{m=-a}^{-1} z^{-m}\ f[m]\color{blue}{+}z^{-a}\sum_{m=-a}^{-1} z^{-m}\ f[m]}^{\text{=0}}= \nonumber \\[10mu]
&\ z^{-a}\,\underbrace{\sum_{m=0}^\infty z^{-m}\ f[m]}_{\color{blue}{=F(z)}}+z^{-a}\sum_{m=1}^{a} z^{m}\ f[-m])
\end{align}
$$
The unilateral Z-transform of this positive time delay follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
f[n-a]\,\color{grey}{\gamma[n]}\, \ztransform z^{-a}\left(F(z)+\sum_{m=1}^{a} z^m\ f[-m]\right)
}
\label{eq:timedelay2}
$$
Time Advance
Consider a sequence advanced by \(a\) samples, where \(a\gt0\), and then truncated at time \(0\)
$$
f[n+a]\,\color{grey}{\gamma[n]}\label{eq:timeadv_def}
$$
The unilateral Z-transforms of \(\eqref{eq:timeadv_def}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n+a]\,\gamma[n]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n+a]\,\gamma[n]
$$
Apply the definition of the unit step function \(\eqref{eq:unitstep_def}\): \(\gamma[n]=0,\ \ \forall_{n\lt 0}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n+a]\,\gamma[n]\,
\ztransform
\sum_{n=0}^\infty z^{-n}\ f[n+a]
$$
Substitute \(m=n+a\) to remove the \(+a\) offset from \(f[n+a]\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n+a]\,\gamma[n]\, \ztransform & \sum_{\color{blue}{m=a}}^\infty z^{-(\color{blue}{m-a})}\ f[\color{blue}{m}] \nonumber \\
\ztransform & z^a\,\sum_{m=a}^\infty z^{-m}\ f[m]
\end{align}
$$
To make the summation start at \(0\), add the first \(a\) terms, and then subtract these points again.
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n+a]\,\gamma[n]\, \ztransform & \ z^a\sum_{m=a}^\infty z^{-m}\ f[m]\ \overbrace{\color{blue}{+}\,z^a\sum_{m=0}^{a-1} z^{-m}\ f[m]\,\color{red}{-}\,z^a\sum_{m=0}^{a-1} z^{-m}\ f[m]}^{\text{=0}} \nonumber \\
\ztransform &\ z^a\Big(\underbrace{\sum_{m=0}^\infty z^{-m}\ f[m]}_{\color{blue}{=F(z)}}-\sum_{m=0}^{a-1} z^{-m}\ f[m])\Big)
\end{align}
$$
The unilateral Z-transform of the positive time advance follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
f[n+a]\,\color{grey}{\gamma[n]}\, \ztransform z^a\Big( F(z)-\sum_{m=0}^{a-1} z^{-m}\ f[m]\Big)
}
\label{eq:timeadvance}
$$
Time Multiply
Consider multiplied by the sample number \(n\), and truncated at \(n=0\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
n\,f[n]\,\gamma[n]
$$
Sorry, proof missing. See “n scaled” pair. Have the proof? Please share in the comments.
The unilateral Z-transform
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
n\,f[n]\,\color{grey}{\gamma[n]} \ztransform -z\frac{\text{d}F(z)}{\text{d}z}
}
$$
Modulation
Consider a sequence multiplied with complex scalar \(a^n\), and truncated at \(n=0\)
$$
a^n\,f[n]\,\color{grey}{\gamma[n]}
$$
Take the unilateral Z-transforms
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
a^n\,f[n]\,\gamma[n]
\ztransform
&\sum_{n=0}^{\infty}z^{-n}a^nf[n]\nonumber\\
\ztransform
&\sum_{n=0}^{\infty}f[n]\left(\underbrace{a^{-1}z}_{\color{blue}{=r}}\right)^{-n}\label{eq:scaling_def}
\end{align}
$$
Recall the power series
$$
\begin{align}
\sum_{n=0}^{\infty}r^n = \frac{1}{1-r},&&|r|\lt1\nonumber
\end{align}
\nonumber
$$
Apply the power series to equation \(\eqref{eq:scaling_def}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
a^n\,f[n]\,\gamma[n] \ztransform & \sum_{n=0}^{\infty}\,f[n]\,\left(za^{-1}\right)^{-n},&|za^{-1}|\gt 1
\end{align}
$$
The unilateral Z-transform of the modulation follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\lfzraised#1{\raise{10mu}{#1}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}}
\newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}}
\begin{align}
\shaded{
a^n\,f[n]\,\color{grey}{\gamma[n]} \ztransform F\left(a^{-1}z\right)
}
\end{align}
$$
In the \(z\)-domain, \(F\left(a^{-1}z\right)\) has a zero at \(z=0\) and pole at \(z=a\).
Note that scaling will affect the region of convergence and all the pole-zero locations will be scaled by a factor of \(a\).
Convolution
Convolution is used to calculate an output signal when the input signal and transfer function are known in the time domain. Convolution is related to autocorrelation that we used in Arduino Pitch Detector.
Consider a convolution of two sequences truncated at the origin (\(n=0\))
$$
\begin{align}
(f\ast g)[n]\,\color{grey}{\gamma[n]} &=\sum_{m=-\infty}^{\infty}f[m]\,g[n-m]\,\gamma[n]\nonumber\\
&=\ \sum_{m=0}^{\infty}\ f[m]\,g[n-m]\\
\end{align}
$$
Apply the unilateral Z-transforms definition
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
(f\ast g)[n]\,\gamma[n]
\ztransform
&\color{purple}{\sum_{n=0}^{\infty}}\left(z^{-n}\color{blue}{\sum_{m=0}^{\infty}}f[m]\,g[n-m]\right)&\text{reverse } \tiny\sum\nonumber\\[8mu]
\ztransform
&\color{blue}{\sum_{m=0}^{\infty}}\left(\color{purple}{\sum_{n=0}^{\infty}}z^{-n}f[m]\,g[n-m]\right)&f[m]\text{ indep.}\nonumber\\[8mu]
\ztransform
&\sum_{m=0}^{\infty}\left(f[m]\,\underbrace{\sum_{n=0}^{\infty}z^{-n}\,g[n-m]}_{\color{green}{z^{-m}G(z)}}\right)
\label{eq:convolution0}
\end{align}
$$
Recall the equations \((\ref{eq:timedelay1b}, \ref{eq:timedelay})\) from the Delay Property
$$
\sum_{n=0}^\infty z^{-n}\ g[n-m]=\color{green}{z^{-m}G(z)}
\label{eq:convolution1}
$$
Apply \(\eqref{eq:convolution1}\) to \(\eqref{eq:convolution0}\)
$$
\begin{align}
(f\ast g)[n]\,\gamma[n] \ztransform & \sum_{m=0}^{\infty}f[m]\ \color{green}{z^{-m}G(z)}&\overset{G(z)\text{ indep.}}{\Rightarrow} \nonumber \\
\ztransform & \ G(z)\underbrace{\sum_{m=0}^{\infty}f[m]\ z^{-m}}_{F(z)}
\end{align}
$$
The unilateral Z-transform of the convolution in the time-domain simplifies to a a multiplication in the \(z\)-domain.
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
(f\ast g)[n]\,\color{grey}{\gamma[n]} \ztransform F(z)\,G(z)
}
\label{eq:convolution}
$$
Conjugation
Consider conjugation
$$
f^\star[n]
$$
Take the unilateral Z-transforms
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f^\star[n] \ztransform & \sum_{n=0}^{\infty}z^{-n}f^*[n] \nonumber \\
f^\star[n] \ztransform & \sum_{n=0}^{\infty}\left(\left(z^\star\right)^{-n} f[n]\right)^\star \nonumber \\
f^\star[n] \ztransform &\left(\sum_{n=0}^{\infty}\left(z^\star\right)^{-n} f[n]\right)^\star
\end{align}
$$
The unilateral Z-transform follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
f^\star[n] \ztransform F^{\star}(z^{\star})
}
$$
First Difference
Differencing in the discrete time domain is analogous to differentiation in the continuous time domain.
Consider the difference
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n]-f[n-1]\ztransform F(z)-z^{-1}F(z)
$$
The unilateral \(z\)-transform of the first difference follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
f[n]-f[n-1]\ztransform \left(1-z^{-1}\right)F(z)
}
$$
Accumulation
Accumulation in the discrete time domain is analogous to integration in the continuous time domain.
Consider the accumulation
$$
\sum_{k=-\infty}^{n}x[k]\label{eq:accum_def}
$$
Take the unilateral Z-transforms of equation \(\eqref{eq:accum_def}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\sum_{k=-\infty}^{n}x[k]
\ztransform
&\sum_{n=0}^{\infty}\left(z^{-n}\sum_{k=-\infty}^{n}x[k]\right)\nonumber\\
\ztransform
&\sum_{n=0}^{\infty}z^{-n}\left(f[-\infty]+\cdots+f[n-2]+f[n-1]+f[n]\right)\nonumber\\
\ztransform
&\sum_{n=0}^{\infty}\left(z^{-n}f[n]+z^{-n}f[n-1]+z^{-n}f[n-2]+\cdots+z^{-n}f[-\infty]\right)\label{eq:accum1}
\end{align}
$$
Take the delay property from equation \(\eqref{eq:timedelay}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-a]\,\color{grey}{\gamma[n}-a\color{grey}{]}\, \ztransform \,z^{-a}F(z) \nonumber
$$
Apply the delay property to \(\eqref{eq:accum1}\)
$$
\begin{align}
\sum_{k=-\infty}^{n}x[k] \ztransform &\sum_{n=0}^{\infty}\left(z^{-n}f[n]+z^{-n}f[n-1]+z^{-n}f[n-2]+\ldots\right) \\[6mu]
\ztransform & \left(F(z)+z^{-1}F(z)+z^{-2}F(z)+\ldots\right) \nonumber \\[4mu]
\ztransform & F(z)\left(1+z^{-1}+z^{-2}+\ldots\right)=F(z)\sum_{k=0}^{\infty}z^{-k}
\end{align}
$$
The unilateral Z-transform of accumulation follows from applying thepower series
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
\sum_{i=0}^{n}x[i] \ztransform F(z)\,\frac{z}{z-1}
}
$$
Double poles
We will take a different approach for this proof.
Recall the Quotient Rule from calculus
$$
\frac{\text{d}}{\text{d}x}\left(\frac{u}{v}\right)=\frac{v\frac{\text{d}u}{\text{d}x}-u\frac{\text{d}v}{\text{d}x}}{v^2}\nonumber
$$
Take the first derivative in \(z^{-1}\) of \(\frac{1}{{1-pz^{-1}}}\)
$$
\frac{\text{d}}{\text{d}y}\left(\frac{1}{{1-pz^{-1}}}\right)=\frac{(1-pz^{-1})0-1(-p)}{(1-pz^{-1})^2}=\frac{p}{(1-pz^{-1})^2}
$$
Apply the power series and differentiate
$$
\begin{align}
\frac{1}{(1-pz^{-1})^2}
&=\frac{1}{p}\ \frac{\text{d}}{\text{d}z^{-1}}\left(\frac{1}{{1-pz^{-1}}}\right), \quad \text{power series} \nonumber \\
&=\frac{1}{p}\ \frac{\text{d}}{\text{d}z^{-1}}\left(\sum_{n=0}^{\infty}\left(pz^{-z}\right)^n\right) \nonumber \\
&=\frac{1}{p}\ \frac{\text{d}}{\text{d}z^{-1}}\left(1+pz^{-1}+p^2z^{-2}+p^3z^{-3}+\cdots\right),\quad \tfrac{\mathrm{d}}{\mathrm{d}z^{-1}} \nonumber \\
&=\frac{1}{p}\ \left(\cancel{0}+p+2p^2z^{-1}+3p^3z^{-2}+\cdots\right)\nonumber\\
&=1+2p^1z^{-1}+3p^2z^{-2}+\cdots
\end{align}
$$
We recognize this as the Z-transform of \((n+1)p^n\)
$$
\begin{align}
\frac{1}{(1-pz^{-1})^2}
&=\sum_{n=0}^{\infty}\underbrace{(n+1)p^n}_{\color{blue}{=f[n]}}\,z^{-n}\\
\end{align}
$$
The unilateral Z-transform of \((n+1)p^n\) follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\shaded{
(n+1)p^n\,\color{grey}{\gamma[n]} \ztransform \frac{1}{(1-pz^{-1})^2}
}
$$
Your blog has been really helpful! I really appreciate what you’re doing! I just noticed that for the Z transform proofs there are a few typos. Nothing serious, as everything is understandable. On the development of equation 98 for the cosine function there are a few Ts missing and there’s an n on the first exp at the beginning. Also on equation 99 you’re missing the T for the cosine’s argument on the numerator. Keep it up! Your blog is amazing!