# Z-transform Proofs

$$\require{AMSsymbols} \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\lfzraised#1{\raise{10mu}{#1}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \require{cancel} \newcommand\ccancel[2][black] {\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black] {\color{#1}{\cancelto{#2}{\color{black}{#3}}}}$$These are the proofs for the Z-transforms 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
\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
$$\shaded{ a\,f[n]+b\,g[n] \ztransform a\,F(z) + b\,G(z) } \label{eq:linearity}$$

### 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
$$f[n-a]\,\gamma[n-a]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n-a]\,\gamma[n-a] \label{eq:timedelay1}$$

Apply $$\gamma[n-a]=0$$ for all $$n\lt a$$ by changing the start value of the summation
$$f[n-a]\,\gamma[n-a]\, \ztransform \sum_{\color{blue}{n=\mathbf{a}}}^{\color{blue}{\infty}} z^{-n}\ f[n-a]\label{eq:timedelay1b}$$

Make the summation start at $$0$$ by subtracting $$a$$ on both sides of the summation start value: introduce $$m=n-a$$
\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
$$\shaded{f[n-a]\,\color{grey}{\gamma[n}-a\color{grey}{]}\, \ztransform z^{-a}F(z)} \label{eq:timedelay}$$

### 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
$$f[n-a]\,\gamma[n]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n-a]\,\gamma[n]$$

Apply the definition of the unit step function $$\eqref{eq:unitstep_def}$$: $$\gamma[n]=0,\ \ \forall_{n\lt 0}$$
$$f[n-a]\,\gamma[n]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n-a]$$

Substitute $$m=n-a$$ to remove the $$-a$$ offset from $$f[n-a]$$
\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.
\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
$$\shaded{f[n-a]\,\color{grey}{\gamma[n]}\, \ztransform z^{-a}\left(F(z)+\sum_{m=1}^{a} z^m\ f[-m]\right)} \label{eq:timedelay2}$$

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 $$\eqref{eq:timeadv_def}$$
$$f[n+a]\,\gamma[n]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n+a]\,\gamma[n]$$

Apply the definition of the unit step function $$\eqref{eq:unitstep_def}$$: $$\gamma[n]=0,\ \ \forall_{n\lt 0}$$
$$f[n+a]\,\gamma[n]\, \ztransform \sum_{n=0}^\infty z^{-n}\ f[n+a]$$

Substitute $$m=n+a$$ to remove the $$+a$$ offset from $$f[n+a]$$
\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.
\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
$$\shaded{f[n+a]\,\color{grey}{\gamma[n]}\, \ztransform z^a\Big( F(z)-\sum_{m=0}^{a-1} z^{-m}\ f[m]\Big)} \label{eq:timeadvance}$$

### Time Multiply

Consider multiplied by the sample number $$n$$, and truncated at $$n=0$$
$$n\,f[n]\,\gamma[n]$$

… proof missing .. see “n scaled” pair

The unilateral Z-transform
$$\shaded{n\,f[n]\,\color{grey}{\gamma[n]} \ztransform -z\frac{\text{d}F(z)}{\text{d}z}}$$

### 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
\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 $$\eqref{eq:scaling_def}$$
\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
\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
\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.
$$\shaded{(f\ast g)[n]\,\color{grey}{\gamma[n]} \ztransform F(z)\,G(z)}\label{eq:convolution}$$

### Conjugation

Consider conjugation
$$f^\star[n]$$

Take the unilateral Z-transforms
\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
$$\shaded{f^\star[n] \ztransform F^{\star}(z^{\star})}$$

### First Difference

Differencing in the discrete time domain is analogous to differentiation in the continuous time domain.

Consider the difference
$$f[n]-f[n-1]\ztransform F(z)-z^{-1}F(z)$$

The unilateral $$z$$-transform of the first difference follows as
$$\shaded{f[n]-f[n-1]\ztransform \left(1-z^{-1}\right)F(z)}$$

### 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 $$\eqref{eq:accum_def}$$
\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 $$\eqref{eq:timedelay}$$

$$f[n-a]\,\color{grey}{\gamma[n}-a\color{grey}{]}\, \ztransform \,z^{-a}F(z)$$

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
$$\shaded{\sum_{i=0}^{n}x[i] \ztransform F(z)\,\frac{z}{z-1}}$$

### 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)&\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)&\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
$$\shaded{(n+1)p^n\,\color{grey}{\gamma[n]} \ztransform \frac{1}{(1-pz^{-1})^2}}$$

## Proofs for pairs

### Impulse

The discrete impulse function $$\delta[t]$$ is different from the continuous impulse function. The impulse function is commonly used as an theoretical input signal to study the filter’s behavior.

The definition is
$$\delta[n]=\begin{cases} 1, & n=0 \\ 0, & n\neq0 \end{cases}\label{eq:impuls_def1}$$

Apply the unilateral Z-transforms definition to equation $$\eqref{eq:impuls_def1}$$
$$\delta[n] \ztransform \Delta(z)=\sum_{n=0}^{\infty}z^{-n}\ \delta[n]$$

Since the impulse is $$0$$ everywhere but at $$n=0$$, the summation simplifies to
$$\delta[n] \ztransform \Delta(z)= \cancelto{1}{z^{-0}}\ \cancelto{1}{\delta[0]}$$

The unilateral Z-Transform of the Unit Impulse Function follows as
\begin{align} \shaded{\delta[n] \ztransform 1\triangleq\Delta(z) }, &&\text{all }z,\text{ including }\infty \end{align} \label{eq:impulse}

This is very similar to the Laplace transform of the continuous impulse function.

### Delayed Impulse

Consider the discrete delayed impulse function $$\delta[n-a]$$
$$\delta[n-a]=\begin{cases} 1, & n=a \\ 0, & n\neq0 \end{cases}\label{eq:delayedimpuls_def1}$$

…proof missing… similar to delay proof?

The unilateral Z-Transform of the Delayed Unit Impulse Function follows as
\begin{align} \shaded{\delta[n-a]\,\ztransform\, \shaded{\begin{cases}z^{-a},&a\geq0\\0,&a\lt0\end{cases}}},&&z\neq0 \end{align} \label{eq:delayedimpulse}

### Unit Step

The discrete unit or Heaviside step function, denoted with $$\gamma[n]$$ is defined as
$$\gamma[n]=\begin{cases} 0, & n\lt 0 \\ 1, & n\geq 0 \end{cases} \label{eq:unitstep_def}$$

The unilateral Z-transforms of $$\eqref{eq:unitstep_def}$$ follows
\begin{align} \gamma[n] \ztransform \Gamma(z) &=\sum_{n=0}^\infty z^{-n}\,\cancelto{1}{\gamma[t] } =\sum_{n=0}^\infty\,z^{-n}\nonumber\\ &=\sum_{n=0}^\infty\,\underbrace{\left(z^{-1}\right)^n}_{\color{blue}{r^n}}\label{eq:unitstep0} \end{align}

Apply the power series

\begin{align} \sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1\nonumber \end{align}\nonumber

to $$\eqref{eq:unitstep0}$$
\begin{align} \gamma[n] \ztransform \Gamma(z) = \frac{1}{1-z^{-1}},&&|z^{-n}|\lt1 \end{align}

The unilateral Z-transform for the step function follows as
\begin{align} \shaded{ \gamma[n] \ztransform \frac{z}{z-1}\triangleq\Gamma(z) },&&|z|\gt1 \end{align} \label{eq:step}

### Scaled

Consider the discrete power function starting at $$n=0$$
$$f[n]=\begin{cases} 0, & n\lt0 \\ a^n, & n\geq 0 \end{cases}\label{eq:scaled_def}$$

Apply the unilateral Z-transforms and the power series
\begin{align} a^n\,\color{grey}{\gamma[n]} \ztransform &\sum_{n=0}^\infty z^{-n}\,a^n\nonumber\\ \ztransform &\sum_{n=0}^\infty (az^{-1})^n\nonumber\\ \ztransform &\frac{1}{1-az^{-1}},&|z|\gt|a| \end{align}

The unilateral Z-Transform of the scaled function follows
\begin{align} \shaded{a^n\,\color{grey}{\gamma[n]} \ztransform {\frac{z}{z-a}}},&&|z|\gt|a|\label{eq:scaled} \end{align}

The Binomial scaled proof comes to this same transform.

### Scaled delayed

Recall the delay property, and the scaled pair

\begin{align} f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\, &\ztransform z^{-b}F(z)\nonumber\\ a^n\ \color{grey}{\gamma[n]} &\ztransform \frac{z}{z-a},&|z|\gt|a|\nonumber \end{align}\nonumber

The unilateral Z-Transform of the scaled delayed function follows
\begin{align} \shaded{a^{n-1}\gamma[n-1] \ztransform \dfrac{1}{z-1}},&&|z|\gt|a| \end{align}

### $$n$$ scaled

Consider the discrete ramp function starting at $$n=0$$
$$\begin{cases} 0, & n\lt0 \\ n\,a^n, & n\geq 0 \end{cases}\label{eq:nscaled_def}$$

Apply the unilateral Z-transforms definition and expand
\begin{align} n\,a^n \ztransform F(z)=&\sum_{n=0}^\infty z^{-n}\ n\,a^n\nonumber\\ \ztransform &\left[\cancel{0}+\cancel{1}az^{-1}+2a^2z^{-2}+3a^3z^{-3}+\cdots\right]\nonumber\\ \ztransform &\left[az^{-1}+2a^2z^{-2}+3a^3z^{-3}+\cdots\right]\label{eq:ramp1} \end{align}

To counter the infinite series, introduce $$az^{-1}F(z)$$
$$az^{-1}\,F(z)=\left[a^2z^{-2}+2a^3z^{-3}+3a^4z^{-4}+\cdots\right]\label{eq:ramp2}$$

Subtract $$\eqref{eq:ramp2}$$ from $$\eqref{eq:ramp1}$$
\begin{align} F(z)-az^{-1}\,F(z) &=\left(az^{-1}+\ccancel[red]{2}a^2z^{-2}+\ccancel[green]{3}a^3z^{-3}+\cdots\right)-\\ &\quad\quad\left(\ccancel[red]{a^2z^{-2}}+\ccancel[green]{2a^3z^{-3}}+\ccancel[orange]{3a^4z^{-4}}+\cdots\right)\Rightarrow\nonumber\\[12mu] (1-az^{-1})\,F(z) &=\left(az^{-1}+a^2z^{-2}+a^3z^{-3}+\cdots\right)\nonumber\\[6mu] &=az^{-1}\left(1+az^{-1}+a^2z^{-2}+\cdots\right)\nonumber\\[6mu] &=az^{-1}\sum_{n=0}^{\infty}\underbrace{\left(az^{-1}\right)^n}_{\color{blue}{r^n}}\label{eq:ramp3} \end{align}

Apply the power series

\begin{align} \sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1\nonumber \end{align}\nonumber

to $$\eqref{eq:ramp3}$$ where $$r=az^{-1}$$
\begin{align} (1-az^{-1})\,F(z) &=az^{-1}\,\frac{1}{1-az^{-1}},&|z|\gt1\nonumber\\ F(z)&=\frac{az^{-1}}{(1-az^{-1})^2},&|z|\gt1 \end{align}

The unilateral Z-Transform of the $$n$$ scaled function follows
\begin{align} \shaded{n\,a^n\ \color{grey}{\gamma[n]} \ztransform \frac{az}{\left(z-a\right)^2}},&&|z|\gt1\label{eq:timescaled} \end{align}

The Binomial scaled proof comes to this same transform.

### Ramp

This is a special case for the $$n$$ scaled $$z$$-transform, where $$a=1$$

Substitute $$a=1$$ in

\begin{align} n\,a^n\ \color{grey}{\gamma[n]} \ztransform \frac{az}{\left(z-a\right)^2},&&|z|\gt1\nonumber \end{align}\nonumber

The unilateral Z-Transform of the discrete ramp function follows
\begin{align} \shaded{n\ \color{grey}{\gamma[n]} \ztransform \frac{z}{\left(z-1\right)^2}},&&|z|\gt1 \end{align} \label{eq:ramp}

### Binomial scaled, $$|z|\gt |a|$$

We will do this proof starting from the $$z$$-domain
$$F(z)=\frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}$$ where $$n$$ is an integer, and $$a$$ is a constant possibly complex.

Recall the Negative Binomial Series

\begin{align} (1-x)^{-m}&=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,x^n,&|x|\lt1\nonumber\\ \end{align}\nonumber

Work towards the form $$(1-x)^{-m}$$
\begin{align} F(z)&=(1-\underbrace{az^{-1}}_{\color{blue}{=x}})^{-m} \end{align}

Use the Binomial Series where $$x=az^{-1}$$
\begin{align} F(z)=\left(1-(az^{-1})\right)^{-m}&=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,(az^{-1})^k,&|az^{-1}|\lt1\nonumber\\ &=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,a^n\,z^{-n},&|z|\gt|a| \end{align}

Apply the Symmetry Rule for Binomial Coefficients $${n \choose k}={n \choose n−k}$$
\begin{align} F(z)&=\sum_{n=0}^{\infty}\underbrace{{n+m-1 \choose m-1}\,a^n}_{\color{blue}{=f[n]}}\,z^{-n},&|z|\gt|a| \end{align}

The unilateral Z-Transform follows
\begin{align} \shaded{{n+m-1 \choose m-1}\,a^n\,\color{grey}{\gamma[n]} \ztransform \frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}}&&|z|\gt |a|\label{eq:binomialscaled} \end{align}

Recall the delay property

$$f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\,\ztransform\,z^{-b}F(z)\nonumber$$

Combining equation $$\eqref{eq:binomialscaled}$$ with the delay property
\begin{align} {n+m-1 \choose m-1}\,a^{n-m+1}\,\color{grey}{\gamma[n-m+1]} \,\ztransform\, &z^{-(m-1)}\frac{z^m}{(z-a)^m}&|z|\gt |a| \nonumber\\ \end{align}

So that in general
\begin{align} \shaded{n+m-1 \choose m-1}\,a^{n}\,\color{grey}{\gamma[n-m+1] \,\ztransform\,\frac{a^{m-1}z}{(z-a)^m}}&&|z|\gt |a| \nonumber\\ \end{align}

For the case where $$m=1$$
\begin{align} {n \choose 0}\,a^n\,\gamma[n] \,\ztransform\, &\frac{z}{z-a}&|z|\gt |a|\nonumber\\ \frac{\ccancel[red]{n!}}{0!\ccancel[red]{(n-0)!}}\,a^n\,\gamma[n] \,\ztransform \nonumber\\ a^n\,\gamma[n] \,\ztransform \end{align} so that \begin{align} \shaded{a^n\,\gamma[n] \,\ztransform \frac{z}{z-a}}&&|z|\gt |a| \end{align}

For the case where $$m=2$$
\begin{align} {n \choose 1}\,a^{n-1}\,\gamma[n-1] \,\ztransform\, &\frac{z}{(z-a)^2}&|z|\gt |a|\nonumber\\ \frac{\ccancelto[red]{n}{n!}}{1!\ccancel[red]{(n-1)!}}\,a^{n-1}\,\gamma[n] \,\ztransform\nonumber\\ n\,a^{n-1}\,\gamma[n] \,\ztransform \end{align} so that \begin{align} \shaded{n\,a^n\,\gamma[n] \,\ztransform \frac{az}{(z-a)^2}}&&|z|\gt |a| \end{align}

For the case where $$m=3$$
\begin{align} {n \choose 3}\,a^{n-2}\,\gamma[n-2] \,\ztransform\, &\frac{z}{(z-a)^3}&|z|\gt |a|\nonumber\\ \frac{\ccancelto[red]{n(n-1)}{n!}}{2!\ccancel[red]{(n-2)!}}\,a^{n-2}\,\gamma[n] \,\ztransform\nonumber\\[6mu] \frac{n(n-1)}{2}\,a^{n-2}\,\gamma[n] \,\ztransform \end{align} so that \begin{align} \shaded{\tfrac{1}{2}{n(n-1)}\,a^n\,\gamma[n] \,\ztransform \frac{a^2z}{(z-a)^3}}&&|z|\gt |a| \end{align}

### Binomial scaled, $$|z|\lt |a|$$

Similar to the previous proof, we will do this starting from the $$z$$-domain
$$F(z)=\frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}$$ where $$n$$ is an integer, and $$a$$ is a constant possibly complex.

Recall the Binomial Series

\begin{align} (1+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,x^k,&|x|\lt1\nonumber\\ \end{align}\nonumber

Work towards the form $$(1-x)^{-m}$$
\begin{align} F(z)&=(1\underbrace{-az^{-1}}_{\color{blue}{=x}})^{-m} \end{align}

Use the Binomial Series where $$x=-az^{-1}$$
\begin{align} F(z)=(1-az^{-1})^{-m}&=\sum_{n=0}^{\infty}{-m \choose n}\,(-az^{-1})^n\nonumber\\ &=\sum_{n=0}^{\infty}{-m \choose n}\,(-1)^n\,a^nz^{-n}\nonumber\\ \end{align}

Apply the Moving Top Index to Bottom in Binomial Coefficient $${n \choose m}=(-1)^{n-m}\,{-(m+1) \choose n-m}$$
\begin{align} F(z)=(1-az^{-1})^{-m} &=\sum_{n=0}^{\infty}{-(n+1) \choose -m-n}\,(-1)^{-m-n}\,(-1)^n\,a^nz^{-n}\nonumber\\ &=\sum_{n=0}^{\infty}{-n-1 \choose -m-n}\,(-1)^{-m}\,a^nz^{-n}\nonumber\\ \end{align}

Apply the Symmetry Rule for Binomial Coefficients $${n \choose k}={n \choose n−k}$$
\begin{align} F(z)=(1-az^{-1})^{-m} &=\sum_{n=0}^{\infty}{-n-1 \choose \cancel{-n}-1+m\cancel{+n})}\,(-1)^{-m}\,a^nz^{-n}\nonumber\\ &=\sum_{n=0}^{\infty}\,\underbrace{(-1)^{-m}\,{-n-1 \choose m-1}\,a^n}_\color{blue}{=f[n]}\,z^{-n}\nonumber\\ \end{align}

The unilateral Z-Transform follows
\begin{align} \shaded{(-1)^{-m}\,{-n-1 \choose m-1}\,a^n\,\color{grey}{\gamma[n]} \ztransform \frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}}&&|z|\lt |a| \end{align}

### Exponential

Consider the discrete exponential function starting at $$n=0$$
$$f[n]=\begin{cases} 0, & n<0 \\ \mathrm{e}^{-anT}, & n\geq 0 \end{cases}\label{eq:exponential_def}$$

Apply the unilateral Z-transforms definition
\begin{align} \mathrm{e}^{-anT}\ \gamma[n] \ztransform &\sum_{n=0}^\infty z^{-n}\ \mathrm{e}^{-anT}\nonumber=\\ &\sum_{n=0}^\infty {\underbrace{\left(z^{-1}\ \mathrm{e}^{-aT}\right)}_a}^n\label{eq:exponential0} \end{align}

Apply the power series

\begin{align} \sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1\nonumber \end{align}\nonumber

to $$\eqref{eq:exponential0}$$, where $$a=z^{-1}\ \mathrm{e}^{-aT}$$
\begin{align} \mathrm{e}^{-anT}\ \gamma[n] \ztransform &\frac{1}{1-z^{-1}\ \mathrm{e}^{-aT}}\ ,&|\mathrm{e}^{-aT}|\lt z \end{align}

The unilateral Z-Transform of the exponential function follows
\begin{align} \shaded{\mathrm{e}^{-anT}\ \gamma[n] \ztransform \frac{z}{z-\mathrm{e}^{-aT}}},\ \ &&{|\mathrm{e}^{-aT}|\lt |z|} \end{align} \label{eq:exponential}

### Sine

The Z-transforms of the sine is similar to that of the cosine function, except that it uses the Euler identity for sine

\begin{align} \shaded{\sin(\omega n)\ztransform\frac{z\sin(\omega T)}{z^2-2z\cos(\omega T)+1}},&& |z|\gt1\\ \end{align} \label{eq:sine}

### Cosine

A common notation is to use $$\Omega$$ to represent frequency in the $$z$$-domain, and $$\omega$$ for frequency in the $$s$$-domain. Here we use $$\omega$$ to represent both types of frequency. Another notation that you may encounter is $$\omega_0$$ to represent $$\omega T$$.

Consider the cosine function starting at $$n=0$$

$$f[n]=\cos(\omega nT)\,\gamma[n]\label{eq:cos_def}$$

The unilateral Z-transforms of the cosine is
$$\cos(\omega nT)\ \gamma[n] \ztransform \sum_{n=0}^\infty z^{-n}\ \cos(\omega nT)$$

Recall the Euler identity for cosine

$$\cos\varphi=\frac{\mathrm{e}^{j\varphi}+\mathrm{e}^{-j\varphi}}{2}\nonumber$$

Apply the identify for cosine
\begin{align} \cos(\omega nT)\ \gamma[n] \ztransform &\frac{1}{2}\left(\frac{1}{1-\left(z^{-1}\ e^{j\omega T}\right)}+\frac{1}{1-\left(z^{-1}\ e^{-j\omega T}\right)}\right),& |z|\leq |e^{j\omega T}|=1\nonumber\\ =\,&\frac{1}{2}\left(\frac{z}{z-e^{j\omega T}}+\frac{z}{z-e^{-j\omega T} }\right),& |z|\lt1 \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 geometric power series
\begin{align} \cos(\omega nT)\ \gamma[n] \ztransform &\frac{1}{2}\left(\frac{1}{1-\left(z^{-1}\ e^{j\omega T}\right)}+\frac{1}{1-\left(z^{-1}\ e^{-j\omega T}\right)}\right),& |z|\leq |e^{j\omega T}|=1\nonumber\\ =\,&\frac{1}{2}\left(\frac{z}{z-e^{j\omega T}}+\frac{z}{z-e^{-j\omega T}}\right),& |z|\lt1 \end{align}

Bring over a common denominator and regroup
\begin{align} F(z)&=\frac{1}{2}\left(\frac{z(z-e^{-j\omega})}{(z-e^{j\omega n})(z-e^{-j\omega})}+\frac{z(z-e^{j\omega})}{(z-e^{-j\omega})(z-e^{-j\omega})}\right),& |z|<1\nonumber\\ &=\frac{1}{2}\left(\frac{z(z-e^{-j\omega})+z(z-e^{j\omega})}{z^2-ze^{-j\omega}+ze^{j\omega}+e^{j\omega}e^{-j\omega}}\right),& |z|<1\nonumber\\ &=\frac{1}{2}\left(\frac{z^2-ze^{-j\omega}+z^2-ze^{j\omega}}{z^2-z\left(e^{-j\omega}+e^{j\omega}\right)+e^0}\right),& |z|<1\nonumber\\ &=\frac{z^2-z\frac{1}{2}\left(e^{-j\omega}+e^{j\omega}\right)}{z^2-z\left(e^{-j\omega}+e^{j\omega}\right)+1},& |z|<1\\ \end{align}

Once more, apply the Euler identity for cosine
\begin{align} \shaded{\cos(\omega n)\ztransform\frac{z^2-z\cos(\omega)}{z^2-2z\cos(\omega T)+1}},&& |z|\gt1 \end{align} \label{eq:cosine}

### Decaying Sine

Consider a decaying sine function for $$n\geq0$$
$$a^n \sin(\omega n)\,\color{grey}{\gamma(n)}$$

…proof missing

The unilateral Z-Transform of the decaying sine follows as
\begin{align} \shaded{a^n \sin(\omega n)\,\color{grey}{\gamma(n)} \ztransform \dfrac{az\sin(\omega)}{z^2-2az\cos(\omega)+a^2}},&&|z|\gt|a| \end{align}

### Decaying Cosine

Consider a decaying cosine function for $$n\geq0$$
$$a^n \cos(\omega n)\,\color{grey}{\gamma(n)}$$

…proof missing

The unilateral Z-Transform of the decaying cosine follows as
\begin{align} \shaded{a^n \cos(\omega n)\,\color{grey}{\gamma(n)} \ztransform \dfrac{1-az\cos(\omega)}{z^2-2az\cos(\omega)+a^2}},&&|z|\gt|a| \end{align}

## Proofs for Initial and Final Values

### 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
\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
\begin{align} \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)=\nonumber\\ &-f[0] + \lim_{n\to\infty}f[n]\label{eq:final1} \end{align}

2) Apply the Z-transform to $$\eqref{eq:final0}$$ using the time advance property $$\eqref{eq:timeadvance}$$ and take the limit for $$z\to1$$
\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
$$-\ccancel[red]{f[0]} + \lim_{n\to\infty}f[n]=-\ccancel[red]{f[0]}+\lim_{z\to1}(z-1)F(z)$$

The final value theorem, for when $$\lim_{n\to\infty}f[n]$$ exists, follows as
$$\shaded{\lim_{n\to\infty}f[n]=\lim_{z\to1}(z-1)F(z)}$$

## What’s next

Suggested next reading is Discrete Transfer Functions.

Embedded software developer
Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

## One Reply to “Z-transform Proofs”

1. Execatl Moreno says:

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!

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