transform

The article on Z-Transforms introduced a difference equation for discrete stable causal Linear Time Invariant (LTI) systems, that from here on we will refer to as a LTI system, or system for short \(\)

$$ \begin{align} \sum_{k=0}^N a_k\,y[n-k]&= \sum_{k=0}^M b_k\,x[n-k]\quad\Rightarrow \nonumber \\[10mu] a_0y[n]+a_1y[n-1]+a_2y[n-2]+\ldots &= b_0x[n]+b_1x[n-1]+b_2x[n-2]+\ldots \label{eq:diffequation} \end{align}\nonumber $$

The article Discrete Transfer Functions showed us the discrete transfer function \(H(z)\) for causal LTI systems. $$ \begin{align} H(z) &= \frac{b_0+b_1z^{-1}+b_2z^{-2}+\cdots+b_Mz^{-M}}{a_0+a_1z^{-1}+a_2z^{-2}+\cdots+a_Nz^{-N}} \label{eq:tf_polynomial} \\[10mu] &= K\,z^{\small N-\small M}\frac{(z-q_1)(z-q_2)\cdots(z-q_{\small M})}{(z-p_1)(z-p_2)\cdots(z-p_{\small N})},&K=\frac{b_M}{a_N} \label{eq:tf_factors} \end{align} $$

Here we will evaluate the response of discrete transfer functions to sinusoidal inputs, introduce stability criteria and give methods to transfer a response back to the time-domain.

We will following the notation used in our piece on Z Transforms, where \( \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \ztransform\) denotes a unilateral Z-transform, equivalent to the more common notation \(\mathfrak{Z}\left\{\,f[n]\,\right\}\), and \(f[n]\) is defined as the sample taken at time \(nT\) or \(f(nT)\). The terms filter and system will be used interchangeably.

Filter Types

In general, the term electronic filters refers to circuits that perform signal processing to remove unwanted frequency components from a signal and/or to enhance wanted components. Examples include enhancing X-ray images at airports and extracting radio signals from far away space probes. An electronic control system typically refers to a circuit that processes one signal into another to give the desired system response. E.g. modern thermostats learn the characteristics of your house, and chemical plants measure fluid levels to control flow pumps. In the context of this writing we refer to filters and systems interchangeably.

In equation \(\eqref{eq:diffequation}\), the \(b_i\) coefficients are called feedforward coefficients, and the \(a_i\) coefficients are called feedback coefficients.

We classify filters based on whether or not they use any previous value of the output, in what case we say they have feedback. Based on this we classify filters into two groups: finite impulse response and infinite impulse response filters.

Finite Impulse Response

As the name implies, Finite Impulse Response (FIR) filters have a finite response to an input. If the filter order is \(M\), then the maximum delay for the input to the output will be \(M\) samples. In other words, given an impulse input, the output will return to \(0\) after \(M\) samples.

In FIR filters, the filter output does not depend on any previous value of the output and the coefficients \(a_i=0\) for all \(i\gt 0\) in equation \(\eqref{eq:diffequation}\).

The figure below shows the signal flow of a FIR filter, where the \(z^{-1}\) block represents an one sample delay.

Output \(y[n]\)depends on	\(x[n-M]\cdots x[n]\)
Impulse response	Has a finite duration
Coefficients	\(b_i\)

Infinite Impulse Response

A filter is said to be recursive when \(a_i\neq 0\) for some \(i\gt 0\). Recursive filters are also called Infinite Impulse Response (IIR) filters.

Output \(y[n]\) depends on	\(x[n-M]\ldots x[n],\ y[n-N]\ldots y[n]\)
Impulse response	Duration depends on feedback
Coefficients	\(a_i, b_i\)

We may view an IIR filter \(H(z)\) as a series combination of two subsystems \(H_1(z)\) and \(H_2(z)\).

The commutative property of multiplication, allows the order of the subsystems to be reversed. If we draw such a circuit, it becomes apparent that each delay element \(z^{-1}\) is next to another delay element with the same input. We may then replace each such set of delay elements by one delay element with the same input. The resulting signal flow graph is called “Direct Form II” as depicted below.

Impulse response

The response of a system to any input can be calculated by the time convolution or the frequency product of the impulse response of the system and the input signal.

We derived the Z-Transform for the impulse function \(\delta[n]\) as

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \delta[n] \ztransform 1 \triangleq \Delta(z)\nonumber \label{eq:impulse} $$

When applying this input \(X(z)\) to the filter with impulse response \(H(z)\), the output \(Y(z)\) is

In other words, the response to an impulse input, is simply the transfer function \(H(z)\) itself. That is not very surprising considering that \(h[n]\) was defined as the response to an impulse input function.

Since the transfer function \(H(z)\) equals the impulse response \(Y(z)\) of the transfer function, these terms are often used interchangeably.

Frequency response

Discrete Transfer Functions introduced the concept of poles and zeros and their effect on the transfer function. Here we will take it a step further by evaluating specific \(z\) values.

The frequency response of a linear time invariant system is defined as the steady state response to a sinusoidal input. In other words the output after all transients have died out.

As part of the Z-transform, we defined:

$$ z\triangleq\mathrm{e}^{sT} \nonumber $$ where \(s=\sigma+j\omega\)

To find the frequency response, we follow the same methodology as we did for the Continuous Frequency Response and evaluate the expression \(F(z)\) along \(s=j\omega\)

Substitute \(\eqref{eq:zunitcircle}\) in \(\eqref{eq:tf_factors}\)

Expressing equation \(\eqref{eq:tf_unitcircle}\) in polar form, helps us distinguish between the amplitude and phase response

Remember, that in the \(z\)-plane, angular frequency are shown in normalized form, where the normalized angular frequency \(\omega T\) is the angle with the positive horizontal axis.

Amplitude response

We find the amplitude response as the magnitude \(|H(\mathrm{e}^{j\omega T})|\) when substituting equation \(\eqref{eq:tf_unitcircle}\) in \(\eqref{eq:tf_polarform}\)

With \(\left|\mathrm{e}^{j(\small N-\small M)\omega T}\right|=1\), according to Euler’s formula, the amplitude response follows as

The amplitude response can be visualized with the length of vectors from the poles and zeros to point \(z\) on the unit circle that corresponding to the natural frequency for which the function is evaluated.

The product of the vector lengths from the zeros divided by that of the poles times \(K\) represents the amplitude response for that normalized angular frequency.

Phase response

Along the same line of thought, the phase response is the angle \(\angle\left(H(\mathrm{e}^{j\omega T})\right)\) when substituting equation \(\eqref{eq:tf_unitcircle}\) in \(\eqref{eq:tf_polarform}\)

With \(K\) a real valued scalar and \(\angle\mathrm{e}^{j\phi}=\phi\), according to Euler’s formula

The phase response \(\angle H(\mathrm{e}^{j\omega T})\) follows as

The phase response can be visualized using the angle of vectors from the poles and zeros to point \(z\) compared to a horizontal line. Where \(z\) is a point on the unit circle (\(|z|=1\)) for which the function is evaluated. The phase response follows as is the sum of the angles from the zeroes minus that of the poles plus \((N-M)\omega T\).

Inverse Z-transform

Eventually there comes a time to return to the time-domain using an inverse Z-transform. The article Z-Transforms metioned some of the techniques for the inverse Z-Transform:

• using the binomial theorem
• using the convolution theorem
• performing long division
• using the initial-value theorem
• partial fractions expansion

We already used the binomial theorem to proof the binomial scaled pair. Here we use long division to reduce the order of the numerator, and use partial fraction expansion to split up the remaining fraction.

Partial Fraction Expansion (and long division)

As we have seen in equation \(\eqref{eq:tf_polynomial}\), the impulse response \(Y(z)=H(z)\) is a rational fraction with \(N\) poles and \(M\) zeroes

$$ \begin{align} Y(z) &= \frac{b_0+b_1z^{-1}+b_2z^{-2}+\ldots+b_Mz^{-M}}{a_0+a_1z^{-1}+a_2z^{-2}+\ldots+a_Nz^{-N}},&a_0=1 \nonumber \end{align} \nonumber $$

This rational fraction is proper when the degree of the numerator polynomial is less than the degree of the denominator polynomial. To make the function proper, we use long division of the denominator/nominator until the order of the numerator is less than that of the denominator.

Let’s call the quotient from the long division \(F(z)\), and the ratio of the remainder/denominator \(G(z)\)

Determine \(F(z)\), the FIR part

The \(F(z)\) part will be a polynomial in \(z^{-1}\) of the order \(M-N\).

Recall the delay from the Z-transform pairs

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \delta[n-a]\,\ztransform\, z^{-a}\nonumber $$

Using this transform, \(F(z)\) transforms to a parallel combination of delayed impulses in the time-domain

Determine \(G(z)\), the IIR part

The proper rational function \(G(z)\) is the remainder/denominator of the long division. Note that the numerator uses \(\dot{b}_i\) coefficients and the constant \(\dot{K}\) brings the numerator and denominator in unity form

This proper fraction can be split up into a sum of simpler fractions as introduced by Oliver Heaviside and described in Partial Fraction Expansion (PFE).

At this point in time we need to decide on the format of the time-domain function \(g[n]\). We will consider \(G(z)\) with singular poles and show you three forms. The same can be done with multiple poles, but it is a bit more involved.

While all formats lead to the same output sequence, some may be more intuitive than others. For example, if you expect a exponentially decaying response, you may want to work towards that and see how well it matches.

Choice 1: The obvious

When \(G(z)\) only has single poles, PFE gives a summation of partial fractions in the form \(\frac{c}{z-a}\)

Thanks to the linearity property of the Z-transform, these the simpler fraction can each be transformed to and summed up in the time-domain. The Z-transform for the scaled delay pair is then found in the table of Z-transform pairs as

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} a^{n-1}\gamma[n-1] \ztransform \dfrac{1}{z-a}, & & |z|\gt|a| \nonumber \end{align} \nonumber $$

Therefore, \(G(z)\) transforms to a parallel combination of delayed scaled step functions in the time-domain

Choice 2: Work towards \(\frac{z}{z-a}\) partial fractions

This time, we decide to work towards partial fractions in the form \(\frac{z}{z-a}\) that transform to \(a^n\,\gamma[n]\) in the time-domain.

Once more, start with equation \(\eqref{eq:gfactors}\), but this time preserve a power of \(z\) by divide both sides by \(z\).

When \(\frac{G(z)}{z}\) has only single poles, according to Heaviside, it expands to the summation

The Z-transform for the constant and the the scaled pair are found in the table of Z-transform pairs as

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} \small{a\delta[n]\triangleq\begin{cases}a,&n=0\\0,&n\neq0\end{cases}} \ztransform a \nonumber \\ a^{n}\gamma[n] \ztransform \dfrac{z}{z-a},&&|z|\gt|a| \nonumber \end{align} \nonumber $$

Thus \(G(z)\) transforms to a parallel combination of impulse and scaled step functions in the time-domain

Choice 3: delayed form

In this variation, the IRR part begins after the FIR part has finished. This can be more accurate in signal modeling applications, as the IIR part may be delayed so that its impulse response begins where that of the FIR part died out.

Start at the beginning with equation \(\eqref{eq:gfactors}\), and multiply the numerator with the highest power of \(z^{-1}\): \(z^{\small M}\)

Similar to before, split the expression in \(F(z)\) and \(G(z)\) parts using long division. Call the quotient from the long division \(F(z)\), and the ratio of the remainder/denominator \(G(z)\)

The \(F(z)\) part will be a polynomial in \(z\) of the order \(M-N\).

Determine the \(G(z)\) part, by first bringing it back to a polynomial in \(z^{−1}\)

Split up the proper fraction into a sum of simpler fractions using PFE, using the “obvious” choice and assuming \(G(z)\) only has single poles

Therefore, \(G(z)\) transforms to a parallel combination of delayed scaled step functions in the time-domain

Examples for these forms are given in the appendix.

Stability

As we have seen in Z-transforms, the convergence of a transfer function depends on its magnitude, while its phase has no effect. A system is stable if the magnitude of its impulse response \(h[n]\) decays to \(0\) as \(t\to\infty\).

As we have seen, every finite-order LTI filter can be expressed as FIR and IIR parts. We will now examine how these parts contribute to the stability of the system.

FIR part

The FIR part \(F(z)\) from equation \(\eqref{eq:firpart5}\) is a finite-order polynomial in \(z^{-1}\)

In the time-domain this transforms to \(\eqref{eq:firpart}\)

$$ f[n]=\sum_{k=0}^{M-N}f_k\ \delta[n-k]\nonumber $$

IIR part

As shown in Choice 1 above, the IIR part \(\eqref{eq:choice2}\) can be expressed as a summation of \(\frac{1}{z-a}\) terms \(\eqref{eq:choise2b}\). These terms transform to the time-domain as \((\ref{eq:iir1},\ref{eq:iir0})\).

If all the poles are inside the unit circle in the \(z\)-plane, then the IIR part is stable and consequently the transfer function \(H(z)\) is stable. More formally:

An irreducible transfer function \(H(z)\) is stable if and only if all its poles have a magnitude less than one.

Appendix

Example 1

This example shows an Inverse Z-Transform for a rational function where the numerator and denominator have the same degree \(N=M=3\).

This conveniently matches Example 2 in the Partial Fraction Expansion article.

$$ \frac{\color{green}{2}x^3+\color{green}{x}^2\color{green}{-1}x+\color{green}{4}}{(x-2)^3}=\color{blue}{-\frac{1}{2}}+\color{blue}{11}\frac{x}{(x-2)^3}+\color{blue}{8}\frac{x}{(x-2)^2}+\color{blue}{\frac{5}{2}}\frac{x}{x-2}\nonumber $$

This implies that \(Y(z)\) can be expressed in partial fractions as

These terms are readily found in the Z-transform pairs table

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} \delta[n] &\,\ztransform\, 1 \nonumber \\ a^n\,\gamma[n] &\,\ztransform \frac{z}{z-a},&|z|\gt |a| \nonumber\\[6mu] n\,a^n\,\gamma[n] &\,\ztransform \frac{az}{(z-a)^2},&|z|\gt |a| \nonumber \\ \tfrac{1}{2}{n(n-1)}\,a^n\,\gamma[n] &\,\ztransform \frac{a^2z}{(z-a)},&|z|\gt |a| \nonumber \end{align} \nonumber $$

Using these transform pairs, the time-domain response is

Example 2

This example shows an Inverse Z-Transform for a rational function where the degree of the numerator is 1 more as that of the denominator \(M=3, N=2\). [CCRMA]

Examine the impulse response of a filter with transfer function

Solve it by using long division to bring the order of the numerator down to \(N-1\), so we can use partial fraction expansion on the remaining IIR part.

To help with notation, define \(d\triangleq z^{-1}\) Do the long division

This bought the order of the numerator down to one less than that of the denominator (\(M=1, N=2\))

Examine the IIR part, \(G(z)\)

This matches Example 1 in the partial fraction expansion article.

$$ \frac{\color{green}{-8}+\color{green}{24}x}{1-2x+x^2}=\frac{\color{blue}{-24}}{1-x}+\frac{\color{blue}{16}}{\left(1-x\right)^{2}}\nonumber $$

Analogue to that example, fraction \(G(z)\) can be expressed in partial fractions as

With \(H(z) = F(z)+G(z)\), and the impulse response \(Y(z) = \Delta(z)\,H(z)=H(z)\)

These terms are readily found in the Z-transform pairs table

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} \delta[n] &\ztransform 1 \nonumber \\ \gamma[n-a]\,\color{grey}{\gamma[n]} &\ztransform \small{\begin{cases}z^{-a},&a\geq0\\0,&a\lt0\end{cases}} &z\neq0 \nonumber \\ a^n\,\color{grey}{\gamma[n]} &\ztransform {\frac{1}{1-az^{-1}}},&|z|\gt|a| \nonumber \\ \small{\left(\begin{array}{c}n+m-1\\m-1\end{array}\right)}\,a^n\,\color{grey}{\gamma[n]} &\ztransform\dfrac{1}{(1-az^{-1})^m}&|z|\gt|a| \nonumber \\ n\,\color{grey}{\gamma[n]} &\ztransform \frac{z^{-1}}{\left(1-z^{-1}\right)^2},&|z|\gt1 \nonumber \end{align}\nonumber $$

The impulse response \(y[n]\) follows as

Delayed version

The delayed version is found by first multiplying numerator and denominator of equation \(\eqref{eq:example2_def}\) with \(z^3\), to make them a polynomial in \(z\) instead of \(z^{-1}\)

Do a long division to reduce the order of the numerator

This bought the order of the numerator down to less than that of the denominator (M=1,N=3)

Examine the IIR part \(G(z)\), by bringing it back to a polynomial in \(z^{-1}\)

Partial fraction expansion, left as an exercise to the reader

With \(H(z)=F(z)+G(z)\), and the impulse response \(Y(z)=\Delta(z)\,H(z)=H(z)\)

Using GNU/Octave

GNU/Octave residuez function returns the FIR part as \(\color{purple}{f}\), the filter-pole residues as \(\color{blue}{r}\), the filter poles as \(\color{brown}{p}\) and the pole multiplicities as \(\color{magenta}{m}\) pkg load signal B=[2 6 6 2]; A=[1 -2 1]; [r,p,f,m] = residuez(B,A) r = -24 16 p = 1 1 f = 10 2 m = 1 2

In other words

We can also use GNU/Octave to determine the delayed form of the IIR

For the same example, the residued function returns [r,p,f,m] = residued(B,A) r = 8 16 p = 1 1 f = 2 10 m = 1 2

In other words

The article on Z-Transforms introduced a difference equation for discrete stable causal Linear Time Invariant (LTI) systems, that from here on we will refer to as a LTI system, or system for short\(\)

$$ \begin{align} \sum_{k=0}^N a_k\,y[n-k] &= \sum_{k=0}^M b_k\,x[n-k]\quad\Rightarrow \nonumber \\[10mu] a_0y[n]+a_1y[n-1]+a_2y[n-2]+\ldots &= b_0x[n]+b_1x[n-1]+b_2x[n-2]+\ldots \nonumber \end{align} \nonumber $$

Here we will focus on the black box with discrete input signal \(x[n]\) and output \(y[n]\), where \(n\) is the sample number, sampled every \(T\) seconds.

We will following the notation from Z-transforms, where \(\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \ztransform\) is equivalent to the more common notation \(\mathfrak{Z}\left\{\,f[n]\,\right\}\), and \(f[n]\) is defined as the sample taken at time \(nT\). The terms filter and system will be used interchangeably.

Linear Time Invariant System

The blackbox that we will examine is a Stable Causal Linear Time Invariant System (LTI). Most of the practical systems can be modeled as LTI systems or at least approximated by one around nominal operating point. Examples are include stereo engineering to counter the effect of a stadium on the music, or process control engineering in chemical plants.

The relation between the causal LTI system \(L\), its input \(x[n]\) and output \(y[n]\) can be expressed as

Definition

Before we start exploring the properties of stable causal LTI systems, let’s take a moment to define what a stable causal LTI system is.

A Causal system:

Is one in which changes in output do not precede changes in input.

The Linearity property states:

If input \(x_{1}(t)\), produces response \(L\big(x_{1}(t)\big)\), and input \(x_{2}(t)\), produces response \(L\big(x_{2}(t)\big)\), and \(a_i\) are real scalars, then a scaled and summed input produces the scaled and summed response

The Time Invariance property states:

When we apply an input to the system now or \(\tau\) seconds from now, the output will be identical except for a time delay of \(a\) samples. That is, if the output due to input \(x(t)\) is \(y(t)\), then the output due to input \(x(t-\tau)\) is

Further more, the system needs to be stable:

LTI system is bounded-input bounded-output stable if all bounded inputs result in bounded outputs.

From here on we will refer to a stable causal linear time invariant system as a LTI system, or system for short.

Combining systems

The two common configurations when combining filters are: series and parallel.

The figure below shows a series, or cascade, connection of filter \(H_1(z)\) and \(H_2(z)\), where the output from the first filter feeds the input of the next filter.

The transfer function for the series circuit is

The other common configuration is called parallel as shown below. In a parallel circuit, both filters get the same input signal and their outputs are summed.

The transfer function for the parallel circuit is

Output of a causal LTI system

In the time-domain, we find the output of a Causal LTI system system by passing the function of the input signal as a parameter to the system equation as shown in equation \(\eqref{eq:ynLx}\)

$$ y[n] = \color{purple}{L\Big(\color{black}{x[n]}\Big)} \nonumber $$

Convolution in the time-domain

The article on Z-transforms showed how any discrete input signal \(x[n]\) can be expressed as a summation of scaled impulses.

$$ x[n] = \sum_{k=0}^{\infty}{\color{green}{x[k]}\ \color{blue}{\delta[n-k]}} \nonumber $$

Consider this \(x[n]\) to be the input to LTI system \(L\). The output of the system follows from substituting the equation for signal \(x[n]\) in \(\eqref{eq:ynLx}\)

Equation \(\eqref{eq:yn}\) matches the general form of a LTI system with a scalar and variable as a function of time \(t=\color{blue}{n}T\). This implies that we can use the part of the linearity property \(\eqref{eq:linear}\) that states:

$$ \color{purple}{L\Big(\color{green}{a}\ \color{blue}{x[n]}\Big)}=\color{green}{a}\ \color{purple}{L\Big(\color{blue}{x[n]}\Big)} $$

Apply the linearity property to bring the time independent part out of the system function \(L\) in equation \(\eqref{eq:yn}\)

Let’s examine the expression \(L\big(\delta[n-k]\big)\). Suppose we know how \(L\) acts on one impulse function \(\delta[n]\), and define it as

Applying the time invariance property \(\eqref{eq:timeinvariance}\) to \(\eqref{eq:hn}\) yields

Substituting \(\eqref{eq:ldelta}\) in \(\eqref{eq:yn2}\), gives the convolution sum for LTI systems

In summary, when we model the transfer function of a LTI black box as \(h[n]\), in the time-domain the output signal \(y[n]\) is a convolution ‘\(\ast\)’ of input \(x[n]\) and the system impulse response \(h[n]\).

Multiplication in the \(z\)-domain

Likewise, in the \(z\)-domain, the transfer function fully describes how the output signal \(Y(z)\) responds to an arbitrary input signal \(X(z)\).

As we have seen in Z-Transforms, the convolution in the time-domain transforms to a multiplication in the \(z\)-domain.

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} (f\ast g)[n]\,\color{grey}{\gamma[n]} \ztransform F(z)\,G(z) \nonumber $$

That implies that the output \(Y(z)\) is the result of the input signal \(X(z)\) multiplied with the impulse response \(H(z)\) of the filter.

Transfer function

Time to take a closer look at the transfer function of the LTI system. We will express the transfer function as a ratio of polynomials and show it in its factorized form.

The Z-Transforms article opened with a generic form of Linear Constant-Coefficient Difference Equation (LCCDE) that expresses the relation between input \(x[n]\) and output \(y[n]\)

$$ \begin{align} \sum_{k=0}^N a_k\,\color{blue}{y[n-k]} &= \sum_{k=0}^M b_k\,\color{blue}{x[n-k]} \label{eq:lccde} \end{align} $$

Recall the time delay property

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} f[n-a]\, \color{grey}{\gamma[n} – a \color{grey}{]} \,\ztransform z^{-a}F(z) \nonumber $$

Apply the time delay property to transform both sides of equation \(\eqref{eq:lccde}\) to the \(z\)-domain

Bring \(X(z)\) and \(Y(z)\) to the left and the summations to the right, yields the generic form of the transfer function \(H(z)\)

The \(b_k\) coefficients are called feedforward coefficients, and the \(a_k\) coefficients are called feedback coefficients.

Factorized form

It is often convenient to factor the polynomials of the transfer function \(\eqref{eq:tf_polynominal}\), and write the function in terms of those factors.

The first step towards a factorized form, is to rewrite \(H(z)\) in a standard from, so that the highest order term of the numerator and denominator are unity.

Factorize the polynomials

The \(q_i\)’s are the roots of the equation \(N(z)=0\) and are called the system zeros. The \(p_i\)’s are the roots of the equation \(D(z)=0\) and are defined as the system poles. The filter order equals the number of poles or zeros in the transfer function, whichever is greater.

If one would make a 3D plot with the \(z\)-plane being the base and \(|H(z)|\) on the vertical axis, then the poles will show as thin “poles” pointing up and the zeros will show as dips pointing down.

The (complex) poles and zeros are properties of the transfer function, and therefore of the difference equation. Together with the gain constant \(K\) and delay \(z^{-(\small N-M})\) give a complete description of the filter.

Visualization

The article Z-transforms introduced the normalized angular frequency \(\omega T\) and the \(z\)-plane. Here we will visualize the poles, zeroes and their evaluation for complex variable \(z\).

Poles and zeros

As we have seen in equation \(\eqref{eq:tf_prod}\), the factorized transfer function can be written as

The system may be represented graphically by plotting the poles and zeros in the complex \(z\)-plane. In these so called pole-zero plots, it is customary to mark a zero location with a circle (\(\circ\)) and a pole location with a cross (\(\times\)).

Evaluating Poles and zeros

We will run ahead of ourselves and describe how the poles and zeroes affect the system response, later we will come back to this subject and explore it further.

Each of the poles \((z-p_i)\) and zeroes \((z-q_i)\) have a unique contribution to the transfer function. To give you an insight, we will we evaluate the factor \((z-p_i)\), but the same applies to zeroes.

this factor can be visualized with a vector drawn from \(r\) to \(z\). Each of the factors in the numerator and denominator may be interpreted as a vector in the z-plane, originating from the zero \(z_i\) or pole \(p_i\) and directed to the point \(z\) at which the function is evaluated.

The length and angle of these factors represent their contribution to the transfer function. For example for a pole \(p_i=\Re(p_i)+j\,\Im(p_i)\), the magnitude and angle of the vector to the variable \(z=\Re(z)+j\,\Im(z)\) are

Multiplication and division

While we are on the subject, recall multiplication and division of complex numbers as it is most easily done in polar form

Applying \(|K|=K\) and \(\angle{K}=\mathrm{atan2}(0,K)=0\), the magnitude and angle of the complete transfer function \(H(z)\) may be written as

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 \(n\lt0\)

The initial value follows as

Final Value Theorem

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

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

write out the summation to find common terms

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

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

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

Proofs for Z-transform properties, presented in the Z-Transforms article.\(\)

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

Apply the unilateral Z-transforms definition to equation \(\eqref{eq:impuls_def1}\)

Since the impulse is \(0\) everywhere but at \(n=0\), the summation simplifies to

The unilateral Z-Transform of the Unit Impulse Function follows as

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

Delayed Impulse

Consider the discrete delayed impulse function \(\delta[n-a]\)

Sorry proof missing. Similar to delay proof? Have the proof, please share it in the comments.

The unilateral Z-Transform of the Delayed Unit Impulse Function follows as

Unit Step

The discrete unit or Heaviside step function, denoted with \(\gamma[n]\) is defined as

The unilateral Z-transforms of \(\eqref{eq:unitstep_def}\) follows

Apply the power series

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

to \(\eqref{eq:unitstep0}\)

The unilateral Z-transform for the step function follows as

Scaled

Consider the discrete power function starting at \(n=0\)

Apply the unilateral Z-transforms and the power series

The unilateral Z-Transform of the scaled function follows

The Binomial scaled proof comes to this same transform.

Scaled delayed

Recall the delay property, and the scaled pair

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\, & \ztransform z^{-b}F(z) \nonumber \\ a^n\ \color{grey}{\gamma[n]} & \ztransform \frac{z}{z-a},&|z|\gt|a| \nonumber \end{align} \nonumber $$

The unilateral Z-Transform of the scaled delayed function follows

\(n\) scaled

Consider the discrete ramp function starting at \(n=0\)

Apply the unilateral Z-transforms definition and expand

To counter the infinite series, introduce \(az^{-1}F(z)\)

Subtract \(\eqref{eq:ramp2}\) from \(\eqref{eq:ramp1}\)

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 \(r=az^{-1}\)

The unilateral Z-Transform of the \(n\) scaled function follows

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

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} n\,a^n\ \color{grey}{\gamma[n]} \ztransform \frac{az}{\left(z-a\right)^2}, && |z|\gt1 \nonumber \end{align} \nonumber $$

The unilateral Z-Transform of the discrete ramp function follows

Binomial scaled, \(|z|\gt |a|\)

We will do this proof starting from the \(z\)-domain

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 \((1-x)^{-m}\)

Use the Binomial Series where \(x=az^{-1}\)

Apply the Symmetry Rule for Binomial Coefficients \({n \choose k}={n \choose n-k}\)

The unilateral Z-Transform follows

Recall the delay property

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\,\ztransform\,z^{-b}F(z)\nonumber $$

Combining equation \(\eqref{eq:binomialscaled}\) with the delay property

So that in general

For the case where \(m=1\)

so that

For the case where \(m=2\)

so that

For the case where \(m=3\)

so that

Binomial scaled, \(|z|\lt |a|\)

Similar to the previous proof, we will do this starting from the \(z\)-domain

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 \((1-x)^{-m}\)

Use the Binomial Series where \(x=-az^{-1}\)

Apply the Moving Top Index to Bottom in Binomial Coefficient \({n \choose m}=(-1)^{n-m}\,{-(m+1) \choose n-m}\)

Apply the Symmetry Rule for Binomial Coefficients \({n \choose k}={n \choose n-k}\)

The unilateral Z-Transform follows

Exponential

Consider the discrete exponential function starting at \(n=0\)

Apply the unilateral Z-transforms definition

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 \(a=z^{-1}\ \mathrm{e}^{-aT}\)

The unilateral Z-Transform of the exponential function follows

Sine

The Z-transforms of the sine is similar to that of the cosine function, except that it uses the Euler identity for 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 \(n=0\)

The unilateral Z-transforms of the cosine is

Recall the Euler identity for cosine

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

Apply the identify for cosine

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

Bring over a common denominator and regroup

Once more, apply the Euler identity for cosine

Decaying Sine

Consider a decaying sine function for \(n\geq0\)

Sorry proof is missing. If you have the proof, please help make this list complete and share it in the comments.

The unilateral Z-Transform of the decaying sine follows as

Decaying Cosine

Consider a decaying cosine function for \(n\geq0\)

Sorry the proof is missing. If you have the proof, please share it in the comments.

The unilateral Z-Transform of the decaying cosine follows as

Proofs for Z-transform properties, presented in the Z-Transforms article.\(\)

Proofs for Properties

Linearity

Consider the time-domain function

This function transforms to the \(z\)-domain as

From which follows

Time Delay

Consider a sequence truncated at n=0, and delayed by \(a\) samples, where \(a\gt0\)

Where the delayed step function \(\gamma[n-a]\) is defined as

The unilateral Z-transforms of \(\eqref{eq:delay0}\) is

Apply \(\gamma[n-a]=0\) for all \(n\lt a\) by changing the start value of the summation

Make the summation start at \(0\) by subtracting \(a\) on both sides of the summation start value: introduce \(m=n-a\)

The unilateral Z-transform of the positive delay property follows as

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 \(a\gt0\)

The unilateral Z-transforms is

Apply the definition of the unit step function \(\eqref{eq:unitstep_def}\): \(\gamma[n]=0,\ \ \forall_{n\lt 0}\)

Substitute \(m=n-a\) to remove the \(-a\) offset from \(f[n-a]\)

To make the summation start at \(0\), subtract the first \(a\) terms, and then add these points back again.

The unilateral Z-transform of this positive time delay follows as

Time Advance

Consider a sequence advanced by \(a\) samples, where \(a\gt0\), and then truncated at time \(0\)

The unilateral Z-transforms of \(\eqref{eq:timeadv_def}\)

Apply the definition of the unit step function \(\eqref{eq:unitstep_def}\): \(\gamma[n]=0,\ \ \forall_{n\lt 0}\)

Substitute \(m=n+a\) to remove the \(+a\) offset from \(f[n+a]\)

To make the summation start at \(0\), add the first \(a\) terms, and then subtract these points again.

The unilateral Z-transform of the positive time advance follows as

Time Multiply

Consider multiplied by the sample number \(n\), and truncated at \(n=0\)

Sorry, proof missing. See “n scaled” pair. Have the proof? Please share in the comments.

The unilateral Z-transform

Modulation

Consider a sequence multiplied with complex scalar \(a^n\), and truncated at \(n=0\)

Take the unilateral Z-transforms

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 \(\eqref{eq:scaling_def}\)

The unilateral Z-transform of the modulation follows as

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 (\(n=0\))

Apply the unilateral Z-transforms definition

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 \(\eqref{eq:convolution0}\)

The unilateral Z-transform of the convolution in the time-domain simplifies to a a multiplication in the \(z\)-domain.

Conjugation

Consider conjugation

Take the unilateral Z-transforms

The unilateral Z-transform follows as

First Difference

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

Consider the difference

The unilateral \(z\)-transform of the first difference follows as

Accumulation

Accumulation in the discrete time domain is analogous to integration in the continuous time domain.

Consider the accumulation

Take the unilateral Z-transforms of equation \(\eqref{eq:accum_def}\)

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 \(\eqref{eq:accum1}\)

The unilateral Z-transform of accumulation follows from applying thepower series

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 \(\frac{1}{{1-pz^{-1}}}\)

Apply the power series and differentiate

We recognize this as the Z-transform of \((n+1)p^n\)

The unilateral Z-transform of \((n+1)p^n\) follows as

Overview of the Unilateral Z-transform properties, pairs and initial/final theorem. Includes links to the the proofs.\(\)

The tables below introduce commonly used properties, common input functions and initial/final value theorems that I collected over time.

The time-domain function is usually given in terms of a discrete index \(n\), rather than time. Since \(t=nT\), we may replace \(f[n]\) with \(f(nT)\) where \(T\) is the sampling period.

Unilateral Z-Transform properties

Unilateral Z-transform pairs

The binomial coefficient, used in the table above, is defined as

$$ \left(\begin{array}{c}a\\ b\end{array}\right)\triangleq\frac{a!}{b!(a-b)!} \nonumber $$

Initial and final value theorem

The proof for these transforms can be found in the post Z-Transforms Proofs.

What is now called the Z-transform (named in honor of Lotfi Zadeh) was known to, mathematician and astronomer, Pierre-Simon Laplace around 1785. With the introduction of digitally sampled-data, the transform was re-discovered by Hurewicz in 1947, and developed by Lotfi Zadeh and John Ragazzinie around 1952, as a way to solve linear, constant-coefficient difference equations.\(\)

As we will see, the convolution property makes the Z-transform a powerful tool in analyzing sampled-data systems.

Introduction

Just as causal continuous systems are controlled by differential equations

Causal discrete systems operate in accordance with difference equations.

From the difference equation we can derive the system characteristics such as the impulse response, step response and frequency response.

Unlike the continuous-time case, causal difference equations can be iterated just like a computer program would do. All one needs to do, is to rewrite the difference equation so that the term \(y[n]\) is on the left and then iterating forward in time. This will give each value of the output sequence without ever obtaining a general expression for \(y[n]\). In this article, we however will look for a general analytical expression for \(y[n]\) using the Z-transform.

The Z-transform can be thought of as an operator that transforms a discontinuous sequence to a continuous algebraic function of complex variable \(z\). As we will see, one of the nice feature of this transform is that a convolution in time, transforms to a simple multiplication in the \(z\)-domain.

Unilateral Z-Transform

We solve the difference equations, by taking the Z-transform on both sides of the difference equation, and solve the resulting algebraic equation for output \(Y(z)\), and then do the inverse transform to obtain \(y[n]\).

Assuming causal filters, the output of the filter will be zero for \(t\lt 0\).

Sampling creates a discontinuous function

For digital systems, time is not continuous but passes at discrete intervals. When it measures a continuous-time signal every \(T\) seconds, it is said to be discrete with sampling period \(T\).

To help understand the sampling process, assume a continuous function \(x_c(t)\) as shown below

To work toward a mathematical representation of the sampling process, consider a train of evenly spaced impulse functions starting at \(t=0\). This so called Dirac comb, \(s(t)\), has a spacing of \(T\gt0\) and contains \(t=0\).

The Dirac comb \(s(t)\) can be expressed as

When multiplying the Dirac comb \(s(t)\) with a continuous-time signal \(x_c(t)\), that signal will scale the comb by that \(x_c(t)\)

The resulting signal \(x_s(t)\) follows from substituting \(s(t)\) from equation \(\eqref{eq:combcond}\) as

The impulse function \(\delta(t-nT)\) is \(0\) everywhere but at \(t=nT\), the so called sifting property, so we can replace \(x_c(t)\)

Work towards a continuous function

The goal is to form a continuous algebraic expression, so we can use algebra to manipulate the difference equations.

Start with the Laplace transform of sampled signal \(x_s(t)\) from equation \(\eqref{eq:fstarnT}\)

Once more, the impulse function \(\delta(t-nT)\) is \(0\) everywhere but at \(t=nT\), the “sifting property”, so we can replace \(\mathrm{e}^{-st}\) with

After substituting \(\eqref{eq:est}\) in \(\eqref{eq:laplace0}\), the terms \(\mathrm{e}^{-snT}\) and \(x(nT)\) are independent of \(t\) and can be taken outside of the integration.

The Z-transform follows

Define \(z\) and \(x[n]\) as

The scaling with sample period \(T\) in the form \(x[n]\) matches the notation for computer arrays. Anytime you see an \([n]\) you can translate to seconds replace it with \((nT)\). Be careful with integer expressions, such as \([n – k]\), which stands for \(((n-k)T)\) seconds, not \((nT – k)\).

From equation \(\eqref{eq:z}\) follows

Substitute \((\ref{eq:z},\ref{eq:s})\) in \(\eqref{eq:Fs}\) and apply the power rule \(x^{ab}=(x^a)^b\). Call the function \(F(z)\) because \(z\) is the only variable after the substitution \(s=\tfrac{1}{T}\ln z\)

The unilateral Z-transform of the discrete function \(x[n]\) follows as

Note that we use the notation \(\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \ztransform\) as equivalent to the more common Z-transform notation \(\mathfrak{Z}\left\{\,x[n]\,\right\}\).

In review: the Z-transform maps a sequence \(x[n]\) to a continuous polynomial \(X(z)\) of the complex variable \(z\).

Normalized frequency

The discrete signal only exists at time \(t=nT\) where \(n={0,1,2,\ldots}\) By normalizing time \(t\) with the sampling interval \(T\), using definition \(\eqref{eq:z}\), we get the natural time, measured in “samples”, on the time-axis

$$ x[n]\triangleq x(nT) \nonumber $$ we highlight the normalized time, by using the \([n]\) notation

When using normalized time, other time-dependent variables should be normalized as well. The angular frequency \(\omega\), measured in [rad/s], normalizes to the normalized angular frequency with units of [rad/sample] by multiplying it with the sample period \(T\) [s/sample]

Using normalized frequency allows an author to present concepts independent of sample rate, but it comes at a loss of clarity as \(T\) and \(f_s\) are omitted from expressions.

When visualizing variable \(z\), the normalized angular frequency \(\omega T\) corresponds to the angle with the positive horizontal axis.

Formulas expressed in terms of \(f_{s}\) and/or \(T\) are not normalized and can be readily converted to normalized frequency by setting those parameters to \(1\). The inverse is accomplished by replacing instances of the angular frequency parameter \(\omega\), with \(\omega T\).

Note that some authors use \(\omega\) for normalized angular frequency in [rad/sample], and \(\Omega\) for angular frequency in [rad/s]. Here we avoid using \(\omega\) for natural frequencies. Instead we use the product \(\omega T\) to refer to natural angular frequency. By doing so, it is clear that the angular frequency \(\omega\) is scaled by the sample time \(T\).

Nyquist–Shannon sampling theorem

The frequency-domain representation of a sampled signal teaches us about the limitations of using a discrete signal. We will show this by doing a Fourier transform on the sampled signal \(x_s(t)\).

Recall equation \(\eqref{eq:fstarnT}\) and \(\eqref{eq:fstar0}\), but this time bilateral

$$ \begin{align} x_s(t)=x_c(t)\,s(t) &= \sum_{n=-\infty}^{\infty}{x_c(nT)\ \delta(t-nT)} \nonumber \\ &= \underbrace{\sum_{n=-\infty}^{\infty}x_c(nT)}_{f(t)}\ \ \underbrace{\sum_{n=-\infty}^{\infty}\delta(t-nT)}_{g(t)} \nonumber \end{align} \nonumber $$

The convolution theorem states that multiplication in the time-domain corresponds to convolution in the frequency-domain

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathcal{F}}} f(t)\,g(t)\fourier \frac{1}{2\pi}{\Large(}F(\omega)*G(\omega){\Large)} \nonumber $$ where \(*\) is the convolution sign

Apply the Fourier transform of a product to \(x_s(t)\) and call it \(X_s(\omega)\)

Recall the Fourier transform of the Dirac comb \(s(t)\)

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\fourier{\lfz{\mathcal{F}}} s(t)=\sum _{n=-\infty }^{\infty }\delta(t-nT) \fourier {\frac {2\pi }{T}}\sum _{k=-\infty }^{\infty }\delta \left(\omega -{\frac {2\pi k}{T}}\right)\triangleq S(\omega) \nonumber $$

Substituting the Fourier transform of the Dirac comb in \(\eqref{eq:fstar2}\) where \(\frac{2\pi}{T}=2\pi f_s=\omega_s\), the angular sample frequency

Recall the convolution with a delayed impulse function \(\delta(\omega-a)\)

$$ F(\omega)*\delta(\omega-a) = F(\omega-a)\nonumber $$

Apply the convolution to \(\eqref{eq:XsjOmega0}\)

Equation \(\eqref{eq:XsjOmega}\) implies that the Fourier transform of \(x_s(t)\) consists of periodically repeated copies of the Fourier transform of \(x_c(t)\). The copies of \(\color{purple}{X_c(\omega)}\) are shifted by integer multiples of the sampling frequency and then superimposed as depicted below.

Plot (a) represents the frequency spectrum of the continuous signal \(\color{purple}{X_c(\omega)}\) where \(\omega_{\small N}\) is the highest frequency component. Plot (b) shows the frequency spectrum of the Dirac comb \(S(\omega)\). Finally, plot (c) shows \(X_s(\omega)\), the result of the convolution between \(\color{purple}{X_c(\omega)}\) and the \(S(\omega)\).

From (c), we see that the replicas of \(\color{purple}{X_c(\omega)}\) do not overlap when

Consequently, \(x_c(t)\) can be recovered from \(x_s(t)\) with an ideal low-pass filter \(H_r\).

Inequality \(\eqref{eq:ineq}\) is captured in the Nyquist–Shannon sampling theorem

Let \(x_c(t)\) be a bandlimited signal with $$ \begin{align} X_c(\omega)&=0,&|\omega|\gt \omega_{\small N}\nonumber \end{align} \nonumber $$ then \(x_c(t)\) is uniquely determined by its samples $$ \begin{align} x[n] &= x_c(nT),&n\in\mathbb{Z}\nonumber \end{align} \nonumber $$ if $$ \omega_s = \frac{2\pi}{T}\geq2\,\omega_{\small N} \nonumber $$ where \(\omega_{\small N}\) is called the Nyquist frequency and \(2\omega_{\small N}\) is called the Nyquist rate.

When the inequality of \(\eqref{eq:ineq}\) does not hold, e.g. when the sampling frequency \(\omega_s\) is less than twice the maximum frequency \(\omega_{\small N}\), the copies of \(X_c(\omega)\) overlap, and \(X_c(\omega)\) is no longer recoverable by low-pass filtering as shown below. The resulting distortion is called aliasing.

\(z\)-plane

We introduced the normalized angular frequency \(\omega \) and how it maps to a polar representation of the complex variable \(z\). Here we will extent this to include the magnitude \(|z|\).

Let’s start with the definition of \(z\) from equation \(\eqref{eq:z}\) and split \(s\) in its real and imaginary parts \(s=\sigma+j\omega\)

The polar notation for the complex variable \(z\) is a function of the natural angular frequency \(\omega T\) and \(\mathrm{e}^{\sigma T}\)

This conveniently matches the topology of the \(z\)-plane, where the modules \(|z|\) corresponds to the length of a vector from the origin to \(z\), and \(\omega T\) corresponds to the angle of that vector with the positive horizontal axis.

According to the Nyquist–Shannon sampling theorem, a discrete signal can only have frequencies \(|\omega|\) between \(0\) and half the sampling frequency

The maximum frequency in the discrete signal, the so called Nyquist frequency corresponds to \(\pi\) radians, because with sample period \(T=\frac{1}{f_s}=\frac{2\pi}{\omega_s}\) and the Nyquist frequency \(\omega_{\small N}=\frac{\omega_s}{2}\), the natural angular frequency for the Nyquist frequency follows as

For those familiar with the Laplace transform, we will map specific features between the \(s\) and \(z\)-domain:

• The origin \(s=0\) of the \(s\)-plane is mapped to \(z=e^0=1\) on the real axis in \(z\)-plane.
• Each vertical line \(\sigma=\sigma_0\) in \(s\)-plane is mapped to a circle \(|z|=e^{\sigma_0}\) centered about the origin in \(z\)-plane. E.g.
  • Leftmost vertical line \(\sigma\to-\infty\) is mapped as the origin, where \(|z|=\mathrm{e}^{-\infty}=0\)
  • The imaginary axis \(\sigma=0\) is mapped as the unit circle, where \(|z|=\mathrm{e}^0=1\)
  • Rightmost vertical line \(\sigma\to\infty\) is mapped as a circle with an infinite radius, where \(|z|=\mathrm{e}^{\infty}=\infty\).
• Each horizontal line \(j\omega=j\omega_0\) in \(s\)-plane is mapped to an angle from the origin in \(z\)-plane of angle \(\omega_0 T\) with respect to the positive horizontal direction.

Region of convergence

Recall the definition of the Z-transform from equations \((\ref{eq:ztransform},\ref{eq:zalt},\ref{eq:zphi})\)

$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} f[n]\ztransform\sum_{n=0}^\infty z^{-n}\ f[n]\\ \text{where } z\triangleq|z|\,\mathrm{e}^{j\varphi}\text{, } |z|\triangleq \mathrm{e}^{\sigma T}\text{, } \varphi\triangleq \omega T $$

This converges depending on the duration and magnitude of \(f[n]\) as well as on the magnitude \(|z|\). The phase \(\varphi\) has no effect on the convergence.

The power series for the Z-transform is called a Laurent series. The Laurent series, represents an analytic function at every point inside the region of convergence. Therefore, the Z-transform and all its derivatives must be continuous function of \(z\) inside the region of convergence.

Laurent series converge in an annular (=ring shaped) region of the \(z\)-plane, bounded by poles. The set of values of \(z\) for which the Z-transform converges is called the region of convergence (ROC). This means that anytime we use the Z-transform, we need to keep the region of convergence in mind.