The z-Transform

4.1  Two things the DTFT cannot do

Lecture 2 built four Fourier transforms and Lecture 3 explained where sequences come from. It is fair to ask what is left to want. The answer is specific: the DTFT fails at exactly two jobs, and both of them are jobs an engineer needs done.

Both are fixed by the same small change. Instead of evaluating the transform only on the unit circle, evaluate it on every circle. Write \(z=re^{j\omega}\) and define

Now separate the modulus from the angle. Since \(z^{-n}=r^{-n}e^{-j\omega n}\),

A one-sample delay transforms to \(z^{-1}X(z)\). Since a difference equation is a weighted sum of delayed inputs and outputs, the \(z\)-transform converts it into a polynomial equation in \(z^{-1}\). Factoring the resulting numerator and denominator reveals the system's zeros and poles.

The parallel with continuous time is exact, and worth putting side by side because it is the last rung of Lecture 2's ladder:

One question remains, and it is the good one: why a circle? In continuous time the Fourier case sits on a straight line; here it sits on a closed curve. Lecture 3 has already supplied the answer.

4.2  Definition, and where the unit circle comes from: \(z=e^{sT}\)

Lecture 1 established that complex exponentials are the eigenfunctions of LTI systems, and Lecture 2 built every transform out of them. So ask what happens to one when it passes through the sampler of Lecture 3. Take \(x(t)=e^{st}\) and sample at \(t=nT\):

Sampling the continuous-time exponential \(e^{st}\) at \(t=nT\) gives \((e^{sT})^n\). The discrete-time variable is therefore \(z=e^{sT}\). Writing \(s=\sigma+j\Omega\) shows how locations in the \(s\)-plane map into the \(z\)-plane:

Read off the consequences one at a time. They are the three facts Lecture 2 could only tabulate.

• \(\sigma<0 \Rightarrow r<1\). The left half-plane maps inside the unit circle. Decaying continuous exponentials sample into decaying discrete ones.
• \(\sigma=0 \Rightarrow r=1\). The \(j\Omega\)-axis — the home of the Fourier transform — maps onto the unit circle. That is the answer to the question at the end of Section 4.1: the circle is the image of a straight line under \(e^{sT}\).
• \(\omega=\Omega T\). The same exchange rate as Eq. (3.13), arrived at by a different route. Consistency here is not a coincidence — both come from sampling.

The map is periodic in \(\Omega\) with period \(2\pi/T\). Thus, \(s\) and \(s+j2\pi k/T\) map to the same point \(z\). This many-to-one mapping is another view of aliasing and explains why distinct discrete-time frequencies occupy only one \(2\pi\)-radian interval.

4.3  The region of convergence is half the answer

Because Eq. (4.1) is an infinite sum, it converges only in part of the \(z\)-plane. This region of convergence (ROC) is part of the transform. Without it, the algebraic expression \(X(z)\) may correspond to more than one sequence.

A causal sequence. Let \(x[n]=a^nu[n]\). Then only \(n\ge0\) contributes, and the sum is geometric with ratio \(az^{-1}\):

An anti-causal sequence. Now let \(x[n]=-a^nu[-n-1]\), which is zero for \(n\ge0\). Substituting \(k=-n\) turns the sum into a geometric series in \(a^{-1}z\):

4.3.1  Properties of the ROC, and how to infer it

The good news is that you rarely have to derive an ROC. Its shape is heavily constrained, and the constraints let you read it off from a single qualitative fact about the sequence.

• It is always an annulus centred on the origin (possibly degenerate): \(r_{\min}<|z|<r_{\max}\). Convergence depends on \(r\) alone — Eq. (4.2) — never on \(\omega\), so the region cannot be any other shape.
• It never contains a pole. By definition the sum blows up there. The boundaries of the ROC are therefore always pole radii.
• Right-sided \(\Rightarrow\) outward. If \(x[n]=0\) for \(n<N_1\), the ROC is \(|z|>r_{\max}\): everything outside the outermost pole. Large \(r\) suppresses large positive \(n\), which is where the sequence lives.
• Left-sided \(\Rightarrow\) inward. If \(x[n]=0\) for \(n>N_2\), the ROC is \(|z|<r_{\min}\): everything inside the innermost pole.
• Two-sided \(\Rightarrow\) an annulus between two poles — or nothing at all, if the inward and outward requirements are incompatible.
• Finite-length \(\Rightarrow\) the entire plane, except possibly \(z=0\) or \(z=\infty\). A finite sum always converges.

In practice, the ROC is often inferred from physical information. If a problem states that a system is causal, the ROC must extend outward beyond the outermost pole. If it states that the system is stable, the ROC must include the unit circle.

EXAMPLE 4.1 · A right-sided sequence, and its ROC

Find \(X(z)\) and its ROC for \(x[n]=\left(\tfrac12\right)^nu[n]\), and say whether a system with this impulse response is stable.

{{ l4Ex1Label }}

Step 1. Only \(n\ge0\) contributes, so by Eq. (4.5) with \(a=\tfrac12\):

$$X(z)=\frac{z}{z-\tfrac12},\qquad \text{ROC: } |z|>\tfrac12.$$

Step 2 \middot; Stability. Right-sided \(\Rightarrow\) the ROC extends outward from the pole, and it already contains \(|z|=1\). So the DTFT exists (Section 4.1) and the system is BIBO stable — this is the ordinary causal, decaying case.

EXAMPLE 4.2 · A left-sided sequence with the identical X(z)

Find \(X(z)\) and its ROC for \(x[n]=-\left(\tfrac12\right)^nu[-n-1]\), and say whether a system with this impulse response is stable.

{{ l4Ex2Label }}

Step 1. Only \(n\le-1\) contributes, so by Eq. (4.6) with \(a=\tfrac12\):

$$X(z)=\frac{z}{z-\tfrac12},\qquad \text{ROC: } |z|<\tfrac12.$$

Step 2 \middot; Stability. The algebra is identical to Example 4.1 — same pole, same \(X(z)\). But this ROC extends inward and excludes \(|z|=1\), so the DTFT does not exist and the system is unstable. Only the shaded region distinguishes the two sequences.

EXAMPLE 4.3 · A two-sided sequence, and why its ROC is an annulus

Find \(X(z)\) and its ROC for \(x[n]=\left(\tfrac12\right)^nu[n]-2^{\,n}u[-n-1]\), and say whether a system with this impulse response is stable.

{{ l4Ex3Label }}

Step 1. Transform each part separately and intersect the regions — that is all the linearity property allows.

Step 2 \middot; Right-sided part. By Eq. (4.5) with \(a=\tfrac12\):

$$X_1(z)=\frac{z}{z-\tfrac12},\qquad |z|>\tfrac12.$$

Step 3 \middot; Left-sided part. This is \(-a^nu[-n-1]\) with \(a=2\), so by Eq. (4.6):

$$X_2(z)=\frac{z}{z-2},\qquad |z|<2.$$

Step 4 \middot; Combine. The sum exists only where both do:

$$X(z)=\frac{z}{z-\tfrac12}+\frac{z}{z-2}, \qquad \text{ROC: } \tfrac12<|z|<2.$$

Step 5 \middot; Stability. The annulus contains \(|z|=1\), so the sequence is absolutely summable there and a system with this impulse response is BIBO stable — even though one pole (at \(z=2\)) lies outside the unit circle. Section 4.4 makes this precise as Eq. (4.7): stability is about the ROC, not about a blanket rule that poles must sit inside the circle — that shortcut (Eq. (4.8)) only holds for causal systems, and this sequence is two-sided. Such a system is stable but not realisable in real time, since it needs future inputs.

4.4  Stability and causality, read off the z-plane

A discrete-time LTI system is BIBO stable when its impulse response is absolutely summable. This is also the condition for its DTFT to converge. Since the DTFT is the \(z\)-transform evaluated at \(r=1\), stability requires the ROC to include the unit circle:

For an LTI system, BIBO stability, absolute summability of \(h[n]\), and existence of the ordinary frequency response are therefore linked. If the ROC excludes the unit circle, the DTFT of the impulse response does not converge.

Combine Eq. (4.7) with the causality rule of Section 4.3.1 and the everyday test appears. Causal means the ROC is \(|z|>r_{\max}\); for that outward region to include \(|z|=1\) we need \(r_{\max}<1\):

This is the single most-used sentence in digital filter design, and it is worth noticing how weak its hypotheses are. It says nothing about zeros: zeros may sit anywhere, inside, on or outside the circle, without affecting stability. Lecture 7 exploits that freedom heavily.

Section 4.3's Example 4.3 already worked through a two-sided case in full — the same sequence used there, \(x[n]=\left(\tfrac12\right)^nu[n]-2^{\,n}u[-n-1]\), has one pole inside the unit circle and one outside, yet the annulus \(\tfrac12<|z|<2\) still contains \(|z|=1\). By Eq. (4.7) that system is BIBO stable, even though the causal rule of Eq. (4.8) alone would have said otherwise — a reminder that Eq. (4.8) presumes causality, while Eq. (4.7) is the general criterion.

4.5  Properties, and the standard pairs

The properties mirror Lecture 2's Fourier properties almost line for line, so rather than re-derive them the table below states each with the one thing Fourier did not have to worry about: what happens to the ROC.

Two of these deserve a sentence more than a table row gives them.

Convolution becomes multiplication: \(Y(z)=H(z)X(z)\). For cascaded filters, the individual system functions multiply, so their poles and zeros combine. This allows high-order filters to be implemented as a cascade of simpler second-order sections, or biquads.

Multiplication by \(a^n\) moves poles and zeros geometrically. Their radii are scaled by \(|a|\), and their angles are shifted by \(\arg a\). When \(a=e^{j\omega_0}\), the movement is a pure rotation, which is useful for shifting a response to a different centre frequency.

Rows 4 and 5 have the same algebraic expression but different regions of convergence, so they represent different sequences. A single complex pole at \(re^{j\omega_0}\) produces a complex exponential. For a real signal, poles occur as the conjugate pair \(re^{\pm j\omega_0}\); adding their contributions gives a real sinusoid. The angle \(\omega_0\) sets the oscillation frequency, and the radius \(r\) sets the factor \(r^n\). As \(r\) moves closer to 1, the oscillation decays more slowly and the system rings for longer.

4.6  Getting back: three ways to invert

There is a formal inversion formula — a contour integral counter-clockwise around a closed path inside the ROC,

evaluated by residues. The contour must lie inside the ROC, which shows mathematically why a different ROC gives a different time sequence. For routine engineering problems, inspection, partial fractions and power-series expansion are usually more convenient.

4.6.1  Inspection, and partial fractions

If \(X(z)\) is rational — and for any LCCDE system it is — split it into first-order pieces and read each off the table. Two mechanical cautions cause most of the lost marks:

• Mind the extra \(z\). The table's pairs have a \(z\) in the numerator: \(z/(z-a)\), not \(1/(z-a)\). A term \(A/(z-a)\) is \(Az^{-1}\cdot z/(z-a)\), and the \(z^{-1}\) is a one-sample delay by the shift property — so the answer is \(Aa^{n-1}u[n-1]\), not \(Aa^nu[n]\). Alternatively, do the partial fractions in \(z^{-1}\) from the start and the problem does not arise.
• Repeated poles need every order. A double pole at \(a\) needs both \(A/(z-a)\) and \(B/(z-a)^2\); the second inverts to \(n a^n u[n]\) by the differentiation property, giving the characteristic \(n\,r^n\) growth-then-decay.

Then — and only then — use the ROC to decide the direction of each term. This is the step students skip, and Example 4.5 is entirely about it.

MATLAB corner for Example 4.5

A rational expression in \(z\) does not identify a unique sequence unless its region of convergence is also known. MATLAB's standard impulse-response functions assume a causal system, so the script constructs the anti-causal sequence separately. Comparing the two results makes the role of the ROC explicit.

Goal: plot the causal and anti-causal sequences that share one \(X(z)\), and confirm which one is absolutely summable, then verify Example 4.5's three inversions numerically.

%% Same X(z), two ROCs, two sequences  (Section 4.3)
% X(z) = 1/(1 - 0.5 z^-1).  MATLAB's impz assumes CAUSAL, i.e. |z|>0.5.
b = 1; a = [1 -0.5];
n = 0:12;  h_causal = impz(b, a, numel(n));
disp([n(1:6)', h_causal(1:6)]);           % (0.5)^n : 1, .5, .25, ...
 
% The anti-causal partner, -(0.5)^n u[-n-1], has the SAME b and a.
% No toolbox function can return it - the coefficients do not encode
% the ROC. You must construct it by hand:
nn = -12:-1;  h_anti = -(0.5).^nn;
figure('Color','w');
subplot(2,1,1); stem(n, h_causal, 'filled'); grid on;
  title('ROC |z|>0.5 : decays forward, absolutely summable -> STABLE');
subplot(2,1,2); stem(nn, h_anti, 'filled'); grid on;
  title('ROC |z|<0.5 : grows backward, not summable -> UNSTABLE');
 
fprintf('sum|h| causal = %.4f   anti-causal = %.1f\n', ...
        sum(abs(h_causal)), sum(abs(h_anti)));
 
%% Inversion by partial fractions - Example 4.5, all three ROCs
bb = [0 1]; aa = poly([0.5 2]);   % X(z) = z / ((z-0.5)(z-2))
[res, pol, kdir] = residuez(bb, aa);
fprintf('residue %+.4f at pole %+.4f\n', [res(:) pol(:)]');
% residuez gives the residues; YOU choose each term's direction from
% the ROC: outside a pole -> causal term, inside it -> anti-causal term.
 
%% Power series when there is nothing to factor (Section 4.6.2)
% X(z) = exp(z^-1) is not rational - but its series is immediate.
N = 8;  x = 1./factorial(0:N-1);
fprintf('x[n] = 1/n! : '); fprintf('%.4f ', x); fprintf('\n');
% Long division does the same job numerically for rational X(z):
disp(deconv([1 0 0 0 0 0], [1 -0.5]));   % -> 1, .5, .25, .125, ...

4.6.2  Power series, when the transform is not rational

Eq. (4.1) says \(X(z)\) is a power series in \(z^{-1}\) whose coefficients are the samples. So any expansion valid in the ROC identifies \(x[n]\) directly by matching coefficients — no factoring, no table.

The ROC decides which expansion is legal, and this is where the method earns its keep. For \(X(z)=1/(1-\tfrac12z^{-1})\):

Power-series expansion also works for non-rational transforms. For example, \(e^{z^{-1}}=\sum_k z^{-k}/k!\) immediately gives \(x[n]=1/n!\) for \(n\ge0\). The method may produce only a sequence of coefficients rather than a compact formula, but that is often enough when only the first few impulse-response samples are required.

4.7  Difference equations become algebra: LCCDEs, FIR and IIR

Here is the payoff that motivated the transform in the first place. A great many discrete systems — every digital filter, every discrete controller, every recursive average — are described by a linear constant-coefficient difference equation:

Solved for the present output, it is a recipe a processor can execute literally:

Now transform Eq. (4.10). Every delay \(y[n-k]\) becomes \(z^{-k}Y(z)\), the sums come out of the transform by linearity, and what was a recursion in time is a single algebraic equation. Dividing:

Two coefficient lists, and the whole system is specified. Factor numerator and denominator and you have the form design actually works in:

The \(q_m\) are the zeros: frequencies the filter kills. The \(p_k\) are the poles: frequencies it emphasises. Designing a filter is choosing where to put these points, and Eq. (4.8) constrains the poles to the interior of the unit circle.

4.7.1  Recursive or not: IIR and FIR

Eq. (4.11) splits the world in two, on one question: does the output depend on past outputs?

That trade — guaranteed stability and linear phase against efficiency — is the central engineering decision of the second half of this module, and it is visible entirely in \(N\).

4.7.2  Zero-input and zero-state response

A recursive system has memory, so its output has two independent causes, and the transform keeps them apart:

The zero-input response is caused by the system's initial stored state when the external input is zero. It is the homogeneous solution and is built from the natural modes \(p_k^n\). The zero-state response is the response to the input when all initial conditions are zero: \(y_{\text{ZS}}=x*h\), or \(Y_{\text{ZS}}(z)=H(z)X(z)\).

Note which one the poles govern: both. The homogeneous solution's terms are exactly \(p_k^{\,n}\), so if every \(|p_k|<1\) any initial-condition transient dies out on its own. That is the time-domain meaning of Eq. (4.8) — a stable filter forgets its own startup state.

4.7.3  Realization: three blocks, and that is all

Eq. (4.11) needs an adder, a constant multiplier and a one-sample delay. Nothing else. Every filter structure in this module is those three components wired differently, and the wiring diagram is a faithful picture of the arithmetic a processor performs per sample.

The number of delays is the filter's order, and it is what costs memory; the multipliers are what cost arithmetic. Reading a structure for those two counts is how implementations are compared.

4.8  The geometric method: seeing \(|H(e^{j\omega})|\) in the z-plane

The pole–zero plot gives a visual way to estimate a filter's frequency response. It also provides a design guide: place poles near frequencies to be emphasized and zeros near frequencies to be suppressed.

Start from the factored form, Eq. (4.13), and evaluate it on the unit circle:

Every factor has the form \((e^{j\omega}-c)\), which is a vector in the complex plane: it points from the fixed point \(c\) to the moving point \(e^{j\omega}\) on the unit circle. Taking magnitudes and angles of Eq. (4.15) separates cleanly:

So: walk the point \(e^{j\omega}\) once around the unit circle and watch the rubber bands stretch. Where a pole vector gets short, \(|H|\) rises. Where a zero vector gets short, \(|H|\) falls. Nothing needs to be computed to know the shape.

Three consequences do most of the work in practice.

• A pole at radius \(r\) and angle \(\theta\) makes a peak at \(\omega=\theta\), and its sharpness is governed by \(1-r\). At the closest approach the pole vector has length \(1-r\), so the peak height goes as \(1/(1-r)\): a pole at \(r=0.99\) gives a peak ten times taller and far narrower than one at \(r=0.9\). Resonance is proximity to the circle.
• A zero on the circle at \(e^{j\omega_0}\) forces \(|H(e^{j\omega_0})|=0\) exactly — a perfect null. That is how a notch filter removes mains hum: place a zero pair at \(\omega_0=2\pi f_0/f_s\). A zero merely near the circle gives a finite dip.
• Distance mutes influence. A pole or zero near the origin is roughly the same distance from every point on the circle, so it contributes an almost constant factor and hardly shapes the response at all. Only features near the circle matter — which is why pole–zero plots are read radially outward.

MATLAB corner for Example 4.5

This script makes Section 4.8 tangible: drag a pole outward and watch the resonance peak sharpen, then builds the Example 4.7 notch and shows it three ways — pole–zero plot, magnitude response, and difference-equation output — from the same coefficients.

Goal: sweep the pole radius \(r\) toward the unit circle to see the peak height and width track \(1/(1-r)\), then reproduce Example 4.5's notch from its pole–zero placement.

%% L4_2_fixed.m
%% Part 1 - resonance IS proximity to the unit circle  (Eq. 4.16)
w0 = pi/3;
figure('Color','w'); hold on; grid on;
for r = [0.5 0.8 0.95 0.99]
    b = 1;
    a = [1, -2*r*cos(w0), r^2];
    [H,w] = freqz(b, a, 1024);
    plot(w/pi, 20*log10(abs(H)), 'LineWidth',1.5, ...
         'DisplayName', sprintf('r = %.2f', r));
end
xline(w0/pi, 'k:', '\omega_0'); legend show;
xlabel('\omega / \pi'); ylabel('|H| (dB)');
title('Peak height ~ 1/(1-r): the pole vector is shortest at \omega=\omega_0');
 
%% Part 2 - the Example 4.7 notch
fs = 500; f0 = 50;  w0 = 2*pi*f0/fs;  r = 0.95;
b = [1, -2*cos(w0), 1];
a = [1, -2*r*cos(w0), r^2];
figure('Color','w');
subplot(1,3,1); zplane(b,a); title('pole-zero');
subplot(1,3,2); [H,f] = freqz(b,a,1024,fs);
  plot(f, 20*log10(abs(H)),'LineWidth',1.4); grid on; xlim([0 150]);
  xlabel('Hz'); ylabel('dB'); title('50 Hz is annihilated');
subplot(1,3,3); impz(b,a,80); title('impulse response rings at 0.95^n');
ylim([-0.5 0.5]);
grid on
t = (0:1/fs:2)';
sig = sin(2*pi*3*t) + 0.5*sin(2*pi*50*t);
out = filter(b, a, sig);
figure('Color','w'); plot(t,sig,'Color',[.7 .7 .7]); hold on;
plot(t,out,'LineWidth',1.4); legend('with hum','notched'); grid on;
 
%% Part 3 - stability test (no ternary operator in MATLAB)
a_bad = [1, -1.2, 0.85];
pmax = max(abs(roots(a_bad)));
if pmax < 1
    status = 'stable';
else
    status = 'UNSTABLE';
end
fprintf('max |pole| = %.4f  ->  %s\n', pmax, status);
isstable(tf(b, a_bad, 1/fs));   % requires Control System Toolbox

4.9  Closing the ladder: what each transform was for

The transforms introduced so far form a connected family rather than six unrelated formulas. The Laplace transform is the general continuous-time transform, and the \(z\)-transform is its discrete-time counterpart. The Fourier transforms are obtained by evaluating these general transforms on particular boundaries.

The choice of transform depends on whether the signal is continuous or discrete and whether the task is theoretical analysis, frequency-response measurement or finite computation.

The next lectures use this toolkit for filter design. Lecture 5 develops analog prototypes, Lecture 7 designs FIR filters, and Lecture 8 maps analog designs into digital IIR filters. In each case, the goal is to place poles and zeros so that selected frequencies pass while others are attenuated.

?Problems with solutions

Summary and key formulas

The reference sheet: definition, geometry, stability and the route from a difference equation to a frequency response.

• The \(z\)-transform is the DTFT of \(x[n]r^{-n}\). That one sentence explains why it exists where the DTFT does not, and why the DTFT is the special case \(r=1\) — the unit circle.
• \(z=e^{sT}\), straight from sampling \(e^{st}\). Hence \(\sigma<0\mapsto|z|<1\), the \(j\Omega\)-axis \(\mapsto\) the unit circle, \(\omega=\Omega T\), and — because \(e^{sT}\) is periodic — a many-to-one map, which is aliasing restated.
• The ROC is half the answer. \(a^nu[n]\) and \(-a^nu[-n-1]\) have identical \(X(z)\) and identical poles; only the region distinguishes them. Right-sided \(\Rightarrow\) outward, left-sided \(\Rightarrow\) inward, two-sided \(\Rightarrow\) an annulus, finite \(\Rightarrow\) everywhere.
• Stability, absolute summability and existence of a frequency response are one condition: the ROC contains \(|z|=1\). Add causality and it becomes the familiar all poles inside the unit circle. Zeros are unconstrained.
• Because a delay is \(z^{-1}\), an LCCDE becomes a rational \(H(z)\). \(N=0\) means FIR — poles only at the origin, unconditionally stable; \(N\ge1\) means IIR — efficient, but stability must be checked.
• Invert by inspection and partial fractions when \(X(z)\) is rational (watch the numerator \(z\), and let the ROC choose each term's direction), and by power series when it is not.
• The geometric method is the skill to keep: \(|H|\) is the product of zero-vector lengths over pole-vector lengths. Poles near the circle make peaks of height \(\sim1/(1-r)\); zeros on it make exact nulls; anything near the origin barely matters.
• Next, in Lecture 5: classical analog filter design — Butterworth, Chebyshev and elliptic prototypes, and the specification language (passband ripple, stopband attenuation, transition width) that Lectures 7 and 8 will translate into pole and zero positions on the plane you have just learned to read.