Thinking in Waves Episode 31 ／ Part XII · Moving the dimension

From the $s$ plane to the $z$ plane

Building the order digitally ──
integer orders are FIR, fractional orders are IIR Expand $D^\alpha=(1-z^{-1})^\alpha$ and you get the Grünwald–Letnikov fractional derivative itself.
★ Only at integer orders do the taps terminate; otherwise they run on forever, decaying as $k^{-\alpha-1}$.
★★ And last episode's "dimension" was exactly the same count ── the dimension of space is a filter's tap count.

Prerequisites: Episodes 1, 2, 5, 26, 27, 30Computation: kenshou/dejitaru.js

For thirty episodes we have stayed analogue ── differentiation is $\times(i\omega)$, the order is $(i\omega)^\alpha$, stability is "poles in the left half-plane".
Sample it into discrete time and the stage moves.

Concept	Analogue ($s$ plane)	Digital ($z=e^{sT}$ plane)
Frequency axis	imaginary axis $s=i\omega$	★ unit circle $z=e^{i\omega T}$
Stability	$\operatorname{Re}s<0$	★ $\|z\|<1$
Differentiation	$s$	★ $(1-z^{-1})/T$
Integration	$1/s$	★ $1/(1-z^{-1})$ ＝ accumulator
Delay $\tau$	$e^{-s\tau}$	$z^{-n}$ ── ★ integers only
Constants (Ep. 5)	$\omega=0$	★ a pole at $z=1$
Fractional order	$(i\omega)^\alpha$	★ $(1-z^{-1})^\alpha$ ← this episode

01★ The digital fractional derivative ── Grünwald–Letnikov

Just a binomial expansion $$(1-z^{-1})^\alpha=\sum_k w_k\,z^{-k},\qquad w_k=(-1)^k\binom{\alpha}{k}$$

By recursion, $w_0=1,\ w_k=w_{k-1}\left(1-\dfrac{\alpha+1}{k}\right)$. The $w_k$ are the filter's taps

$\alpha$	$w_0\dots w_5$	Taps
−1	1, 1, 1, 1, 1, 1	★ infinite (accumulator)
−1/2	1, 0.5, 0.375, 0.3125, 0.2734, 0.2461	★ infinite
0	1, 0, 0, 0, 0, 0	★ ends after 1
1/2	1, −0.5, −0.125, −0.0625, −0.0391, −0.0273	★ infinite
1	1, −1, 0, 0, 0, 0	★ ends after 2
2	1, −2, 1, 0, 0, 0	★ ends after 3
3	1, −3, 3, −1, 0, 0	★ ends after 4

The heart of §01

★★ Only when $\alpha$ is a non-negative integer do the taps terminate (FIR).

★ $\alpha=-1$ is all ones ── an accumulator: whatever goes in stays forever.
★ $\alpha=0$ passes straight through ($\delta$). ★ $\alpha=1$ is the difference $[1,-1]$, two taps.
★ $\alpha=\pm1/2$ never ends.

02The tail decays as a power law ── the same shape as Episodes 26 and 27

Asymptotics $$w_k\ \approx\ \frac{k^{-\alpha-1}}{\Gamma(-\alpha)}\qquad(\text{large }k)$$

$\alpha$	$k$	$w_k$	$k^{-\alpha-1}/\Gamma(-\alpha)$	Ratio
−0.50	$10^{5}$	1.7841e−3	1.7841e−3	0.999999
0.50	$10^{5}$	−8.9207e−9	−8.9206e−9	1.000004
1.50	$10^{5}$	1.3381e−13	1.3381e−13	1.000019
−0.25	$10^{5}$	4.9048e−5	4.9048e−5	0.999999

The heart of §02

A fractional-order digital filter has coefficients that decay as a power law ── memory that never ends. This is the same shape as Episodes 26 and 27 ──
Episode 26: fractional order ⇐ time constants distributed as $g(\tau)\propto\tau^{\alpha-1}$
★ Here　　: fractional order ⇐ taps distributed as $w_k\propto k^{-\alpha-1}$
Continuous versus discrete is the only difference; they say the same thing.

03Does applying the half order twice give the first?

The standard check for a fractional order ── by convolution $$D^{1/2}\!\circ D^{1/2}=D^1\quad\Longleftrightarrow\quad w^{(1/2)}*w^{(1/2)}=w^{(1)}=[1,-1]$$

$n$	$(w^{½}*w^{½})_n$	$w^{(1)}_n$
0	1.000000000000	1
1	−1.000000000000	−1
2–7	0.000000000000	0

★ Pass a signal through an infinite filter twice and you get a filter that ends after two taps ── the half order really is the square root of the first.

04In the frequency domain ── which discretization preserves the order?

There is more than one way to replace $s$ by $z$. Put them on the unit circle $z=e^{i\theta}$ ($\theta=\omega T$) and look at the phase.

Backward difference $$1-e^{-i\theta}=2\sin\frac\theta2\cdot e^{i(\pi/2-\theta/2)}\quad\Longrightarrow\quad\text{phase}=\alpha\left(90^\circ-\frac\theta2\right)$$

Bilinear (Tustin) $$\frac{1-e^{-i\theta}}{1+e^{-i\theta}}=i\tan\frac\theta2\quad\Longrightarrow\quad\text{phase}=90\alpha^\circ\ \text{(independent of frequency)}$$

$\alpha$	$\omega/\omega_N$	Backward diff.	Bilinear	Analogue $90\alpha^\circ$	Error of backward diff.
1.0	0.01	89.100	90.000	90.000	−0.900°
1.0	0.30	63.000	90.000	90.000	−27.000°
1.0	0.90	9.000	90.000	90.000	−81.000°
0.5	0.50	22.500	45.000	45.000	−22.500°
−0.5	0.50	−22.500	−45.000	−45.000	+22.500°

The heart of §04

★★ The bilinear transform holds the phase at exactly $90\alpha^\circ$ at every frequency.
Episode 1's "the order is the phase" has become the criterion for choosing a discretization.
★ The backward difference tips toward zero phase near Nyquist: even at first order, staying within 1° means staying below 1 % of Nyquist (0.6 % at $\alpha=2$). The higher the order, the tighter.

05★★ The link to last episode ── the dimension was a tap count

Episode 30: the order of the $d$-dimensional Green's function is $\alpha=(d-3)/2$; odd dimensions are sharp, even ones leave a tail. This episode: only non-negative integer $\alpha$ gives finite taps. Overlay them.

$d$	$\alpha$	GL taps	Episode 30's tail	In digital terms
1	−1	infinite (never decays ── all ones)	$\theta(s)$, a step that never leaves	★ IIR
2	−1/2	infinite ($k^{-1/2}$)	tail $s^{-1/2}$	★ IIR
3	0	★ 1 tap [1]	a bare $\delta$	★ FIR
4	+1/2	infinite ($k^{-3/2}$)	tail $s^{-3/2}$	★ IIR
5	+1	★ 2 taps [1, −1]	$\delta'$ (differentiated)	★ FIR
7	+2	★ 3 taps [1, −2, 1]	$\delta''$	★ FIR

The heart of §05

★★★ The dimension of space can be counted as a filter's tap count.

★ Odd dimensions ＝ integer order ＝ FIR (taps end after $(d-1)/2$)
★ Even dimensions ＝ half-integer order ＝ IIR (taps run on as a power law)
── ★★ "Huygens' principle holds" ＝ "that filter is FIR".

Read one by one ── $d=1$ is an all-ones accumulator so sound stays forever ($\theta(s)$); $d=3$ is a single-tap pass-through, undistorted and gone once past ($\delta$); $d=5$ is a two-tap difference, sharp but differentiated ($\delta'$).

06Actually ringing the 2D reverberation

What do the $d=2$ taps ($\alpha=-1/2$) sound like as reverberation? Here is the step response ── how much piles up while you keep sounding.

Samples $n$	$d=1$ ($\alpha=-1$)	$d=2$ ($\alpha=-1/2$)	$d=3$ ($\alpha=0$)	Ratio to $\sqrt n$
$10^{2}$	101.0	11.3260	1.0	1.132604
$10^{3}$	1001.0	35.6959	1.0	1.128802
$10^{5}$	100001.0	356.8262	1.0	1.128383

The heart of §06

$d=3$ stays at 1 ── nothing piles up. $d=1$ grows as $n$. ★ $d=2$ grows as $\sqrt n$ ── in between.
The ratio converges to $2/\sqrt\pi=1.128379$.
★ "In a two-dimensional room, sound piles up as $\sqrt{\text{time}}$" ── an hour of talking leaves 7.7 times what a minute does.

07A problem that only appears digitally ── fractional delay

In the analogue world $e^{-s\tau}$ lets $\tau$ be as fine as you like. Digitally, delay is $z^{-n}$ with $n$ an integer. So what does a "half-sample delay" $z^{-1/2}$ look like?

The ideal fractional delay $$h_k=\operatorname{sinc}(k-D),\qquad D=\text{delay in samples}$$

$k$	$D=0$ (integer)	$D=0.5$ (half sample)	$D=1$ (integer)
−1	0	−0.212207	0
0	1.000000	0.636620	0
1	0	0.636620	1.000000
2	0	−0.212207	0
3	0	0.127324	0

The heart of §07

An integer delay is one tap. A half-sample delay runs on forever, and decays only as $1/k$.
★ Exactly the pattern of §01 ── integer → finite (FIR), fractional → infinite (IIR).
★★ In the order, in the delay, and in the dimension: the moment you leave the integers, a tail grows.

Exercises

You want a half-order filter in an audio device. Backward difference or bilinear?
Show the answer
Bilinear. The backward difference gives a phase $\alpha(90^\circ-\theta/2)$, so past 1 % of Nyquist it is already off by more than a degree. At the top of the audible band (20 kHz against a 24 kHz Nyquist, i.e. 0.83) the backward-difference phase has essentially collapsed.
★ Bilinear keeps the phase at exactly $90\alpha^\circ$ at every frequency because $\tan(\theta/2)$ is purely imaginary. The price is frequency warping, $\omega\to(2/T)\tan(\omega T/2)$ (pre-warping is the standard fix).
★ Episode 1's "the order is the phase" becomes a design criterion.
The $\alpha=-1/2$ taps are all positive; the $\alpha=+1/2$ taps are negative from the second onward. Why?
Show the answer
Because the sign of the asymptote $w_k\approx k^{-\alpha-1}/\Gamma(-\alpha)$ is set by $\Gamma(-\alpha)$. At $\alpha=-1/2$, $\Gamma(1/2)=\sqrt\pi>0$; at $\alpha=+1/2$, $\Gamma(-1/2)=-2\sqrt\pi<0$.
★ In meaning ── the integrating side ($\alpha<0$) accumulates the past, so every tap is positive; the differentiating side ($\alpha>0$) subtracts it, so they go negative.
★ The extreme case is $\alpha=-1$ (pure integration), all ones ── exactly Episode 30's $d=1$, "a step that never leaves".
If "Huygens holds ＝ FIR", what would creatures in a non-FIR dimension struggle with?
Show the answer
They could not separate moments in time, because the past keeps mixing in. In $d=2$ the reverberation piles up as $\sqrt n$ (§06), so the background rises the more you speak.
★ To hear "when" something sounded, the impulse response must be finite ── a different constraint from Episode 9's bandwidth theorem ($\Delta t\Delta\omega\ge1/2$). However wide your bandwidth, an IIR medium muddies timing.
★ Only 3 dimensions has $\alpha=0$ (one tap), so conversation is a privilege of three dimensions (Episode 30).

Summary　The order becomes visible as a number of taps

We moved the stage from the $s$ plane to the $z$ plane. Expanding $D^\alpha=(1-z^{-1})^\alpha$ binomially gives the Grünwald–Letnikov fractional derivative, and the coefficients $w_k$ are the filter's taps.

★ Only for non-negative integer $\alpha$ do the taps terminate (FIR): 1 tap at $\alpha=0$, $[1,-1]$ at $\alpha=1$, $[1,-2,1]$ at $\alpha=2$. Otherwise they continue forever, decaying as $w_k\approx k^{-\alpha-1}/\Gamma(-\alpha)$ (agreeing to 6 digits out to $k=10^5$). ★ The same shape as Episode 26's $g(\tau)\propto\tau^{\alpha-1}$ ── continuous versus discrete is the only difference.

We checked it too ── convolving the $D^{1/2}$ taps with themselves gives exactly $[1,-1]$. An infinite filter applied twice ends after two taps. The half order really is the square root of the first.

The order also decides how to discretize: the bilinear (Tustin) transform holds the phase at exactly $90\alpha^\circ$ at every frequency (because $\tan(\theta/2)$ is purely imaginary), while the backward difference is already off by a degree past 1 % of Nyquist even at first order. ★ Episode 1's "the order is the phase" became a design criterion.

★★★ And last episode's "dimension" was exactly the same count.
Odd dimensions ＝ integer order ＝ FIR ($(d-1)/2$ taps); even dimensions ＝ half-integer ＝ IIR.
The dimension of space can be counted as a filter's tap count.

$d=1$ is the all-ones accumulator ($\theta(s)$), $d=3$ the single-tap pass-through ($\delta$), $d=5$ the two-tap difference ($\delta'$) ── matching Episode 30's table line for line. And the $d=2$ reverberation piles up as $\sqrt n$, with the ratio converging to $2/\sqrt\pi=1.1284$.

Finally, the problem that only appears digitally has the same shape ── an integer delay is one tap; a half-sample delay is an infinite tail decaying only as $1/k$. In the order, in the delay, and in the dimension: leave the integers and a tail grows.

Next episode The tools are all in place. ★ From next time, Part XIII ── looking at frontier physics thoroughly as waves.
★ First, the Higgs. A negative mass term is an imaginary cutoff frequency (Episodes 19 and 28): the field does not oscillate but grows exponentially, and where nonlinearity (Episode 23) stops it is the vacuum.
★★ And ── the Higgs boson is amplitude modulation; the Nambu–Goldstone boson is phase modulation. AM and FM, exactly as in audio.

This document is Episode 31 of the "Thinking in Waves" series. The computations are in kenshou/dejitaru.js. On the numbers here: the GL coefficients and their asymptotics, the self-convolution of the half order, the phase of each discretization, the step responses, and the fractional-delay sinc coefficients were all computed for this article. The Grünwald–Letnikov definition, the coefficient asymptotics $w_k\sim k^{-\alpha-1}/\Gamma(-\alpha)$, the fact that the bilinear transform maps the imaginary axis onto the unit circle, and that the ideal fractional delay is a sinc are all well-known results and not discoveries of this article ── what this article did was line them up against Episode 30's "dimension = order" table row by row and make the correspondence "odd dimensions = FIR, even dimensions = IIR" explicit. That correspondence means the Green's functions have the same shape; it is not a claim that the $d$-dimensional wave equation is literally a GL filter (one is an order in time, the other a discrete-time operator). The phase-error table in §04 is for the first-order backward difference; higher-order discretizations do better. The "$\sqrt n$ pile-up" in §06 is a statement about the step response; real room acoustics also involve absorption. ── To print, use your browser's Print → Save as PDF.

Concept	Analogue (\(s\) plane)	Digital (\(z=e^{sT}\) plane)
Frequency axis	imaginary axis \(s=i\omega\)	★ unit circle \(z=e^{i\omega T}\)
Stability	\(\operatorname{Re}s<0\)	★ \(\|z\|<1\)
Differentiation	\(s\)	★ \((1-z^{-1})/T\)
Integration	\(1/s\)	★ \(1/(1-z^{-1})\) ＝ accumulator
Delay \(\tau\)	\(e^{-s\tau}\)	\(z^{-n}\) ── ★ integers only
Constants (Ep. 5)	\(\omega=0\)	★ a pole at \(z=1\)
Fractional order	\((i\omega)^\alpha\)	★ \((1-z^{-1})^\alpha\) ← this episode