Global attraction in the sigma-delta modulator piecewise isometry J.H.B. Deane School of Electronics, Computing and Mathematics, University of Surrey, Guildford, GU2 7XH, UK e-mail:
[email protected] January 6, 2006
Abstract This paper presents a geometrical proof of a conjecture in [1], that the map that describes the behaviour of the sigma-delta modulator [2] has a global, polygonal attractor. The map in question is given in equation (1), and is a two-dimensional piecewise isometry, that is, a map whose action is to cut the phase plane into several pieces each one of which is then rigidly rotated within the plane. The lines along which the cuts are made are the discontinuities of the map, and global attraction takes place for all points in the plane except those whose iterates eventually fall on a discontinuity.
1
Introduction
In this paper we consider the one-parameter piecewise isometry f : R 2 7→ R2 , introduced by Feely and Fitzgerald [2], and simplified and discussed in [1]: − cot θ (1) xn+1 = f(xn ) = R(θ)xn + σ(xn ) 1 where xn = [xn , yn ]T , σ(xn ) = sgn(xn sin θ + yn cos θ) − 2 cos θ sgn yn and R(θ) is the 2 × 2 matrix representing a clockwise rotation by θ. This mapping arises as a description of an electronic circuit, the bandpass sigma-delta modulator, in the case of zero damping. In [1] we showed that the effect of f, for θ ∈ (π/3, 2π/3) ∪ (4π/3, 5π/3), can be reduced to a rearrangement of four kite-shaped quadrilaterals M A , MB , MC , MD , whose geometry we determined, in such a way that M = MA ∪ MB ∪ MC ∪ MD = f (MA ) ∪ f (MB ) ∪ f (MC ) ∪ f (MD ) apart from a set of zero measure. See figure 1. A result in [3] implies that if this holds for an area preserving map, then it must be invertible on M except possibly on a set of zero measure. The purpose of the present paper is to prove geometrically that, in the Lebesgue sense almost all (a.a.) points in R2 eventually enter the invariant set M under iteration of f , for θ ∈ [π/3, 2π/3] ∪ [4π/3, 5π/3] — that is, we show that M is a global attractor. The proof is simplified by the fact that K, the closure of the set 1
f(MA)
MA MB MC
f(MD) MD
f(MB)
f(MC)
Figure 1: Left: the set M = MA ∪ MB ∪ MC ∪ MD on which the mapping f is invertible. Right: the effect of f on each of MA , MB , MC and MD leaves their union M unchanged. Hence, f is invertible on M . In both diagrams, θ = 1.8.
x ∈ R2 | f (f (x)) ∈ M
i.e. the union of M and its first and second preimages under f , is a rhombus.
2
Definitions
The map f splits the plane into four infinite triangular sectors, open on all boundaries and divided from each other by the lines of discontinuity of f, which are y = 0 and y = −x tan θ. For brevity, we refer to these lines just as discontinuities. In each sector σ(x) is constant. We name the sectors A, B, C, D, where A = {x | x > −y/ tan θ, y > 0}
and
D = {x | x > −y/ tan θ, y < 0}
(2)
and B and C are obtained from D and A by x 7→ −x (rotation by 180 ◦ ) respectively. We also define sets KT = K ∩ T where T is one of A, B, C, D. All these sets are illustrated in figure 2. To prove global attraction to M , we need to define a measure of distance d(x) valid for all x ∈ R2 and such that we can show something like d(f I (x)) < d(x) for some finite integer I and a.a. x ∈ R2 \ E. Here E is some subset of R2 that can be excluded while simultaneously simplifying, but not invalidating, the proof. For instance, we can certainly exclude x ∈ M with impunity since we are only interested in attraction to, and not dynamics within, M . The function d(x) is roughly analogous to a discrete Lyapunov function. A simple proof of global attraction results if we choose d(x) = kxk 2 (the squared Euclidean distance from the origin) along with I = 3 and E = K; numerical evidence for this choice is presented in figure 3. Accordingly, we define ∆T1 T2 T3 (x) = kxk2 − kf (f (f (x)))k2 with x ∈ T1 , f (x) ∈ T2 , f 2 (x) ∈ T3 ,
(3)
where T1 , T2 and T3 each are one of the sectors A–D. Note that even though ∆ is a squared distance, the fact that ∆(x) is greater than, equal to, or less than zero, implies that the actual distance of x from the origin is respectively decreased, unchanged or increased by f 3 . Two subsets of [0, 2π] are required: S 1 = [π/2, 2π/3] and S = [π/3, 2π/3] ∪ [4π/3, 5π/3]. 2
θ y=− x tan
B σ = −1− 2 cos θ
A σ = 1− 2 cos θ
KB
KA
θ
y=0
KC
KD
C
D
σ = −1+ 2 cos θ
σ = 1+ 2 cos θ
Figure 2: Illustration of the sets defined, with θ = 1.8. The discontinuities are shown as dotted lines.
4
4
2
2
0
0
-2
-2 θ = 1.2
-4 -4
-2
θ = 1.4 0
2
-4 4 -4
4
4
2
2
0
0
-2
-2
-2
0
-2
4
θ = 2π/3
θ = 1.8 -4 -4
2
0
2
-4 4 -4
-2
0
2
4
Figure 3: Numerical evidence for the conjecture that the set of points that move away from the origin under f 3 (grey region) is a subset of K (the rhombus) for θ ∈ [π/3, 2π/3].
3
3
Symmetries
In this section we write f(x) = f θ [x, y]T in order to emphasise the dependence of the mapping on θ. We state two symmetry properties of f(x) under transformations of θ, and one under a transformation of x, that leave ∆ unchanged. Symmetry 1 f π−θ [x, y]
T
= Hf θ [x, −y]
T
where H =
−1 0 0 1
Symmetry 2 f 2π−θ [x, y]T = Hf θ [−x, y]T Symmetry 3 f θ [−x, −y]T = −f θ [x, y]T All these can be verified by direct calculation. These symmetries imply that kf π−θ [x, y]T k2 = kf θ [x, −y]T k2 , kf 2π−θ [x, y]T k2 = kf θ [−x, y]T k2 and kf θ [x, y]T k2 = kf θ [−x, −y]T k2
so that ∆ is unchanged by the transformations θ 7→ π − θ, θ 7→ 2π − θ and x 7→ −x. The last of these implies that we need only consider y > 0.
4
The construction of K
In order to construct K, we first need the vertices of M , which follow from those of M A – MD . These were obtained in [1] from which we quote that ±1 MA = conv [0, 0] , ,0 , MB sin θ ±(1 + cos 2θ) − cos θ ±1 + cos 3θ , −1 − 2 cos 2θ , , 1 ∓ 2 cos θ sin θ sin θ where the upper signs apply to MA , the lower to MB and conv(p1 , . . . p4 ) is the convex hull of the points p1 , . . . , p4 . MC and MD are obtained by applying Symmetry 3 to M A and MB respectively. (Technically, the lower left vertex of M A should be [0+ , 0+ ] etc. to exclude the discontinuities from M .) To find K, we first define L = {x ∈ R2 | f (x) ∈ M }, whose construction entails the inversion of f . This can be done provided that the sector in which the result is to lie is specified (and valid). In that case,
4
w=f
−1
− cot θ (x) = R(−θ) x − σ(w) 1
(4)
We now determine the valid sector(s) for preimage(s) of points in A or D. In fact, there is a semi-infinite strip JAB in A ∪ D such that x ∈ JAB has two preimages, one in A and one in B; otherwise it has one preimage. To define J AB , observe that f(A) is the open set x sin θ > −y cos θ and y cos 2θ > −x sin 2θ + 2 cos θ − 1. Similarly, f (B) is defined by x sin θ > −y sin θ and y cos 2θ < −x sin 2θ + 2 cos θ + 1. Then J AB = f(A) ∩ f (B) and JA = (A ∪ D) \ f(B), JB = (A ∪ D) \ f(A). These results are obtained simply by applying f to the boundaries of A and B. We summarise this in Lemma 1 (i) If x ∈ A ∪ D, then x has 1, 2, 1 preimages, which are in A, A and B, B, for x ∈ JA , JAB and JB respectively, where JA = {(x, y) | x sin θ > −y cos θ, y cos 2θ > −x sin 2θ + 2 cos θ + 1} JAB
= {(x, y) | x sin θ > −y cos θ, y cos 2θ > −x sin 2θ + 2 cos θ − 1 and y cos 2θ < −x sin 2θ + 2 cos θ + 1}
JB = {(x, y) | x sin θ > −y cos θ, y cos 2θ < −x sin 2θ + 2 cos θ − 1} (ii) If x ∈ B ∪ C, then x has 1, 2, 1 preimages, which are in C, C and D, D, for x ∈ J C , JCD and JD respectively, where JC = {(x, y) | x sin θ < −y cos θ, y cos 2θ < −x sin 2θ − 2 cos θ − 1} JCD = {(x, y) | x sin θ < −y cos θ, y cos 2θ > −x sin 2θ − 2 cos θ − 1 and y cos 2θ < −x sin 2θ − 2 cos θ + 1}
JD = {(x, y) | x sin θ < −y cos θ, y cos 2θ > −x sin 2θ − 2 cos θ + 1} Part (ii) is obtained from part (i) by Symmetry 3. The construction of K is simplified if we can represent M and L as unions of their subsets, such that the preimages of each union are entirely within a single sector. The various subsets needed are: LA , LB = f −1 (MA ∪ MD ) = f −1 (MAD ) LC , LD = f −1 (MB ∪ MC ) = f −1 (MBC ) KA , KB = f −1 (LA ∪ LD ) = f −1 (LAD ) KC , KD = f −1 (LB ∪ LC ) = f −1 (LBC ) . 5
The subscripts on the set names (K, L, M ) denote the sector of which the set is a subset. Some of these are illustrated in figure 4. It is easily verified that M AD , LAD ⊂ JAB and MBC , LBC ⊂ JCD . Hence, these sets all have two preimages. Thus, M AD is given by 1 + cos 3θ 1 − cos θ + cos 2θ , 1 − 2 cos θ , , −1 − 2 cos 2θ , MAD = conv sin θ sin θ 1 + cos θ + cos 2θ 1 − cos 3θ , 1 + 2 cos 2θ , , −1 − 2 cos θ sin θ sin θ and its preimages LA , LB are ±2 ±1 − cos θ − cos 2θ ±1 − cos θ + cos 2θ conv [0, 0] , ,0 , , 1 + 2 cos θ , , 1 − 2 cos θ sin θ sin θ sin θ where the upper/lower signs apply to L A /LB respectively. The preimages of MBC are just the reflection (Symmetry 3) of those of M AD . Now, the preimages of LAD , denoted KA and KB , are ±2 ±2 − 2 cos θ ,2 , ,0 conv [0, 0] , [−2 cot θ, 2] , sin θ sin θ with the upper/lower signs applying to K A /KB . The vertices of KC and KD follow by Symmetry 3; hence, we have proved Lemma 2 K, the closure of KA ∪ KB ∪ KC ∪ KD , is the rhombus with vertices ±2 2 2 ±2 − ,2 , + , −2 sin θ tan θ sin θ tan θ
MB MC
MA
LA
LB
MD
LC
LD
Figure 4: The sets MA . . . MD (left) and LA . . . LD (right) illustrated for θ = 1.8.
6
[−2/s − 2/t, 2]
[−2/t, 2]
KB
[2/s − 2/t, 2]
KA
−2/s
2/s
KC
KD
[−2/s + 2/t, −2]
[2/s + 2/t, −2]
Figure 5: The set K, the closure of KA ∪ KB ∪ KC ∪ KD , illustrated for θ = 1.8. In the figure, s = sin θ, t = tan θ.
5
Global attraction to M
We can now prove that a.a. points in R 2 are attracted to M under iteration of f for θ ∈ S 1 . We first observe that ∆ as defined in equation 3 is a linear scalar field, i.e. it is of the form xAx + yAy + A0 where Ax , Ay and A0 are functions of θ only. That this must be so is a direct result of the fact that f is a piecewise isometry. Hence, the ‘equipotentials’ of ∆ will be parallel straight lines. See table 1 for the explicit expressions for ∆. We establish global attraction by 1. proving that ∆ increases as we move away from K; 2. proving that ∆ ≥ 0 for a.a x ∈ R2 \ K; 3. proving attraction even in the special case ∆ = 0; for y > 0 (sectors A and B) and θ ∈ [π/2, 2π/3] = S 1 only. Symmetries 1–3 then allow us to extend these results immediately to θ ∈ S and a.a. R 2 \ K. To prove items 1 and 2, refer to figure 6. The upper half plane is split into various subsectors labelled A > D > B, A > D > C etc. according to the itinerary of x in each subsector. So, for instance, for x in the subsector labelled A > D > B, x ∈ A, f(x) ∈ D and f 2 (x) ∈ B. The subsectors are separated from one another by the discontinuities and lines named, for example, ADCDB etc., which forms the boundary between a subsector that is mapped from A to D then C, and one which goes to D then B. The equations of these dividing lines are found by calculating the relevant first and second preimages of the discontinuities, and are as follows: ADCDB y cos 2θ = x sin 2θ + 2 cos 2θ ∓ 4 cos θ + 1 BADAA AADDC y cos θ = x sin θ + 2 cos θ ∓ 1 BADDC where the upper signs apply to ADCDB , AADDC and the lower to BADAA , BADDC . 7
B>A>D
B>D>C
g
A>D>C
f
e
DC AD
B ADDC
A
B ADA
AA
A>A>D
c
A ADDC
DC AD
h
AD
d
DB
A DC
b
DC DB
A>D>B B>A>A
i
j
O
k
a
Figure 6: The subsectors of the upper half plane, labelled according to the itinerary of points in them under f, for θ = 1.8. The dotted lines show the boundaries between these subsectors. Points a–i are on the boundary of K and are used in the global attraction proof. To establish item 1, first note that unit normals to the edges of K, directed outwards, are ur = [sin θ, cos θ] (normal to ac); ut = [0, 1] (normal to cg); and ul = [− sin θ, − cos θ] (normal to ig). Now define the two-dimensional gradient operator as ∇ = [∂/∂x, ∂/∂y]. Then, from the definition of ∇ and the dot product, ∆ increases with increasing distance from K if the direction inequality, u · ∇(∆) > 0, is satisfied. Since ∆ is always a linear scalar field, u · ∇(∆) is independent of x and y within a subsector. ∆(x)
Expression for ∆(x) sin2 θ
∆ADB
s −8c3 + 2 c + 4 x + 8 c4 − 10 c2 + 2 y − 16 c4 + 16 c3 + 8 c2 − 4 c − 5
∆ADC ∆AAD ∆BDC ∆BAD ∆BAA
s 8c3 − 6c + 4 x + −8 c4 + 6 c2 + 2 y + 16 c4 − 16 c3 − 16 c2 + 12 c − 5 s 8c3 − 6c x + −8 c4 + 6 c2 + 2 y + 16 c4 − 16 c2 − 1 ∆AAD
s 8c3 − 6c − 4 x + −8 c4 + 6 c2 + 2 y + 16 c4 + 16 c3 − 16 c2 − 12 c − 5 s −8c3 + 2c − 4 x + 8 c4 − 10 c2 + 2 y − 16 c4 − 16 c3 + 8 c2 + 4 c − 5
Table 1: Expressions for ∆(x) sin2 θ in the upper half plane. Note that x = [x, y], c = cos θ, s = sin θ. √ Now, for θ ∈ S1 , cos θ ∈ [−1/2, 0] and sin θ ∈ [ 3/2, 1]. The direction inequalities are given in table 2 and their positivity is easily proved. It turns out in practice that we can prove positivity by using the simple lower bound: 8
∆
∆ sin2 θ
L.b.
Direction inequality: ∇(∆) · u
L.b.
∆ADB (a)
−16c4 + 8c2 + 3
2
−16c3 + 4c + 4
2
∆ADB (b)
−4c2 − 4c + 5
4
∆ADC (b)
= ∆ADB (b)
4
16c3 − 4c + 4
2
∆ADC (c)
−16c4 + 8c2 − 8c + 7
6
∆ADC (d)
(3 − 4c2 )(1 + 2c)
0
8c2 + 2
2
∆AAD (d)
8c3 − 4c2 − 6c + 3
1
8c2 + 2
2
∆AAD (e)
= ∆ADB (a)
2
∆BDC (e)
= ∆ADB (a)
2
8c2 + 2
2
∆BDC (f )
= ∆ADC (d)
0
∆BAD (f )
= ∆AAD (d)
1
8c2 + 2
2
∆BAD (g)
−16c4 + 8c2 + 8c + 7
2
∆BAD (h)
−4c2 + 4c + 5
2
−16c3 + 4c + 4
2
∆BAA (h)
= ∆BAD (h)
2
16c3 − 4c + 4
2
∆BAA (i)
= ∆ADB (a)
2
Table 2: Expressions for the various ∆ at points on the boundary of K (see figure 6), their lower bounds, the direction inequalities, and their lower bounds, for θ ∈ [π/2, 2π/3]. Note that c = cos θ.
9
f ADB
8 c4 − 8 c 3 − 6 c 2 + 6 c + 1 β , −γs x + α y − 2 (c − 1) (2 c + 1) (2 c − 1) αx − y + s s
f ADC
8 c4 − 8 c3 − 10 c2 + 6 c + 1 β , −γs x + α y − 8 c3 + 8 c2 + 6 c − 2 αx − y + s s
f AAD
β 8 c4 − 10 c2 + 1 αx − y + , −γs x + α y − 2α s s
= f AAD
f BDC
f BAD
f BAA
8 c4 + 8 c3 − 10 c2 − 6 c + 1 β , −γs x + α y − 8 c3 − 8 c2 + 6 c + 2 αx − y + s s
8 c4 + 8 c 3 − 6 c 2 − 6 c + 1 β , −γs x + α y − 2 (2 c + 1) (2 c − 1) (c + 1) αx − y + s s
Table 3: Expressions for f 3 (x) in the various subsectors of the upper half plane. Note that c = cos θ, s = sin θ, α = c(4c2 − 3), β = (c2 − 1)(4c2 − 1) and γ = (4c2 − 1).
min θ∈T
N X i=1
gi (θ)
!
≥
N X i=1
min gi (θ) θ∈T
(5)
where the gi are functions of θ and T is a set of values of θ. Hence, a lower bound for −16 cos3 θ + 4 cos θ + 4 is 0 − 2 + 4 = 2 > 0; similarly, a lower bound for 16 cos 3 θ − 4 cos θ + 4 is −2 + 0 + 4 = 2 > 0; and the lower bound for 8 cos 2 θ + 2 = 2 > 0. Since we merely need to prove positivity, no more sophisticated lower bounds are needed. The positivity of the various ∆ — see figure 8 — can be proved in the same way, except for ∆ADC (d) = ∆BDC (f ), which we consider separately. In fact, we can easily prove that ∆ADC (d) is a monotonically decreasing function of θ ∈ S 1 , and hence that θ = 2π/3 is the only zero of this function in S1 . Using c = cos θ, s = sin θ, we have ∆ADC (d) = s−2 (−8c3 − −3 4 2 4c2 +6c+3). The derivative of √ this with respect to θ is s (−8c +18c +2c−6). By observing bounds of this as that c ∈ [−1/2, 0] and s ∈ [ 3/2, 1], we can find simple lower and upper √ before: the reciprocal of the denominator is positive and lies in [1, 3 3/8]. The four terms in the numerator are bounded by [−1/2, 0], [0, 9/2], [−1, 0] and −6. Hence, the numerator is in [−15/2, −3/2], and so the derivative is always negative and ∆ ADC (d) is a monotonically decreasing function of θ. Therefore, the minimum value of ∆ ADC (d) is zero, and can only occur for θ = 2π/3. All the expressions for ∆/ sin2 θ, evaluated at the named points on the boundary of K (see figure 6), are given in table 2, as well as their lower bounds. Note that sin 2 θ ∈ [3/4, 1], so ∆ > 0 if ∆ sin2 θ > 0. Thus, we can conclude from table 2 that ∆ ≥ 0 and that ∆ increases with increasing distance from K, for θ ∈ S 1 : hence, items 1 and 2 are proved. It only remains to consider the situation at the two points d and f , where ∆ ADC (d) = ∆BDC (f ) = (3−4 cos 2 θ)(1+2 cos θ) can equal zero. This can only happen when cos θ = −1/2 10
so that θ = 2π/3. The results so far obtained do not exclude the possibility that in this instance, a non-zero measure set of points from R 2 could accumulate just outside the boundary of K — these points would never be attracted to M , which would therefore not be a global attractor for a.a. x ∈ R2 . We now show that this is not the case. B>D>C
4 x-
1/2
x
1/2
3
x
x+
y=
x+
1/2
1/2
-3
1/2
-3
3
c
y=
y=
y=
d
3
e
4
f
y=
g
B>A>D
A>D>C
4
i
O
a
A>D>B
Figure 7: Figure analogous to figure 6 when θ = 2π/3. Line A DCDB is superimposed on ac and BADAA is superimposed on gi. √ According√to the results in table 2, ∆ = 0 only when θ = 2π/3, at points d = [2/ 3, 2] and f = [−2/ 3, 2]. Considering point d first, we take from table 3 the expression√for f ADC (x) = f 3 (x) for x ∈ subsector ADC. Substituting θ = 2π/3 and x = d 0 = [2/ 3 + δx, 2 + δy] √ 0 where δx is small and ADC), √ positive, and 3δx > δy > 0 (which ensures that d ∈ subsector 3 0 gives f (d ) = [−4/ 3 + δx, δy]. The inequalities obeyed by δx, δy ensure that f 3 (d0 ) ∈ K, and hence, although d itself is not mapped into K by f 3 , all points in its neighbourhood in subsector ADC, are. Now, ∆ADC is a linear scalar field and, since ∆ADC (c) = 16 when θ = 2π/3, it has non-zero gradient. Thus, even when θ = 2π/3, ∆ ADC > 0 for a.a. x ∈ subsector ADC. Around point f , a similar argument applies. Taking from table 3 the expression for f BDC (f 0 ), √ 0 with θ = 2π/3 and point [−2/√ 3 + δx, 2 + δy], now with δy small and positive, √ where f is the 3 0 and δx > −δy/ 3, we obtain f (f ) = [−4/ 3 + δx, δy]. That is, f 3 translates the subsector DBC so that f coincides with i in figure 7. Hence, points in the neighbourhood of f are either mapped into K (those that lie to the right of line if produced), or have ∆ ≥ 0 (those between Of produced and if produced), with equality only at f . We have thus shown that ∆(x) ≥ 0 for a.a. x ∈ R 2 \K. We have further shown that ∆(x) ≡ 0 at two points only, d and f , on the boundary of K, and this happens when θ = 2π/3 but for no other θ ∈ S1 . See figure 8. This figure also shows that all the ∆ are bounded below by ∆ADC (d) which is monotonic. The results of this section have only been proved for the upper half plane, y > 0, and for θ ∈ S1 . They can, however, be immediately extended by Symmetries 1–3: first, Symmetry 3 extends the results to the whole plane; then Symmetry 1 extends the range of θ to [π/3, 2π/3]; and finally, Symmetry 2 extends the extended range of θ to the whole of S = [π/3, 2π/3] ∪ [4π/3, 5π/3]. Putting together all the results of this section, we can now state Theorem 1 Almost all x ∈ R2 \ K are moved closer to the origin by f 3 . Since K surrounds 11
∆ADC(c)
16 14 12
∆
10 8
∆ADB(b)
6
∆BAD(g)
∆AAD(d) ∆ADB(a)
4
∆BAD(h)
2 0 0.50
∆ADC(d)
0.55
θ/π
0.60
0.65
Figure 8: The distinct ∆ as a function of θ ∈ [π/2, 2π/3].
12
the origin, a.a. x ∈ / K are moved closer to K by f 3 ; and since K consists of points a.a. of which enter M after at most two iterations of f, M is a global attractor for a.a. x ∈ R 2 , for θ ∈ [π/3, 2π/3] ∪ [4π/3, 5π/3].
6
Conclusions
We have presented a proof of global attraction of the invariant set M for the piecewise isometry that describes a system known as the sigma-delta modulator. The proof is geometrical and applies to almost all points in the two-dimensional phase space of the system. The proof in this particular case was made easier by the fact that the closure of the second preimage of M is a rhombus and that the inequalities required can be proved positive by equation (5).
Figure 9: Up to the first 20 preimages of M for θ = 5π/12. The preimages are coloured black (M itself), grey (first preimage), white (second preimage) and so on, cyclically. The rhombus K is clearly visible. That the proof is of the ‘almost all’ variety follows from the fact that the mapping is not defined on a subset of the phase plane, this subset being the discontinuities of the mapping. One interesting consequence of the global attraction property is that the entire phase plane, with the exception of a set of measure zero, is divided up into polygons, each one being a 13
preimage of some subset of M . These preimages must, by global attraction, cover almost all the plane, and in special cases form a simple periodic tiling of the plane. Numerical computations suggest the well-known tilings of the plane by equilateral triangles and by squares occur when θ = 2π/3 and π/2 respectively, but it is an open question as to whether tilings other than these can be exhibited by this dynamical system. Figure 9 is an attempt to show the first twenty preimages for θ = 5π/12.
References [1] P.Ashwin, J.H.B. Deane and X-C. Fu, Dynamics of a bandpass sigma-delta modulator as a piecewise isometry, Proceedings, ISCAS 2001, (Proceedings of the IEEE International Symposium on Circuits and Systems, 2001), pp III-811 – III-814 (Sydney, Australia, 2001) [2] O. Feely and D. Fitzgerald, Non-ideal and chaotic behaviour in bandpass sigma-delta modulators Proceedings of NDES 1996, Sevilla, Spain, pp. 399–404 (1996); O. Feely, Nonlinear dynamics of bandpass sigma-delta modulation, Proceedings of NDES, Dublin, pp 33–36 (1995); O. Feely and D. Fitzgerald, Bandpass sigma-delta modulation: an analysis from the perspective of nonlinear dynamics, Proceedings of ISCAS 1996, Atlanta, Ga, May 1996. A.C. Davies, Periodic Non-linear Oscillations from Bandpass sigma-delta modulators’, Proceedings of ISCAS 1996, Atlanta, Ga, pp. 469–472 (1996). [3] P. Ashwin and X-C. Fu, On the geometry of planar piecewise isometries (to appear in Journal of Nonlinear Science, 2002) P. Ashwin and X-C. Fu, Tangencies in invariant disk packings for certain planar piecewise isometries are rare, Dynamical Systems 16, 4 (2001) 333–345; P. Ashwin, Elliptic behaviour in the sawtooth standard map, Physics Letters A, 232, p. 409 (1997).
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