Advanced Topics on Multidimensional Signals

Abstract

The theory of the previous chapters is applicable to multidimensional signals, but to arrive at their full formulation, some topics, mainly multidimensional groups, must be further developed. In this chapter this will be done for gratings and especially for lattices, first with an efficient representation of gratings and then that of lattices, and then we will consider the problem of generating all subgroups of a given group. Other topics that are further investigated are those of cells, with their multiplicity of representations, and the evaluation of sum and intersection, which is formulated in the framework of integer matrices.

The final advanced topic is change of signal dimensionality, which is encountered, e.g., in television scanning, where a time-varying image (a 3D signal) is converted to a 1D signal. This topic has not been developed in the literature in operator form and represents an original contribution of the Unified Signal Theory.

Appendix: Condition for Getting a Subgroup (Theorem 16.1)

We investigate when the “convenient” basis given by (16.38) generates a subgroup J=J K of G. To this end, we decompose the candidate subgroup J into its 1D components (see (16.8))

$$J=\mathbf{j}_1 K_1+\mathbf{j}_2 K_2+\cdots+\mathbf{j}_m K_m , $$

(16.104)

where K _i are the factors of the signature K=K ₁×⋅⋅⋅×K _m . Now, a necessary and sufficient condition for J be a subgroup of G is that all the 1D components j _r K _r be subgroups of G, that is,

$$\mathbf{j}_1 K_1,\ldots,\mathbf{j}_m K_m \subset G \quad {\Longleftrightarrow}\quad J\subset G . $$

(16.105)

Lemma 16.1

Let j=Gk, k∈ℝ^p×ℤ^q be a point of a group G=Gℝ^p×ℤ^q, and let jℤ or jℝ be the 1D group generated by j. Then:

• (1) jℤ is always a subgroup of G,

• (2) jℝ is a subgroup of G if and only if it belongs to the subspace V(G) of G, that is,

  $$\mathbf{j} \in\mathbf{G} \mathbb{R}^p \times\mathbb{O}^q=V(G)\quad {\Longleftrightarrow}\quad \mathbf{h} \in\mathbb{R}^p \times\mathbb{O}^q . $$

  (16.106)

Proof

(1) We have to prove that n j∈G for ∀n∈ℤ. This is easily done by induction, using the group properties: j∈G, then 2j=j+j∈G, 3j=2j+j∈G, etc. and also −jℤ∈G, −2jℤ∈G, etc.

(2) We have to prove condition (16.106) that r j∈G for all r∈ℝ if and only if the point j belongs to the subspace V(G). The reason of this condition is due to the fact that jℝ is a line through the origin, specifically

$$\mathbf{j}\mathbb{R}=\bigl\{r(h_1 \mathbf{g}_1+\cdots+h_p\mathbf{g}_p+h_{p+1}\mathbf{g}_{p+1}+\cdots+h_{m}\mathbf{g}_m)\bigm{|} r \in\mathbb{R}\bigr\} ,$$

and G consists of hyperplanes, and only one of them crosses the origin, namely

$$\mathbf{g}_1 \mathbb{R}+\cdots+\mathbf{g}_p\mathbb{R}=V(G) .$$

Now, in order that jℝ⊂G, the continuum jℝ must lie in the continuum V(G), and this implies, and is implied by, the condition that jℝ does not receive contributions from the vectors g _p+1,…,g _m . This happens when

$$ h_{p + 1}= 0, \ldots ,h_m= 0. $$

(16.107)

 □

We illustrate this condition in the 2D and 3D case. In the 2D case, consider the grating G=Gℝ×ℤ, which is formed by equally distant parallel lines, and g ₁ℝ is the line passing through the origin and containing g ₁ (see Fig. 16.2). But, also jℝ={r(h ₁ g ₁+h ₂ g ₂)∣r∈ℝ} is a line passing through the origin, which belongs to the grating G if and only if h ₂=0. If this is the case, we have jℝ={rh ₁ g ₁∣r∈ℝ}=g ₁ℝ, provided that h ₁≠0.

Next, suppose that G is a 3D grating G=Gℝ²×ℤ, which consists of equidistant parallel planes, and V(G)=g ₁ℝ+g ₂ℝ is the plane passing through the origin determined by g ₁ and g ₂. Now

$$\mathbf{j} \mathbb{R}=\bigl\{r(h_1 g_1+h_2 g_2+h_3 g_3) \bigm{|} r\in\mathbb{R}\bigr\}$$

is a line through the origin, and it belongs to the grating G if it lies on the plane V(G). This implies that h ₃=0. In other words, jℝ must not receive a contribution from the vector g ₃. If G is the grating G=Gℝ×ℤ², the space V(G) becomes $V(G)=\mathbf{G}\mathbb{R} \times\mathbb{O}^{2}=\mathbf{g}_{1} \mathbb {R}$, that is, a line through the origin, and the line jℝ must coincide with the line V(G). This implies that h ₂=h ₃=0.

We have seen how to generate a basis J=[j ₁⋅⋅⋅j _m ] from m independent points j _s =Gh _s of a group G and the construction of a subgroup in the form J=j ₁ K ₁+⋅⋅⋅+j _m K _m . The subgroup condition j _s K _s ⊂G is that the last q entries of h _s be zero when K _s =ℝ (see (16.107)). When the basis is written in the form J=GK, the above condition requires that some entries of the matrix K be zero.

We illustrate this in the 3D case; starting from the grating G=Gℝ²×ℤ, we suppose to generate three subgroups with the signatures (1) $\mathbb{K}=\mathbb{Z}^{3}$, (2) $\mathbb{K}=\mathbb {R} \times\mathbb{Z}^{2}$, and (3) ℝ²×ℤ. In the matrix K we have the constraints

$$ {\bf K} = [{\bf k}_1 {\bf k}_1 {\bf k}_3 ] = \left[ {\begin{array}{*{20}c} {r_1 } & {r_2 } & {r_3 }\\ {s_1 } & {s_2 } & {s_3 }\\ {n_1 } & {n_2 } & {n_3 }\\ \end{array}} \right]\begin{array}{*{20}c} {{\rm real},}\\ {{\rm real},}\\ {{\rm integer}.}\\ \end{array} $$

In case (1), J=j ₁ℤ+j ₂ℤ+j ₃ℤ, and we have no further constraints. In case (2), J=j ₁ℝ+j ₂ℤ+j ₃ℤ, and the constraints is on j ₁=Gk ₁, where the last entry of k ₁ must be zero, that is, n ₁=0. In case (3), J=j ₁ℝ+j ₂ℝ+j ₃ℤ, and the constraint is on both j ₁=Gk ₁ and j ₃=Gk ₃, specifically, n ₁=n ₂=0. In conclusion, for the matrix K, we find in the three cases the constraints indicated in (16.37).