Unified Theory: Signal Domain Analysis

Abstract

The preceding chapter has dealt with the class of LCA groups, which will represent the domain/periodicity of signals (and also those of Fourier transforms). This chapter deals with the Haar integral of signals specified on LCA groups, which represents the fundamental functional of the Unified Signal Theory. Using the Haar integral, the classes of signals are formulated in a Hilbert space, where also a theory of symmetries is developed. The convolution operation is then introduced and its basic properties are developed in a unified form. Associated with convolution operation is the impulse, which extends the well known properties of the “delta function” to every signal class.

Appendices

Appendix A: Haar Integral Induced by an Isomorphism

If we know the Haar integral over an LCA group H, then we can derive the Haar integral over every G∼H, by using the isomorphism.

Let α:H→G be the isomorphism map. Let s(t),t∈G, and let $\tilde{s}(h), h \in H$, be the corresponding signal defined on H, which is given by (see (3.62))

$$\tilde{s}(h)=\tilde{s}(h),\quad h \in H.$$

Theorem 4.10

The integral defined by

$$\fbox{$\displaystyle\int_G \mathrm{d}t \,s(t)= \mu_G \int_H \mathrm{d} h \, \tilde{s}(h)$,} $$

(4.126)

where μ _G is an arbitrary positive constant, is a correct Haar integral over G.

Proof

We have to show that the integral over G defined by (4.126) has the five identification properties listed as Properties 1–5 in Sect. 4.1. We now see that those properties follow from the Haar integral properties on H and isomorphism rules. Properties 1, 2 and 3 are evident. To prove Property 4, that is, that s ₋(t)=s(−t) and s(t) have the same integral, it is sufficient to note that α(−h)=−α(h). To prove Property 5, that is, that $s_{t_{0}}(t)=s(t-t_{0})$ and s(t) have the same integral, let u=β(t) the inverse mapping and u ₀=β(t ₀) and note that $s(t-t_{0})=s(\alpha(u)-t_{0})=s(\alpha(u)-\alpha(u_{0}))=\tilde{s}(u-u_{0})$, where the last equality is obtained from the separability of the isomorphism map. But, from the shift-invariance of the Haar integral on H, we know that $\tilde{s}(u-u_{0})$ and $\tilde{s}(u)$ have the same integral. □

Appendix B: Integral Independence of a Group Representation

1. We want to prove that the integral defined by (4.18) is independent of the group representation $(\mathbf{G},H) \longmapsto G$, and this should be done for the three kinds of ordinary groups of $\mathcal{G}(\mathbb{R}^{m})$, that is, G=ℝ^m, G= lattice and G= grating.

When G=ℝ^m, (4.18) gives

$$\int_{\mathbb{R}^m} \mathrm{d}\mathbf{t} \,s(\mathbf{t}) = \mathrm{d}(\mathbf{G}) \int_{\mathbb{R}^m} \mathrm{d}\mathbf{h}\, s(\mathbf{G}\mathbf{h}), $$

(4.127)

where G is an arbitrary regular matrix. The first is the ordinary integral of s(t), evaluated with respect to an orthogonal coordinate system (with basis given by the identity matrix). In the second integral, we have the coordinate change t=Gh, which can be done without changing the result, provided that a multiplication by the absolute value of the Jacobian determinant is introduced. But, this factor is just d(G).

When G is a lattice, we have the sum of all signal values on the lattice, which are independent of the lattice representation. On the other hand, also the lattice determinant d(G) is independent of the basis G.

The proof of independence when G is a grating is less trivial [4] and is omitted. It is based on the idea that, starting from an arbitrary representation (G,ℝ^p×ℤ^q), we finally obtain (working with a matrix partitioning) that the result is the same as with a canonical representation (see (16.14)).

2. The integral on a quotient group G/P is obtained by restricting the integration over a cell [G/P). Now, suppose that we have evaluated the integral over a particular cell C, namely

$$\int_{G/P}\mathrm{d}t \, s(\mathbf{t}) = \int_C \mathrm{d}\mathbf{t} \,s(\mathbf{t}),$$

and we compare the result obtained with another cell $\tilde{C}$. As we shall see in Chap. 16, $\tilde{C}$ is related to C by the partition

$$\tilde{C}=\bigcup_{\mathbf{p} \in P_0} \bigl[C(\mathbf{p})+\mathbf{p}\bigr]$$

where {C(p), p∈P ₀} is a suitable partition of C and P ₀⊂P. So, we have

$$\int_{\tilde{C}} \mathrm{d}\mathbf{t} \,s(\mathbf{t})=\sum_{\mathbf{p} \in P_0} \int_{C(\mathbf{p})+\mathbf{p}} \mathrm{d}\mathbf{t} \,s(\mathbf{t})=\sum_{\mathbf{p} \in P_0} \int_{C(\mathbf{p})}\mathrm{d}t \, s(\mathbf{t}-\mathbf{p}),$$

where s(t−p)=s(t) for the periodicity of s(t). Hence

$$\int_{\tilde{C}} \mathrm{d}\mathbf{t} \,s(\mathbf{t})=\sum_{\mathbf{p} \in P_0} \int_{C(\mathbf{p})}\mathrm{d}\mathbf{t} \,s(\mathbf{t})=\int_C \mathrm{d}\mathbf{t} \, s(\mathbf{t}),$$

where we have considered that C(p) is a partition of C.

Appendix C: Proof of Theorem 4.2 on Coordinate Change in ℝ^m

Suppose that G is an ordinary group of ℝ^m with representation (G,H). After the coordinate change, the group becomes

$$G_{\mathbf{a}}=\bigl\{\mathbf{a}^{-1} \mathbf{u} \mid\mathbf{u} \in G\bigr\}= \bigl\{\mathbf{a}^{-1} \mathbf{Gh} \mid\mathbf{h} \in H \bigr\},$$

which states that a representation of G _a is (G _a ,H) with G _a =a ⁻¹ G.

Now, we can apply the general formula (4.18) to derive the Haar integral on G _a , namely

$$\int_{G_{\mathbf{a}}} d\mathbf{t} \,s_{\mathbf{a}}(\mathbf{t})=\mathrm{d}(\mathbf{G}_{\mathbf{a}}) \int_{H} \mathrm{d}\mathbf{h} \, s_{\mathbf {a}}(\mathbf{G}_{\mathbf{a}} \mathbf{h}) $$

(4.128)

where

$$s_{\mathbf{a}}(\mathbf{G}_{\mathbf{a}} \mathbf{h})=s(\mathbf{aG}_{\mathbf{a}} \mathbf{h})=s(\mathbf{G} \mathbf{h}).$$

On the other hand, the integral on G is

$$\int_{G} \mathrm{d}\mathbf{u} \,s(\mathbf{u})=\mathrm{d}(\mathbf{G}) \int_H d\mathbf{h} \,s(\mathbf{Gh}). $$

(4.129)

Then, comparing (4.128) with (4.129) and considering that d(G _a )=d(a ⁻¹)d(G), the conclusion follows.

Appendix D: Proof that L _p (I) Is a Vector Space

It is sufficient to show that

• 1. L _p (I) is closed with respect to the sum;

• 2. L _p (I) is closed with respect to the multiplication by a scalar.

We begin by showing property 2, so let x∈L _p (I), α∈C and y=αx. Then

$$\int_I\mathrm{d}t\,\bigl|y(t)\bigr|^p = \int_I\mathrm{d}t\,|\alpha|^p\bigl|x(t)\bigr|^p= |\alpha|^p \int_I\mathrm{d}t\,\bigl|x(t)\bigr|^p,$$

which exists and is finite. Hence property 2 is proved. To show property 1, let x,y∈L _p (I), z=x+y. Then, by defining A={t∈I:|x(t)|≥|y(t)|}, we can write

$$\bigl|z(t)\bigr|^p=\bigl|x(t)+y(t)\bigr|^p \leq\bigl(\bigl|x(t)\bigr|+\bigl|y(t)\bigr|\bigr)^p$$

and

$$\bigl|x(t)\bigr|+\bigl|y(t)\bigr| \leq\begin{cases}2|x(t)|, & \hbox{if}\ t\in A; \cr 2|y(t)|, &\hbox{if}\ t\notin A.\end{cases}$$

Therefore, we get

$$\int_I\mathrm{d}t\,\bigl|z(t)\bigr|^p \leq\int_I\mathrm{d}t\,\bigl(\bigl|x(t)\bigr|+\bigl|y(t)\bigr|\bigr)^p\leq\int_A\mathrm{d}t\,2^p\bigl|x(t)\bigr|^p +\int_A\mathrm{d}t\,2^p\bigl|y(t)\bigr|^p,$$

which exist and are finite by property 2, and hence property 1 is proved.

Appendix E: Proof of Theorem 4.6 on Periodic Convolution

With the language of transformations, the theorem claims that (i) the periodic repetitions (or up-periodization) x(t),y(t) of a(t), b(t), t∈G, followed by (ii) the convolution c(t)=x∗y(t) is equivalent to (iii) the convolution s(t)=a∗b(t) followed by (iv) the periodic repetition of c(t), as shown in the top part of Fig 4.25.

Fig. 4.25

Diagrams for the proof of Theorem 4.6

The proof is carried out in the frequency domain, where the up-periodization G⟶U=G/P becomes the $\widehat{G}\longrightarrow\widehat{U}$ down-sampling and the convolution becomes a product, as shown in bottom part of Fig 4.13. Then we have to prove that (i′) the down-sampling $\widehat{G} \longrightarrow\widehat{U}$ of A(f), B(f), with equations

$$X(f)=A(f),\qquad Y(f)=B(f),\quad f\in\widehat{U}$$

followed by (ii′) the product S(f)=X(f)Y(f), is equivalent to (iii′) the product C(f)=A(f)B(f), followed by (iv′) the $\widehat{G}\longrightarrow\widehat{U}$ down-sampling, with equation S(f)=C(f), $f \in\widehat{U}$.

Now, the global relation of (i′) and (ii′) is

$$S(f)=A(f)B(f),\quad f \in\widehat{U},$$

and the global relation of (iii′) and (iv′) is just the same. This states the equivalence.

Appendix F: Proof of the Noble Identity on Impulse (Theorem 4.8)

We have already observed that if the pair (I ₁,I ₂) is ordered, that is, I ₁⊂I ₂ or I ₁⊃I ₂, the identity is trivial (see (4.88)). If one of the group is a continuum, then the pair is always ordered. Therefore, it remains to prove the identity in the case of nonordered lattices and nonordered finite groups. The proof is not easy and will be articulated in several steps with the main points illustrated by examples.

The main preliminaries are the determinant identity (3.77), that is,

$$\mathrm{d}(J\cap K)\,\mathrm{d}(J+K)=\mathrm{d}(J)\,\mathrm{d}(K), $$

(4.130)

and the following:

Lemma 4.1

Let (J,K) be a pair of rationally comparable lattices, then for the sum J+K the following partition holds

$$J+p,\quad p\in[K/(J\cap K)) \stackrel{ \varDelta }{=}P. $$

(4.131)

Proof

We start from the partitions of K modulo J∩K, that is,

$$K=J\cap K +\big[K/(J\cap K)\big) =J\cap K +P,$$

which allows writing the sum in the form

$$J+K= J+ J\cap K + P = J+P. $$

(4.132)

 □

Here, we have considered that J∩K is a sublattice of J and then J+J∩K=J. Now, (4.132) assures that partition (4.131) gives the covering of J+K, but not that the cosets J+p are pairwise disjoint. To prove that this property holds, we evaluate the cardinality of P. In the partition of J+K modulo J, given by

$$J+q,\quad q\in\big[(J+K)/J\big) \stackrel{ \varDelta }{=}Q, $$

(4.133)

the cardinality of Q is d(J)/d(J+K), whereas the cardinality of P is d(J∩K)/d(K). But, by identity (4.130), these cardinalities coincide. This proves that (4.131) is itself a partition of J+K.

Example 4.10

Let

$$J=\mathbb{Z}(25),\qquad K=\mathbb{Z}(40),\qquad J+K=\mathbb{Z}(5),\qquad J\cap K=\mathbb{Z}(200). $$

(4.134)

The determinant identity gives 200⋅5=25⋅40. By Lemma 4.1 for the sum J+K=ℤ(5), we find the partition

$$\mathbb{Z}(25)+p,\quad p\in\big[\mathbb{Z}(40)/\mathbb{Z}(200)\big)=\{0,40,80,120,160 \}$$

which is equivalent to (4.133)

$$\mathbb{Z}(25)+q,\quad q \in\big[\mathbb{Z}(5)/\mathbb{Z}(25)\big)=\{ 0,5,10,15,20 \}.$$

Now, we realize that the two partitions coincide. In fact,

$$\mathbb{Z}(25)+40=\mathbb{Z}(25)+15,\qquad \mathbb{Z}(25)+80=\mathbb{Z}(25)+5\mbox{,\quad etc.}$$

Identity

Lemma 4.1 provides the following identity for every function f(⋅) defined on J+K

$$\sum_{v\in J+K} f(v) = \sum_{j \in J}\sum_{p\in[K/(J\cap K))}f(j+p). $$

(4.135)

Proof of the Noble Identity for Lattices

If we let I ₁=A ₁ and I ₂=A ₂, we have to prove that

$$ \begin{array}{l} e_0 \left( {xy} \right) = e_0 \left( x \right) \cap e_0 \left( y \right), \\ e_0 \left( {x + y} \right) \subset e_0 \left( x \right) \cup e_0 \left( y \right). \\ \end{array} $$

(4.136)

where, considering that A ₁ and A ₂ are lattices, the Haar integral is explicitly given by (see (4.8))

$$h_a(t,u) =\sum_{s \in A_1 \cap A_2} \mathrm{d}(A_1 \cap A_2)\, \delta _{A_1}(t-s) \delta_{A_2}(s-u) $$

(4.137)

with t∈A ₁ and u∈A ₂ being fixed arguments. We note that (see (4.79))

$$\delta_{A_1}(t-s)=0,\quad t \neq s \quad\mbox{and}\quad\delta_{A_2}(s-u)=0,\quad s \neq u,$$

and therefore $\delta_{A_{1}}(t-s)\, \delta_{A_{2}}(s-u)=0$ for every s and t≠u. Hence, also the sum is zero for t≠u. On the other hand, $\delta_{A_{1}+A_{2}}(t-u)=0$ for t≠u. So, we have proved (4.136) for t≠u.

Next, consider the case t=u noting that this coincidence can be considered only for t=u∈A ₁∩A ₂. Since for s≠t=u the two impulses give a zero contribution, the summation can be limited to the single value s=t=u. Hence, we have to find

$$\mathrm{d}(A_1 \cap A_2) \delta_{A_1}(0) \delta_{A_2}(0)= \delta_{A_1+A_2}(0)$$

which, considering that δ _I (0)=1/d(I) (see (4.79)), is equivalent to

$$\mathrm{d}(A_1 \cap A_2) \, \mathrm{d}(A_1 +A_2)= \mathrm{d}(A_1) \,\mathrm{d}(A_2). $$

(4.138)

But, this is just the determinant identity.

Example 4.11

Consider the case in which

Now, we suggest the reader to check the different steps of the above proof by writing the arguments in the form t=6m, u=10n, s=30k.

Proof of Noble Identity for Finite Groups

Now, we let

$$I_1=A_1/P_1,\qquad I_2=A_2/P_2, $$

(4.139)

and we first deal with the case P ₁=P ₂=P. Considering that (4.136) has been proved, we perform the summation

$$\sum_{r\in P}h_a (t-r,u)=\sum_{r\in P}h_b (t-r,u), $$

(4.140)

which is permitted since t∈A ₁ and P is a sublattice of A ₁, and therefore t−r∈A ₁. On the right-hand side, we obtain

$$\sum_{r\in P}h_b (t-r,u) \sum_{r\in P}\delta_{A_1+A_2}(t-r-u)=\delta_{(A_1+A_2)/P}(t-u),$$

where we have used (4.84). On the left-hand side, we find

$$\sum_{r\in P}\int_{A_1\cap A_2}\mathrm{d} s\,\delta_{A_1}(t-r-s) \delta_{A_2}(s-u)=\int_{A_1\cap A_2}\mathrm{d} s\,\delta_{A_1/P}(t-s) \delta_{A_2}(s-u) .$$

Next, using integration rule (4.12a, 4.12b), we obtain

where we have considered that $\delta_{A_{1}/P}(t-s-p)=\delta _{A_{1}/P}(t-s)$, and we have used identity (4.84) again. At this point we have obtained the identity

$$\int_{A_1 \cap A_2)/P} \mathrm{d} s \delta_{A_1/P}(t-s)\delta _{A_2}(s-u)=\delta_{(A_1+A_2)/P}(t-u), $$

(4.141)

which proves (4.87) in the case I ₁=A ₁/P, I ₂=A ₂/P.

Finally, we develop the general case (4.139). Considering that (4.141) has been proved, we assume P=P ₁∩P ₂ and on the left-hand side we perform the summation

$$\sum_{p\in[P_1/P)} \sum_{q\in[P_2/P)} \int_{(A_1\cap A_2)/P}\mathrm{d}s\,\delta_{A_1/P}(t-p-s)\delta_{A_2}(s-u+q).$$

Next, using identity (4.85) for the first impulse, we obtain

$$\sum_{p\in[P_1/P)} \delta_{A_1/P}(t-p-s) = \delta_{A_1/P_1}(t-s).$$

Analogously, we deal with the second impulse. Therefore, once the summation has been carried out, the left-hand side gives

$$\int_{(A_1\cap A_2)/P}\mathrm{d}s\,\delta_{A_1/P_1}(t-s)\delta_{A_2/P_2}(s-u),\quad P=P_1+P_2. $$

(4.142)

Next, the same summation is carried out on the right-hand side. Thus, we get

where we have used identities (4.85) and (4.84). Finally, we recall that P=P ₁∩P ₂ and then identity (4.135) allows writing

$$\sum_{v\in P_1+P_2} \delta_{A_1+A_2}(t-u+v)=\delta_{(A_1+A_2)/ (P_1+P_2)}(t-u). $$

(4.143)

In conclusion, starting from (4.141), we have carried out the same summation on both sides. So, we have obtained (4.142) for the left-hand side and (4.143) for the right-hand side. The equality of these two expressions proves identity (4.87) over finite groups.

Example 4.12

We suggest that the reader checks the steps leading to (4.142) and (4.143) with the basis groups given by (4.134) and with the moduli P ₁=ℤ(18) and P ₂=ℤ(60).