Unified Theory: Frequency Domain Analysis

Abstract

In this chapter, signals are studied in the frequency domain by means of the Fourier transform, which is introduced in unified form using the Haar integral as

$$S(f) = \int_I \mathrm{d} t\, s(t) \psi^{\ast}(f,t), \quad f\in \widehat{I},$$

where I is the signal domain, $\widehat{I}$ is the frequency domain and ψ ^∗(f,t) is the conjugate of the kernel. A preliminary problem is the identification of the frequency domain $\widehat{I}$ and of the kernel ψ(f,t). In particular, for the groups of ℝ, we shall find that the kernel has the familiar exponential form ψ(f,t)=exp (i2πft). Since the frequency domain $\widehat{I}$ associated to a given LCA group I is still an LCA group (the dual group), all definitions and operations introduced in the signal domain (in particular the Haar integral) are straightforwardly transferred to the frequency domain. For the Fourier transform, several rules will be established in a unified form and then applied to specific classes of signals, one-dimensional as well as multidimensional.

The chapter also contains two unusual topics: the Fourier transform on multiplicative groups and the fractional Fourier transform.

Appendices

Appendix A: Fourier Kernel from Characters

A character of an LCA group I is every continuous complex function on I, ψ _f (t), t∈I, with the properties (see (5.4) and (5.5))

$$ \psi _f \left( {t_1+ t_2 } \right) = \psi _f \left( {t_1 } \right)\psi _f \left( {t_2 } \right),\,\,t_1 ,t_2\in I, $$

(5.130a)

$$ \left| {\psi _f \left( t \right)} \right| = 1,\,t \in I. $$

(5.130b)

The index f identifies a specific character ψ _f ; the collection of all the indexes f identifies a set, which represents the dual group $\widehat{I}$ (the dual of I).⁴ In fact, it can be shown that the index set is always an LCA group [23]. Now, from the set of characters ψ _f (t), with t∈I and $f\in \widehat{I}$, we can define a function of two variables ψ(f,t)=ψ _f (t), which represents the kernel of the FT on I.

From conditions (5.130a), (5.130b) the following important properties for the kernel ψ(f,t), which generalize the properties of the exponential e^i2πft, can be established

$$ \psi \left( {f,t_1+ t_2 } \right) = \psi \left( {f,t_1 } \right)\psi \left( {f,t_2 } \right), $$

(5.131a)

$$ \psi \left( {f,0} \right) = 1,\,\,\psi \left( {0,t} \right) = 1, $$

(5.131b)

$$ \psi \left( {f,t} \right) = \,\psi *\left( {f, - t} \right) = 1/\psi \left( {f, - t,} \right) $$

(5.131c)

$$ \psi \left( {f, - t} \right) = \psi \left( { - f,t} \right), $$

(5.131d)

$$ \psi \left( {f_1+ f_2 ,t} \right) = \psi \left( {f_1 ,t} \right)\psi \left( {f_2 ,t} \right). $$

(5.131e)

For instance, the first of (5.131b) is proved by fixing in (5.130a) t ₁=t ₂=0; and (5.131c) is proved by letting t ₁=−t ₂=t. We see that the separability with respect to t leads to the separability with respect to f. Note that in (5.130a), (5.130b) and (5.131a), (5.131b), (5.131c), (5.131d), (5.131e) the group operation + is not necessarily the same for I and $\widehat{I}$. In (5.131a), + is the operation on I, whereas in (5.131e) + is the operation on $\widehat{I}$. Finally, in (5.131b) the first 0 is the identity element of I and the second 0 is the identity element of $\widehat{I}$.

Fourier Kernel on ℝ

When I=ℝ in (5.130a), the group operation + is the ordinary addition and the only continuous solution to the functional equation ψ(t ₁+t ₂)=ψ(t ₁)ψ(t ₂) has the form ψ(t)=e^pt with p an arbitrary complex number. But, constraint (5.130b) restricts p to be imaginary and this is sufficient to conclude that $\{\mathrm{e}^{\mathrm{i}at} \:|\: a\in \mathbb{R}\}$ is the character set and that ℝ is the dual of ℝ. On the other hand, we can denote the real number a in the form a=2πf with f still real, thus obtaining the form (5.6), that is, ψ _ℝ(f,t)=e^i2πft.

Fourier Kernel on ℝ^m

If a group is separable, I=I ₁×I ₂, the Fourier kernel on I is given by the product of the kernels on I ₁ and I ₂ (see [23]). Since ℝ² is separable, from this rule we find that the kernel on ℝ² is given by

$$\psi_{\mathbb{R}^2}(f_1,f_2;t_1,t_2)=\psi_{\mathbb{R}}(f_1,t_1) \psi_{\mathbb{R}}(f_2,t_2)=\mathrm{e}^{\mathrm{i}2\pi (f_1 t_1+f_2 t_2)}.$$

The same form holds for all the subgroups of ℝ², also not separable, since they have the group operation in common with ℝ². The generalization to ℝ^m is immediate and gives (5.4).

Appendix B: Invertibility of the Fourier Transform

We prove that, from the FT calculated using (5.1), we can recover the signal using (5.2). Denoting the inverse transform by

$$\tilde{s}(t) = \int_{\widehat{I}} \mathrm {d}f\; S(f) \psi^{\ast}(f,t) ,$$

and substituting the expression of S(f) given by (5.1), we get

$$\tilde{s}(t) = \int_{\widehat{I}} \mathrm {d}f\; \int_I \mathrm {d}u\; s(u)\psi^*(f,u) \psi(f,t),$$

where, from properties (5.131a), (5.131b), (5.131c), (5.131d), (5.131e) ψ ^∗(f,u)ψ(f,t)=ψ(f,−u)ψ(f,t)=ψ(f,t−u). Hence

$$\tilde{s}(t) = \int_I \mathrm {d}u\; s(u) \int_{\widehat{I}} \mathrm {d}f \;\psi(f,t-u)= \int_{I} \mathrm {d}u \, s(u) \delta_I(t-u) = s(t) ,$$

where we used the orthogonality condition (5.10a), (5.10b) and the sifting property of the impulse (4.76).

Appendix C: Proof of Theorem 5.1 on the Dual Group

We prove that if the function ψ _f (t)=ψ(f,t) (with fixed f) is defined on I ₀ and has periodicity P, then the function ψ _t (f)=ψ(f,t) (with fixed t) is defined on P ^⋆ and has periodicity $I_{0}^{\star }$. The periodicity of ψ _f (t) yields ψ _f (t+u)=ψ _f (t), u∈P, and, recalling that ψ _f (t+u)=ψ _f (t)ψ _f (u), it follows that

$$ \psi_f(u) = 1 , \quad u \in P .$$

(5.132)

Thus, any frequency f satisfying (5.132) ensures the periodicity P with respect to the time domain u. The set of such frequencies is $P^{\star}=\{f \:|\: \psi_{f}(u) = 1, \ u\in P\}$, which defines the domain of the kernel with respect to the frequency f.

Analogously, the possible periodicity of the function ψ _t (f) with respect to f is expressed by the condition ψ _t (f+v)=ψ _t (f), t∈I ₀, which is equivalent to ψ _t (v)=1, t∈I ₀. The frequencies v that verify this condition form the reciprocal of I ₀. Therefore, the periodicity of the dual group is given by $I_{0}^{\star }$.

Appendix D: Proof of Theorem 5.2 on the Representation of the Dual Group

To find the representation of the reciprocal J ^⋆ of an mD group J, we note that the condition t∈J in (5.31) can be expressed in the form t=Jh, h∈H. Hence, we have

$$J^\star = \{\textbf{f} \mid \mathbf{f}'\mathbf{Jh}\in \mathbb{Z}, \ \textbf{h}\in H\}.$$

Now, if we compare the expression of J ^⋆ with the expression of H ^⋆, given by

$$H^\star = \{\textbf{v} \mid \textbf{vh}\in \mathbb{Z},\ \textbf{h}\in H\},$$

we find the relation between the frequencies v of H ^⋆ and the frequencies f of J ^⋆, namely

$$\textbf{v}'=\textbf{f}' \textbf{J} \rightarrow \textbf{f} = (\textbf{J}')^{-1} \textbf{v} , \quad \textbf{v} \in H^{\star }.$$

This completes the proof.

Appendix E: Proof of Poisson’s Summation Formula

For the proof we use the theory of elementary transformations and particularly the Duality Theorem of Sect. 6.13. Let us consider a signal s(t), t∈I, that is, I→U down-sampled, where I=I ₀/P→U=U ₀/P and U ₀ is a lattice. The relation is

$$y(t) = s(t) , \quad t \in U $$

(5.133a)

where y(t) is the down-sampled signal (Fig. 5.28).

Fig. 5.28

Diagram for the proof of Poisson’s summation formula

In the frequency domain, we have an $\widehat{I} \rightarrow \widehat{U}$ up-periodization, where $\widehat{I}=P^{\star }/I_{0}^{\star }$, $\widehat{U}= P^{\star }/U_{0}^{\star }$. The relation is (see (6.80))

$$ Y(f) = \sum_{\lambda\in U_0^\star /I_0^\star } S(f-\lambda) =\sum_{\lambda\in U_0^\star /I_0^\star } S(f+\lambda),\quad f\in \widehat{I}.$$

(5.133b)

Finally, the application of the rule area (y)=Y(0) gives

$$ \sum_{u\in U_0/P} \mathrm {d}(U_0)s(u) =\sum_{\lambda\in U_0^\star /I_0^\star } S(\lambda) .$$

(5.134)

This complete the proof.