# 3-D generalized analytic signal associated with linear canonical transform
in Clifford biquaternion domain

Zhen Feng Cai  School of Science, Hubei University of Technology
No.28, Nanli Road, Hong-shan District
Wuhan. Hubei
China<EMAIL_ADDRESS>Kit Ian Kou  Faulty of Science and Technology,
University of Macau
Taipa, Macao
China<EMAIL_ADDRESS>
###### Abstract.

The analytic signal is a useful mathematical tool. It separates qualitative
and quantitative information of a signal in form of the local phase and local
amplitude. The Clifford Fourier transform (CFT) plays a vital role in the
representation of multidimensional signals. By generalizing the CFT to the
Clifford linear canonical transform (CLCT), we present a new type of Clifford
biquaternionic analytic signal. Due to the advantages of more freedom, the
envelop detection problems of 3D images, with the help of this new analytic
signal, can get a better visual appearance. Synthesis examples are presented
to demonstrate these advantages.

###### Key words and phrases:

analytic signal, Clifford biquaternion, 3-D images.

###### 1991 Mathematics Subject Classification:

Primary 45P05; Secondary 30H

##  1\. Introduction

The analytic signal, which was introduced by Gabor [1] and Ville [2], is a
powerful tool in various applications such as communication [3], radar-based
object detection [4], processing of oceanic data [5], etc. It is a complex
signal which derives from adding its Hilbert transform to the original real-
valued signal. This complex signal also can be regarded as being constructed
by suppressing all negative frequency components of the original real signal.
Since one can separate the qualitative and quantitative information from the
local phase and local amplitude of analytic signals, the instantaneous
amplitude and phase are the hot topic of analytic signal analysis, and a
critical analysis about it has been carried out by Picibono [6] in 1997.
During the last several decades, many new complex analytic signal theories [7,
8, 9, 10, 11, 30] were well established with the development of Clifford
algebras and the associated Fourier transform theory.

The linear canonical transform (LCT) - a powerful tool for optics and signal
processing - was first introduced in the 1970s by Collins [12] and Moshinsky
[13]. It is a linear integral transform with three free parameters. Many
famous transforms such as Fourier transform (FT), fractional transform (FRFT),
and the Fresnel transform (FST) are all special cases of the LCT [14, 15, 16].
Due to the property that owning more degrees of freedom and needing similar
computation cost to the FT and FRFT, the LCT has many applications such as in
signal synthesis, radar system analysis, filter design and pattern
recognition, etc [15, 17]. Recently, with the development of the LCT, the
analytic signal has been extended into the LCT domain initially by Fu and Li
[18] and to 2D LCT domain by Xu et al [19]. Furthermore, Kou [20] generalized
the analytic signal to the quaternion domain with the help of the two-sided
quaternion linear canonical transform (QLCT) and got a satisfactory result by
using this method to process the envelop detection problems.

In 3-D image processing domain, such as 3-D ultrasound image registration,
analytic signal is a powerful tool. Zhang [21] first used it to process 3-D
ultrasound image with the help of its phase information. Following this, phase
information of analytic signal is widely used in image processing [22, 23, 31,
32]. Since local amplitude is another important part of analytic signal, Wang
[24] introduced 3-D Clifford biquaternionic analytic signal, which is
associated with Clifford Fourier transform. With the help of partial modules,
it overcomes the shortcoming of losing information of the classical 1-D
analytic envelop detection tools. Since the analytic signal in LCT domain
supplies better results in envelop detection than that in FT domain, which was
shown by Kou in [20]. We generalize Wang’s [24] Clifford biquaternionic
analytic signal to LCT domain, and with the help of the local amplitude, the
envelopes of 3-D images are successfully detected. Synthesis examples show
that our approach presents better results than Wang’s method. Furthermore, by
comparing with the amplitude method of monogenic signal which is another kind
of generalization of analytic signal, the powerful ability of our approach is
verified.

The paper is organized as follows. First, we recall the basic knowledge about
the Clifford biquaternion and $n$ dimensional analytic signal in section 2.
Section 3 is dedicated to give the definition and basic properties of CLCT,
which are the keys of using the generalized analytic signal to detect the
envelopes of 3-D images. In section 4, a novel approach of envelop detection
based on CLCT is supplied and synthetic examples are introduced to show the
advantages of this method. Finally we conclude this article in Section 5.

##  2\. Preliminary

###  2.1. Clifford Biquaternion

Quaternion, which was discovered by Hamilton in 1843 and is denoted by
$\mathbb{H}$, is a generalization of complex number. Each quaternion number
$\mathbf{q}$ (denoted by bold letter in this paper) has a form
$\mathbf{q}=q_{0}+q_{1}\textbf{i}+q_{2}\textbf{j}+q_{3}\textbf{k}$, where
$q_{0},q_{1},q_{2},q_{3}$ are real numbers and $\textbf{i},\textbf{j}$, and k
are imaginary units, which satisfy
$\textbf{i}^{2}=\textbf{j}^{2}=\textbf{k}^{2}=-1$ and
$\textbf{ij}=-\textbf{ji}=\textbf{k}$. For every quaternion number
$\mathbf{q}=q_{0}+q_{1}\textbf{i}+q_{2}\textbf{j}+q_{3}\textbf{k}$, the scalar
part and vector part are $Sc(q):=q_{0}$ and
$\underline{\mathbf{q}}:=q_{1}\textbf{i}+q_{2}\textbf{j}+q_{3}\textbf{k}$
respectively. We also use symbols
$\overline{\mathbf{q}}:=q_{0}-\underline{\mathbf{q}}$ and
$|\mathbf{q}|:=\sqrt{\mathbf{q}\overline{\mathbf{q}}}=\sqrt{q_{0}^{2}+q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}$
to denote the conjugate and norm of $\mathbf{q}$. Furthermore, for $p=1$ and
$2$, the quaternion modules $L^{p}(\mathbb{R}^{n},\mathbb{H})$ is defined by

$L^{p}(\mathbb{R}^{n},\mathbb{H}):=\\{f|f:\mathbb{R}^{n}\rightarrow\mathbb{H},\|f\|_{L^{p}(\mathbb{R}^{n},\mathbb{H})}:=\big{(}\int_{\mathbb{R}^{n}}|f(\mathbf{x})|^{p}d\mathbf{x}\big{)}^{\frac{1}{p}}<\infty\\}.$

Clifford algebra, which is a further generalization of quaternion, was
introduced by William Kingdon Clifford in 1878 and is defined as follows.

######  Definition 2.1.

Suppose that $\\{\mathbf{e}_{1},\mathbf{e}_{2},\cdots,\mathbf{e}_{n}\\}$ is an
orthonormal basis of Euclidean space $\mathbb{R}^{n}$, and satisfies the
relations $\mathbf{e}_{i}^{2}=-1$ for $i=1,2,\cdots,n,$ and
$\mathbf{e}_{i}\mathbf{e}_{j}+\mathbf{e}_{j}\mathbf{e}_{i}=0$ for $1\leq i\neq
j\leq n.$ Then the Clifford algebra $Cl(0,n)$ is an algebra constructed over
these elements, i.e.,

$Cl(0,n):=\bigg{\\{}a=\sum\limits_{S}a_{S}\mathbf{e}_{S}:a_{S}\in\mathbb{R},\mathbf{e}_{S}=\mathbf{e}_{j_{1}}\mathbf{e}_{j_{2}}\cdots\mathbf{e}_{j_{k}}\bigg{\\}},$
(2.1)

where $S:=\\{j_{1},j_{2},\cdots,j_{k}\\}\subseteq\\{1,2,\cdots,n\\}$ with
$1\leq j_{1}<j_{2}<\cdots<j_{k}\leq n$; or $S=\emptyset,$ and
$\mathbf{e}_{\emptyset}:=1$.

It can be easily found from the definition that $Cl(0,n)$ is a $2n$
dimensional real linear vector space. Let $|S|$ be the number of elements in
the set $S$. For each $k\in\\{1,2,\cdots,n\\},$ let

$Cl^{(k)}(0,n):=\bigg{\\{}a\in
Cl(0,n):a=\sum\limits_{|S|=k}a_{S}\mathbf{e}_{S}\bigg{\\}},$

to be the $k$-vectors subset of $Cl(0,n)$. Hence we have

$Cl(0,n)=\bigoplus_{k=0}^{n}Cl^{(k)}(0,n).$

For any Clifford number $a=\sum\limits_{S}a_{S}\mathbf{e}_{S}$ in $Cl(0,n)$,
it has a projection $[a]_{k}$ on $Cl^{(k)}(0,n)$, and can be represented by

$a=\sum\limits_{k=0}^{n}[a]_{k}.$

For $k=0$, $[a]_{0}$ is named the scalar part of $a$. Furthermore $[a]_{1}$,
$[a]_{2}$ and $[a]_{n}$, are the vector part, bivector part and pseudoscalar
part of $a$ respectively.

Similar to quaternion numbers, we use symbols
$\overline{a}:=\sum\limits_{S}a_{S}\overline{\mathbf{e}}_{S}$ and
$|a|:=\left(\sum\limits_{S}|a_{S}|^{2}\right)^{1/2}$ to denote the conjugate
and norm of $a$, where
$\overline{\mathbf{e}}_{S}:=(-1)^{\frac{|S|(|S|+1)}{2}}\mathbf{e}_{S}$.
Furthermore the Clifford modules $L^{p}(\mathbb{R}^{n},Cl(0,n))$ is

$L^{p}(\mathbb{R}^{n},Cl(0,n)):=\\{f|f:\mathbb{R}^{n}\rightarrow
Cl(0,n),\|f\|_{L^{p}(\mathbb{R}^{n},Cl(0,n))}:=\big{(}\int_{\mathbb{R}^{n}}|f(\mathbf{x})|^{p}d\mathbf{x}\big{)}^{\frac{1}{p}}<\infty\\}.$

For $p=2$, and let $f,g\in L^{2}(\mathbb{R}^{n},Cl(0,n))$, an inner product
can be equipped as follows

$(f,g)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}:=\int_{\mathbb{R}^{n}}f(\mathbf{x})\overline{g(\mathbf{x})}d^{n}\mathbf{x}.$
(2.2)

Since Clifford algebra is generalized from quaternion algebra, they have a
close relationship as follows,

######  Theorem 2.2.

[25] If $p+q=2m$ ($m$ is integer),the Clifford algebra $Cl(p,q)$ is the tensor
product of $m$ quaternion algebras. If $p+q=2m-1$, the Clifford algebra
$Cl(p,q)$ is the tensor product of $m-1$ quaternion algebra and the algebra
$(1,\epsilon)$, where $\epsilon$ is the product of the $2m-1$ generators
$(\epsilon=\mathbf{e}_{1}\mathbf{e}_{2}\cdots\mathbf{e}_{2m-1})$.

According to the above theorem, one can easily find some special examples of
Clifford algebras such as: complex $\mathbb{C}$ with $p=0,q=1,m=1$ and
$\mathbf{e}_{1}=\mathbf{i}$; quaternions $\mathbb{H}$ with $p=0,q=2,m=1$ and
$\mathbf{e}_{1}=\mathbf{i},\mathbf{e}_{2}=\mathbf{j}$.

Here we should pay an attention to a special case of Clifford algebra,
$Cl(0,3)$, which will be used in this paper to detect the envelop of 3-D
images. According to Theorem 2.2, $Cl(0,3)$ corresponds $p=0,q=3,$ and is a
tensor product of quaternion algebra $\mathbb{H}$ and the algebra
$(1,\epsilon)$, where
$\mathbf{e}_{1}=\epsilon\mathbf{i},\mathbf{e}_{2}=\epsilon\mathbf{j},\mathbf{e}_{3}=\epsilon\mathbf{k}$
are the generators of $Cl(0,3)$,
$\epsilon=\mathbf{e}_{1}\mathbf{e}_{2}\mathbf{e}_{3}$ commuting with
$\mathbf{i,j,k}$ and $\epsilon^{2}=1$. Thus each number in this algebra

$A=a_{0}+a_{1}\mathbf{e}_{1}+a_{2}\mathbf{e}_{2}+a_{3}\mathbf{e}_{3}+a_{4}\mathbf{e}_{1}\mathbf{e}_{2}+a_{5}\mathbf{e}_{3}\mathbf{e}_{1}+a_{6}\mathbf{e}_{2}\mathbf{e}_{3}+a_{7}\mathbf{e}_{1}\mathbf{e}_{2}\mathbf{e}_{3},$

can be expressed by $A=\mathbf{p}+\epsilon\mathbf{q}$, where
$\mathbf{p},\mathbf{q}$ are two quaternions and

$\displaystyle\mathbf{p}$
$\displaystyle=(a_{0}+a_{6}\mathbf{i}+a_{5}\mathbf{j}+a_{4}\mathbf{k}),$
$\displaystyle\mathbf{q}$
$\displaystyle=(a_{7}+a_{1}\mathbf{i}+a_{2}\mathbf{j}+a_{3}\mathbf{k}).$

Here $a_{0}$ is the scalar part of $A$, $a_{1},a_{2},a_{3}$ correspond to the
vector part, and $a_{4},a_{5},a_{6}$ are the bivector parts, while $a_{7}$ is
the pseudoscalar part of $A$.

######  Remark 2.3.

Since $Cl(0,3)$ is a special case of geometric algebra (Clifford algebra) and
isomorphic to $\mathbb{H}\bigoplus\mathbb{H}$, it is also named Clifford
biquaternion[24, 25] (clifbquat for short) by some authors.

Since Clifford biquaternion algebra $Cl(0,3)$ is a special case of geometric
algebra (Clifford algebra), it follows the reflecting properties of geometric
algebra [26], which states that under an orthogonal symmetry with respect to a
plane which is perpendicular to a unit vector $a$, the reflecting of a
Clifford biquaterion $A=\sum\limits_{|S|=0}^{3}a_{S}\mathbf{e}_{S}$ is
$A^{\prime}=\sum\limits_{|S|=0}^{3}(-1)^{|S|+1}a_{S}a\mathbf{e}_{S}a$. In
particular, if $a=\mathbf{e}_{1}$, we can get

$A^{\prime}=K_{1}(A)=(a_{0}+a_{6}\mathbf{i}-a_{5}\mathbf{j}-a_{4}\mathbf{k})+\epsilon(-a_{7}-a_{1}\mathbf{i}+a_{2}\mathbf{j}+a_{3}\mathbf{k});$
(2.3)

if $a=\mathbf{e}_{2}$, it is

$A^{\prime}=K_{2}(A)=(a_{0}-a_{6}\mathbf{i}+a_{5}\mathbf{j}-a_{4}\mathbf{k})+\epsilon(-a_{7}+a_{1}\mathbf{i}-a_{2}\mathbf{j}+a_{3}\mathbf{k});$
(2.4)

if $a=\mathbf{e}_{3}$,

$A^{\prime}=K_{3}(A)=(a_{0}-a_{6}\mathbf{i}-a_{5}\mathbf{j}+a_{4}\mathbf{k})+\epsilon(-a_{7}+a_{1}\mathbf{i}+a_{2}\mathbf{j}-a_{3}\mathbf{k}).$
(2.5)

###  2.2. Analytic Signal in N Dimension

Analytic signal is constructed by suppressing the negative frequency
components of the original signal. So it is closely related with the Fourier
transform of the original signal. In [25], Girard generalized the analytic
signal to $n$ dimension by introducing a new Clifford-Fourier transform as
follows

######  Definition 2.4.

Given a Clifford valued function $f(\mathbf{x})\in
L^{1}(\mathbb{R}^{n},Cl(0,n))$ with $\mathbf{x}=(x_{1},x_{2},\cdots,x_{n})$,
its Clifford Fourier transform $F(u)$ is defined as

$F(\mathbf{u})=\int_{\mathbb{R}^{n}}f(\mathbf{x})\prod\limits_{k=1}^{n}e^{-\mathbf{e}_{k}2\pi
u_{k}x_{k}}d^{n}\mathbf{x}.$ (2.6)

Furthermore, Girard supplied the inverse Clifford Fourier transform as follows

$f(\mathbf{x})=\int_{\mathbb{R}^{n}}F(\mathbf{u})\prod\limits_{k=0}^{n-1}e^{\mathbf{e}_{n-k}2\pi
u_{n-k}x_{n-k}}d^{n}\mathbf{u}.$ (2.7)

With the help of this CFT, Girard introduced the $n$ dimensional analytic
signal $f_{A}(\mathbf{x})$ which corresponds to $f(\mathbf{x})$ by the
following steps

$F_{A}(\mathbf{u})=\prod\limits_{k=1}^{n}[1+sign(u_{k})]F(\mathbf{u}),$ (2.8)
$f_{A}(\mathbf{x})=\int_{\mathbb{R}^{n}}F_{A}(\mathbf{u})\prod\limits_{k=0}^{n-1}e^{\mathbf{e}_{n-k}2\pi
u_{n-k}x_{n-k}}d^{n}\mathbf{u},$ (2.9)

where $sign(u_{k})$ is the classical signum function and $F(\mathbf{u})$ is
the CFT of original function $f(\mathbf{x})$.

It can be easily verified that when $n=1$, CFT degenerates to the classical FT
and the corresponding analytic signal is $f_{A}(t)$. When $n=2$, the CFT turns
to the right-sided quaternion Fourier transform and the analytic signal is
quaternionic valued function $f_{A}(x,y)$.

Since LCT is a generalization of FT, one can generalize the analytic signal by
replacing the FT by LCT, which was introduced by Fu in [18]. One can connect
these two approaches to generalize the analytic signal to $n$ dimensional LCT
domain.

Monogenic signal which was introduced by Felsberg [11] is another
generalization of analytic signal. Since we will compare our method with that
of monogenic signal, the basic knowledge about monogenic signal will be
reviewed in the following.

######  Definition 2.5.

[30](Monogenic signal) For $f\in L^{2}(\mathbb{R}^{n},Cl(0,n))$, the monogenic
signal $f_{M}\in L^{2}(\mathbb{R}^{n},Cl(0,n))$ is defined by

$f_{M}(\underline{x}):=f(\underline{x})+H[f](\underline{x}),$ (2.10)

where $H[f]$ is the isotropic Hilbert transform of $f$ defined by

$\displaystyle H[f](\underline{x}):$
$\displaystyle=p.v.\frac{1}{\omega_{n}}\int_{\mathbf{R}^{n}}\frac{\overline{\underline{x}-\underline{t}}}{|\underline{x}-\underline{t}|^{n+1}}f(\underline{t})d\underline{t}$
(2.11) $\displaystyle=\lim\limits_{\epsilon\rightarrow
0^{+}}\frac{1}{\omega_{n}}\int_{{|\underline{x}-\underline{t}}|>\epsilon}\frac{\overline{\underline{x}-\underline{t}}}{|\underline{x}-\underline{t}|^{n+1}}f(\underline{t})d\underline{t}$
$\displaystyle=-\sum\limits_{j=1}^{n}R_{j}(f)(\underline{x})\mathbf{e}_{j}.$

Furthermore,

$R_{j}(f)(\underline{x}):=\lim\limits_{\epsilon\rightarrow
0^{+}}\frac{1}{\omega_{n}}\int_{{|\underline{x}-\underline{t}}|>\epsilon}\frac{x_{j}-t_{j}}{|\underline{x}-\underline{t}|^{n+1}}f(\underline{t})d\underline{t}$

and $\omega_{n}=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}$.

##  3\. Clifford Linear Canonical Transform

Since LCT is a generalization of many famous integral transform and has more
degrees of freedom, it is natural to generalize the LCT to Clifford domain. In
the last decade, many mathematicians tried different approaches to develop
this kind of generalization. In [27], Kou introduced the CLCT of a function
$f\in L^{1}(\mathbb{R}^{n},Cl(0,n))$ and investigated the maxima of energy
preservation problem. Yang [28] paid attention to the CLCT with the kernel
constituted by the complex unit $I$. Here, we will introduce a new type of
generalization as follows

######  Definition 3.1.

(Right-sided CLCT) Let $\Lambda$:
$A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)\in\mathbb{R}^{2\times 2}$ ($k=1,2,\cdots,n$) be a set of
parameter matrices with $det(A_{k})=1$ and $b_{k}\neq 0$, the right-sided CLCT
of a signal $f\in L^{1}(\mathbb{R}^{n},CL(0,n))$ is a Clifford valued function
$\mathscr{L}^{r}_{\Lambda}(f):\mathbb{R}^{n}\rightarrow CL(0,n)$ defined as
follows

$\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u}):=\int_{\mathbb{R}^{n}}f(\mathbf{x})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x},$
(3.1)

where $K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})$ is the kernel of the CLCT
defined as

$K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k}):=\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}}}e^{\mathbf{e}_{k}(\frac{a_{k}}{2b_{k}}x_{k}^{2}-\frac{1}{b_{k}}x_{k}u_{k}+\frac{d_{k}}{2b_{k}}u_{k}^{2})}.$
(3.2)

Due to the non-commutativity of Clifford algebra, there is a different type of
CLCT, left-sided CLCT, which can be defined as follows

######  Definition 3.2.

(Left-sided CLCT) Let $\Lambda$:
$A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)\in\mathbb{R}^{2\times 2}$ ($k=1,2,\cdots,n$) be a set of
parameter matrices with $det(A_{k})=1$ and $b_{k}\neq 0$, the left-sided CLCT
of a signal $f\in L^{1}(\mathbb{R}^{n},Cl(0,n))$ is a Clifford valued function
$\mathscr{L}^{l}_{\Lambda}(f):\mathbb{R}^{n}\rightarrow Cl(0,n)$ defined as
follows

$\mathscr{L}^{l}_{\Lambda}(f)(\mathbf{u}):=\int_{\mathbb{R}^{n}}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})f(\mathbf{x})d^{n}\mathbf{x}.$
(3.3)

Here we focus on the right-sided CLCT in the following content. Furthermore,
we can simulate the approach Hitzer used in [29] to introduce a two-sided
Clifford LCT with two square roots of $-1$ in $Cl(0,n)$, which is out of our
scope.

######  Remark 3.3.

It can be easily verified that when the dimension $n=1$, the CLCT degenerates
to the classical LCT. When $n=2$, the CLCT turns to the right-sided quaternion
linear canonical transform.

Now let us investigate an example to see how the CLCT works:

######  Example.

Considering the right-sided CLCT of
$\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(x_{n-k},v_{n-k})$.

Since $f(t)$ is product of $n$ LCT kernel functions, we can integrate the
functions separately when taking the CLCT. That is

$\begin{split}&\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})=\int_{\mathbb{R}^{n}}\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(x_{n-k},v_{n-k})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}\\\
&=\int_{\mathbb{R}^{n-1}}\prod\limits_{k=0}^{n-2}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(x_{n-k},v_{n-k})\int_{\mathbb{R}}K^{-\mathbf{e}_{1}}_{A_{1}}(x_{1},v_{1})K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})dx_{1}\\\
&\cdot\prod\limits_{k=2}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n-1}\mathbf{x},\end{split}$
(3.4)

where

$\begin{split}\int_{\mathbb{R}}K^{-\mathbf{e}_{1}}_{A_{1}}(x_{1},v_{1})K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})&dx_{1}=\int_{\mathbb{R}}\frac{1}{\sqrt{-\mathbf{e}_{1}2\pi
b_{1}}}e^{-\mathbf{e}_{1}(\frac{a_{1}}{2b_{1}}x_{1}^{2}-\frac{1}{b_{1}}x_{1}v_{1}+\frac{d_{1}}{2b_{1}}v_{1}^{2})}\\\
&\cdot\frac{1}{\sqrt{\mathbf{e}_{1}2\pi
b_{1}}}e^{\mathbf{e}_{1}(\frac{a_{1}}{2b_{1}}x_{1}^{2}-\frac{1}{b_{1}}x_{1}u_{1}+\frac{d_{1}}{2b_{1}}u_{1}^{2})}dx_{1}\\\
&=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-\mathbf{e}_{1}\frac{x_{1}}{b_{1}}(u_{1}-v_{1})}d(\frac{x_{1}}{b_{1}})e^{-\mathbf{e}_{1}\frac{d_{1}}{2b_{1}}(u_{1}^{2}-v_{1}^{2})}\\\
&=\delta(u_{1}-v_{1}).\end{split}$ (3.5)

The last step of the above equation comes from the relationship of Fourier
transform $\widehat{1}(\omega)=2\pi\delta(\omega)$. Applying result (3.5) to
the right side of equation (3.4), one can derive

$\mathscr{L}^{r}_{\Lambda}\big{(}\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(x_{n-k},v_{n-k})\big{)}(\mathbf{u})=\prod\limits_{k=1}^{n}\delta(u_{k}-v_{k}).$

In the following content, we turn to investigate the main properties of CLCT.

######  Proposition 3.4.

(Left linearity) Let $\alpha\in Cl(0,n),\beta\in Cl(0,n)$, for
$f_{1}(\mathbf{x}),f_{2}(\mathbf{x})\in L^{1}(\mathbb{R}^{n},Cl(0,n))$, we
have $\mathscr{L}^{r}_{\Lambda}(\alpha f_{1}+\beta
f_{2})(\mathbf{u})=\alpha\mathscr{L}^{r}_{\Lambda}(f_{1})(\mathbf{u})+\beta\mathscr{L}^{r}_{\Lambda}(f_{2})(\mathbf{u})$.

This proposition can be easily verified by the linearity of integral.

######  Proposition 3.5.

(Translation) Let $\mathbf{\bm{\alpha}}=(\alpha_{1},0,\cdots,0,\alpha_{n})$,
for $f(\mathbf{x})\in L^{1}(\mathbb{R}^{n},\mathbb{R})$, we have
$\mathscr{L}^{r}_{\Lambda}(f(\mathbf{x}-\bm{\alpha}))(\mathbf{u})=e^{\mathbf{e}_{1}(c_{1}u_{1}\alpha_{1})}e^{-\mathbf{e}_{1}\frac{a_{1}c_{1}}{2}\alpha_{1}^{2}}\mathscr{L}^{r}_{\Lambda}(f)(u_{1}-a_{1}\alpha_{1},u_{2},\cdots,u_{n}-a_{n}\alpha_{n})e^{\mathbf{e}_{n}(c_{n}u_{n}\alpha_{n})}e^{-\mathbf{e}_{n}\frac{a_{n}c_{n}}{2}\alpha_{n}^{2}}.$

###### Proof.

According to definition of right-sided CLCT, one can get

$\mathscr{L}^{r}_{\Lambda}(f(\mathbf{x}-\bm{\alpha}))(\mathbf{u})=\int_{\mathbb{R}^{n}}f(x_{1}-\alpha_{1},x_{2},\cdots,x_{n-1},x_{n}-\alpha_{n})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}.$
(3.6)

Let $t_{1}=x_{1}-\alpha_{1}$, $t_{n}=x_{n}-\alpha_{n}$, the above integral
turns to a new integral about variables $t_{1},x_{2},\cdots,x_{n-1},t_{n}$,
where the first LCT kernel becomes

$\displaystyle K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})$
$\displaystyle=\frac{1}{\sqrt{\mathbf{e}_{1}2\pi
b_{1}}}e^{\mathbf{e}_{1}[\frac{a_{1}}{2b_{1}}(t_{1}+\alpha_{1})^{2}-\frac{1}{b_{1}}(t_{1}+\alpha_{1})u_{1}+\frac{d_{1}}{2b_{1}}u_{1}^{2}]}$
$\displaystyle=\frac{1}{\sqrt{\mathbf{e}_{1}2\pi
b_{1}}}e^{\mathbf{e}_{1}[\frac{a_{1}}{2b_{1}}t_{1}^{2}+\frac{a_{1}}{b_{1}}t_{1}\alpha_{1}+\frac{a_{1}}{2b_{1}}\alpha_{1}^{2}-\frac{t_{1}u_{1}}{b_{1}}-\frac{u_{1}\alpha_{1}}{b_{1}}+\frac{d_{1}}{2b_{1}}u_{1}^{2}]}$
$\displaystyle=e^{\mathbf{e}_{1}(c_{1}u_{1}\alpha_{1})}e^{-\mathbf{e}_{1}\frac{a_{1}c_{1}}{2}\alpha_{1}^{2}}\frac{1}{\sqrt{\mathbf{e}_{1}2\pi
b_{1}}}e^{\mathbf{e}_{1}[\frac{a_{1}}{2b_{1}}t_{1}^{2}-\frac{t_{1}}{b_{1}}(u_{1}-a_{1}\alpha_{1})+\frac{d_{1}}{2b_{1}}(u_{1}-a_{1}\alpha_{1})^{2}]},$

and the last LCT kernel turns to

$K^{\mathbf{e}_{n}}_{A_{n}}(x_{n},u_{n})=e^{\mathbf{e}_{n}(c_{n}u_{n}\alpha_{n})}e^{-\mathbf{e}_{n}\frac{a_{n}c_{n}}{2}\alpha_{n}^{2}}\frac{1}{\sqrt{\mathbf{e}_{n}2\pi
b_{n}}}e^{\mathbf{e}_{n}[\frac{a_{n}}{2b_{n}}t_{n}^{2}-\frac{t_{n}}{b_{n}}(u_{n}-a_{n}\alpha_{n})+\frac{d_{n}}{2b_{n}}(u_{n}-a_{n}\alpha_{n})^{2}]}.$

Since $f(\mathbf{x})$ is real valued function in this case and it can
interchange the position with $K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})$ in
equation (3.6) freely. Hence equation (3.6) derives

$\displaystyle\mathscr{L}^{r}_{\Lambda}(f(\mathbf{x}-\bm{\alpha}))(\mathbf{u})=$
$\displaystyle
e^{\mathbf{e}_{1}(c_{1}u_{1}\alpha_{1})}e^{-\mathbf{e}_{1}\frac{a_{1}c_{1}}{2}\alpha_{1}^{2}}\mathscr{L}^{r}_{\Lambda}(f)(u_{1}-a_{1}\alpha_{1},u_{2},\cdots,u_{n}-a_{n}\alpha_{n})$
$\displaystyle
e^{\mathbf{e}_{n}(c_{n}u_{n}\alpha_{n})}e^{-\mathbf{e}_{n}\frac{a_{n}c_{n}}{2}\alpha_{n}^{2}}.$

∎

######  Proposition 3.6.

(Scaling)Let $\sigma_{1},\sigma_{2},\cdots,\sigma_{n}>0$, for
$f(\mathbf{x})\in L^{1}(\mathbb{R}^{n},Cl(0,n))$, we have

$\mathscr{L}^{r}_{\Lambda}(f(\sigma_{1}x_{1},\sigma_{2}x_{2},\cdots,\sigma_{n}x_{n})(\mathbf{u})=\frac{1}{\sqrt{\prod\limits_{k=1}^{n}\sigma_{k}}}\mathscr{L}^{r}_{\Lambda^{\prime}}(f)(\mathbf{u}),$

where $\Lambda^{\prime}$:
$B_{k}=\left(\begin{array}[]{cc}a_{k}/\sigma_{k}&b_{k}\sigma_{k}\\\
c_{k}/\sigma_{k}&d_{k}\sigma_{k}\\\ \end{array}\right).$

###### Proof.

According to equation (3.1), one can find

$\begin{split}\mathscr{L}^{r}_{\Lambda}(f(\sigma_{1}x_{1},\sigma_{2}x_{2},\cdots,\sigma_{n}x_{n}))(\mathbf{u})=&\int_{\mathbb{R}^{n}}f(\sigma_{1}x_{1},\sigma_{2}x_{2},\cdots,\sigma_{n}x_{n})\\\
&\cdot\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}.\end{split}$
(3.7)

Let $t_{k}=\sigma_{k}x_{k}\quad(k=1,\cdots,n)$, each kernel
$K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})$ turns to

$\displaystyle K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})$
$\displaystyle=\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}}}e^{\mathbf{e}_{k}[\frac{a_{k}}{2b_{k}}(\frac{t_{k}}{\sigma_{k}})^{2}-\frac{1}{b_{k}}(\frac{t_{k}}{\sigma_{k}})u_{k}+\frac{d_{k}}{2b_{k}}u_{k}^{2}]}$
$\displaystyle=\frac{\sqrt{{\sigma_{k}}}}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}{\sigma_{k}}}}e^{\mathbf{e}_{k}[\frac{a_{k}/{\sigma_{k}}}{2b_{k}{\sigma_{k}}}t_{k}^{2}-\frac{1}{b_{k}{\sigma_{k}}}t_{k}u_{k}+\frac{d_{k}{\sigma_{k}}}{2b_{k}{\sigma_{k}}}u_{k}^{2}]}.$

Hence the left side of equation (3.7) becomes

$\begin{split}\mathscr{L}^{r}_{\Lambda}(f(t_{1},t_{2},\cdots,t_{n}))(\mathbf{u})&=\int_{\mathbb{R}^{n}}f(t_{1},t_{2},\cdots,t_{n})\\\
&\cdot\prod\limits_{k=1}^{n}\frac{\sqrt{{\sigma_{k}}}}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}{\sigma_{k}}}}e^{\mathbf{e}_{k}[\frac{a_{k}/{\sigma_{k}}}{2b_{k}{\sigma_{k}}}t_{k}^{2}-\frac{1}{b_{k}{\sigma_{k}}}t_{k}u_{k}+\frac{d_{k}{\sigma_{k}}}{2b_{k}{\sigma_{k}}}u_{k}^{2}]}d{\frac{t_{1}}{\sigma_{1}}}\cdots
d{\frac{t_{n}}{\sigma_{n}}}\\\
&=\frac{1}{\sqrt{\prod\limits_{k=1}^{n}\sigma_{k}}}\int_{\mathbb{R}^{n}}f(t_{1},t_{2},\cdots,t_{n})\prod\limits_{k=1}^{n}\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}{\sigma_{k}}}}\\\ &\cdot
e^{\mathbf{e}_{k}[\frac{a_{k}/{\sigma_{k}}}{2b_{k}{\sigma_{k}}}t_{k}^{2}-\frac{1}{b_{k}{\sigma_{k}}}t_{k}u_{k}+\frac{d_{k}{\sigma_{k}}}{2b_{k}{\sigma_{k}}}u_{k}^{2}]}d^{n}\mathbf{t},\end{split}$

which is the desired result. ∎

######  Proposition 3.7.

(Partial derivative) For $f(\mathbf{x})$ and $\frac{\partial
f(\mathbf{x})}{\partial x_{1}},\frac{\partial f(\mathbf{x})}{\partial
x_{n}}\in L^{1}(\mathbb{R}^{n},\mathbb{R})$, we have

$\mathscr{L}^{r}_{\Lambda}(\frac{\partial f(\mathbf{x})}{\partial
x_{1}})(\mathbf{u})=(a_{1}\frac{\partial}{\partial
u_{1}}-\mathbf{e}_{1}c_{1}u_{1})\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u}),$
(3.8)

and

$\mathscr{L}^{r}_{\Lambda}(\frac{\partial f(\mathbf{x})}{\partial
x_{n}})(\mathbf{u})=\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})(a_{n}\frac{\partial}{\partial
u_{n}}-\mathbf{e}_{n}c_{n}u_{n}).$ (3.9)

###### Proof.

Since $\frac{\partial f(\mathbf{x})}{\partial x_{1}}\in
L^{1}(\mathbb{R}^{n},\mathbb{R})$, it has CLCT as following

$\displaystyle\mathscr{L}^{r}_{\Lambda}(\frac{\partial f(\mathbf{x})}{\partial
x_{1}})(\mathbf{u})$ $\displaystyle=\int_{\mathbb{R}^{n}}\frac{\partial
f(\mathbf{x})}{\partial
x_{1}}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}$
$\displaystyle=\int_{\mathbb{R}^{n-1}}\bigg{[}\int_{\mathbb{R}}\frac{\partial
f(\mathbf{x})}{\partial
x_{1}}K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})dx_{1}\bigg{]}\prod\limits_{k=2}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n-1}\mathbf{x},$

where

$\displaystyle\int_{\mathbb{R}}\frac{\partial f(\mathbf{x})}{\partial x_{1}}$
$\displaystyle
K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})dx_{1}=\big{[}f(\mathbf{x})K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})\big{]}\big{|}_{-\infty}^{+\infty}-\int_{\mathbb{R}}f(\mathbf{x})\frac{\partial}{\partial
x_{1}}K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})dx_{1}$
$\displaystyle=-\int_{\mathbb{R}}f(\mathbf{x})\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{1}}}e^{\mathbf{e}_{1}(\frac{a_{1}}{2b_{1}}x_{1}^{2}-\frac{1}{b_{1}}x_{1}u_{1}+\frac{d_{1}}{2b_{1}}u_{1}^{2})}\mathbf{e}_{1}(\frac{a_{1}}{b_{1}}x_{1}-\frac{u_{1}}{b_{1}})dx_{1}$
$\displaystyle=\int_{\mathbb{R}}(\frac{u_{1}}{b_{1}}-\frac{a_{1}}{b_{1}}x_{1})f(\mathbf{x})\mathbf{e}_{1}K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})dx_{1}.$

Hence

$\mathscr{L}^{r}_{\Lambda}(\frac{\partial f(\mathbf{x})}{\partial
x_{n}})(\mathbf{u})=\int_{\mathbb{R}^{n}}(\frac{u_{1}}{b_{1}}-\frac{a_{1}}{b_{1}}x_{1})f(\mathbf{x})\mathbf{e}_{1}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}.$
(3.10)

While

$\displaystyle a_{1}\frac{\partial}{\partial
u_{1}}\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$
$\displaystyle=a_{1}\frac{\partial}{\partial
u_{1}}\int_{\mathbb{R}^{n}}f(\mathbf{x})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}$
$\displaystyle=\int_{\mathbb{R}^{n}}(\frac{a_{1}d_{1}}{b_{1}}u_{1}-\frac{a_{1}x_{1}}{b_{1}})f(\mathbf{x})\mathbf{e}_{1}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x},$

and

$\displaystyle\mathbf{e}_{1}c_{1}u_{1}\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})=\int_{\mathbb{R}^{n}}c_{1}u_{1}\mathbf{e}_{1}f(\mathbf{x})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}.$

According to equation (3.10), We can easily verified that
$\mathscr{L}^{r}_{\Lambda}(\frac{\partial f(\mathbf{x})}{\partial
x_{1}})(\mathbf{u})=(a_{1}\frac{\partial}{\partial
u_{1}}-\mathbf{e}_{1}c_{1}u_{1})\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$,
where the property $a_{1}d_{1}-b_{1}c_{1}=1$ is used. The proof of equation
(3.9) is similar to that of equation (3.8), and we will omit it here. ∎

######  Theorem 3.8.

(Plancherel) Suppose $f(\mathbf{x})$ and $g(\mathbf{x})\in L^{1}\bigcap
L^{2}(\mathbb{R}^{n},Cl(0,n))$,
$F(\mathbf{u}):=\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$ and
$G(\mathbf{u}):=\mathscr{L}^{r}_{\Lambda}(g)(\mathbf{u})$ are their CLCTs
respectively, then we have

$(f,g)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}=(F,G)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}.$
(3.11)

Furthermore, when $f=g$, they turn to

$(f,f)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}=(F,F)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}.$
(3.12)

###### Proof.

According to equation (2.2),

$(F,G)_{L^{2}(\mathbb{R}^{n},Cl(0,n))}=\int_{\mathbb{R}^{n}}F(\mathbf{u})\overline{G(\mathbf{u})}d^{n}\mathbf{u}.$
(3.13)

Inputting the integral formulaes of $F(\mathbf{u})$ and $G(\mathbf{u})$ into
the above equation, one can derive

$\displaystyle(F,G)$
$\displaystyle=\int_{\mathbb{R}^{n}}\bigg{[}\int_{\mathbb{R}^{n}}f(\mathbf{x})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})d^{n}\mathbf{x}\bigg{]}\bigg{[}\overline{\int_{\mathbb{R}^{n}}g(\mathbf{y})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(y_{k},u_{k})d^{n}\mathbf{y}}\bigg{]}d^{n}\mathbf{u}$
$\displaystyle=\int_{\mathbb{R}^{n}}\bigg{[}\int_{\mathbb{R}^{2n}}f(\mathbf{x})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(y_{n-k},u_{n-k})\overline{g(\mathbf{y})}d^{n}\mathbf{x}d^{n}\mathbf{y}\bigg{]}d^{n}\mathbf{u}$
$\displaystyle=\int_{\mathbb{R}^{2n}}f(\mathbf{x})\bigg{[}\int_{\mathbb{R}^{n}}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(y_{n-k},u_{n-k})d^{n}\mathbf{u}\bigg{]}\overline{g(\mathbf{y})}d^{n}\mathbf{x}d^{n}\mathbf{y}.$

The interchanging of the integral in the above equation is due to the Fubini
theorem which is insured by $f(\mathbf{x})$ and $g(\mathbf{x})\in
L^{2}(\mathbb{R}^{n},Cl(0,n))$. Furthermore, the integral in bracket can be
turned into

$\displaystyle\int_{\mathbb{R}^{n}}$
$\displaystyle\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(y_{n-k},u_{n-k})d^{n}\mathbf{u}=\int_{\mathbb{R}^{n-1}}\prod\limits_{k=1}^{n-1}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})$
$\displaystyle\bigg{[}\int_{\mathbb{R}}K^{\mathbf{e}_{n}}_{A_{n}}(x_{n},u_{n})K^{-\mathbf{e}_{n}}_{A_{n}}(y_{n},u_{n})du_{n}\bigg{]}\prod\limits_{k=1}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(y_{n-k},u_{n-k})d^{n-1}\mathbf{u},$

where
$\int_{\mathbb{R}}K^{\mathbf{e}_{n}}_{A_{n}}(x_{n},u_{n})K^{-\mathbf{e}_{n}}_{A_{n}}(y_{n},u_{n})du_{n}$
can be calculated as equation (3.5) and equals $\delta(x_{n}-y_{n})$. Hence,
the above equation is

$\int_{\mathbb{R}^{n}}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k})\prod\limits_{k=0}^{n-1}K^{-\mathbf{e}_{n-k}}_{A_{n-k}}(y_{n-k},u_{n-k})d^{n}\mathbf{u}=\prod\limits_{k=1}^{n}\delta(x_{k}-y_{k}),$

and $(F,G)$ turns to

$\displaystyle(F,G)$
$\displaystyle=\int_{\mathbb{R}^{n}}f(\mathbf{x})\bigg{[}\int_{\mathbb{R}^{n}}\prod\limits_{k=1}^{n}\delta(x_{k}-y_{k})\overline{g(\mathbf{y})}d^{n}\mathbf{y}\bigg{]}d^{n}\mathbf{x}$
$\displaystyle=\int_{\mathbb{R}^{n}}f(\mathbf{x})\overline{g(\mathbf{x})}d^{n}\mathbf{x}=(f,g).$

Let $f=g$, the Parseval theorem (3.12) is derived. ∎

######  Theorem 3.9.

(Inverse Theorem) Let $\Lambda$:
$A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)$ ($k=1,2,\cdots,n$), suppose $f(\mathbf{x})\in
L^{1}(\mathbb{R}^{n},Cl(0,n))$. Then the CLCT of $f$,
$F(\mathbf{u}):=\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$, is an invertible
transform and its inverse is

$f(\mathbf{x})=\mathscr{L}^{-1}_{\Lambda}(F)(\mathbf{x})=\int_{\mathbb{R}^{n}}F(\mathbf{u})\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u},$
(3.14)

where $A^{-1}_{k}=\left(\begin{array}[]{cc}d_{k}&-b_{k}\\\ -c_{k}&a_{k}\\\
\end{array}\right),$ ($k=1,2,\cdots,n$).

###### Proof.

$f(\mathbf{x})\in L^{1}(\mathbb{R}^{n},Cl(0,n))$, by straightforward
computation, one can find

$\displaystyle\int_{\mathbb{R}^{n}}F(\mathbf{u})\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u}=\int_{\mathbb{R}^{n}}\bigg{[}\int_{\mathbb{R}^{n}}f(\mathbf{y})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(y_{k},u_{k})d^{n}\mathbf{y}\bigg{]}$
(3.15)
$\displaystyle\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u}$
$\displaystyle=\int_{\mathbb{R}^{2n}}f(\mathbf{y})\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(y_{k},u_{k})d^{n}\mathbf{y}\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{y}d^{n}\mathbf{u},$
$\displaystyle=\int_{\mathbb{R}^{n}}f(\mathbf{y})\bigg{[}\int_{\mathbb{R}^{n}}\prod\limits_{k=1}^{n}K^{\mathbf{e}_{k}}_{A_{k}}(y_{k},u_{k})\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u}\bigg{]}d^{n}\mathbf{y}$

where
$K^{\mathbf{e}_{k}}_{{A^{-1}_{k}}}(u_{k},x_{k})=\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}}}e^{\mathbf{e}_{k}(-\frac{a_{k}}{2b_{k}}x_{k}^{2}+\frac{1}{b_{k}}x_{k}u_{k}-\frac{d_{k}}{2b_{k}}u_{k}^{2})}$,
similar to equation (3.5), the integral in bracket of above equation is equal
to $\prod\limits_{k=1}^{n}\delta(x_{k}-y_{k})$. Thus equation (3.15) becomes

$\int_{\mathbb{R}^{n}}F(\mathbf{u})\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u}=\int_{\mathbb{R}^{n}}f(\mathbf{y})\prod\limits_{k=1}^{n}\delta(x_{k}-y_{k})d^{n}\mathbf{y}=f(\mathbf{x}),$

which completes the proof. ∎

##  4\. Analytic Signal in CLCT Domain and Envelop Detection

###  4.1. Analytic Signal in CLCT Domain

Following the approach of generalizing the analytic signal to LCT domain,
which was introduced by Fu and Li [18] and extended to quaternion algebra by
Kou [20], we supply the definition of generalized analytic signal which is
associated with CLCT.

######  Definition 4.1.

Given a Clifford valued function $f(\mathbf{x})$ with
$\mathbf{x}=(x_{1},x_{2},\cdots,x_{n})$, its corresponding analytic signal
$f_{A}(\mathbf{x})$ is defined as

$f_{A}(\mathbf{x})=\int_{\mathbb{R}^{n}}F_{A}(\mathbf{u})\prod\limits_{k=0}^{n-1}K^{\mathbf{e}_{n-k}}_{{A^{-1}_{n-k}}}(u_{n-k},x_{n-k})d^{n}\mathbf{u},$
(4.1)

where $F_{A}(\mathbf{u})$ is

$F_{A}(\mathbf{u})=\prod\limits_{k=1}^{n}[1+sign(\frac{u_{k}}{b_{k}})]F(\mathbf{u}),$
(4.2)

and $F(\mathbf{u}):=\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$ is the CLCT of
original function $f(\mathbf{x})$, $\Lambda:$
$A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)\in\mathbb{R}^{2\times 2}$ ($k=1,2,\cdots,n$) be a set of
parameter matrices with $det(A_{k})=1$ and $b_{k}\neq 0$.

When the dimension $n=1$, the analytic signal $f_{A}(t)$ is just what Fu
introduced in [18]. While $n=2$, this analytic signal $f_{A}(x,y)$ is
different from Kou’s generalized quaternionic analytic signal, which is
associated with two-sided QLCT.

Since we intend to investigate the envelop detection problems of 3-D images,
we will focus on $n=3$ case. Due to the advantages of Clifford algebra
$Cl(0,3)$, we will adopt this special Clifford algebra in the following part.

######  Definition 4.2.

Let $\Lambda$: $A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)\in\mathbb{R}^{2\times 2}$ ($k=1,2,3$) be a set of parameter
matrices with $det(A_{k})=1$ and $b_{k}\neq 0$, the right-sided CLCT of a
signal $f\in L^{1}(\mathbb{R}^{3},Cl(0,3))$ is a Clifford valued function
$\mathscr{L}^{r}_{\Lambda}(f):\mathbb{R}^{3}\rightarrow Cl(0,3)$ defined as
follows

$\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u}):=\int_{\mathbb{R}^{3}}f(\mathbf{x})K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})K^{\mathbf{e}_{2}}_{A_{2}}(x_{2},u_{2})K^{\mathbf{e}_{3}}_{A_{3}}(x_{3},u_{3})d^{3}\mathbf{x},$
(4.3)

where $K^{\mathbf{e}_{1}}_{A_{1}}(x_{1},u_{1})$,
$K^{\mathbf{e}_{2}}_{A_{2}}(x_{2},u_{2})$ and
$K^{\mathbf{e}_{3}}_{A_{3}}(x_{3},u_{3})$ are kernels of the CLCT and defined
by

$K^{\mathbf{e}_{k}}_{A_{k}}(x_{k},u_{k}):=\frac{1}{\sqrt{\mathbf{e}_{k}2\pi
b_{k}}}e^{\mathbf{e}_{k}(\frac{a_{k}}{2b_{k}}x_{k}^{2}-\frac{1}{b_{k}}x_{k}u_{k}+\frac{d_{k}}{2b_{k}}u_{k}^{2})}.$

######  Definition 4.3.

Given a Clifford biquaternion valued function $f(\mathbf{x})$ with
$\mathbf{x}=(x_{1},x_{2},x_{3})$, its corresponding Clifford biquaternion
analytic signal $f_{A}(\mathbf{x})$ is defined as

$f_{A}(\mathbf{x})=\int_{\mathbb{R}^{3}}F_{A}(\mathbf{u})K^{\mathbf{e}_{3}}_{{A^{-1}_{3}}}(u_{3},x_{3})K^{\mathbf{e}_{2}}_{{A^{-1}_{2}}}(u_{2},x_{2})K^{\mathbf{e}_{1}}_{{A^{-1}_{1}}}(u_{1},x_{1})du_{1}du_{2}du_{3},$
(4.4)

where $F_{A}(\mathbf{u})$ is

$F_{A}(\mathbf{u})=[1+sign(\frac{u_{1}}{b_{1}})][1+sign(\frac{u_{2}}{b_{2}})][1+sign(\frac{u_{3}}{b_{3}})]F(\mathbf{u}),$
(4.5)

and $F(\mathbf{u}):=\mathscr{L}^{r}_{\Lambda}(f)(\mathbf{u})$ is the CLCT of
original function $f(\mathbf{x})$, $\Lambda:$
$A_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\ c_{k}&d_{k}\\\
\end{array}\right)\in\mathbb{R}^{2\times 2}$ ($k=1,2,3$) be a set of parameter
matrices with $det(A_{k})=1$ and $b_{k}\neq 0$.

The analytic signal $f_{A}(\mathbf{x})$ defined by Definition 4.3 is a
Clifford biquaternion valued function, which can be written as

$f_{A}(\mathbf{x})=(p_{0}+\mathbf{i}p_{1}+\mathbf{j}p_{2}+\mathbf{k}p_{3})+\epsilon(q_{0}+\mathbf{i}q_{1}+\mathbf{j}q_{2}+\mathbf{k}q_{3}).$

From this function, one can derive three quaternions:

$f_{A_{\mathbf{e}_{1}\mathbf{e}_{2}}}=p_{0}+q_{1}\mathbf{e}_{1}+q_{2}\mathbf{e}_{2}+p_{3}\mathbf{e}_{1}\mathbf{e}_{2},$
$f_{A_{\mathbf{e}_{2}\mathbf{e}_{3}}}=p_{0}+q_{2}\mathbf{e}_{2}+q_{3}\mathbf{e}_{3}+p_{1}\mathbf{e}_{2}\mathbf{e}_{3},$

and

$f_{A_{\mathbf{e}_{3}\mathbf{e}_{1}}}=p_{0}+q_{3}\mathbf{e}_{3}+q_{1}\mathbf{e}_{1}+p_{2}\mathbf{e}_{3}\mathbf{e}_{1}.$

For each of this quaternionic valued signals, one can write them into polar
form representations, and obtain their modules which named the partial modules
of the original Clifford biquaternion valued analytic signal
$f_{A}(\mathbf{x})$ as following

$mod_{\mathbf{e}_{1}\mathbf{e}_{2}}=\sqrt{f_{A_{\mathbf{e}_{1}\mathbf{e}_{2}}}\overline{(f_{A_{\mathbf{e}_{1}\mathbf{e}_{2}}})}},$
$mod_{\mathbf{e}_{2}\mathbf{e}_{3}}=\sqrt{f_{A_{\mathbf{e}_{2}\mathbf{e}_{3}}}\overline{(f_{A_{\mathbf{e}_{2}\mathbf{e}_{3}}})}},$
$mod_{\mathbf{e}_{3}\mathbf{e}_{1}}=\sqrt{f_{A_{\mathbf{e}_{3}\mathbf{e}_{1}}}\overline{(f_{A_{\mathbf{e}_{3}\mathbf{e}_{1}}})}}.$

###  4.2. Envelop Detection of 3-D Images

In [24], Wang constructed a experiment platform for acquiring radio frequency
(RF) ultrasound volume with a biopsy needle in it. The volume is a $128\times
1280\times 33$ pixels in lateral ($x_{1}$ axes), elevation ($x_{2}$ axes) and
axial ($x_{3}$ axes) directions respectively. By using classical 1-D analytic
signal and novel 3-D analytic signal approaches, different resolution appears
in ultrasound image envelop detection. Since 3-D analytic signal takes into
account the information of the neighbouring scan lines, this novel approaches
supplies better result than classical 1-D analytic signal approach. In this
paper, we will take the same experiment platform and show the shortcoming of
the analytic signal associated with CFT. We also propose the analytic signal
associated with CLCT in Clifford biquaternion domain to this problem and show
that by modifying the matrix parameters, one can obtain a satisfactory result.

Figure 1. original envelop of 3-D image.

Here we modify the volume to a $128\times 128\times 128$ pixels, and in the
plane of $x_{3}=64$ there is a biopsy needle located at $x_{2}=64$ with $0\leq
x_{1}\leq 32$, which can be easily found in Figure 1. The flow-process diagram
of using the amplitude method of analytic signals is supplied in Figure 2.

Inputting original data $f(x_{1},x_{2},x_{3})$Obtain analytic signal
$f_{A}(x_{1},x_{2},x_{3})$Derive the polar formObtain the modulus
$|f_{A}(x_{1},x_{2},x_{3})|$draw the picture of $|f_{A}(x_{1},x_{2},x_{3})|$
Figure 2.  The flow-process diagram of using analytic signal’s amplitude to
envelop detection.

Figure 3 shows the result by using Wang’s analytic signal to detect the
envelop of biopsy needle which locates at plane of $x_{3}=64$. Thank to take
into account the information of the neighbouring scan lines, one can find the
biopsy needle in the nearest planes besides in plane $x_{3}=64$, which are
planes $x_{3}=63$ and $x_{3}=65$. Since it will induce many noise and lose
information in converting RF ultrasound signal to the Brightness mode (B-mode)
signal, this property will insure one can find a clear envelop in B-mode
images. But unfortunately, the envelops in these three planes distort badly,
which are much longer than the true needle. This shortcoming can be well
overcame by our approach based on the analytic signal associated with CLCT,
which can be found in Figure 4. The envelop also can be found in planes
$x_{3}=63$,$x_{3}=64$ and $x_{3}=65$, and the length is very close to that of
true needle. Here the parameter matrices are
$A_{1}=\left(\begin{array}[]{cc}1&10\\\ -0.5&0.5\\\ \end{array}\right)$,
$A_{2}=\left(\begin{array}[]{cc}1&10\\\ -0.5&0.5\\\ \end{array}\right)$ and
$A_{3}=\left(\begin{array}[]{cc}1000&1\\\ -0.1&0.0009\\\ \end{array}\right)$,
and they are found by two steps: first, to the 2-D image, plane $x_{3}=64$, we
try several times and get good result. Furthermore, we slightly modify the
parameters of $A_{3}$ and find a better result. Since there are two many
parameters in these three matrices, we can not insure these three matrices are
the best parameters. Due to the fact, Wang’s method is a special case of our
approach, we can insure that our approach will supply a better result by
choosing proper parameter matrices.

Furthermore, we also use the envelop method of monogenic signal, which was
introduced by Yang [30] and is another higher dimensional generalization of
analytic signal, to detect the envelop of this image. Figure 4 shows that it
supplies an accurate envelop in every planes but one can find this envelop in
every $x_{3}$ plane, which means that the image will be contaminated by the
background signal badly.

Figure 3. Wang’s analytic signal associated with CFT [24].

Figure 4. This paper’s method: analytic signal associated with CLCT, where
$A_{1}=(1,10;-0.05,0.5)$, $A_{2}=(1,10;-0.05,0.5)$ and
$A_{3}=(1000,1;-0.1,0.0009)$.

Figure 5. Monogenic signal approach of [30].

##  5\. Conclusion

Inspired by the success of 3-D biquaternionic analytic signal in detecting 3-D
ultrasound images envelop, we generalize this approach to LCT domain. Thanks
to the special structure of the Clifford biquaternion algebra, this
generalized analytic signal has three partial modules where
$mod_{\mathbf{e}_{1}\mathbf{e}_{2}}(f_{A})$ corresponds the envelop of 3-D
images in $x_{3}$ planes. Real valued synthetic 3-D image is introduced to
test the valid of this novel envelop detection method. Comparing with the
classical 1-D analytic signal method and Wang’s generalized biquaternionic
analytic signal method even the envelop method of monogenic signal, our
approach supplies a best result. Furthermore, our method can derive a better
resolution by modifying the parameter matrices and adding post processing such
as average filter processing.

### Acknowledgment

The first author acknowledges financial support from the PhD research startup
foundation of Hubei University of Technology No. BSQD2019052. Partial support
by the Foundation for Science and Technology from Department of Education of
Hubei province (B2019047). Partial support by the Macao Science and Technology
Development Fund 0085/2018/A2.

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