A New Form of Convolution Theorem for One-Dimensional Quaternion Linear Canonical Transform and Application

Abstract

In this research work, we focus on the one-dimensional quaternion linear canonical transform (1-D QLCT). Under certain conditions, we first derive the symmetry property of the 1-D QLCT for real signals. The new form of the convolution theorem related to this transformation is proposed. We develop this convolution definition to derive the correlation theorem for the 1-D QLCT. We then show that the direct connection between the quaternion convolution and quaternion correlation definitions permits us to provide a different way for proving the correlation theorem concerning the 1-D QLCT. Finally, we present a simple application of the convolution theorem to the study of quaternion swept-frequency filter analysis.

1. Introduction

Due to the importance of the linear canonical transform [1,2], a series of studies have been conducted on the generalization of the classical linear canonical transform using the quaternion algebra, called the quaternion linear canonical transform (QLCT). Recently, the two-dimensional quaternion linear canonical transform (2-D QLCT) was studied theoretically and proved to be a useful tool in several areas, like signal processing, optics, quantum physics, and statistics [3,4,5,6]. It is an expansion of popular generalized transformations like the two-dimensional quaternion Fourier quaternion [7,8,9,10,11,12,13,14] and the two-dimensional quaternion fractional Fourier transform [15,16,17]. More recently, the authors of [18,19,20] proposed the offset quaternion linear canonical transform, which is the expansion of the 2-D QLCT. With the rigorous research of the 2-D QLCT, a number of useful properties of the 2-D QLCT have been investigated, including shifting, modulation, correlation, differentiation, energy conservation, inequalities, and so on. In this research paper, our work is focused on the one-dimensional quaternion linear canonical transform (1-D QLCT), which was recently studied by the authors in [21]. As we know, the authors of [21] discussed a convolution definition of the 1-D QLCT and obtained the convolution theorem, which is the multiplication of their 1-D QLCT and a phase factor. The form of this convolution theorem contains an extra chirp factor and is not exactly parallel to the convolution theorem for the one-dimensional quaternion Fourier transform [22].

Therefore, we first propose a new form of the convolution definition related to the 1-D QLCT, which is quite different from the one proposed in [21]. The convolution theorem is found to be equal to a simple multiplication of the 1-D QLCT and the quaternion Fourier transform. In addition, based on the convolution definition, we propose a correlation definition for the 1-D QLCT. We then provide direct proof of the correlation theorem. By leveraging the basic connection between the convolution definition and correlation definition, we provide further proof of the correlation theorem related to this transformation. We finally provide a simple application of the convolution theorem to quaternion swept-frequency filter analysis.

The remainder of this paper is organized as follows. In Section 2, we focus on the basic properties of quaternion algebra used in the next section. In Section 3, we recall the definition of the one-dimensional quaternion Fourier transform and collect its useful properties, such as the convolution and correlation theorems. In Section 4, we provide the definition of the 1-D QLCT and propose a new form of its convolution theorem. The derivation of the correlation theorem related to this transformation is also provided in this section. For the illustration of the proposed transformation, we provide in Section 5 a simple application of the convolution theorem concerning the 1-D QLCT. Finally, in Section 6, we provide a simple conclusion.

2. Preliminaries

This part collects a few well-known facts about quaternions, which will be important for the rest of paper. Real quaternion is a generalization of the complex numbers over real number $\mathbb{R}$ and is denoted by $\mathbb{H}$. Every element of $\mathbb{H}$ may be expressed as [23]

$$\mathbb{H}=\{r=r_0+\mathbf{i}{r}_1+\mathbf{j}{r}_2+\mathbf{k}{r}_3:r_0,r_1,r_2,r_3\in \mathbb{R}\},$$

(1)

which follows the following multiplication laws:

$$\begin{array}{ll}\mathbf{ij}\hfill & =-\mathbf{ji}=\mathbf{k},\mathbf{jk}=-\mathbf{kj}=\mathbf{i},\mathbf{ki}=-\mathbf{ik}=\mathbf{j},\hfill \\ {\mathbf{i}}^2\hfill & ={\mathbf{j}}^2={\mathbf{k}}^2=\mathbf{ijk}=-1.\hfill \end{array}$$

(2)

For a quaternion $r=r_0+\mathbf{i}{r}_1+\mathbf{j}{r}_2+\mathbf{k}{r}_3$, $r_0=S\left(r\right)$ is called scalar part and $V\left(r\right)=\mathbf{r}=\mathbf{i}{r}_1+\mathbf{j}{r}_2+\mathbf{k}{r}_3$ is called the vector part or the pure quaternion. This will lead to

$$\begin{array}{c}\hfill r=S\left(r\right)+V\left(r\right).\end{array}$$

(3)

According to relation (2), we get the quaternion product $rz$ in the form

$$rz=S\left(r\right)S\left(z\right)-V\left(r\right)·V\left(z\right)+S\left(z\right)V\left(r\right)+S\left(r\right)V\left(z\right)+V\left(r\right)\times V\left(z\right),$$

(4)

for which

$$\begin{array}{ll}\hfill V\left(r\right)·V\left(z\right)& =r_1{z}_1+r_2{z}_2+r_3{z}_3\hfill \\ \hfill V\left(r\right)\times V\left(z\right)& =\mathbf{i}(r_2{z}_3-r_3{z}_2)+\mathbf{j}(r_3{z}_1-r_1{z}_3)+\mathbf{k}(r_1{z}_2-r_2{z}_3).\hfill \end{array}$$

(5)

We introduce to each quaternion r its conjugate one; that is,

$$\overline{r}=r_0-\mathbf{i}{r}_1-\mathbf{j}{r}_2-\mathbf{k}{r}_3,$$

(6)

with the property

$$\overline{rz}=\overline{z}\overline{r},\overline{\overline{r}}=r.$$

(7)

Note that, in contrast to complex conjugate, the conjugate quaternion changes the order of multiplication. From (6), we can get the norm or modulus of $r\in \mathbb{H}$ as

$$\left|r\right|=\sqrt{r\overline{r}}=\sqrt{r_0^2+r_1^2+r_2^2+r_3^2}.$$

(8)

It is routine to verify that

$$\begin{array}{c}\hfill \left|rz\right|=\left|r\right|\left|z\right|,|r+z|\le \left|r\right|+\left|z\right|,|r^2{|=|r|}^2,\end{array}$$

(9)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{r+\overline{r}}{2}}=S\left(r\right),{\displaystyle \frac{r-\overline{r}}{2}}=V\left(r\right),S\left(r\right)\le \left|r\right|,V\left(r\right)\le \left|r\right|,\forall r,z\in \mathbb{H}.\end{array}$$

(10)

The following aspects are satisfied via

$$\begin{array}{c}\hfill S\left(rz\right)=S\left(zr\right),S\left(\overline{r}z\right)=S\left(r\overline{z}\right),\end{array}$$

(11)

but

$$\begin{array}{c}\hfill S\left(\overline{r}z\right)\ne S\left(rz\right),\forall r,z\in \mathbb{H}.\end{array}$$

(12)

Based on quaternion conjugate (6) and the modulus of r, we define the inverse of a non-zero quaternion $r\in \mathbb{H}$ as

$$r^{-1}={\displaystyle \frac{\overline{r}}{{\left|r\right|}^2}}.$$

(13)

Utilizing (13) above, one can get

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{p}{r}}\right|={\displaystyle \frac{\left|p\right|}{\left|r\right|}},\forall p\in \mathbb{H}.\end{array}$$

(14)

Like the complex case, we may introduce the inner product for two continuous functions $f,g:\mathbb{R}⟶\mathbb{H}$ as

$${(f,g)}_{L^2(\mathbb{R};\mathbb{H})}={\int }_{-\infty }^{\infty }f\left(x\right)\overline{g\left(x\right)}dx.$$

(15)

Especially, we get

$${\parallel f\parallel }_{L^2(\mathbb{R};\mathbb{H})}={\left({\int }_{-\infty }^{\infty }{\left|f\left(x\right)\right|}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}.$$

(16)

Using the basic properties of quaternion explained above, we can derive the following result.

Theorem 1. 

For any function $f\in L^2(\mathbb{R};\mathbb{H})$, the following inequality is satisfied

$$\begin{array}{c}\hfill \left|{\int }_{-\infty }^{\infty }f\left(x\right)dx\right|\le 2{\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx.\end{array}$$

(17)

Proof. 

In view of Equation (8), we have

$$\begin{array}{cc}\hfill & |{\int }_{-\infty }^{\infty }f\left(x\right)dx{|}^2\hfill \\ \hfill & ={\left|{\int }_{-\infty }^{\infty }(f_0\left(x\right)+\mathbf{i}{f}_1\left(x\right)+\mathbf{j}{f}_2\left(x\right)+\mathbf{k}{f}_3\left(x\right))dx\right|}^2\hfill \\ \hfill & ={\left|{\int }_{-\infty }^{\infty }{f}_0\left(x\right)dx+\mathbf{i}{\int }_{-\infty }^{\infty }{f}_1\left(x\right)dx+\mathbf{j}{\int }_{-\infty }^{\infty }{f}_2\left(x\right)dx+\mathbf{k}{\int }_{-\infty }^{\infty }{f}_3\left(x\right)dx\right|}^2\hfill \\ \hfill & =\left({\int }_{-\infty }^{\infty }{f}_0\left(x\right)dx+\mathbf{i}{\int }_{-\infty }^{\infty }{f}_1\left(x\right)dx+\mathbf{j}{\int }_{-\infty }^{\infty }{f}_2\left(x\right)dx+\mathbf{k}{\int }_{-\infty }^{\infty }{f}_3\left(x\right)dx\right)\hfill \\ \hfill & \times \left({\int }_{-\infty }^{\infty }{f}_0\left(x\right)dx-\mathbf{i}{\int }_{-\infty }^{\infty }{f}_1\left(x\right)dx-\mathbf{j}{\int }_{-\infty }^{\infty }{f}_2\left(x\right)dx-\mathbf{k}{\int }_{-\infty }^{\infty }{f}_3\left(x\right)dx\right)\hfill \\ \hfill & ={\left({\int }_{-\infty }^{\infty }{f}_0\left(x\right)dx\right)}^2+{\left({\int }_{-\infty }^{\infty }{f}_1\left(x\right)dx\right)}^2+{\left({\int }_{-\infty }^{\infty }{f}_2\left(x\right)dx\right)}^2+{\left({\int }_{-\infty }^{\infty }{f}_3\left(x\right)dx\right)}^2.\hfill \end{array}$$

(18)

By using (10), relation (18) above changes to

$$\begin{array}{cc}\hfill & |{\int }_{-\infty }^{\infty }f\left(x\right)dx{|}^2\hfill \\ \hfill & \le {\left({\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx\right)}^2+{\left({\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx\right)}^2+{\left({\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx\right)}^2+{\left({\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx\right)}^2\hfill \\ \hfill & =4{\left({\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx\right)}^2.\hfill \end{array}$$

Hence,

$$\left|{\int }_{-\infty }^{\infty }f\left(x\right)dx\right|\le 2{\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx,$$

(19)

and proof is complete. □

Remark 1. 

If a complex function $f\in L^2\left(\mathbb{R}\right)$, Equation (17) will lead to

$$\begin{array}{c}\hfill \left|{\int }_{-\infty }^{\infty }f\left(x\right)dx\right|\le \sqrt{2}{\int }_{-\infty }^{\infty }\left|f\left(x\right)\right|dx.\end{array}$$

(20)

3. The 1-D QLCT with Properties

In this part, we briefly review the definition of the one-dimensional quaternion Fourier transform (1-D QFT) and main properties. More details have been presented in [22].

Definition 1. 

The definition of the one-dimensional quaternion Fourier transform (1-D QFT) for the quaternion signal $f\in L^2(\mathbb{R};\mathbb{H})$ is given by

$${\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)={\int }_{-\infty }^{\infty }f\left(x\right)e^{-\mathbf{i}\omega x}dx,$$

(21)

where $x,\omega \in \mathbb{R}$.

It is observed that, if the quaternion function $f\left(x\right)$ is decomposed into

$$\begin{array}{c}\hfill f\left(x\right)=f_0\left(x\right)+\mathbf{i}{f}_1\left(x\right)+\mathbf{j}{f}_2\left(x\right)+\mathbf{k}{f}_3,\end{array}$$

(22)

we get

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)=R_0\left(\omega \right)+\mathbf{i}{R}_1\left(\omega \right)+\mathbf{j}{R}_2\left(\omega \right)+\mathbf{k}{R}_3\left(\omega \right),\end{array}$$

(23)

where

$$\begin{array}{ll}\hfill R_0\left(\omega \right)& ={\int }_{-\infty }^{\infty }\left(f_0\left(x\right)cos\omega x+f_1\left(x\right)sin\omega x\right)dx\hfill \\ \hfill R_1\left(\omega \right)& ={\int }_{-\infty }^{\infty }\left(f_1\left(x\right)cos\omega x-f_0\left(x\right)sin\omega x\right)dx\hfill \\ \hfill R_2\left(\omega \right)& ={\int }_{-\infty }^{\infty }\left(f_2\left(x\right)cos\omega x-f_3\left(x\right)sin\omega x\right)dx\hfill \\ \hfill R_3\left(\omega \right)& ={\int }_{-\infty }^{\infty }\left(f_3\left(x\right)cos\omega x+f_2\left(x\right)sin\omega x\right)dx.\hfill \end{array}$$

(24)

If $f\left(x\right)$ in (21) is a real-valued function, then relation (23) changes to

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)=R_0\left(\omega \right)+\mathbf{i}{R}_1\left(\omega \right),\end{array}$$

(25)

for which

$$\begin{array}{c}\hfill R_0\left(\omega \right)={\int }_{-\infty }^{\infty }f\left(x\right)cos\omega xdx,\mathrm{and}{R}_1\left(\omega \right)={\int }_{-\infty }^{\infty }f\left(x\right)sin\omega xdx.\end{array}$$

(26)

Now Equation (21) can be rewritten as

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)={\int }_{-\infty }^{\infty }\overline{\overline{f\left(x\right)}}\overline{e^{\mathbf{i}\omega x}}dx.\end{array}$$

(27)

In view of Equation (7), we obtain

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)={\int }_{-\infty }^{\infty }\overline{e^{\mathbf{i}\omega x}\overline{f\left(x\right)}}dx.\end{array}$$

(28)

Definition 2. 

If $f\in L^1(\mathbb{R};\mathbb{H})$ with ${\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\in L^1(\mathbb{R};\mathbb{H})$, then the inverse transform of the 1-D QFT is computed by

$$f\left(x\right)={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right)e^{\mathbf{i}\omega x}d\omega .$$

(29)

Let us introduce the quaternion convolution for two quaternion functions $f,g\in L^2(\mathbb{R};\mathbb{H})$ through the following definition:

$$(f\star g)\left(x\right)={\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)dy.$$

(30)

The next result provides the convolution theorem that describes how the 1-D QFT interacts with the quaternion convolution.

Theorem 2. 

Suppose $f\in L^2(\mathbb{R};\mathbb{H})$ and $g\in L^2(\mathbb{R};\mathbb{H})$. Then, the following holds

$$\begin{array}{ll}\hfill {\mathcal{F}}_{\mathbf{i}}\{f\star g\}\left(\omega \right)& ={\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left(\omega \right)+\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left(\omega \right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left(\omega \right)+\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left(\omega \right).\hfill \end{array}$$

(31)

Moreover,

$$\begin{array}{ll}\hfill (f\star g)\left(x\right)& ={\mathcal{F}}_{\mu }^{-1}[{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left(\omega \right)+\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left(\omega \right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left(\omega \right)+\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left(\omega \right)]\left(x\right).\hfill \end{array}$$

(32)

Definition 3. 

The correlation definition for the 1-D QFT of two quaternion functions $f,g\in L^1(\mathbb{R};\mathbb{H})$ is given by the integral

$$(f\circ g)\left(x\right)={\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)dy.$$

(33)

With the above definition, one gets (see [22])

Theorem 3. 

Suppose that $f,g$ belong to $\in L^2(\mathbb{R};\mathbb{H})$; we have

$$\begin{array}{ll}\hfill {\mathcal{F}}_{\mathbf{i}}\{f\circ g\}\left(\omega \right)& ={\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}(-\omega )-\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}(-\omega )\hfill \\ \hfill & -\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}(-\omega )-\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}(-\omega ).\hfill \end{array}$$

(34)

4. Convolution and Correlation Theorems for the I-D QLCT

Based on the quaternion algebra, we introduce the 1-D QLCT in this part. It will be shown that the convolution theorem for the 1-D QLCT is a general form of the convolution theorem for the 1-D QFT. Let us now introduce the definition of the 1-D QLCT.

Definition 4. 

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in {\mathbb{R}}^2$ be a matrix parameter satisfying $det\left(A\right)=1$. The 1-D QLCT of the function $f\in L^2(\mathbb{R};\mathbb{H})$ is defined by

$${\mathcal{L}}_A\left\{f\right\}\left(\omega \right)={\int }_{-\infty }^{\infty }f\left(x\right)K_A^{\mathbf{i}}(x,\omega )dx,$$

(35)

and kernel $K_A^{\mathbf{i}}(x,\omega )$ is given by

$$K_A^{\mathbf{i}}(x,\omega )=\left\{\begin{array}{cc}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})},\hfill & b\ne 0\hfill \\ \sqrt{d}{e}^{\mathbf{i}{\displaystyle \frac{cd}{2}}{\omega }^2}\delta (x-d\omega ),\hfill & b=0.\hfill \end{array}\right.$$

(36)

It should be clear that, for $b=0$, the 1-D QLCT definition reduces to chirp multiplication operator and is of no particular interest for the objective in this research. Therefore, we always consider the case of $b\ne 0$ in the work unless stated otherwise.

Now observe that

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\left\{f\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }f\left(x\right)K_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }f\left(x\right)e^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dx\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }\left(f_0\left(x\right)+\mathbf{i}{f}_1\left(x\right)+\mathbf{j}{f}_2\left(x\right)+\mathbf{k}{f}_3\left(x\right)\right)e^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dx\hfill \\ \hfill & ={\mathcal{L}}_A\left\{f_0\right\}\left(\omega \right)+\mathbf{i}{\mathcal{L}}_A\left\{f_1\right\}\left(\omega \right)+\mathbf{j}{\mathcal{L}}_A\left\{f_2\right\}\left(\omega \right)+\mathbf{k}{\mathcal{L}}_A\left\{f_3\right\}\left(\omega \right).\hfill \end{array}$$

(37)

This equation implies that

$$\begin{array}{cc}\hfill & \overline{{\mathcal{L}}_A\left\{f\right\}(-\omega )}\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }\overline{K_A^{\mathbf{i}}(x,-\omega )}\overline{f\left(x\right)}dx\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2+{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\overline{f\left(x\right)}dx\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left(f_0\left(x\right)-\mathbf{i}{f}_1\left(x\right)-\mathbf{j}{f}_2\left(x\right)-\mathbf{k}{f}_3\left(x\right)\right)dx.\hfill \end{array}$$

(38)

In particular, when the function $f\left(x\right)$ is real-valued and the matrix parameter $A=\left[\begin{array}{cc}-a& b\\ c& -d\end{array}\right]$, relation (38) changes to

$$\begin{array}{cc}\hfill & \overline{{\mathcal{L}}_A\left\{f\right\}(-\omega )}={\mathcal{L}}_A\left\{f\right\}\left(\omega \right).\hfill \end{array}$$

(39)

Relation (38) is often called the symmetry property of the 1-D QLCT.

Again, from (37) above, it is straightforward to verify that

$$\begin{array}{c}\hfill \left|{\mathcal{L}}_A\left\{f_i\right\}\left(\omega \right)\right|\le \left|{\mathcal{L}}_A\left\{f\right\}\left(\omega \right)\right|,i=0,1,2,3.\end{array}$$

(40)

In this case, we may define

$$\begin{array}{c}\hfill \left|{\mathcal{L}}_A\left\{f\right\}\left(\omega \right)\right|=\sqrt{|{\mathcal{L}}_A\left\{f_0\right\}\left(\omega \right){|}^2+|{\mathcal{L}}_A\left\{f_1\right\}\left(\omega \right){|}^2+|{\mathcal{L}}_A\left\{f_2\right\}\left(\omega \right){|}^2+|{\mathcal{L}}_A\left\{f_3\right\}\left(\omega \right){|}^2}.\end{array}$$

(41)

Definition 5. 

If $f\in L^1(\mathbb{R};\mathbb{H})$ with ${\mathcal{L}}_A\left\{f\right\}\in L^1(\mathbb{R};\mathbb{H})$, then the inverse transform of the 1-D QLCT is calculated via

$$\begin{array}{cc}\hfill f\left(x\right)& ={\int }_{-\infty }^{\infty }{\mathcal{L}}_A\left\{f\right\}\left(\omega \right)\overline{K_A^{\mathbf{i}}(x,\omega )}d\omega \hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\mathcal{L}}_A\left\{f\right\}\left(\omega \right)e^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}d\omega .\hfill \end{array}$$

(42)

Now we present Parseval’s formula for the 1-D QLCT, which is stated as follows.

Theorem 4 

([22]). For any functions $f,g$ belonging to $L^1(\mathbb{R};\mathbb{H})\cap L^2(\mathbb{R};\mathbb{H})$, the following identity holds:

$${(f,g)}_{L^2(\mathbb{R};\mathbb{H})}={({\mathcal{L}}_A\left\{f\right\},{\mathcal{L}}_A\left\{g\right\})}_{L^2(\mathbb{R};\mathbb{H})},$$

(43)

and

$$\begin{array}{c}\hfill {\parallel f\parallel }_{L^2(\mathbb{R};\mathbb{H})}={\parallel {\mathcal{L}}_A\left\{f\right\}\parallel }_{L^2(\mathbb{R};\mathbb{H})}.\end{array}$$

(44)

From relations (21) and (35), it follows that

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\left\{f\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }f\left(x\right)K_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}f\left(x\right)dx\hfill \\ \hfill & ={\displaystyle \frac{e^{-\mathbf{i}\frac{\pi }{4}}}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}{\displaystyle \frac{d}{2b}}{\omega }^2}{\int }_{-\infty }^{\infty }{e}^{-\mathbf{i}{\displaystyle \frac{x\omega }{b}}}{e}^{\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}f\left(x\right)dx\hfill \\ \hfill & ={\displaystyle \frac{e^{-\mathbf{i}\frac{\pi }{4}}}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}{\displaystyle \frac{d}{2b}}{\omega }^2}{\int }_{-\infty }^{\infty }{\mathcal{F}}_{\mathbf{i}}\left\{h\right\}({\displaystyle \frac{\omega }{b}})dx,\hfill \end{array}$$

(45)

where

$$\begin{array}{c}\hfill h\left(x\right)=e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}f\left(x\right).\end{array}$$

(46)

Equation (45) provides the direct relationship between the 1-D QFT and 1-D QLCT. This relation is very useful to derive some properties of the 1-D QLCT.

Remark 2. 

It should be observed that, when $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$, then, from Equation (45), we get

$$\begin{array}{c}\hfill {\mathcal{L}}_A\left\{f\right\}\left(\omega \right)={\displaystyle \frac{e^{-\mathbf{i}\frac{\pi }{4}}}{\sqrt{2\pi }}}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\omega \right).\end{array}$$

(47)

Let us now introduce the definition of the quaternion convolution in the 1-D QLCT domain. It is an important technical tool in the area of image processing. The quaternion convolution for the 1-D QLCT plays a crucial role in the sequel.

Definition 6. 

The convolution operation of the 1-D QLCT for signals $f,g\in L^2(\mathbb{R};\mathbb{H})$ is given by

$$\begin{array}{c}\hfill (f\ast g)\left(x\right)={\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{(x-y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}.\end{array}$$

(48)

It is easy to check that, for $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$, Equation (48) becomes the quaternion convolution definition (30). With the above definition, we obtain the following important convolution theorem.

Theorem 5. 

Suppose that $f,g\in L^2(\mathbb{R};\mathbb{H})$; then, the 1-D QLCT of the convolution of f and g is described by

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{f\ast g\}\left(\omega \right)& ={\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right).\hfill \end{array}$$

(49)

In addition,

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{\overline{f}\ast g\}\left(\omega \right)& ={\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)-\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)-\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)-\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right).\hfill \end{array}$$

(50)

Proof. 

According to the definition of the 1-D QLCT (35), we can obtain

$$\begin{array}{cc}\hfill & {\mathcal{L}}_A\{f\ast g\}\left(\omega \right)\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }(f\ast g)\left(x\right)K_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)e^{\mathbf{i}{\displaystyle \frac{a}{b}}{(x-y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{K}_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)e^{\mathbf{i}{\displaystyle \frac{a}{b}}{(x-y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dxdy\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)e^{\mathbf{i}{\displaystyle \frac{a}{b}}{(x-y)}^2}{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dxdy\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g(x-y)e^{\mathbf{i}{\displaystyle \frac{a}{b}}{(x-y)}^2}{e}^{-\mathbf{i}({\displaystyle \frac{x\omega }{b}}-{\displaystyle \frac{d}{2b}}{\omega }^2+{\displaystyle \frac{\pi }{4}})}dydx.\hfill \end{array}$$

Using the change of the variables $z=x-y$, we obtain

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{f\ast g\}\left(\omega \right)=& {\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g\left(z\right)e^{\mathbf{i}{\displaystyle \frac{a}{b}}{z}^2}{e}^{-\mathbf{i}({\displaystyle \frac{(z+y)\omega }{b}}-{\displaystyle \frac{d}{2b}}{\omega }^2+{\displaystyle \frac{\pi }{4}})}dydz\hfill \\ \hfill =& {\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g\left(z\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dydz.\hfill \end{array}$$

The decomposition of $f\left(y\right)=f_0\left(y\right)+\mathbf{i}{f}_1\left(y\right)+\mathbf{j}{f}_2\left(y\right)+\mathbf{k}{f}_3\left(y\right)$ will lead to

$$\begin{array}{cc}\hfill & {\mathcal{L}}_A\{f\ast g\}\left(\omega \right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(z\right)e^{\mathbf{i}({\displaystyle \frac{a}{b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_0\left(y\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)dz\hfill \\ \hfill & +\mathbf{i}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(z\right)e^{\mathbf{i}({\displaystyle \frac{a}{b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_1\left(y\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)dz\hfill \\ \hfill & +\mathbf{j}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(z\right)e^{\mathbf{i}({\displaystyle \frac{a}{b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_2\left(y\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)dz\hfill \\ \hfill & +\mathbf{k}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(z\right)e^{\mathbf{i}({\displaystyle \frac{a}{b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_3\left(y\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)dz.\hfill \end{array}$$

The proof is complete. □

The alternative form of the convolution theorem for the 1-D QLCT was discussed in [21]. Their convolution theorem is the product of the 1-D QLCT and phase factor, which is not similar to the convolution theorem for the 1-D QFT in Theorem 2.

Let us now collect some findings regarding Theorem 5 in the following remark.

Remark 3. 

• If f and g are real-valued functions, Equation (49) becomes

  $$\begin{array}{c}\hfill {\mathcal{L}}_A\{f\ast g\}\left(\omega \right)={\mathcal{L}}_A\left\{f\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{g\right\}\left({\displaystyle \frac{\omega }{b}}\right),\end{array}$$

  (51)
  which corresponds to the convolution theorem for the classical linear canonical transform (compare to [24]).
• Generalization of convolution theorem for the 1-D QLCT to the 2-D QLCT may be considered in future work (compare to [25,26])

Further, we obtain the inequality related the convolution theorem in the form

$$\begin{array}{cc}\hfill & |{\mathcal{L}}_A\{f\ast g\}\left(\omega \right){|}^2\hfill \\ \hfill & \le |{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^2\left(|{\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\right).\hfill \end{array}$$

(52)

To see this, we observe that

$$\begin{array}{cc}\hfill |{\mathcal{L}}_A\{f\ast g\}\left(\omega \right){|}^2& =|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2.\hfill \end{array}$$

(53)

The triangle inequality for quaternion yields

$$\begin{array}{cc}\hfill |{\mathcal{L}}_A\{f\ast g\}\left(\omega \right){|}^2& \le |{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\hfill \\ \hfill & +|\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\hfill \\ \hfill & =|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\hfill \\ \hfill & +|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\hfill \\ \hfill & =\left|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^2\right(|{\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2\hfill \\ \hfill & +|{\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2+|{\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^2).\hfill \end{array}$$

(54)

Remark 4. 

It should be observed that, when $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$, Equation (49) changes to the convolution theorem for the 1-D QFT (31).

Below, we define the quaternion correlation concerning the 1-D QLCT.

Definition 7. 

The correlation related to the 1-D QLCT of two quaternion functions $f,g\in L^2(\mathbb{R};\mathbb{H})$ is expressed as

$$(f•g)\left(x\right)={\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)e^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{(x+y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}.$$

(55)

With the above definition, we get the following result on the 1-D QLCT of a correlation of two quaternion functions.

Theorem 6. 

Assume that $f,g\in L^2(\mathbb{R};\mathbb{H})$; then, the correlation for f and g in terms of the 1-D QLCT is provided by

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{f•g\}\left(\omega \right)=& {\mathcal{L}}_{A^{\prime }}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)-\mathbf{i}{\mathcal{L}}_{A^{\prime }}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)+\mathbf{j}{\mathcal{L}}_{A^{\prime }}\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & \times {\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)-\mathbf{k}{\mathcal{L}}_{A^{\prime }}\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left(-{\displaystyle \frac{\omega }{b}}\right),\hfill \end{array}$$

(56)

where $A^{\prime }=\left[\begin{array}{cc}-a& b\\ c& d\end{array}\right]$.

Proof. 

We first provide direct proof of the result. By virtue of relations (35) and (55), we see that

$$\begin{array}{cc}\hfill & {\mathcal{L}}_A\{f•g\}\left(\omega \right)\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }(f•g)\left(x\right)K_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)e^{-\mathbf{i}{\displaystyle \frac{a}{b}}{(x+y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{K}_A^{\mathbf{i}}(x,\omega )dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)e^{-\mathbf{i}{\displaystyle \frac{a}{b}}{(x+y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dxdy\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)e^{-\mathbf{i}{\displaystyle \frac{a}{b}}{(x+y)}^2}dy{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dxdy\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g(x+y)e^{-\mathbf{i}{\displaystyle \frac{a}{b}}{(x+y)}^2}{e}^{-\mathbf{i}({\displaystyle \frac{x\omega }{b}}-{\displaystyle \frac{d}{2b}}{\omega }^2+{\displaystyle \frac{\pi }{4}})}dydx.\hfill \end{array}$$

By substituting, $u=x+y$ in the above identity, we arrive at

$$\begin{array}{cc}\hfill & {\mathcal{L}}_A\{f•g\}\left(\omega \right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g\left(u\right)e^{-\mathbf{i}{\displaystyle \frac{a}{b}}{u}^2}{e}^{-\mathbf{i}({\displaystyle \frac{(u-y)\omega }{b}}-{\displaystyle \frac{d}{2b}}{\omega }^2+{\displaystyle \frac{\pi }{4}})}dydu\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\overline{f\left(y\right)}g\left(u\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}{e}^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dydu\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\left(f_0\left(y\right)-\mathbf{i}{f}_1\left(y\right)-\mathbf{j}{f}_2\left(y\right)-\mathbf{k}{f}_3\left(y\right)\right)g\left(u\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}\hfill \\ \hfill & e^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dydu\hfill \end{array}$$

This equation may be rewritten as

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{f•g\}\left(\omega \right)& ={\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(u\right)e^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_0\left(y\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)du\hfill \\ \hfill & -\mathbf{i}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(u\right)e^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_1\left(y\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)du\hfill \\ \hfill & -\mathbf{j}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(u\right)e^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_2\left(y\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)du\hfill \\ \hfill & -\mathbf{k}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(u\right)e^{\mathbf{i}(-{\displaystyle \frac{a}{b}}{u}^2-{\displaystyle \frac{u\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}\left({\int }_{-\infty }^{\infty }{f}_3\left(y\right)e^{\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\right)du.\hfill \end{array}$$

Thus, the proof is complete. □

Let us provide further proof of correlation theorem for the 1-D QLCT using the interaction between the definition of convolution and the definition of correlation.

Alternative Proof of Theorem 6. 

Putting $y=-z$ and setting $f(-z)=h\left(z\right)$ in relation (55) yields

$$\begin{array}{cc}\hfill (f•g)\left(x\right)=& {\int }_{-\infty }^{\infty }\overline{f(-z)}g(x-z)e^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{(x-z)}^2}dz{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}\hfill \\ \hfill =& {\int }_{-\infty }^{\infty }\overline{h\left(z\right)}g(x-z)e^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{(x-z)}^2}dz{e}^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2)}\hfill \\ \hfill =& (\overline{h}\ast g)\left(x\right).\hfill \end{array}$$

(57)

Applying (50) yields

$$\begin{array}{cc}\hfill (f•g)\left(x\right)=& {\mathcal{L}}_A^{-1}[{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{h_0\right\}({\displaystyle \frac{\omega }{b}})-\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{h_1\right\}({\displaystyle \frac{\omega }{b}})\hfill \\ \hfill & -\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{h_2\right\}({\displaystyle \frac{\omega }{b}})-\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{h_3\right\}({\displaystyle \frac{\omega }{b}})]\left(x\right).\hfill \end{array}$$

(58)

Now we observe that

$$\begin{array}{cc}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{h_0\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }{f}_0(-z)e^{-\mathbf{i}{\displaystyle \frac{\omega z}{b}}}dz\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{f}_0\left(z\right)e^{\mathbf{i}{\displaystyle \frac{\omega z}{b}}}dz\hfill \\ \hfill & ={\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}(-{\displaystyle \frac{\omega }{b}}).\hfill \end{array}$$

(59)

Similarly, we get

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{h_i\right\}\left(\omega \right)={\mathcal{F}}_{\mathbf{i}}\left\{f_i\right\}(-{\displaystyle \frac{\omega }{b}}),i=1,2,3.\end{array}$$

(60)

Substituting Equations (59) and (60) into Equation (58) results in

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\{f•g\}\left(\omega \right)=& {\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)-\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)-\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left(-{\displaystyle \frac{\omega }{b}}\right)-\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left(-{\displaystyle \frac{\omega }{b}}\right),\hfill \end{array}$$

(61)

and the proof is complete. □

5. Application of Convolution Theorem

As a simple application of the quaternion convolution theorem (48) above, we get the following result, which is the main result of this section.

Theorem 7. 

Consider a system $g\left(x\right)$ such that its response satisfies

$$\begin{array}{c}\hfill h\left(x_0\right)={\int }_{-\infty }^{\infty }f\left(x\right)g(x_0-x)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{(x_0-x)}^2}dx{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2},\end{array}$$

(62)

for each fixed $x_0$. If the energy of the system $g\left(x\right)$ is specified, that is,

$$\begin{array}{c}\hfill E={\int }_{-\infty }^{\infty }|g\left(x\right){|}^{2}dx,\end{array}$$

(63)

then, for every input signal $f\left(x\right)$, the following is satisfied:

$$\begin{array}{cc}\hfill |h\left(x_0\right){|}^2\le & {\displaystyle \frac{E}{2\pi \left|b\right|}}[{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ].\hfill \end{array}$$

(64)

Proof. 

From Equations (29) and (71), it can be deduced that

$$\begin{array}{cc}\hfill h\left(x_0\right)=& {\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }({\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right))e^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}_0^2-{\displaystyle \frac{x_0\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}d\omega .\hfill \end{array}$$

(65)

We further get

$$\begin{array}{cc}\hfill & |h\left(x_0\right){|}^2\hfill \\ \hfill & =\left|{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }\right({\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right))e^{-\mathbf{i}({\displaystyle \frac{a}{2b}}{x}_0^2-{\displaystyle \frac{x_0\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}d\omega {|}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2\pi \left|b\right|}}{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right)\hfill \\ \hfill & {\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi \left|b\right|}}\left[{\int }_{-\infty }^{\infty }\right|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +|\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +|\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +|\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ].\hfill \end{array}$$

(66)

With the help of Cauchy–Schwartz inequality and Parseval’s Formula (44), we see that

$$\begin{array}{cc}\hfill |h\left(x_0\right){|}^2& \le {\displaystyle \frac{1}{2\pi \left|b\right|}}[{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^{2}d\omega {\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^{2}d\omega {\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^{2}d\omega {\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^{2}d\omega {\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ].\hfill \end{array}$$

(67)

This relation is simplified to

$$\begin{array}{cc}\hfill & |h\left(x_0\right){|}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2\pi \left|b\right|}}{\int }_{-\infty }^{\infty }|{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){|}^{2}d\omega [{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ]\hfill \\ \hfill & ={\displaystyle \frac{E}{2\pi \left|b\right|}}[{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega \hfill \\ \hfill & +{\int }_{-\infty }^{\infty }|{\mathcal{F}}_i\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ],\hfill \end{array}$$

(68)

and the proof is complete. □

Remark 5. 

If $f\in L^2\left(\mathbb{R}\right)$ is a real-valued function, relation (64) will lead to

$$\begin{array}{c}\hfill |h\left(x_0\right){|}^2\le {\displaystyle \frac{E}{2\pi \left|b\right|}}{\int }_{-\infty }^{\infty }|{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega .\end{array}$$

(69)

To verify the correctness of Theorem 7, we provide a simple example as below.

Example 1. 

Consider the signals of the form

$$\begin{array}{c}\hfill f\left(x\right)=g\left(x\right)=e^{-x^2}.\end{array}$$

(70)

Thus, we obtain

$$\begin{array}{cc}\hfill h\left(x_0\right)& ={\int }_{-\infty }^{\infty }f\left(x\right)g(x_0-x)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{(x_0-x)}^2}dx{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2}\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{e}^{-x^2}{e}^{-{(x_0-x)}^2}{e}^{\mathbf{i}{\displaystyle \frac{a}{2b}}{(x_0-x)}^2}dx{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2}\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{e}^{-x^2}{e}^{-(x_0^2-2x_{0}x+x^2)}{e}^{\mathbf{i}{\displaystyle \frac{a}{2b}}(x_0^2-2x_{0}x+x^2)dx{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2}}\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{e}^{-x^2}{e}^{-x_0^2+2x_{0}x-x^2}{e}^{\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2}{e}^{-\mathbf{i}{\displaystyle \frac{a}{b}}{x}_{0}x}{e}^{\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}dx{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}_0^2}.\hfill \end{array}$$

(71)

This equation yields

$$\begin{array}{c}\hfill |h\left(x_0\right)|\le {\int }_{-\infty }^{\infty }{e}^{-2x^2+2x_{0}x-x_0^2}dx=\sqrt{{\displaystyle \frac{\pi }{2}}}{e}^{-{\displaystyle \frac{x_0^2}{2}}}.\end{array}$$

(72)

On the other side, we have

$$\begin{array}{c}\hfill E={\int }_{-\infty }^{\infty }|g\left(x\right){|}^{2}dx={\int }_{-\infty }^{\infty }{e}^{-x^2}dx=\sqrt{\pi },\end{array}$$

(73)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{E}{2\pi \left|b\right|}}{\int }_{-\infty }^{\infty }|{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left({\displaystyle \frac{\omega }{b}}\right){|}^{2}d\omega ={\displaystyle \frac{\sqrt{\pi }}{2\pi \left|b\right|}}{\int }_{-\infty }^{\infty }\sqrt{\pi }{e}^{-{\displaystyle \frac{{\omega }^2}{4b^2}}}d\omega ={\displaystyle \frac{1}{2\left|b\right|}}\sqrt{4\pi b^2}=\sqrt{\pi }.\end{array}$$

(74)

Equations (72) and (74) show that Equation (64) is valid.

Now, setting $x-y=z$ in Equation (48), we immediately obtain

$$\begin{array}{c}\hfill r\left(x\right)={\int }_{-\infty }^{\infty }f(x-z)g\left(z\right)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{z}^2}dz{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}.\end{array}$$

(75)

Equation (75) is known as the output of generalized swept-frequency filters. Taking the 1-D QLCT of the above identity, we obtain

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\left\{r\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x-z)g\left(z\right)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{z}^2}dz{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dx\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x-z)g\left(z\right)e^{\mathbf{i}{\displaystyle \frac{a}{2b}}{z}^2}{e}^{-\mathbf{i}{\displaystyle \frac{a}{2b}}{x}^2}{\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{x}^2-{\displaystyle \frac{x\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dxdz.\hfill \end{array}$$

(76)

This equation is simplified to

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\left\{r\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g\left(z\right){\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{z}^2-{\displaystyle \frac{(y+z)\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dzdy\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(y\right)g\left(z\right){\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{e}^{\mathbf{i}({\displaystyle \frac{a}{2b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}{e}^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dzdy\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }f\left(y\right)\left({\displaystyle \frac{1}{\sqrt{2\pi \left|b\right|}}}{\int }_{-\infty }^{\infty }g\left(z\right)e^{\mathbf{i}({\displaystyle \frac{a}{2b}}{z}^2-{\displaystyle \frac{z\omega }{b}}+{\displaystyle \frac{d}{2b}}{\omega }^2-{\displaystyle \frac{\pi }{4}})}dz\right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }f\left(y\right){\mathcal{L}}_A\left\{g\right\}\left(\omega \right)e^{-\mathbf{i}{\displaystyle \frac{u\omega }{b}}}dy.\hfill \end{array}$$

(77)

Note that, if ${\mathcal{L}}_A\left\{g\right\}\left(\omega \right)$ is a real-valued function, then we immediately obtain

$$\begin{array}{c}\hfill {\mathcal{L}}_A\left\{r\right\}\left(\omega \right)={\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left({\displaystyle \frac{\omega }{b}}\right){\mathcal{L}}_A\left\{g\right\}\left(\omega \right).\end{array}$$

Further, when ${\mathcal{L}}_A\left\{g\right\}\left(\omega \right)$ is a quaternion-valued function, then we decompose $f\left(y\right)=f_0\left(y\right)+\mathbf{i}{f}_1\left(y\right)+\mathbf{j}{f}_2\left(y\right)+\mathbf{k}{f}_3\left(y\right)$ and obtain

$$\begin{array}{cc}\hfill {\mathcal{L}}_A\left\{r\right\}\left(\omega \right)& ={\int }_{-\infty }^{\infty }f\left(y\right){\mathcal{L}}_A\left\{g\right\}\left(\omega \right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }\left(f_0\left(y\right)+\mathbf{i}{f}_1\left(y\right)+\mathbf{j}{f}_2\left(y\right)+\mathbf{k}{f}_3\left(y\right)\right){\mathcal{L}}_A\left\{g\right\}\left(\omega \right)e^{-\mathbf{i}{\displaystyle \frac{y\omega }{b}}}dy\hfill \\ \hfill & ={\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_0\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{i}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_1\right\}\left({\displaystyle \frac{\omega }{b}}\right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_2\right\}\left({\displaystyle \frac{\omega }{b}}\right)+\mathbf{k}{\mathcal{L}}_A\left\{g\right\}\left(\omega \right){\mathcal{F}}_{\mathbf{i}}\left\{f_3\right\}\left({\displaystyle \frac{\omega }{b}}\right).\hfill \end{array}$$

6. Conclusions

In this work, we introduced the 1-D QLCT. The convolution theorem related to this transformation was derived in detail. We found that the convolution theorem is equivalent to simple multiplication of the 1-D QLCT and the quaternion Fourier transform. We proposed a simple application of the convolution theorem regarding the study of quaternion swept-frequency filter analysis. In the future, we will concentrate on investigating the convolution and correlation theorems associated with the 2-D QLCT, whose proof is more complicated, and discuss several applications, such as quaternion swept-frequency filter analysis, image denoising, and so on.