The identities of n-dimensional s-transform and applications

Abstract

The Stockwell transform is a signal processing tool based on Fourier transform (FT) with a specific window function. The important properties of 1-dimensional (1-D) and 2-dimensional (2-D) S-transform (ST) are available in the literature. This paper includes the generalized case of these ST as n-dimensional (n-D) ST. First, a definition of n-D ST is given followed by the definitions of its important properties. This definition of n-D ST utilizes the approach of separable function to achieve its closed-form definition starting from the conventional definition for 1-D ST. Thereafter, these analytical concepts are being utilized in filtering the unwanted component and the performance of the proposed technique is compared with the discrete cosine transform (DCT) based filtering.

Appendices

Appendix A

Proof of Linearity Property:

$$ S\left\{{\alpha}_1{S}_1+{\alpha}_2{S}_2\right\}\left(t,\nu \right)={\alpha}_1S\left\{{S}_1\left(t,\nu \right)\right\}+{\alpha}_2S\left\{{S}_2\left(t,\nu \right)\right\} $$

(25)

$$ ={\int}_{-\infty}^{\infty }{\alpha}_1{S}_1(t)\frac{\mid \nu \mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left(\tau -t\right)}^2{\nu}^2}{2}\right)\exp \left(-i2\pi \nu t\right) dt+{\int}_{-\infty}^{\infty }{\alpha}_2{S}_2(t)\frac{\mid \nu \mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left(\tau -t\right)}^2{\nu}^2}{2}\right)\exp \left(-i2\pi \nu t\right) dt $$

(26)

$$ ={\alpha}_1{\int}_{-\infty}^{\infty }{s}_1(t)\frac{\mid \nu \mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left(\tau -t\right)}^2{\nu}^2}{2}\right)\exp \left(-i2\pi \nu t\right) dt+{\alpha}_2{\int}_{-\infty}^{\infty }{s}_2(t)\frac{\mid \nu \mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left(\tau -t\right)}^2{\nu}^2}{2}\right)\exp \left(-i2\pi \nu t\right) dt $$

(27)

$$ ={\alpha}_1S\left\{{S}_1\left(t,\nu \right)\right\}+{\alpha}_2S\left\{{S}_2\left(t,\nu \right)\right\} $$

(28)

In general, for n-dimension ST can be defined as

$$ S\left\{{\alpha}_1{S}_1+{\alpha}_2{S}_2+\dots +{\alpha}_n{S}_n\right\}\left(t,\nu \right)={\alpha}_1S\left\{{S}_1\left(t,\nu \right)\right\}+{\alpha}_2S\left\{{S}_2\left(t,\nu \right)\right\}+\dots +{\alpha}_nS\left\{{S}_n\left(t,\nu \right)\right\} $$

(29)

Appendix B

Proof of Shifting property:

$$ S\left\{{t}_1,{t}_2,{\nu}_1,{\nu}_2\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left({t}_1,{t}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{t}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right){dt}_1{dt}_2 $$

(30)

Suppose, S(t) function is shifted by ζ then,

$$ S\left\{{\tau}_1,{\tau}_2,{\nu}_1,{\nu}_2\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left({t}_1+{\zeta}_1,{t}_2+{\zeta}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{t}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right){dt}_1{dt}_2 $$

(31)

$$ ={\int}_{-\infty}^{\infty}\frac{\mid {\nu}_1\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2}{2}\right)\exp \left(-j2\pi {t}_1{\nu}_1\right){\int}_{-\infty}^{\infty }S\left({t}_1+{\zeta}_1,{t}_2+{\zeta}_2\right)\frac{\mid {\nu}_2\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_2-{t}_2\right)}^2{\nu}_2^2}{2}\right)\exp \left(-j2\pi {t}_2{\nu}_2\right){dt}_2{dt}_1 $$

(32)

Suppose, t₂ + ζ₂ = y and dt₂ = dy then

$$ ={\int}_{-\infty}^{\infty}\frac{\mid {\nu}_1\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2}{2}\right)\exp \left(-j2\pi {t}_1{\nu}_1\right){\int}_{-\infty}^{\infty }S\left({t}_1+{\zeta}_1,y\right)\frac{\mid {\nu}_2\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_2-y+{\zeta}_2\right)}^2{\nu}_2^2}{2}\right)\exp \left(-j2\pi \left(y-{\zeta}_2\right){\nu}_2\right){dydt}_1 $$

(33)

Suppose, t₁ + ζ₁ = x and dt₂ = dx then

$$ {\displaystyle \begin{array}{l}={\int}_{-\infty}^{\infty}\frac{\mid {\nu}_1\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_1-x+{\zeta}_1\right)}^2{\nu}_1^2}{2}\right)\exp \left(-j2\pi \left(x-{\zeta}_1\right){\nu}_1\right){\int}_{-\infty}^{\infty }S\left(x,y\right)\frac{\mid {\nu}_2\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_2-y+{\zeta}_2\right)}^2{\nu}_2^2}{2}\right)\\ {}\kern2em \exp \left(-j2\pi \left(y-{\zeta}_2\right){\nu}_2\right) dydx\end{array}} $$

(34)

$$ {\displaystyle \begin{array}{l}=\exp \left(j2\pi \left({\zeta}_1{\nu}_1+{\zeta}_2{\nu}_2\right)\right){\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left(x,y\right)\frac{\mid {\nu}_1\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_1+{\zeta}_1-x\right)}^2{\nu}_1^2}{2}\right)\exp \left(-j2\pi x{\nu}_1\right)\frac{\mid {\nu}_2\mid }{\sqrt{2\pi }}\exp \left(\frac{-{\left({\tau}_2+{\zeta}_2-y\right)}^2{\nu}_2^2}{2}\right)\\ {}\kern2em \exp \left(-j2\pi y{\nu}_2\right) dydx\end{array}} $$

(35)

$$ {\displaystyle \begin{array}{l}=\exp \left(j2\pi \left({\zeta}_1{\nu}_1+{\zeta}_2{\nu}_2\right)\right){\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left(x,y\right)\frac{\mid {\nu}_1\mid }{\sqrt{2\pi }}\frac{\mid {\nu}_2\mid }{\sqrt{2\pi }}\exp \left(\frac{-\left\{{\left({\tau}_1+{\zeta}_1-x\right)}^2{\nu}_1^2+{\left({\tau}_2+{\zeta}_2-y\right)}^2{\nu}_2^2\right\}}{2}\right)\\ {}\kern2em \exp \left(-j2\pi \left({\nu}_1x+{\nu}_2y\right)\right) dydx\end{array}} $$

(36)

$$ =\exp \left(j2\pi \left({\zeta}_1{\nu}_1+{\zeta}_2{\nu}_2\right)\right)S\left\{x,y,{\nu}_1,{\nu}_2\right\} $$

(37)

Similarly, for n-dimensional ST can be defined as

$$ S\left\{{t}_1\pm {\zeta}_1,{t}_2\pm {\zeta}_2,\dots, {t}_n\pm {\zeta}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right\}=\exp \left(\pm j2\pi \left({\zeta}_1{\nu}_1+{\zeta}_2{\nu}_2+\dots +{\zeta}_n{\nu}_n\right)\right)\mathcal{S}\left\{h\left({t}_1,{t}_2,\dots, {t}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right)\right\} $$

(38)

Appendix C

Proof of Scaling Property:

$$ S\left\{{t}_1,{t}_2,{\nu}_1,{\nu}_2\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left({t}_1,{t}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{t}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right){dt}_1{dt}_2 $$

(39)

$$ S\left\{{a}_1{t}_1,{a}_2{t}_2,{\nu}_1,{\nu}_2\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left({a}_1{t}_1,{a}_2{t}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{t}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{t}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right){dt}_1{dt}_2 $$

(40)

Suppose, a₁t₁ = pand a₂t₂ = q, hence ${dt}_1=\frac{dp}{\mid {a}_1\mid }$ and ${dt}_2=\frac{dq}{\mid {a}_2\mid }$, then

$$ =\frac{1}{\mid {a}_1\Big\Vert {a}_2\mid }{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }S\left(p,q\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left({\left({\tau}_1-\frac{p}{a_1}\right)}^2{\nu}_1^2+{\left({\tau}_2-\frac{q}{a_2}\right)}^2{\nu}_2^2\right)}{2}\right)\exp \left(-j2\pi \left(\frac{p{\nu}_1}{a_1}+\frac{q{\nu}_2}{a_2}\right)\right) dpdq $$

(41)

$$ =\frac{1}{\mid {a}_1\Big\Vert {a}_2\mid }{S}_h\left\{{a}_1{\tau}_1,{a}_2{\tau}_2,\frac{\nu_1}{a_1},\frac{\nu_2}{a_2}\right\} $$

(42)

$$ =\frac{1}{\mid \mathrm{detM}\mid }{S}_h\left\{\left({a}_1{\tau}_1,{a}_2{\tau}_2,{M}^{-T}\nu \right)\right\} $$

(43)

where, $M=\left[\begin{array}{cc}{a}_1& 0\\ {}0& {a}_2\end{array}\right]$and detM = a₁a₂ ≠ 0

$$ {M}^T=\left[\begin{array}{cc}{a}_1& 0\\ {}0& {a}_2\end{array}\right] $$

(44)

$$ \mathrm{Or},\kern0.5em {M}^{-T}=\left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${a}_1$}\right.& 0\\ {}0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${a}_2$}\right.\end{array}\right] $$

(45)

where T is the transpose of the matrix

$$ \mathcal{S}\left\{h\left({a}_1{\tau}_1,{a}_2{\tau}_2,{\nu}_1,{\nu}_2\right)\right\}=\frac{1}{\mid \mathrm{detM}\mid }{S}_h\left\{\left({a}_1{\tau}_1,{a}_2{\tau}_2,{M}^{-T}\nu \right)\right\} $$

(46)

Similarly, for n-dimensional ST can be defined as

$$ \mathcal{S}\left\{h\left({a}_1{t}_1,{a}_2{t}_2,\dots, {a}_n{t}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right)\right\}=\frac{1}{\mid {a}_1\mid, \mid {a}_2\mid, \dots, \mid {a}_n\mid }{S}_h\left\{\left({a}_1{\tau}_1,{a}_1{\tau}_2,\dots, {a}_1{\tau}_n,\frac{\nu_1}{a_1},\frac{\nu_2}{a_2},\dots, \frac{\nu_n}{a_n}\right)\right\} $$

(47)

Appendix D

Proof of Complex Conjugate Property:

$$ S\left\{h\left({t}_1,{t}_2\right)\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty}\left\{{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{S}_h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)d{t}_1d{t}_2\right\}\exp \left(j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(48)

Taking complex conjugate represented as ‘*’ in both sides, give

$$ {S}^{\ast}\left\{h\left({t}_1,{t}_2\right)\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty}\left\{{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{S}_h^{\ast}\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)d{t}_1d{t}_2\right\}\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(49)

Substituting ν₁ =  − ν₁ and ν₂ =  − ν₂ then

$$ ={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty}\left\{{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{S}_h^{\ast}\left({t}_1,{t}_2,-{\nu}_1,-{\nu}_2\right)d{t}_1d{t}_2\right\}\exp \left(j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(50)

$$ ={S}^{-1}\left\{{H}^{\ast}\left({t}_1,{t}_2,-{\nu}_1,-{\nu}_2\right)\right\} $$

(51)

In general, n-dimensional ST can be defined as

$$ {S}^{\ast}\left\{h\left({t}_1,{t}_2,\dots, {t}_n\right)\right\}={S}^{-1}\left\{{H}^{\ast}\left({t}_1,{t}_2,\dots, {t}_n,-{\nu}_1,-{\nu}_2,\dots, -{\nu}_n\right)\right\} $$

(52)

Similar to the 1-dimension case, it S(t)is even then S{S(t, ν)}is even and if S(t)is real and even then S{S(t, ν)}is real and even.

Appendix E

Proof of Time Reversal Property:

By using the definition of ST, the function h(−t₁, −t₂) can be expressed as

$$ S\left\{h\left(-{t}_1,-{t}_2\right)\right\}={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }h\left(-{t}_1,-{t}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1+{t}_1\right)}^2{\nu}_1^2+{\left({\tau}_2+{t}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left({t}_1{\nu}_1+{t}_2{\nu}_2\right)\right)d{t}_1d{t}_2 $$

(53)

Substituting−t₁ = u₁ − t₂ = u₂, anddt₁ =  − du₁, dt₂ =  − du₂ in (53),

$$ ={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }h\left({u}_1,{u}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{u}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{u}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(-j2\pi \left(-{u}_1{\nu}_1-{u}_2{\nu}_2\right)\right)d{u}_1d{u}_2 $$

(54)

$$ ={\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }h\left({u}_1,{u}_2\right)\frac{\mid {\nu}_1\Big\Vert {\nu}_2\mid }{2\pi}\exp \left(\frac{-\left\{{\left({\tau}_1-{u}_1\right)}^2{\nu}_1^2+{\left({\tau}_2-{u}_2\right)}^2{\nu}_2^2\right\}}{2}\right)\exp \left(j2\pi \left({u}_1{\nu}_1+{u}_2{\nu}_2\right)\right)d{u}_1d{u}_2 $$

(55)

Therefore,

$$ S\left\{h\left(-{t}_1,-{t}_2\right)\right\}\overset{ST}{\leftrightarrow }{S}_h^{\ast}\left(-{u}_1,-{u}_2,{\nu}_1,{\nu}_2\right) $$

(56)

Similarly, for n-dimensional, the ST can be defined as

$$ S\left\{h\left(-{t}_1,-{t}_2,\dots, {t}_n\right)\right\}\overset{ST}{\leftrightarrow }{S}_h^{\ast}\left(-{t}_1,-{t}_2,\dots, {t}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right) $$

(57)

Appendix F

Proof of Time-derivatives Property:

Considering the definition of n-dimensional IST of h(t)can be written as

$$ h\left(t,\nu \right)=\underset{\Re^n}{\int}\left(\underset{\Re^n}{\int }{S}_h\left(t,\nu \right) d t\right)\exp \left(j2\pi t\nu \right) d\nu $$

(58)

For n = 2, the IST of h(t)can be defined as

$$ h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)=\underset{\Re^2}{\int}\left(\underset{\Re^2}{\int }{S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right){du}_1{du}_2\right)\exp \left(j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(59)

Comparing (59) with the definition of FT, after that differentiating it with respect to time in both sides as

$$ \frac{d^2}{dt_1{dt}_2}\left\{h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right\}=\frac{d^2}{dt_1{dt}_2}\left\{\underset{\Re^2}{\int}\left(\underset{\Re^2}{\int }{S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right){du}_1{du}_2\right)\exp \left(j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2\right\} $$

(60)

$$ =\underset{\Re^2}{\int}\left(\underset{\Re^2}{\int }{S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right){du}_1{du}_2\right)\frac{d^2}{dt_1{dt}_2}\exp \left(j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(61)

$$ =\underset{\Re^2}{\int}\left(\underset{\Re^2}{\int }{S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right){du}_1{du}_2\right)\left\{j2\pi \left({\nu}_1,{\nu}_2\right)\right\}\exp \left(j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(62)

$$ =j2\pi \left({\nu}_1,{\nu}_2\right){S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right) $$

(63)

Therefore,

$$ \frac{d^2}{dt_1{dt}_2}\left\{h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right\}\overset{ST}{\leftrightarrow }j2\pi \left({\nu}_1,{\nu}_2\right){S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right) $$

(64)

Similarly, for n-dimensional ST can be defined as

$$ \frac{d^n}{d{t}_1d{t}_2\dots d{t}_n}\left\{h\left({t}_1,{t}_2,\dots, {t}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right)\right\}\overset{ST}{\leftrightarrow }j2\pi \left({\nu}_1,{\nu}_2,\dots, {\nu}_n\right){S}_h\left({u}_1,{u}_2,\dots, {u}_n,{\nu}_1,{\nu}_2,\dots, {\nu}_n\right) $$

(65)

Appendix G

Proof of Parseval’s theorem:

Using the expression of inverse ST for two dimensional h(t) [2] as

$$ h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)=\underset{{\mathfrak{R}}^2}{\int}\left(\underset{{\mathfrak{R}}^2}{\int }{S}_h\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right)d{u}_1d{u}_2\right)\exp \left(j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(66)

Similarly, taking the complex conjugate of (66),

$$ {h}^{\ast}\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)=\underset{{\mathfrak{R}}^2}{\int}\left(\underset{{\mathfrak{R}}^2}{\int }{S}_h^{\ast}\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right)d{u}_1d{u}_2\right)\exp \left(-j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2 $$

(67)

Now, the LHS of Parseval’s theorem is written as

$$ {\displaystyle \begin{array}{c}\underset{{\mathfrak{R}}^2}{\int }h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right){h}^{\ast}\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)d{t}_1d{t}_2=\underset{{\mathfrak{R}}^2}{\int }h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\underset{{\mathfrak{R}}^2}{\int}\left(\underset{{\mathfrak{R}}^2}{\int }{S}_h^{\ast}\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right)d{u}_1d{u}_2\right)\\ {}\kern3em \exp \left(-j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{\nu}_1d{\nu}_2d{t}_1d{t}_2\end{array}} $$

(68)

Now, by the convergence property of ST into FT, once the ST is integrated foru₁, u₂ variable, that is

$$ {H}^{\ast}\left({\nu}_1,{\nu}_2\right)=\underset{{\mathfrak{R}}^2}{\int }{S}_h^{\ast}\left({u}_1,{u}_2,{\nu}_1,{\nu}_2\right)d{u}_1d{u}_2 $$

(69)

Substituting (69) into (68) and rearranging the terms

$$ =\underset{{\mathfrak{R}}^2}{\int }{H}^{\ast}\left({\nu}_1,{\nu}_2\right)\underset{{\mathfrak{R}}^2}{\int}\left\{h\left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\exp \left(-j2\pi \left({t}_1,{t}_2,{\nu}_1,{\nu}_2\right)\right)d{t}_1d{t}_2\right\}\kern0.1em d{\nu}_1d{\nu}_2 $$

(70)

$$ =\underset{{\mathfrak{R}}^2}{\int }H\left({\nu}_1,{\nu}_2\right){H}^{\ast}\left({\nu}_1,{\nu}_2\right)d{\nu}_1d{\nu}_2 $$

(71)

Therefore, after transformation, the energy is preserved. Hence, it is written as

$$ \underset{{\mathfrak{R}}^2}{\int }{\left|h\left({t}_1,{t}_2\right)\right|}^2d{t}_1d{t}_2=\underset{{\mathfrak{R}}^2}{\int }{\left|H\Big({\nu}_1,{\nu}_2\Big)\right|}^2d{\nu}_1d{\nu}_2 $$

(72)

Similarly, for n-dimensional, the ST can be defined as

$$ \underset{{\mathfrak{R}}^n}{\int }{\left|h\left({t}_1,{t}_2,\dots, {t}_n\right)\right|}^2d{t}_1d{t}_2\dots d{t}_n=\underset{{\mathfrak{R}}^n}{\int }{\left|H\Big({\nu}_1,{\nu}_2,\dots, {\nu}_n\Big)\right|}^2d{\nu}_1d{\nu}_2\dots d{\nu}_n $$

(73)