This note emphasizes a point of view on the Bilinear Hilbert Transform (BHT) in the Bedrosian identity, which is originally based on the conventional Hilbert Transform in the theory of analytic signals. We show that the generalized sinc functions from Möbius play a crucial role. We have demonstrated some properties of BHT: Riesz representation theorem, boundedness, and Bedrosian identity for BHT. The generalized sinc function is a special solution of the Bedrosian identity. After that, we consider some of the nonlinear sinc functions, which are also the solutions of the Bedrosian identity. Lastly, the nonlinear sinc function system is orthonormal. In addition, we propose a framework for analytical decision-making applications that leverages the properties of BHT. incorporating this framework into existing real-world systems further enhances the adaptability and responsiveness of decision-making models, positioning the BHT as a critical tool for optimizing processes in dynamic, data-rich environments.
Denote the translation and modulation of a signal f by
ibt
Taf(t) = f(t − a), Mbf(t) = e f(t), t ∈ R .
The vector-valued Hilbert transform and the bilinear Hilbert transform(BHT)
Hf(t) = p.v. f(t − x) dx π R x
and Z 1 dx H(f, g)(t) = p.v. f(t − x)g(t + x) .
π R x
Theorem 1. BHT is bilinear transform.
Proof. Z 1 dx H(αf + βg, h)(t) = p.v. [αf(t − x) + βg(t − x)]h(t + x) π R x 1 dx 1 dx = p.v. αf(t − x)h(t + x) + p.v. βg(t − x)h(t + x) π R x π R x 1 dx 1 dx = α p.v. f(t − x)h(t + x) + β p.v. g(t − x)h(t + x) π R x π R x
= αH(f, h)(t) + βH(g, h)(t).
Let’s verify H(f, g)(t) = H(g, f)(t)
1 dx H(f, g)(t) = p.v. f(t − x)g(t + x) π R x 1 dx = − p.v. f(x)g(2t − x) π R t − x
1 dx H(g, f)(t) = p.v. g(t − x)f(t + x) π R x 1 dx = p.v. f(x)g(2t − x) π R x − t
H(f, g)(t) = H(g, f)(t),BHT is bilinear transform. □
We give an alternative characterization of BHT by the Fouerier transform
1 dx < w ˆ , ψ >=< w , ψ > ˆ = lim ψ ˆ(x) π ε→ 0 x
|x|≥ ε
1 − 2πixξ dx = lim ψ(ξ)e dξ π ε→ 0 1 x
≥| x|≥ ε R
1 − 2πixξ dx = lim ψ(ξ)[ e ]dξ ε→ 0 π 1 x
R ≥| x|≥ ε
−i dx = lim ψ(ξ)[ sin(2πxξ) ]dξ ε→ 0 π 1 x
R ≥| x|≥ ε
−i dx = lim ψ(ξ)[ sgn(ξ) sin(x|ξ|) ]dξ ε→ 0 π 1 ε x
R ≥| x|≥
2 πε 2 π
The Lebesgue dominated convergence theorem allows the passage of the limit inside the integral.We obtain that
< w ˆ0, ψ >= ψ(ξ)(−isgn(ξ))dξ This implies that
w c 0 (x) = −isgn(x)
We use identity< w ˆ0, ψ >= ψ(ξ)(−isgn(ξ))dξ to write
H(f, g)(t) = f (x −\t) g (x + t) (−isgn(x)) (t)
From the above we can see that the Riesz representation theorem applies to BHT.
For all 1 < p < 2 ,we give the L boundedness of BHT
Lemma 2. For BHT given by
1 f (t − x) g (t + x) H (f, g) = p.v. dx π R x
the inequality
∥H (f, g)∥p ⩽ Kp p ∥f∥p ∥g∥p
3 1 2 1 2
1 1 1
exsits, provided 2 < p1, p2 < ∞, = + and 1 < p3 < 2.
p3 p1 p2
The following lemma establishes a relationship betweeen Hilbert transform and BHT by modulation operator.
i<ξ,·>
Lemma 3. Set eξ(·) = e . For any ξ ∈ R , we have
H(Mξf)(t) = −M2ξH(f, e− ξ)(t), t ∈ R . (1)
Proof. By the definition of BHT and a change of variable in the integral, we get
1 dy H(f, eξ)(t) = p.v. f(t − y)eξ(t + y) π R y 1 dy = − p.v. f(y)eξ(2t − y) π R t − y
i<ξ,2t> 1 − i<ξ,y> dy = −e p.v. f(y)e .
π R t − y
By recalling the definition of modulation and Hilbert transform, we obtain
i<2ξ,t>
H(f, eξ)(t) = −e H(M− ξf)(t),
that is
H(M− ξf)(t) = −M− 2ξH(f, eξ)(t).
We therefore conclude that this is an alternative form of (1) by setting ξ instead of −ξ. The proof of this lemma is finished. □
The theorem below states that Bedrosian identity
Theorem 4. Suppose thatg ∈ L [(0, 2π)] is2π-periodic function of real variable. Then the Bedrosian identity
H(f g)(t) = f(t)Hg(t), t ∈ R (2)
is equivalent to
H(f, g(2t − ·))(t) = f(t)H(1, g(2t − ·))(t), t ∈ R . (3) 2 ikt
Proof. Since g ∈ L [(0, 2π)], g has a convergent Fourier expansion g(t) = ck(g)e . One hand, by linearity
k∈ Z
of Hilbert transform, we have
i<k,·>
H(f g)(t) = H f(·) ck(g)e (t)
k∈ Z i<k,·>
= ck(g)H f(·)e (t)
k∈ Z
= ck(g)H(Mkf)(t).
k∈ Z
Using the equation (1) and the linearity of BHT implies
H(f g)(t) = − ck(g)M2kH(f, e− k)(t)
k∈ Z i2kt
= − ck(g)e H(f, e− k)(t)
k∈ Z
i2kt
= −H f, ck(g)e e− k (t)
k∈ Z
i<k,2t−· >
= −H f, ck(g)e (t)
k∈ Z
= −H (f, g(2t − ·)) (t).
One the other hand, by the similar technique, we have that
f(t)Hg(t) = −f(t)H(1, g(2t − ·))(t).
We therefore conclude that the two equations are equivalent each other. The proof of this theorem is completed.
This section is focusing to investigate the Bedrosian identity of BHT type in the case that g is nonlinear Fourier atom
H (ρ, cos θa(2t − ·)) (t) = ρ(t)H (1, cos θa(2t − ·)) (t), t ∈ R . (4)
The following theorem shows that the generalized Sinc function is a special solution of above equation.
sint sinθa (t)
Theorem 5. The generalized Sinc function Sinc a(t) := pa(t) = satisfies the equation (4), that is,
sin θa(t) H(Sinca(·), cos θa(2t − ·))(t) = Sinca(t)H(1, cos θa(2t − ·))(t) = , t ∈ R . (5)
To prove this theorem, we need some preparations. The first fact is that the so called one-sided ladder shape
it
+ ⌊|t|⌋ √ 1 1 1− e
filter H (t) = a χR (t) and the function r(t) = it form a pair of Fourier transform, that is
a 2π 1− ae − it
Lemma 6.
− iξ
+ ∧ 1 1 1 − e (Ha ) (ξ) = r(−ξ) = √
− iξ
2π 1 − ae iξ
and
i· ∧
1 1 1 − e + r ˆ(ξ) = √ (ξ) = Ha (ξ).
2π 1 − ae −i· Proof. The detail of calculation is below
+ ∧ 1 + − iξt (Ha ) (ξ) = √ Ha (t)e dt 2π −∞ Z ∞ ∞ 1 − a X
k− 1 − iξt
= √ a χ[0,k)(t)e dt 2π −∞ ∞ Z k ∞ X X − ikξ 1 − a k− 1 − iξt 1 − a k− 1 1 − e = √ a e dt = √ a 2π 0 2π iξ
k=1 k=1 − iξ − iξ − iξ
1 − a 1 e 1 1 − ae − (1 − a)e = √ − /(iξ) = √
− iξ − iξ
2π 1 − a 1 − ae 2π iξ(1 − ae )
− iξ
1 1 1 − e = √ .
− iξ
2π 1 − ae iξ
The proof of this lemma is finished. □
The second fact is the representation of cos θa(t) and sin θa(t).
Lemma 7. For real number a ∈ (−1, 1), the following equations hold
(1 − a ) sin t sin θa(t) = pa(t) sin t = , t ∈ R , 1 − 2a cos t + a (1 + a ) cos t − 2a cos θa(t) = , t ∈ R .
1 − 2a cos t + a
Proof. The boundary on the unit circle of Möbius transform can conclude the two formulae. The detail is as follows
it it − it
e − a (e − a)(1 − ae ) τa(z)|z= eit = =
it 2
1 − ae 1 − 2a cos t + a
it 2 − it 2 2
e − 2a + a e (1 + a ) cos t − 2a (1 − a ) sin t = = + i .
2 2 2
1 − 2a cos t + a 1 − 2a cos t + a 1 − 2a cos t + a
The proof of this lemma is finished. □
The third is the formula of the Hilbert transform of generalized Sinc function
Lemma 8. The Hilbert transform of Sinc a is
1 + a 1 − cos t 1 − cos θa(t) H(Sinca)(t) = pa(t) = , t ∈ R . (6) 1 − a t t
it
√ 1 1 e − 1
Proof. We have shown that the Fourier transform of r(t) = it is the one-sided ladder shape function
2π 1− ae it
Ha (ξ). Since r has no negative frequency, it is an analytic signal and therefore the real part and imaginary part form a pair of Hilbert transform
H(Rer)(t) = Imr(t), t ∈ R .
Note that
1 (1 − a) sin t 1 1 (1 + a)(1 − cos t) 1 r(t) = √ + i√ 2π 1 − 2a cos t + a t 2π 1 − 2a cos t + a t 1 sin t 1 1 − cos t = √ pa(t) + i√ pa(t) .
2π(1 + a) t 2π(1 − a) t
We get
sin(·) 1 + a 1 − cos t H pa(·) (t) = pa(t) , · 1 − a t which is equivalent to
1 + a 1 − cos t H (Sinca(·)) (t) = pa(t) .
1 − a t
We are left to show 1 + a 1 − cos θa(t) = pa(t)(1 − cos t), t ∈ R .
1 − a In fact, by the representation of cos θa(t) (see Lemma 7), we have
2 2 2
(1 + a ) cos t − 2a (1 + a) − (1 + a) cos t 1 − cos θa(t) = 1 − = 1 − 2a cos t + a 1 − 2a cos t + a (1 + a) (1 − cos t) 1 + a = = pa(t)(1 − cos t).
1 − 2a cos t + a 1 − a
The proof of this lemma is completed. □
Remark 1 : This lemma indicates an interesting formula: pseudo-Bedrosian formula
1 + a H(pa(·)Sinc(·))(t) = pa(t)H(Sinc(·))(t), t ∈ R .
1 − a
Remark 2 : There is an alternative approach to prove this lemma. By using Qian’s result
1 sin θa(y) dy H (Sinca(·)) (t) = π −∞ y t − y 1 1 1 1 = sin θa(y) ( + )dy π −∞ t y t − y Z ∞ Z ∞ 1 1 sin θa(y) 1 sin θa(y) = dy + dy t π −∞ y π −∞ t − y
= (−H(sin θa)(0) + H(sin θa)(t)) = (−(− cos θa(0)) − cos θa(t))
1 − cos θa(t)
sin(2θa (t))
The fourth is the Hilbert transform of the function (different from sinca(2t)).
Lemma 9. The following formula holds
sin(2θa(·)) sin θa(t) H (t) = , t ∈ R . (7) 2· t
Proof. Direct calculation leads to
sin(2θa(·)) H (t)
1 sin(2θa(y)) dy
π −∞ 2y t − y 1 1 1 1 = sin(2θa(y)) ( + )dy 2π −∞ t y t − y Z ∞ Z ∞ 1 1 sin(2θa(y)) 1 sin(2θa(y)) = − dy + dy 2t π −∞ 0 − y π −∞ t − y
= (−H(sin(2θa(·)))(0) + H(sin(2θa(·)))(t)) .
Utilizing Qian’s result
sin(2θa(·)) 1 1 − cos(2θa(t)) H (t) = (−(− cos(2θa(0))) − cos(2θa(t))) = 2· 2t 2t
sin θa(t)
This completes the proof of this lemma. □
Now it it time to prove Theorem 5.
Proof. Set ρ = Sinca. One hand, the left side of (4) equals to
Left = H (ρ, cos θa(2t − ·)) (t)
1 sin θa(t − y) dy = cos θa(2t − (t + y)) π −∞ t − y y 1 sin θa(y) dy = cos θa(y) π −∞ y t − y 1 sin(2θa(y)) dy sin(2θa(·)) = = H (t).
π −∞ 2y t − y 2·
By (7), we conclude that sin θa(t) Left = , t ∈ R .
One the other hand, the right side of (4) is
Right = ρ(t)H (1, cos θa(2t − ·)) (t)
sin θa(t) 1 dy sin θa(t) = cos θa(2t − (t + y)) = H cos(θa(·))(t).
t π −∞ y t
Using Qian’s result
sin θa(t) Right = , t ∈ R .
The proof of this theorem is finished. □
The following theorem offers us a more generalized result.
Theorem 10. For any l sequence c = {ck}, the function
X sin θ (t − 2kπ) X ρ(t) = ck = ckSinca(t − 2kπ), t ∈ R t − 2kπ
k∈ Z k∈ Z
satisfies the equation (4)
Proof. We first establish an important equation. By (5), i.e. (rewrite it for convenience)
sin θa(t) H(Sinca(·), cos θa(2t − ·))(t) = Sinca(t)H(1, cos θa(2t − ·))(t) = ,
we know that, for any integer k, the following equation holds
H(Sinca(·), cos θa(2(t − 2kπ) − ·))(t − 2kπ) = Sinca(t − 2kπ)H(1, cos θa(2(t − 2kπ) − ·))(t − 2kπ)
sin θa(t − 2kπ)
t − 2kπ By the definition of ρ, the linearity of BHT and the 2π-periodicity of cos θa(·), we get
H(ρ(·), cos θ(2t − ·))(t)
= ckH (Sinca(· − 2kπ), cos θa(2t − ·)) (t)
k∈ Z
X 1 dy = ck Sinca(t − y − 2kπ) cos θa(2t − (t + y)) π −∞ y
k∈ Z
X 1 dy = ck Sinca((t − 2kπ) − y) cos θa(t − y) π −∞ y
k∈ Z
X 1 dy = ck Sinca((t − 2kπ) − y) cos θa(2(t − 2kπ) − ((t − 2kπ) + y)) π −∞ y
k∈ Z
= ckH(Sinca(·), cos θa(2(t − 2kπ) − ·))(t − 2kπ).
k∈ Z
By using the important equation we established at beginning, we get
H(ρ(·), cos θ(2t − ·))(t)
= ckSinca(t − 2kπ)H(1, cos θa(2(t − 2kπ) − ·))(t − 2kπ)
k∈ Z
X 1 dy = ckSinca(t − 2kπ) cos θa (2(t − 2kπ) − ((t − 2kπ) + y) π −∞ y
k∈ Z
X 1 dy = ckSinca(t − 2kπ) cos θa(2t − (t + y)) π −∞ y
k∈ Z
= ckSinca(t − 2kπ)H(1, cos θa(t − ·))(t)
k∈ Z
= ρ(t)H(1, cos θa(2t − ·))(t).
The proof of this theorem is completed. □
Lemma 11. The following equation holds
H [(HSinca) (·) cos θa(·)] (t) = (HSinca) (t) sin θa(t), t ∈ R . (8)
1− cos θa (t)
Proof. The fact HSinca(t) = implies that we can write the equation (8) as
1 − cos θa(·) 1 − cos θa(t) H cos θa(·) (t) = sin θa(t), t ∈ R .
Noting that −H equals to identity operator, we are left to show the equivalent equation
1 − cos θa(·) 1 − cos θa(t) H sin θa(·) (t) = − cos θa(t), t ∈ R .
This can be checked by direct calculation combing with the equations (6) and (7)
1 − cos θa(·) H sin θa(·) (t)
sin(2θa(·)) = HSinca(t) − H (t)
1 − cos θa(t) sin θa(t)
1 − cos θa(t) = − cos θa(t).
This completes the proof of this lemma. □
Theorem 12. The functionρ(t) = HSinca(t) is a solution of the equation (4), that is,
H(HSinca(·), cos θa(2t − ·))(t) = HSinca(t)H(1, cos θa(2t − ·))(t), t ∈ R . (9)
Proof. By the definition of BHT and a suitable variable changing, we get
1 1 − cos θa(t − y) dy H(HSinca(·), cos θa(2t − ·))(t) = cos θa(t − y) π −∞ t − y y = H [(HSinca) (·) cos θa(·)] (t).
Using (8) and the identity H(1, cos θa(2t − ·))(t) = sin θa(t) shows that
H(HSinca(·), cos θa(2t − ·))(t) = (HSinca(t)) sin θa(t) = HSinca(t)H(1, cos θa(2t − ·))(t).
The proof of this theorem is completed. □
Theorem 13. For any l sequence d = {dk}, the function
X 1 − cos θ (t − 2kπ) X ρ(t) = dk = dk(HSinca)(t − 2kπ), t ∈ R t − 2kπ
k∈ Z k∈ Z
satisfies the equation (4)
Proof. By (9), we first claim that, for any integer k,
H(HSinca(·), cos θa(2(t − 2kπ) − ·))(t − 2kπ)
= HSinca(t − 2kπ)H(1, cos θa(2(t − 2kπ) − ·))(t − 2kπ), t ∈ R .
By the definition of ρ, the linearity of BHT and the 2π-periodicity of cos θa(·), we get
H(ρ(·), cos θ(2t − ·))(t)
= dkH ((HSinca)(· − 2kπ), cos θa(2t − ·)) (t)
k∈ Z
X 1 dy = dk (HSinca)(t − y − 2kπ) cos θa(2t − (t + y)) π −∞ y
k∈ Z
X 1 dy = dk (HSinca)((t − 2kπ) − y) cos θa(t − y) π −∞ y
k∈ Z
X 1 dy = dk (HSinca)((t − 2kπ) − y) cos θa(2(t − 2kπ) − ((t − 2kπ) + y)) π −∞ y
k∈ Z
= dkH((HSinca)(·), cos θa(2(t − 2kπ) − ·))(t − 2kπ).
k∈ Z
Therefore, we obtain
H(ρ(·), cos θ(2t − ·))(t)
= dk(HSinca)(t − 2kπ)H(1, cos θa(2(t − 2kπ) − ·))(t − 2kπ)
k∈ Z
= dk(HSinca)(t − 2kπ)H(1, cos θa(2t − ·))(t)
k∈ Z
= ρ(t)H(1, cos θa(2t − ·))(t).
This completes the proof of this theorem. □ Theorem 14. For any sequence pairsc and d in l (Z ), the function
ρ(t) = ckSinca(t − 2kπ) + dk(HSinca)(t − 2kπ)
k∈ Z k∈ Z
satisfies the equation (4).
Proof. This is a direct conclusion of previous two theorems by utilizing the linearity of bilinear Hilbert transform.
Lemma 15. For any real number a with |a| < 1, the following identity holds
X X 1
l− 1 m− 1
a a min {l, m } = .
(1 − a) (1 + a)
l=1 m=1
Proof.
l− 1 m− 1
a a min {l, m }
l=1 m=1 ∞ m ∞ l− 1
X X X X X X
l− 1 m− 1 l− 1 m − 1 m − 1 l− 1
= ( + )a a min {l, m } = a a l+ a a m
m≥ l m<l m = 1 l= 1 l= 2 m = 1
Note that
m m m+1 m m+1
X d X d a − a 1 − (m + 1)a + ma
l− 1 l
la = ( a ) = ( ) = da da 1 − a (1 − a)
l=1 l=1
and
l− 1 l− 1 l l− 1 l
X d X d a − a 1 − la + (l − 1)a
m− 1 m
ma = ( a ) = ( ) = .
da da 1 − a (1 − a)
m=1 m=1 Thus
l− 1 m− 1
a a min {l, m }
l=1 m=1 ∞ m m+1 ∞ l− 1 l
X 1 − (m + 1) a + ma X 1 − la + ( l − 1)a
m− 1 l− 1
= a + a (1− a) (1− a)
m=1 l=2 ∞ m m+1 ∞ m m+1
X 1 − (m + 1) a + ma X 1 − (m + 1) a + ma
m− 1 m
= a + a (1− a) (1− a)
m=1 m=1
m− 1 2m− 1 2m m 2m 2m+1
= a − (m + 1) a + ma + a − (m + 1) a + ma (1− a)
1 X 1 X
m− 1 m 2m 2m+1 2m− 1
= a + a − a + ma − (m + 1) a (1− a) (1− a)
m=1 m=1 ∞ ∞ ∞
1 X 1 X 1 X
m− 1 m 2m 2m+1 2m+1
= a + a − a + ma − (m + 2) a
2 2 2
(1− a) (1− a) (1− a)
m=1 m=1 m=0
1 X 2a 2 X
m− 1 m 2m 2m+1
= a + a − a − − a
2 2 2
(1− a) (1− a) (1− a)
m=1 m=1
1 1 a a 2a 2 a = + − − −
2 2 2 2 2
(1− a) 1 − a 1 − a 1 − a (1− a) (1− a) 1 − a 1 2 3 = (1 + a) + a(1 + a)− a − 2a(1 + a)(1− a)− 2a (1− a) (1 + a) 1 2 2 3 3 1 = 1 + a + a + a − a − 2a + 2 a − 2a = .
(1− a) (1 + a) (1− a) (1 + a)
The following theorem offers us the result of orthogonality
1− a
Theorem 16. The system { Sinca( · − nπ) :n ∈ Z } is an orthonormal system.
π(1+ a)
Proof. We know that X sinlπt
2 l− 1
Sinca (t) = pa (πt)Sinc(πt )) = (1 − a ) a .
πt
l= 1
The Fourier transform of Sinca is )
∧ 2 l− 1
(Sinca ) (ξ) = (1 − a ) a √ χ[− lπ,lπ](ξ).
We therefore get that, for any integer n, Z r r 1 − a 1 − a Sinca (t) Sinc (a (t− n)dt −∞ 1 + a 1++ a Z ∞ ∞ ∞ 2 2 X X 1 − a (1− a ) l− 1 m− 1 inξ = a a χ[− lπ,lπ](ξ)χ[− mπ,mπ](ξ)e dξ 1 + a 2π −∞
l=1 m=1
∞ ∞ Z 3 X X (1− a) (1 + a) l− 1 m− 1 inξ = a a e dξ ,0= δn,0. 2π [− lπ,lπ]∩ [− mπ,mπ]
l=1 m=1
3 l− 1 m− 1
= (1 − a) (1 + a) a a min {l, m }δn ,0
l=1 m=1
= δn,0.
We therefore get that, for any integer n,
The Bilinear Hilbert Transform (BHT) offers a powerful framework for enhancing decision-making processes in environments where nonlinear signals
The BHT’s ability to decompose nonlinear signals allows decision-makers to uncover hidden patterns in data, leading to better predictions and responses. For example, in predictive maintenance, the BHT helps detect early signs of equipment failure by analyzing complex signal behaviors. Similarly, in finance, it reveals trends and volatility in stock prices, improving investment strategies.
The Bedrosian identity plays a key role in combining different signal components while maintaining their integrity. This is especially useful in models involving multiple interacting variables, such as economic forecasting, where inflation, interest rates, and currency values need to be harmonized for accurate predictions.
Generalized Sinc functions enhance the precision of signal decomposition in BHT, improving the accuracy of models in fields such as biomedical signal processing or industrial control. These functions allow for more reliable interpretations of sensor data, leading to optimized processes and improved decision outcomes.
The BHT’s real-time processing capability makes it ideal for applications requiring immediate action, such as telecommunications or supply chain management. Its adaptability extends to multiple fields, including healthcare, finance, and environmental monitoring, where it aids in optimizing resource use and improving response times.
By integrating BHT with machine learning, decision-making processes can become more automated and adaptive. For example, in autonomous systems, BHT enhances real-time sensor data processing, improving machine learning models’ performance. As industries move toward Industry 5.0, this framework provides a way to combine human insight with machine efficiency, particularly in sustainable practices.
With the Hilbert transform, we define the bilinear Hilbert transform. We have demonstrated some properties of BHT.The generalized Sinc function is a special solution of Hρ, ↼ cos θ a↼2 t − ·⇒⇒ ⇐ t↽ / ρ↼ t↽ H↼1 ,cosθa↼2 t − ·⇒⇒ ⇐ t↽ .After that, we consider some the nonlinear Sinc function, which is also the
1− a
solution of this equation.Last,The nonlinear Sinc function system { Sinca(· − nπ↽ . n ∈ Z } is an
π(1+ a)
orthonormal system.
Additionally, we have introduced a framework that extends BHT’s utility into analytical decision-making applications.This framework leverages the decomposition properties of the BHT to support real-time decision-making, precision modeling, and integration with AI technologies across multiple fields. The BHT enables more accurate and responsive decision-making, particularly in environments that require the handling of complex nonlinear data and real-time signal processing. Moreover, by applying this framework to various industries such as finance, healthcare, and industrial optimization, we highlight the versatility of the BHT in improving efficiency and predictive accuracy. As industries continue to adopt AI-driven and real-time decision-making tools, the BHT framework offers an adaptable and scalable solution for future technological advancements, aligning with the principles of Industry 5.0 and sustainable practices.
The authors would like to express their sincere gratitude to the editors and anonymous reviewers for their invaluable comments and constructive feedback, which significantly contributed to the enhancement of this paper.
All authors have equal contributions to this project. All authors have read and agreed to the published version of the manuscript.
The authors declare that no external funding or support was received for the research presented in this paper, including administrative, technical, or in-kind contributions. Data Availability All data supporting the reported findings in this research paper are provided within the manuscript.
The authors declare that there is no conflict of interest concerning the reported research findings. Funders played no role in the study’s design, in the collection, analysis, or interpretation of the data, in the writing of the manuscript, or in the decision to publish the results.