Int. Journal of Math. Analysis, Vol. 2, 2008, no. 20, 977 - 986
Analyticity and Operation Transform on
Generalized Fractional Hartley Transform
*P. K. Sontakke and A. S. Gudadhe
* H.V.P.M’s College of Engineering and Technology, Amravati-444601 (M.S), India Govt. Vidarbha Institute of science and Humanities, Amravati-444604 (M.S), India
Abstract
In this paper we have discussed the analyticity theorem and inversion formula for the generalized fractional Hartley transform and using that we have proved uniqueness theorem. Also we have discussed fractional Hartley transform of selected functions and obtained operation transform formulae for this transform.
Keywords: Fractional Fourier transform, Hartley transform, Testing function space, Generalized function
1. Introduction: Now a days, fractional integral transforms play an important role in signal processing, image reconstruction, pattern recognition, accostic signal processing [8], [9]. Fourier analysis is one of the most frequently used tools in signal processing and many other scientific disciplines. Beside the Fourier transform for time frequency representation of signal, Wigner distribution, the ambiguity function, the short time fourier transform and the spectrogram are often used. e.g in speech processing radar, quantum physics. In mathematics literature a generalization of the Fourier transform to the fractional Fourier transform is as under. Namias [10] introduced the concept of Fourier transform of fractional order, which depends on a continuous parameterα . The generalization of ordinary fourier transform and it’s properties were discussed in Cariolaro et.al [3] Zayed [1] Dragoman [4] etc. Fractional fourier transform is further generalized to the integral with respect to new measure dρ and a new generalized integral transform was obtained by Zayed [1]. Bhosale and chaudhary [2] had extended fractional fourier transform to the distribution of compact support. The fractional Fourier transform with α = 1 corresponds to the classical Fourier transform and fraction al Fourier transform with α = 0 corresponds to the identity operator. In [5] other integral transforms of Fourier class, that is cosine transform, sine 978 P. K. Sontakke and A. S. Gudadhe
transform and Hartley transform, are also generalized to the corresponding fractional integral transform and studied by different mathematicians. The Hartley transform is an attractive alternative and convenient real replacement for the well known complex Fourier transform. Hartley transform is getting greater importance in several applications. In [6] Bracewell suggests that the Hartley transform is as fast or faster than the Fourier transform and can be used as a replacement for the Fourier transform. Using the eigenvalue function, as used in fractional Fourier transform, different integral transform in Fourier class are generalized to fractional transform by Pei [5]. Hartley transform is also generalized to fractional Hartley transform by −t 2 2 him. He had shown that for all non negative integer m, e H m (t) is the eigen function of the Hartley transform and had given the formula for fractional Hartley transform as, ∞ Hα f (t) (s) = f (t)K (t,s)dt, {} ∫ α −∞ where s 2 t 2 1− i cot φ i cot φ i cot φ 1 K (t, s) = e 2 e 2 [](1− ie iφ )cas(cscφ.st) + (1+ ie iφ )cas(− cscφ.st) . α 2π 2
In this paper first we define generalized fractional Hartley transform in section 2 and prove its analyticity. The function f (t) is then recovered from H α {}f (t) (s) by means of the inversion formula in section 3. Using inversion formula we have proved that fractional Hartley transformable generalized function having the same strip of definition and the same transform must be identical, which is named as uniqueness theorem in section 3. Fractional Hartley transforms of some selected functions and operation transform formulae are obtained in section 4 and section 5 respectively. Lastly conclusion is given in section 6.
2 Analyticity Of Fractional Hartley Transform : 2.1 The Generalized Fractional Hartley Transform on E' :
n n Let Sα = {t ∈ R , t ≤ a, a > 0 }. Let s ∈ R and t ∈ Sα and 0 ≤ α ≤ 1, if s2 t 2 1− icotφ i cotφ i cotφ 1 K (t,s) = e 2 e 2 [](1− ieiφ )cas(cscφ.st) + (1+ ieiφ )cas(−cscφ.st) ., α 2π 2 απ where φ = , 2 n k then Kα (t, s) ∈ E(R ) if γ k {Kα (t, s)}= sup D Kα (t, s) < ∞ . −∞ α H {}f (t) (s) = f (t), Kα (t, s) , (2.1.1) where E ' (Rn ) is the dual space of the testing function space. 2.2 Analyticity Theorem : Theorem : Let f ∈ E ' (R n ) and its fractional Hartley transform is defined as (2.1.1). Then H α {f (t)}(s) is analytic on R n if the α Supp. f ⊂ S a = {t : t ∈ R, t ≤ a, a > 0} then H {f (t)}(s) is differentiable and k α k Ds []H {}f (t) (s) = f (t), D Kα (t, s) . n Proof : Let s : (s1 , s2 ...... sn ) ∈ R , we first prove that ∂ α ∂ . []H {}f (t) (s) = f (t), K α (t, s) ∂s j ∂s j We prove the result for k = 1, the general result follows by induction. For some 1 s j ≠ 0 , choose two concentric circles C and C with centre at s j and radii r and r1 respectively, such that 0.< r < r1 < s j Let Δ s j be a increment, satisfying 0.< Δs j < r Consider, H α s s H α (s ) ()j + Δ j − j ∂ , 2.2.1) − f (t ), K α (t, s) = f (t ), ΨΔ s j (t ) Δs j ∂s j 1 ∂ where, ΨΔs j (t) = []Kα (t, s1 , s2 .....s j + Δs j .....sn ) − Kα (t, s) − Kα (t, s) Δs j ∂s j For any fixed t ∈ R n , D1Kα (t, s) = Cs,φ {(−cscφ.s)K α (t, s) + it cotφ Kα (t, s)}, s2 1− icotφ i cotφ where,C = e 2 and s,φ 2π t 2 i cotφ 2 iφ K α (t, s) = e [sin(cscφ.st) + ie cos(cscφ.st)]. 980 P. K. Sontakke and A. S. Gudadhe Since for any fixed t ∈ R n fixed integer k and α ranging from 0 to 1, 1 D1 Kα (t, s) is analytic inside and on C , we have by Cauchy integral formula, Δs M (t, s ) D ΨΔ s (t) = j dz , 1 j ∫ 2 2π i C 1 (z − s j − Δs j )( z − s j ) where, s =(s1...... sj−1,z,sj+1...... sn ). But for all z ∈ C1 and t restricted to a compact subset of Rn , 0 ≤ α ≤ 1, M (t, s) = D1Kα (t, s) is bounded by a constant K . Therefore we have, K D1ΨΔs j (t) = Δs j . (r1 − r)r1 Thus as Δs j → 0, D1 ΨΔs j (t) tends to zero uniformly on the compact n n subsets of R therefore it follows that ΨΔs j (t) converges in E(R ) to zero. Since H α ∈ E ' we conclude that (2.2.1) also tends to zero therefore H α (s ) is α differentiable with respect to S j . But this is true for all j = 1,2...... n .Hence H (s) is analytic and k α k Ds []H {}f (t) (s) = f (t), D Kα (t, s) . 3 Inverse And Uniqueness Theorem : 3.1 Inverse Of Fractional Hartley Transform: Generalized fractional Hartley transform as defined in (2.1.1) can be written as H α {}f (t) (s) s2 ∞ t2 iφ iφ 1−icotφ i cotφ i cotφ ⎡(1−ie )cas(cscφ.st)+(1+ie )cas(−cscφ.st)⎤ = e 2 ∫e 2 ⎢ ⎥ f (t)dt . 2π −∞ ⎣ 2 ⎦ s 2 2π −i cot φ ∴ 2 e 2 H α {}f (t) (s) 1 − i cot φ t 2 t 2 ⎧ i cot φ ⎫ ⎧ i cot φ ⎫ i ⎪ ⎪ i ⎪ ⎪ = (1 − ie φ )H ⎨e 2 f (t )⎬(csc φ .s) + (1 + ie φ )H ⎨e 2 f (−t)⎬(csc φ .s), ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ Analyticity and operation transform 981 v Put (cscφ.s) = v ∴s = = vsinφ cscφ Taking Hartley inverse of both side which is same as Hartley transform, 2 2 ∞ 1 −iv sin φ.cotφ ∴ 2 ∫ e 2 .H α {}f (t) (vsinφ).casvt dt −∞ 1− icotφ t 2 t 2 ⎧ i ( ) cot φ ⎫ ⎧ i ( ) cot φ ⎫ i ⎪ ⎪ i ⎪ ⎪ = (1 − ie φ )⎨e 2 f (t )⎬ + (1 + ie φ )⎨e 2 f (− t )⎬ . ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ If f (t) is even then f (−t) = f (t) then, we get ∞ t 2 i cotφ 2 ∫ g(v)cas vt dv = 2 f (t)e 2 , −∞ 2 2 1 −iv sin φ.cot φ where g(v) = e 2 .H α {}f (t) (v sin φ) 1 − i cot φ t 2 ∞ −i cotφ ∴ f (t) = e 2 ∫ g(v)cas vt dv , when f (t) is even. −∞ Putting g(v) and solving, we get ∞ f (t) = ∫ K(t, s) H α {}f (t) (s) ds , −∞ t 2 s2 cscφ −i cotφ −i cotφ where K(t,s) = e 2 e 2 cas(cscφ.st) . 1− icotφ Now if f (t ) is odd then f (−t) = − f (t ) . ∞ t 2 t 2 i cotφ i cotφ ∴2 ∫ g(v)cas vt dv =(1− ieiφ )e 2 f (t) + (1+ ieiφ )e 2 (− f (t)) −∞ t 2 ∞ −i(φ + cotφ ) ∴ f (t) = i e 2 ∫ g(v)cas vt dv . −∞ Again putting g(v) and solving, we get 982 P. K. Sontakke and A. S. Gudadhe ∞ f (t) = ∫ K(t, s) H α {}f (t) (s) ds −∞ t 2 π s2 cscφ −i cotφ −i(φ − ) −i cotφ where, K(t,s) = e 2 e 2 e 2 cas(cscφ.st) . 1− i cotφ 3.2 Uniqueness Theorem : Theorem : If H α {}f (t) (s) and H α {g(t)}(s) are fractional Hartley transform of f (t) and g(t ) respectively for o ≤ α ≤ 1 and supp. f ⊂ Sα : Sα = {t : t ∈ R, t ≤ a} and supp. g ⊂ Sα : Sα = {t : t ∈ R, t ≤ a} α α ' if H {}f (t) (s) ={H g(t)}(s) then f = g in the sense of equality of D (I ) Proof : By inversion theorem, N 1 α α f − g = lim Kα (t, s)[]H {}f (t) (s) − H {}g(t) (s) ds . N →∞ ∫ 2π −N ' Thus f = g in D (I ) . 4 Fractional Hartley Transform Of Selected Functions : Fractional Hartley Transform of selected functions are tabulated as follows. Analyticity and operation transform 983 TABLE 1 Sr. Signal Fractional Hartley transform No i − s2 tanφ 1. f (t) = 1 H α {}1 (s) = (tanφ − i) e 2 H α {}δ (t − a) = f (t) = δ (t − a) ⎛ s2 +a 2 ⎞ i⎜ ⎟ cot φ 2. 1− i cotφ ⎜ 2 ⎟ e ⎝ ⎠ []cos(cscφ.sa) − ie iφ sin(cscφ.sa) 2π s2 1− i cotφ i cot φ 3. f (t) = δ (t) H α {}δ (t) = e 2 2π i − (s2 +1) tanφ 4. f (t) = sin t H α {}sint (s) = eiφ (tanφ −i) e 2 sin(s.secφ) i − (s2 +1) tan φ 5. f (t) = cost H α {}cost (s) = (tanφ − i) e 2 cos(s.secφ) 5 Operation Transform Formulae : In this section we prove some operation transform formulae for fractional Hartley transform, for which following two lemmas can be easily proved. 5.1 Lemma : Fractional Hartley transform given in section 2.1, can also be expressed as H α {}f (t) (s) = s2 ∞ t 2 1− i cotφ i cotφ i cotφ e 2 ∫ e 2 []cos(cscφ.st) − ieiφ sin(cscφ.st) f (t)dt . 2π −∞ 5.2 Lemma : H α {}f (−t) (s) = s2 ∞ t2 1− icotφ i cotφ i cotφ e 2 ∫e 2 []cos(cscφ.st) + ieiφ sin(cscφ.st) f (t)dt . 2π −∞ 984 P. K. Sontakke and A. S. Gudadhe 5.3 Lemma : s2 ∞ t2 1− icotφ i cotφ i cotφ e 2 ∫e 2 []sin(cscφ.st) + ieiφ cos(cscφ.st) f (t)dt 2π −∞ =i cosφ Hα {}f (t) −sinφ Hα {f (−t)}. 5.4 Formula : If FrHT{ f (t)}(s) = H α {f (t)}(s) then α ⎧ d ⎫ α α α H ⎨ f (t)⎬(s) = i cotφ[]sH {}f (t) − H {}tf (t) − sH {}f (−t) . ⎩dt ⎭ Proof : Since H α {}f (t) (s) s2 ∞ t 2 1− icotφ i cotφ i cotφ = e 2 ∫e 2 []cos(cscφ.st) − ieiφ sin(cscφ.st) f (t)dt 2π −∞ ∞ t 2 d i cot φ ∴ H α { f (t)} = Csφ ∫ e 2 []cos(csc φ.st) − ie iφ sin(csc φ.st) f ' (t)dt dt −∞ Solving we get =icotφ.[sHα{f (t)}− Hα{t f (t)}]−sHα{f (−t)}. 5.5 Formula: If FrHT{ f (t)}(s) = H α {f (t)}(s) then α d H {}t f (t) (s) = sH α { f (t)}+ i tanφ.sH α { f (−t)}+ i tanφ H α { f (t)} dt α ' α α α Proof : QH {f (t)}(s) =icotφ.[sH {f (t)}− H {t f (t)}]−sH {f (−t)} α d ∴H {t f (t)}(s) = sH α { f (t)}+ i tanφ.sH α { f (−t)}+ i tanφ H α { f (t)}. dt 5.6 Formula: H α {}f (t) (s + c) 2 (2sc+c2 ) (2sc+c ) i cotφ i cotφ = e 2 H α {}cos(cscφ.ct) f (t) − e 2 i cosφ H α {}sin(csc φ.ct) f (t) + sinφ H α {}sin(cscφ.ct) f (−t) Proof : we know that, H α {f (t)}(s + c) (s+c)2 ∞ t 2 1− i cotφ i cotφ i cotφ = e 2 ∫ e 2 {}cos(cscφ.(s + c)t) − ieiφ sin(cscφ.(s + c)t) f (t)dt 2π −∞ (2sc+c2 ) (2sc+c2 ) i cotφ i cotφ = e 2 H α {}cos(cscφ.ct) f (t) − e 2 i cosφ H α {}sin(cscφ.ct) f (t) + sinφ H α {}sin(cscφ.ct) f (−t) . d 5.7 Formula : H α {}f (t) (s) = 2si cotφ H α { f (t)}− sH α { f (−t)} ds Proof : Consider, ∞ s 2 t 2 d d 1 − i cotφ i cot φ i cot φ ∴ H α {}f (t) (s) = ∫ e 2 e 2 {}cos(cscφ.st) − ie iφ sin(cscφ.st) f (t)dt ds ds 2π −∞ Analyticity and operation transform 985 s s = −i .cosφ H α { f (t)}− sinφ H α { f (−t)}+ is cotφ H α { f (t)} sinφ sinφ = isH α { f (t)}{cotφ + cotφ }− sH α { f (−t)} = 2si cotφ H α { f (t)}− sH α { f (−t)}. 5.8 Formula : Let a be a fixed real number.The mapping ∗ ∗ f (t) → f (t − a) is continuous linear mapping on a S a to S a and a2 i cotφ H α {}f (t − a) (s) = e 2 []cos(cscφ.sa) − ie −iφ sin(cscφ.sa) .H α {}eiat cotφ f (t) , Where a is real number. Proof : By the definition of fractional Hartley transform, H α {}f (t − a) (s) s2 ∞ t 2 1− i cotφ i cotφ i cotφ = e 2 ∫e 2 []cos(cscφ.st) − ieiφ sin(cscφ.st) f (t − a)dt . 2π −∞ Putting t − a = x and solving we get a2 i cotφ H α {}f (t − a) (s) = e 2 []cos(cscφ.sa) − ie −iφ sin(cscφ.sa) .H α {}eiat cotφ f (t) . 5.9 Formula: If FrHT{ f (t)}(s) = H α {f (t)}(s) then a 2 −1 i a 2 −1 ⎧ i( )t 2 cot φ ⎫ ( )s 2 cot φ α ⎪ ⎪ 1 2 α ⎛ s ⎞ H ⎨e 2 . f (at)⎬(s) = e 2 a H {}f (t) ⎜ ⎟ . ⎩⎪ ⎭⎪ a ⎝ a ⎠ Proof : By the definition of fractional Hartley transform, a2 −1 ⎧ i( )t 2 cotφ ⎫ α ⎪ ⎪ H ⎨e 2 . f (at)⎬(s) ⎩⎪ ⎭⎪ s2 ∞ t2 a2 −1 1− icotφ i cotφ i cotφ i( )t 2 cotφ = e 2 ∫e 2 []cos(cscφ.st) − ieiφ sin(cscφ.st) e 2 f (at)dt . 2π −∞ Putting at =T and solving we get a 2 −1 i a 2 −1 ⎧ i( )t 2 cot φ ⎫ ( )s 2 cot φ α ⎪ ⎪ 1 2 α ⎛ s ⎞ H ⎨e 2 . f (at)⎬(s) = e 2 a H {}f (t) ⎜ ⎟ . ⎩⎪ ⎭⎪ a ⎝ a ⎠ 986 P. K. Sontakke and A. S. 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