] Time-Domain Synthesis Problem Involving I-Function of Two Variables

Time-Domain Synthesis Problem Involving I-Function of Two Variables

S. S. SRIVASTAVA

and ANSHU SINGH

1,2

Department of Mathematics, Govt. P. G. College, Shahdol, M. P., INDIA.

(Received on: February 17, 2013) ABSTRACT

The object of this paper is to evaluate an integral involving Bessel polynomial and I-function of two variables and employ it to obtain a particular solution of the general solution derived in this paper of the classical problem known as the, "time-domain synthesis problem", occurring in electrical network theory.

Keywords:.

1. INTRODUCTION

The I–function of two variables

introduced by Sharma & Mishra

, will be defined and represented as follows:

I [ ] = I [ |

: [(c

; γ

)

n

], [(c

; γ

) n

p

]; [(e

; E

)

n

], [(e

; E

) n

p

] : [(d

; δ

)

m

], [(d

; δ

) m

q

]; [(f

; F

)

m

], [(f

; F

) m

q

]

= ∫ ∫ φ

(ξ, η) θ

(ξ) θ

(η)x

y

dξ dη, (1) where

Γ ( 1 − a

+ α

ξ + A

η) φ

( ξ, η ) =

p_i, q_i: r : p_i´, q_i´: _{r´ :}p_i´´, q_{i´: r´´} [(b_ji; βji, B_ji)_1, q_i]

]

1 .

(2πω) ² L₁ L₂

Π n j = 1

r pi q_i

Σ [ Π Γ(a

- α

ξ − Α

η) Π Γ (1 − b

+ β

ξ + B

η)

i = 1 ^{j = n+ 1 j = 1} x

y

0, n : m

, n

m

, n

[(a

; α

, A

)

], [(a

; α

, A

)

p

]

Γ ( d

− δ

ξ ) Π Γ (1 − c

+ γ

ξ )

θ

(ξ) =

Γ (f

− F

η) Π Γ (1 − e

+ E

η) θ

(η) =

x and y are not equal to zero, and an empty product is interpreted as unity p

, p

, p

, q

, q

, q

, n, n

, n

, n

and m

are non negative integers such that p

≥ n ≥ 0, p

≥ n

≥ 0, p

≥ n

≥ 0, q

>0, q

>0, q

> 0, (i = 1, …, r; i´

= 1, …, r´; i´´ = 1, …, r´´; k = 1, 2) also all the A’s, α’s, B’s, β’s, γ’s, δ’s, E’s and F’s are assumed to be positive quantities for standardization purpose; the definition of I- function of two variables given above will however, have a meaning even if some of these quantities are zero. The contour L

is in the ξ−plane and runs from – ω∞ to + ω∞,

with loops, if necessary, to ensure that the poles of Γ( d

δ

ξ ) (j = 1, ..., m

) lie to the right, and the poles of Γ (1 − c

 γ

ξ) (j = 1, ..., n

), Γ (1− a

+ α

ξ + A

η) (j = 1, ..., n) to the left of the contour.

The contour L

is in the η− plane and runs from – ω∞ to + ω∞ , with loops, if necessary, to ensure that the poles of Γ (f

- F

η) (j=1,..., n

) lie to the right, and the poles of Γ (1− e

 E

η) (j = 1, ..., m

), Γ ( 1− a

+ α

ξ + A

η ) (j = 1, ..., n) to the left of the contour.

R = Σ α

+ Σ γ

− Σ β

− Σ δ

< 0, S = Σ A

+ Σ E

− Σ B

− Σ F

< 0,

U = Σ α

− Σ β

+ Σ δ

− Σ δ

+ Σ γ

− Σ γ

> 0, (2) Π

m

n

j = 1 j = 1

r ´ q_i´ p_i´

Σ [ Π Γ (1 − d

+ δ

ξ) Π Γ ( c

− γ

ξ) ]

i´ = 1 j = m

1 + 1 ^{j = n}1 + 1

m Π

n

j = 1 j = 1

r ´´ q_i´´ p_i´´

Σ [ Π Γ (1 − f

+ F

η) Π Γ (e

− E

η) ]

i´´ = 1 j = m

2 + 1 ^{j = n}2 + 1

q

j = 1 q

j =1 p

j = 1 p

j = 1 p

j = 1 p

j = 1 q

j =1

q

j = 1 p

j = n +

q

m

n

p_i´

j = n₁ + 1 q_i´

j = m₁ + 1

V = − Σ A

− Σ B

− Σ F

− Σ F

+ Σ E

− Σ E

> 0, (3)

and | arg x | < ½ U π , | arg y | < ½ V π . (4)

2. RESULT REQUIRED

The following results are required in our present investigation:

Bessel polynomial

y

x; a, b

∑





=

F

 x/b. (5) Orthogonality property of Bessel polynomials

x_^∞
e^భ౮y_x; a, 1dx 

^౤ ! 
π

Γ
 Γ 
  π
δ_, ,

(6) where Re a < l − m − n.

3. INTEGRAL

x_^{∞ σ}e^{ భ౮}y_x; a, 1dx ^Γ σΓ  σ

Γ  σ

, (7) where Re

1, 2, 3, ….

The integral (7) can easily established by expressing the Bessel Polynomial in the integrand as its series representation (5), interchainging the order of integration and summation, evaluating the resulting integral with the help of

, using Gauss's theorem

.

 x

e

y

x; a, 1I

 dx

∞

= I

,:;^′,^′:^′;^′′,^′′:´´

,
; ,
;_,
_

|

…...,……:σ
,λ, σ
,λ,……:……,…….

……,…...:…….,σ,λ:……,…….

, (8) provided that Re σ < 0, Re (a − σ) < 2,

  1, 2, 3, …. and U  0,   0, |argξ| "

Uπ, where U and V are given in (2) and (3).

Proof:

To establish (8), expressing the I- function in the integrand as a Mellin-Barnes type integral (1) and interchainging the order of integrations which is justified due to the absolute convergence of the integrals involved in the process, evaluating the inner- integral with the help of (7) and using the definition of I-function, we arrive at (8).

4. SOLUTION OF THE TIME-DOMAIN SYNTHESIS PROBLEM OF

SIGNALS

The classical time-domain synthesis problem occurring in electrical network theory is stated as follows

.

Given an electrical signal described by a real valued conventional function f(t) on 0 < t < ∞ , construct an electrical network consisting of finite number of components

j = n + 1

q

m

j =1

^qi´´

j = m₂ + 1

n

j =1 p_i´´

j = n₂ + 1

R, C and I which are all fixed, linear and positive, such that the output of f

(t), resulting from a delta-function δ(t) approximates f(t) on 0 < t < ∞ in some sense.

In order to obtain a solution of this problem, we expand the function f(t) into a convergent series:

f(t) = ∑

ψ

t (9) of real-valued function ψ

t. Let every partial sum

f

(t) = ∑

ψ

t , N = 0, 1, 2, … (10)

possess the two properties:

(i) f

(t) = 0, for − ∞ < t < 0.

(ii) The laplace transform F

(s) of f

(t) is a rational function having a zero at s = ∞ and all its poles in the left-half s-plane, except possibly for a simple pole at the origin.

After choosing n in (10) sufficiently large to satisfy whatever approximation criterion is being used, an orthonormal series expansion may be employed. The Bessel polynomial transformation and (5) yields an immediate solution in the following form:

f(t) = ∑

c

t

e

y

t; a, 1 , where

c  1 ^Γ
 Γ 
  π

! 
π
ftt^{∞ 
}e^మ౪భy t; a, 1dt



, (11)

where Re a < 1 − 2n.

This case is an example of the use of hypergeometric integral in an orthonormal series expansion.

5. PARTICULAR SOLUTION OF THE PROBLEM

The particular solution of the problem in terms of I-function is as follows:

f(t) =

π

∑ 1

!


∞


t

e

y

t; a, 1I

,:;^′,^′:^′;^′′,^′′:´´

,
; ,
;_,
_

|

ξ

…...,……:σ
,λ, σ
,λ:……,…….

……,…...:……..,σ,λ:……,…….

, (12)

provided that Re σ < 0, Re (a − σ) < 2,   1, 2, 3, …. and U  0,   0, |argξ| "

Uπ, where U and V are given in (1.3.4) and (1.3.5).

Proof:

Let us consider f(t) = t

e

I

ξ^λ



= ∑

c

t

e

y

t; a, 1 . (13)

Equation (13) is valid, since f(t) is continous and of bounded variation in the open interval (0, ∞).

Multiplying both sides of (13) by t

e

y

t; a, 1 and integrating with respect to t from 0 to ∞ , we get

% t

e

y

t; a, 1 I

∞

dt

= ∑

c

% t

e

y

t; a, 1 dt.

Now using (8) and (6), we get c

!π

I

,:;^′,^′:^′;^′′,^′′:´´

,
; ,
;_,
_

|

ξ

…...,……:σ,λ, σ,λ,……:……,…….

……,…...:……..,σ,λ:……,…….

, (14)

From (13) and (14), the solution (12) follows immediately.

REFERENCES

1. Sharma C. K. and Mishra, P. L.: On the I-function of two variables and its certain properties, ACI, 17, 1-4 (1991).

2. Exton, H. On Orthogonal of Bessel Polynomials. Rev. Mat. Univ. Parma (4) 12, 213-215(1986).

3. Erdelyi, A.: Table of integral transform, Vol. I, McGraw-Hill, New York, (1954).

4. Erdelyi, A. et. al. Higher Transcendental

Functions, Vol. 1 McGraw-Hill, New

York, (1953).

5. Exton, H. Handbook of Hypergeometric Integrals, Ellis Horwood Ltd., Chiches- ter, (1978).

6. Srivastava, S. S.: Studies on Some

Contributions of Special Functions in

Various Discipline, Minor Research

Project Submitted to MPCOST, Bhopal,

(2008).