# Asymptotic Nature Of Mellin Transform In Electrical Field

DOI : 10.17577/IJERTV2IS4676

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#### Asymptotic Nature Of Mellin Transform In Electrical Field

Dr. N. A. Patil Vijaya N. Patil

Department Of Applied Sciences and Humanities (Mathematics), Sant Gadge Baba Amravati, University

Abstract

Integral transform techniques widely used for solving linear differential equations in mathematics, especially in Engineering. It is commonly used to solve electrical circuit and systems problems. In this paper we show the relation between the Mellin

2. Generalities on Mellin transform

We recall first the definition of Mellin transform.

Let f(t)denotes a complex-valued function of the real, positive variable t. The Mellin transform for f(t) will be denoted by M [ f ; s]and defined by

transform of the derivative of a function is not simple nature as that of Laplace transform. A transform table will enable to obtain the solution by a method similar

M [ f ; s] F (s)

0

f (t) ts 1 dt

(1)

to the method used in Laplace transforms theory. Also we focus on some properties of Mellin transform and may be used to solve the Euler-Cauchy differential

n d i y(t)

i

i

Wheres is complex. The basic properties of the Mellin transform follows immediately from those of the Laplace transform since these transforms are intimately connected.

equation Ait

i0

dti

x(t) which will simplify

The integral (1) defines the Mellin transform in a vertical strip in the s plane whose boundaries are

the solution of such an equation. The applications will illustrate instrumentation and Network analysis

determined by the analytic structure of f(t) as t tend to 0+ and t tends to +.If we suppose that

problems.

Keywords

0 (t )

f (t)

0 (tb )

as t 0

as t

Mellin transform, Differential Properties, Euler- Cauchy differential equation, Instrumentation, Network analysis.

Where 0 and a b , then the integral (1) converges absolutely and defines an analytic function

1. Introduction

in the strip

a Re(s) b

The first occurrence of the transform is found in a memoir by Riemann in which he used it to study the famous zeta function. However it is the Finnish mathematicians R.H. Mellin (1854-1933) who was the first to give a systematic formulation of the

transformation and its inverse. The Mellin transform is

And the inversion integral formula for (1) follows directly from the corresponding inversion formula for the bilateral Laplace transform. Thus we find the result

1

1

c j

f (t) ts M [ f ; s]ds a c b

a basic tool for analyzing the behaviour of many

2 j c j

(2)

important functions in mathematics and mathematical physics.

Outlines of the paper-

In first section we discuss some generalities on Mellin transform. In Second section we shows how the relation between the Mellin transform of the derivative of a function is not simple nature as that of Laplace transform. In third section we see the application of Mellin to Euler-Cauchy differential equation and in fourth section applications will illustrate Instrumentation problem and then Network analysis problem. Lastly we conclude.

Which is valid at all points t 0 where f(t) is continuous.

1. Nature of Mellin transforms

In common with other integral transforms, the Mellin transform possesses a series of simple translational properties which greatly facilitate the evaluation of transforms of more involved functions.

All these results can be obtained by straight forward manipulation of the definition (1).(See Table-I)

 Table-I SN Original function Mellin Transform 1 f (t),t 0 F (s) f (t) ts 1 dt 0 2 f (t) F (s) 3 Af1 (t) Bf2 (t) AF1 (s) BF2 (s) 4 f (at), a 0 as F (s) 5 f 1 t F (s) 6 f (ta ), a real 0 a 1 F (a1s) 7 (t)a f (t), a 0 F(s a)
 Table-I SN Original function Mellin Transform 1 f (t),t 0 F (s) f (t) ts 1 dt 0 2 f (t) F (s) 3 Af1 (t) Bf2 (t) AF1 (s) BF2 (s) 4 f (at), a 0 as F (s) 5 f 1 t F (s) 6 f (ta ), a real 0 a 1 F (a1s) 7 (t)a f (t), a 0 F(s a)

d n f (t)

d n1 f (t)

0

0

dtn

t s1dt

dtn1

t s1

0

d n1 f (t)

(s 1)

0

dtn1

t s2dt

(5)

So that, if we assumef to be of such a nature that the square bracket vanishes, we have the relation

M [ f n (t); s] (s 1)F n1 (s 1)

(a Re(s 1) b)

(6)

In similar manner we obtain by induction, (Appling this rule until we reach F0(s))the Mellin transform of n

the derivative of f (t) is given by

The Mellin transform of derivative of f(t) can be found by integration by parts to yield

M [ f n (t); s] (1)n s

(s 1)

F (s n)

M [ f (t); s] F (s) f (t) ts 1 dt

0

(a Re(s n) b)

Where, (7)

s 1

(s 1) 1

s (s n)(s n 1)……(s 1)

t f (t)

(s 1) f (t) t dt

0 0

(3)

(s 1)

If f(t) satisfies (1) ,we have

This formula gives the Mellin transform of the

limts1 f (t) 0 for

t0

limts1 f (t) 0 for

t

And hence (3) becomes,

Re(s) a 1 Re(s) b 1

derivative in terms of the Mellin transform of the function itself.

A similar relation is the Mellin transform of the expression:

M [ f (t); s] (s 1)F(s 1) (a Re(s 1) b)

n dn f (t) s dn f (t)

n 1

n 1

M t n t

n dt

M [ f (t); s] F (s) f (t) ts 1 dt

dt 0 dt

0

t sn1

dn1 f (t)

(s n 1)

t sn1

dn1 f (t)

dt

s 1

s 2

dtn1

dtn1

t f (t)

(s 1) f (t) t dt

0 0

(4)

0

0

(8)

And solving same above we get,

M [ f (t); s] (s 1)(s 2)F (s 2)

(a Re(s 2) b)

Where n=0,1,2..

where the function is such that the quantity in the bracket again goes to zero. Repeating the process, we get:

n d n f (t) n

M t

dtn

1

s s 1s 2

But the relation between the Mellin transform of the derivative of a function is not simple nature as that of Laplace transform. For e.g.

….s n 1 F s

(9)

Where F(s) is the Mellin transform of the function f(t).Other simple relations which can be derived in the same way are:

d n

du

1 c j

t s

M t f (t) (1)n sn F (s)

g(u) F (s) ds

dt

0 u 2 j c j

u

And

(10)

f t g(u) du

d

M

M

n

n

t

f (t) (1)n (s n)n F (s)

u

u

0

u (..By(1) and(2)) (14)

dt

 Table-III SN Original function Mellin Transform 1 d n t dt f (t) (1)n sn F(s) 2 d n dt t f (t) (1)n (s n)n F(s) 3 d n f (t) dt n (1)n s F (s n) (s 1) 4 d n f (t) tn dtn 1n (s) F s n 5 f (t)g(t) 1 c j F ()G(s )d 2 j c j 6 t du f u g(u) u 0 F()G(s)
 Table-III SN Original function Mellin Transform 1 d n t dt f (t) (1)n sn F(s) 2 d n dt t f (t) (1)n (s n)n F(s) 3 d n f (t) dt n (1)n s F (s n) (s 1) 4 d n f (t) tn dtn 1n (s) F s n 5 f (t)g(t) 1 c j F ()G(s )d 2 j c j 6 t du f u g(u) u 0 F()G(s)

(11)

 Table-II SN Original function Mellin Transform 1 t 1 s 1 2 t a 1 s a 3 1 ta tb b a 1 s as b 4 eat a s (s) 5 Sin t s sin s 2 6 Cos t s cos s 2

The convolution or faulting theorem for the Mellin transform is derived in the same way as that of for the Laplace Transform. Let us suppose that F(S) and G(S) be the Mellin transforms of the functions f(t) and g(t)respectively; then the Mellin transform of the product f(t) g(t) is defined to be

2. Application on Euler-Cauchy differential equation:

1 c j

f (t)g(t)ts1dt g(t)ts1dt F () t d

0 0 2 j c j

(using equation (2)) (12)

1

1

c j

F ()d g(t)ts 1dt

Certain types of linear systems give rise to Euler- Cauchy differential equations.

Application of a Mellin transform to this type of equation will yield an algebraic equation.

A Eulers Cauchy differential equation is of the form,

2 j c j 0

n d i y(t)

i

i

1 c j

Ait

i0

dti

x(t)

(15)

2 j

F ()G(s )d

Where Ais are constants.

c j

(By equation (1)) (13)

In a similar way the Mellin transform of the product

Apply the Mellin Transform using Table-I& III to obtain the transformed equation.

n i di y(t)

F(s) G(s) is,

Ai M

t i

M x(t)

1 c j

i0

dt

F ()G(s) t sds n

i i

i i

2 j c j

A (1)i (s) Y (s) X (s)

1 c j

i0

F (s) t sds g(u) us1du

2 j c j 0

Where i is a positive integer and

(s)n s(s 1)(s 2)…..(s n 1)

n

n

A 1i s s 1s 2

If the current meter reads, i(t ) 1 1

then

i

i0

t2 t3

….s i 1Y s X s

(16)

the equation by transforming (Table-II)we get,

I (S) 2s 5

Y s

X s

n i

n i

A 1 s s 1s 2….s i 1

(s 2)(s 3)

Therefore equation (19) becomes,

R

(20)

i

i0

(17)

E(s 1)

(2s 5)L

0 s

L

Using the familiar partial fraction method and a Table

(s 2)(s 3)

of transform pair, the inverse transform of (17) is

L (3 a) L a s

R0 a (by using partial

easily obtained. Hence for linear Mellin transform the sum of the inverse transform of each fraction is equal to the inverse transform of the sum.

3. Application of Mellin transforms in the Instrumentation

If the current which is measured by a meter in a circuit

s 2 s 3 L

fraction method) (21)

Taking Inverse Mellin transform of both sides by using Table II, we get

Or

Or

t e(t) (3 a)Lt2 (2 a)Lt3

3

3

e(t) L 3 a 2 a

with varying current i(t ) 1 1

, and the resistance,

t t

(22)

t2 t3

R

R

R 0

t

.what is the driving voltage of the network

which is the voltage necessary to provide the given current.

shown in Fig.1 ?

4. Application of Mellin transforms in Network analysis

When the switch S is closed, current i(t) is measured

by a meter in a circuit is given by

i (t )1 1 , and the

t t2

R

R

Capacitor 1 co ,the resistance, R 0

find the driving

c2 t2 t

voltage e(t ) for the network shown in Fig.2 ?

The differential equation for Fig.1 is:

e(t) L di(t) Ri(t) L di(t) R0 i(t)

dt dt t Or

te(t) Lt di(t) R i(t)

dt 0

(18)

Taking Mellin transform, we get

M{te(t)} L M{t di(t) R M{i(t)}

Fig.2

The differential equation for Fig.2 is:

dt } 0

E(s 1) LSI (S) R0 I (S)

e(t) L di(t) Ri(t) 1 i(t)dt

I (S)(R LS)

dt c

(23)

(using Table -I) (19)

(using Table -I) (19)

e(t) L

e(t) L

0 i(t)

0 i(t)

o

o

i(t)dt

i(t)dt

0 di(t) R c

di(0)

dt t t2

,where i(0) dt 0 OR

te(t) Lt di(t) R i(t) c

i(t) dt

Conclusion

dt 0 0 t

(24)

The use of the Laplace integral transform for some of

Taking Mellin transform (Table-I & II), we get,

E(s 1) L SI (S) R I (S ) C0 I (S)

0 s

0

0

L S R C I (S )

0 s

b

the random variables is mostly used and explained in every advanced Engineering and Science field, now brief theoy of Mellin integral transform for electrical Engineering is given in this paper .It seems for any statisticians, mathematicians and engineers will also take interest in developing Mellin transform.

Here the paper presented some background on Mellin

S a I (S )

s

(25)

transform theory and motivated to compute the Euler-

n d i y(t)

i

i

R C Cauchy differential equation Ait

dti

x(t)

a

0 b 0

i0

Where

i(0) di(0) 0

dt

L L

i (t )1 1

and the application of Mellin transform in different areas of Electrical Engineering. It is a very effective mathematical tool to simplify very complex problems in the area of Instrumentation and Network analysis.

t t2

References

Equation by transforming we get,

I (S) 2s 3

(s 1)(s 2)

Therefore equation (25) becomes,

(2s 3)L S a b

(26)

1. George J. Fikioris Mellin Transform Method for Integral Evaluation

2. BogdanZiemian The Mellin Transformation and Fuchsian Type Partial Differential Equations [3]R. B. Paris, D. Kaminski AsymptoticsandMellin- BarnesIntegrals

[4]Alexander D. PoularikasHandbook of Formulas and Tables for Signal Processing

E(s 1)

(s 1)(s 2)

s

[5]D.ZagierAPPENDIXTHEMELLIN TRANSFORM ANDRELATEDANALYTIC TECHNIQUES

1. Jacqueline Bertrand, Pierre Bertrand, Jean-Philippe Ovarlez The Mellin Transform CRC Press inc, 1995

(by using partial fraction method)

L a b 2

2. Francis R Gerardi Application of Mellin and Hankel Transforms to Networks with Time-Varying Parameters.

L (a b 1)

s 1

2

s 2

(27)

[8]P. A. Martin Some Applications of the Mellin Transform to Asymptotics of Series Peter Lang, Frankfurt,1995

Taking Inverse Mellin transform of both sides by using Table II, we get

t e(t) L (a b 1) t1 L a b 2 2t 2

[9] Christopher C. TisdellEngineering mathematics wookbook2012

[10]A. K. GhoshIntroducton to Instrumentation and Control

1. Peter D. Hiscocks A Laplace Transform Cookbook

Syscomp Electronic Design Limited,March 1, 2008

e(t) L (a b 1) t2

• L a

b 2 t 3

2

2

Or

2. A: R. Forsyth, A Treatise on Differential Equations,

Macmillan and Co., Ltd., London, Eng.; 1885.

e(t)

L a b 2 2

(a b 1)

t2 t

(28)

which is the voltage necessary to provide the given current.