 Open Access
 Total Downloads : 388
 Authors : Dr. N. A. Patil, Vijaya N. Patil
 Paper ID : IJERTV2IS4676
 Volume & Issue : Volume 02, Issue 04 (April 2013)
 Published (First Online): 25042013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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 complexvalued 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 EulerCauchy 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 (18541933) 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 EulerCauchy 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.

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 TableI)
TableI
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)
TableI
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
TableIII
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)
TableIII
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)
TableII
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

Application on EulerCauchy 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 TableI& 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 (TableII)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.

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 ?

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 (TableI & 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.
If the current meter reads,
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)

George J. Fikioris Mellin Transform Method for Integral Evaluation

BogdanZiemian The Mellin Transformation and Fuchsian Type Partial Differential Equations [3]R. B. Paris, D. Kaminski AsymptoticsandMellin BarnesIntegrals
E(s 1)
(s 1)(s 2)
s
[5]D.ZagierAPPENDIXTHEMELLIN TRANSFORM ANDRELATEDANALYTIC TECHNIQUES
Jacqueline Bertrand, Pierre Bertrand, JeanPhilippe Ovarlez The Mellin Transform CRC Press inc, 1995
(by using partial fraction method)
L a b 2

Francis R Gerardi Application of Mellin and Hankel Transforms to Networks with TimeVarying 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,1995Taking 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
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


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.