Progress In Electromagnetics Research, PIER 59, 199–213, 2006

FRACTIONAL CURL OPERATOR IN CHIRAL

MEDIUM AND FRACTIONAL NON-SYMMETRIC
TRANSMISSION LINE

A. Hussain and Q. A. Naqvi

Electronics Department
Quaid-i-Azam University
Islamabad, Pakistan

Abstract—Fractional curl operator has been utilized to wave

propagation in lossless, isotropic, homogeneous and reciprocal chiral
medium when it contains interfaces. The fractional solutions for the
corresponding standing wave solution and transverse impedance are
determined. Equivalent fractional non-symmetric transmission line has
also been analyzed.

1. INTRODUCTION

Fractional calculus is a branch of mathematics that deals with

operators having non-integer and/or complex order, e.g., fractional
derivative and fractional integral [1]. Tools of fractional calculus have
various applications in different disciplines of science and engineering,
e.g., Optics, Control and Mechanics etc. Mathematical recipe
to fractionalize a linear operator is available in [2, 3]. Recently,
while exploring the roles and applications of fractional calculus in
electromagnetics a new fractional operator has been introduced [2].
The new fractional operator is termed as fractional curl operator.
Fractional curl operator has been utilized to find the new set of
solutions to Maxwell’s equations by fractionalizing the principle of
duality [2]. New set of solutions is named as fractional dual solutions
to the Maxwell equations. In electromagnetics, principle of duality
states that if (E, ηH) is one set of solutions (original solutions) to
Maxwell equations, then other set of solutions (dual to the original
solutions) is (ηH, −E), where η is the impedance of the medium. The
solutions which may be regarded as intermediate step between the
original and dual to the original solutions may be obtained using the
200 Hussain and Naqvi

following relations [2]

1
Ef d = (∇×)α E
(ik)α
1
ηH f d = (∇×)α ηH
(ik)α
√
where (∇×)α means fractional curl operator and k = ω µ is the
wavenumber of the medium. It may be noted that f d means fractional
dual solutions. Naqvi et al. [4] afterward extended the work [2] and
discussed the behavior of fractional dual solutions in an unbounded
chiral medium. Lakhtakia [5] derived theorem which shows that a
dyadic operator which commutes with curl operator can be used to
find new solutions of the Faraday and Ampere-Maxwell equations.
Veliev and Engheta [6] utilized the fractional curl operator to a fixed
solution and obtained the fractional fields that represent the solution
of reflection problem from an anisotropic surface. Naqvi and Rizvi [7]
determined the sources corresponding to the fractional dual solutions.
Naqvi and Abbas [8, 9] extended the work for metamaterial and for
complex and higher order fractional curl operator respectively. In
present work, we have determined the fractional solutions when chiral
medium contains interfaces. Equivalent fractional non-symmetric
transmission line has also been studied.

2. UNBOUNDED CHIRAL MEDIUM

Consider a uniform plane wave propagating along z-axis in an

unbounded, lossless, isotropic and reciprocal chiral medium. According
to field decomposition approach [10], field quantities, E and H may
be pictured as consisting of two parts, i.e., (E + , H + ) and (E − , H − ).
Two parts are termed as wavefields. The electric fields corresponding
to two wavefields are
E ± (z) = E ± (0) exp(ik± z) (1)
where k± = k(1 ±κr ) are wavenumbers of the two wavefields. k =
√
ω µ and κr = κ µ0 0 /µ . κ is the chirality parameter. Using the
following relation
η± H ± (z) = ±iE ± (z) (2)
corresponding magnetic field may be obtained. In above expression

µ±
η± = =η
±
Progress In Electromagnetics Research, PIER 59, 2006 201

This means that each wavefield sees chiral medium as achiral

medium with equivalent constitutive parameters ( + , µ+ ) and ( − , µ− ).
Medium parameters of the equivalent isotropic media are related to the
parameters of chiral medium by the following relations

± = (1 ± κr ), µ± = µ(1 ± κr )

Simple expressions for the wavefields can be written as

1
E+ = (E − jηH)
2
1
E− = (E + jηH)
2 
1 j
H+ = H+ E
2 η
 
1 j
H− = H− E
2 η
The total fields in chiral medium are
E(z) = E i+ (0) exp(ik+ z) + E i− (0) exp(ik− z)
  (3)
ηH(z) = i E i+ (0) exp(ik+ z) − E i− (0) exp(ik− z)

Fractionalizing the electric fields E + (z) and E − (z)

1
E f d+ (z) = (∇×)α E + (z)
(ik+ )α
1  dα
= (ẑ×)α E i+ (0) exp(ik+ z)
(ik+ )α dz α
 
απ
= E i+ (0) exp i k+ z +
2
Similarly  
απ
E f d− (z) = i k− z −
E i− (0) exp
2
Fractional dual fields corresponding to the original fields given in
202 Hussain and Naqvi

equation (3) may be written as

 
απ
E f d (z) = E i+ (0) exp i k+ z +
 2 
απ
+E i− (0) exp i k− z −
 2 
(4)
απ
ηH f d (z) = i E i+ (0) exp i k+ z +
 2 
απ
− E i− (0) exp i k− z −
2
It is obvious from equation (4) that for α = 0
E f d (z) = E(z), ηH f d (z) = ηH(z)
and for α = 1
E f d (z) = ηH(z), ηH f d (z) = −E(z)
which is consistent with electromagnetics principle of duality. For
0 < α < 1, solutions may be regarded as intermediate between the
original and dual to the original solutions.

3. STANDING WAVES IN CHIRAL MEDIUM

Assume a plane wave hits normally a chiral-chiral interface located at

z = 0. The intrinsic impedance before the interface is η while after
the interface is η1 . Total electric field of wavefields before the interface
may be written as
E ± (z) = E i± (0) exp(ik± z) + R∓ ± E i± (0) exp(−ik∓ z)

where R+ − is the reflection co-efficient for negative incident wavefield

and positive reflected wavefield while R− + is the reflection coefficient
for positive incident wavefield and negative reflected wave field. For
reciprocal chiral media, the reflection coefficient becomes R∓ ± = R =
η1 −η
η1 +η . So, we can write

E ± (z) = E i± (0) exp(ik± z) − RE i± (0) exp(−ik∓ z)

= E i± (0){exp(ik± z) + R exp(−ik∓ z)}
The corresponding magnetic field is
ηH ± (z) = ±iE i± (0){exp(ik± z) − R exp(−ik∓ z)}
Progress In Electromagnetics Research, PIER 59, 2006 203

Using above expressions, total fields may be written as

E(z) = E i+ (0){exp(ik+ z) + R exp(−ik− z)}
+E i− (0){exp(ik− z) + R exp(−ik+ z)}
= E 1 + E2 + E3 + E4 (5)
ηH(z) = iE i+ (0){exp(ik+ z) − R exp(−ik− z)}
−iE i− (0){exp(ik− z) − R exp(−ik+ z)}
= ηH1 + ηH2 + ηH3 + ηH4 (6)
Let us write fractional dual solution to each of the field
components as
1
E 1f d = (×)α {E i+ (0) exp(ik+ z)}
(ik+ )α
= (i)α E i+ (0) exp(ik+ z)
Similarly other components can be written as
E 2f d = −(i)α (−1)α E i+ (0) exp(−ik− z)
E 3f d = (−i)α E i− (0) exp(ik− z)
E 4f d = −(−i)α (−1)α E i− (0) exp(−ik+ z)
ηH 1f d = i(i)α E i+ (0) exp(ik+ z)
ηH 2f d = i(i)α (−1)α E i+ (0) exp(−ik− z)
ηH 3f d = −i(−i)α E i− (0) exp(ik− z)
ηH 4f d = −i(−i)α (−1)α E i− (0) exp(−ik+ z)

Using these values in equation (5) and equation (6), we can write
fractional dual solutions as
E f d (z) = (i)α E i+ (0) [exp {i (k+ z)} + R(−1)α exp {−i (k− z)}]
+(−i)α E i− (0) [exp {i (k− z)} + R(−1)α exp {−i (k+ z)}]
ηH f d (z) = i(i)α E i+ (0) [exp {i (k+ z)} − R(−1)α exp {−i (k− z)}]
−i(−i)α E i− (0) [exp {i (k− z)} − R(−1)α exp {−i (k+ z)}]
on simplifying, we can write
 
iαπ
E f d (z) = exp
2
204 Hussain and Naqvi
   
απ απ
(i)α E i+ (0) exp i k+ z − +R exp −i k− z−
2 2
   
απ απ
+ (−i)α E i− (0) exp i k− z− +R exp −i k+ z −
2 2
(7)
 
iαπ
ηH f d (z) = i exp
2
   
απ απ
(i)α
E i+ (0)
exp i k+ z− −R exp −i k− z−
2 2
   
απ απ
− (−i) E − (0) exp i k− z−
α i
−R exp −i k+ z−
2 2
(8)
It is obvious that for α = 0
E f d (z) = E(z) ηH f d (z) = ηH(z)
which is original set of solutions given by equations (5) and (6). For
α=1
E f d (z) = ηH(z) ηH f d (z) = −E(z)
which is dual to the original set of solutions. For 0 < α < 1, solutions
set given by (7) and (8) may be regarded as fractional dual solutions
corresponding to the original solutions set given by (5) and (6).

4. TRANSVERSE WAVE IMPEDANCE

Transverse impedance of fractional dual fields is defined as

Ef dx Ef dy
Zf dxy = =− (9)
Hf dy Hf dx
where
iαπ
Ef dx (z) = exp( )
2    
απ απ
(i) α i
E+ (0)
exp i k+ z− +R exp −i k− z−
2 2
   
απ απ
α i
+ (−i) E− (0) exp i k− z− +R exp −i k+ z−
2 2
and
 
iαπ
ηHf dy (z) = exp
2
Progress In Electromagnetics Research, PIER 59, 2006 205
   
απ απ
(i)α E+
i
(0) exp i k+ z− −R exp −i k− z−
2 2
   
απ απ
+ (−i)α E−
i
(0) exp i k− z− −R exp −i k+ z−
2 2
Using relations k+ = k + kκr and k− = k − kκr , above equations take
the following form
     
iαπ απ απ
Ef dx (z) = exp exp i kz− +R exp −i kz−
2 2 2

(i)α E+
i
(0) exp(+ikκr z) + (−i)α E−
i
(0) exp(−ikκr z)
     
1 iαπ απ απ
Hf dy (z) = exp exp i kz− −R exp −i kz−
η 2 2 2

(i)α E+
i
(0) exp(+ikκr z) + (−i)α E−
i
(0) exp(−ikκr z)

Substituting the Ef dx and Hf dy in equation (9), following is obtained

   
απ απ
exp i kz − + R exp −i kz −
 2   2 
Zf dxy =η
απ απ
exp i kz − − R exp −i kz −
2 2
For PEC interface, above yields the following
 
απ
Zf dxy = iηtan kz − (10a)
2
and at z = 0  
απ
Zf dxy = −iηtan (10b)
2
Similarly
Ef dy
Zf dyx = −
Hf dx
 
απ
= iηtan kz − (10c)
2
and at z = 0  
απ
Zf dyx = −iηtan (10d)
2
From equation (10b) and (10d), it is obvious that for α = 0, value of
transverse wave impedance at z = 0 is Zf d = 0, which describes the
206 Hussain and Naqvi

situation as if a PEC interface is placed at z = 0. For α = 1, the

transverse wave impedance is Zf d = ∞, which describes a situation as
if a PMC interface is placed at z = 0. Hence for 0 < α < 1, equation
(10b) and (10d) describe a surface located at z = 0 whose impedance
may be regarded intermediate step of the impedance of a PEC interface
and PMC interface.
It is of interest to extend the above results for the case of multiple
boundary mediums. For this purpose, geometry with two interfaces are
considered. Similarly geometries having more than two interfaces may
be treated. Consider a chiral slab which is sandwiched between two
different chiral media. The width and intrinsic impedance of the chiral
slab are L and η3 respectively. Intrinsic impedance of the medium
before the slab is η1 while after the slab is η2 . Front end of the slab
is located at z = 0 while back end is located at z = −L. Transverse
impedance, for the fractional dual solution corresponding to present
case, before the slab may be written as
   
απ απ
exp i k1 z − + R1 exp −i k1 z −
 2   2 
Z1f dxy = η1
απ απ
exp i k1 z − − R1 exp −i k1 z −
2 2
(11)
where
Z2f dxy − η1
R1 =
Z2f dxy + η1
   
απ απ
exp i k2 z − + R2 exp −i k2 z −
 2   2 
Z2f dxy = η1
απ απ
exp i k2 z − − R2 exp −i k2 z −
2 2
η3 − η2
R2 =
η3 + η2
It is obvious from above expression that for value of fractional
parameter α = 2kπ2 L , the situation behaves as if there is no chiral
slab.
Above result may be easily extended for the case of chiral slab
backed by PEC interface and the result is given by
   
απ απ
exp i k1 z − + R1 exp −i k1 z −
 2   2 
Z1f dxy = η1
απ απ
exp i k1 z − − R1 exp −i k1 z −
2 2
(11a)
Progress In Electromagnetics Research, PIER 59, 2006 207

where
Z2f dxy − η1
R1 =
Z2f dxy + η1
 
απ
Z2f dxy = iη2 tan k2 L −
2
It may be noted that for α = 0, equation (11a) deals with situation of
a slab backed by PEC interface while for α = 1, equation (11a) deals
with situation of a slab backed by PMC interface. For 0 < α < 1,
equation (11a), deals with fractional dual situation.

5. NON-SYMMETRIC TRANSMISSION LINE

When an electromagnetic wave of any polarization propagates through

a bi-isotropic medium, solution of Maxwell equations in bi-isotropic
medium gives rise to two circularly polarized waves. One of the
waves is right circularly polarized (RCP) while the other one is left
circularly polarized (LCP). These RCP and LCP components move
with different phase velocities and may be represented in terms of wave
numbers k+ and k− respectively. A plane wave propagating through
a plane-parallel structure of bi-isotropic medium can be analyzed in
terms of two non-interacting scalar transmission lines with two eigen
waves much in the same way as simple isotropic medium. The main
difference is that in case of bi-isotropic medium re ected wave has
wave number different than the incident wave. This means that a left
circularly polarized wave will become right circularly polarized upon
re ection from the interface and vice versa. Thus incident and re ected
components of the wave will see different effective media and hence the
corresponding transmission line becomes non symmetric with different
parameters for the waves propagating in the opposite directions. For
the two circularly polarized TEM eigen waves depending only upon
z-coordinate, the source free Maxwell equations can be written as
uz × E  ± (z) = −iωµ± H ± (z), uz × H  ± (z) = iω ± E ± (z) (12)
where the primes denote differentiation with respect to z. In terms of
circular polarization (CP) unit vectors u± satisfying uz × u± = ±iu±
we can write
E± = u± E± and H± = u± H± .
Using these values, we may write the scalar form of equation (12) as
E  ± (z) = −iωµ± {∓iH± (z)}, ∓iH  ± (z) = −iω ± E± (z)
208 Hussain and Naqvi

They resemble the transmission line equations

V  (z) = −iωLI(z), I  (z) = −iωCV (z) (13)

If we identify electric field with voltage V± = E± , the current must be

recognized as I± = ∓iH± .

6. FRACTIONAL NON-SYMMETRIC TRANSMISSION

LINE

The positive and negative wave-fields are represented by their

respective voltage and current components. These components are
added to get total voltage and current. For a boundary with reflection
coefficient Γ∓ ± , we can write wave fields as

V± (z) = V± (0) exp(ik± z) + Γ∓ ± V± (0) exp(−ik∓ z)

ZI± (z) = V± (0) exp(ik± z) − Γ∓ ± V± (0) exp(−ik∓ z)

ZL −Z
For a reciprocal chiral medium Γ∓ ± = Γ = ZL +Z . So, we can write

V± (z) = V± (0){exp(ik± z) + Γ exp(−ik∓ z)}

ZI± (z) = V± (0){exp(ik± z) − Γ exp(−ik∓ z)}

Now total voltage and current equations can be written as

V (z) = V+ (0){exp(ik+ z) + Γ exp(−ik− z)}

+V− (0){exp(ik− z) + Γ exp(−ik+ z)}
= V 1 + V2 + V3 + V4 (14a)
ZI(z) = V+ (0){exp(ik+ z) − Γ exp(−ik− z)}
+V− (0){exp(ik− z) − Γ exp(−ik+ z)}
= ZI1 + ZI2 + ZI3 + ZI4 (14b)

Let us write fractional dual solution to each component of

equation(14a) and equation (14b) as

1 d2α
V1f d = V + (0) exp(ik+ z)
(ik+ )2α dz 2α
= exp(iαπ)V+ (0) exp{i(k+ z − απ)}

Similarly other components can be written as

V2f d = exp(iαπ)V+ (0)Γ exp{−i(k− z − απ)}

Progress In Electromagnetics Research, PIER 59, 2006 209

V3f d = exp(iαπ)V− (0) exp{i(k− z − απ)}

V4f d = exp(iαπ)V− (0)Γ exp{−i(k+ z − απ)}
ZI1f d = exp(iαπ)V+ (0) exp{i(k+ z − απ)}
ZI2f d = exp(iαπ)V+ (0)Γ exp{−i(k− z − απ)}
ZI3f d = exp(iαπ)V− (0) exp{i(k− z − απ)}
ZI4f d = exp(iαπ)V− (0)Γ exp{−i(k+ z − απ)}

Using these values in equation (14a) and equation (14b) we have

Vf d (z) = exp(iαπ)V+ (0)[exp{i(k+ z − απ)} + Γ exp{−i(k− z − απ)}]

+ exp(iαπ)V− (0)[exp{i(k+ z−απ)}+Γ exp{−i(k+ z−απ)}] (15a)

ZIf d (z) = exp(iαπ)V+ (0)[exp{i(k+ z − απ)} − Γ exp{−i(k− z − απ)}]

+ exp(iαπ)V− (0)[exp{i(k− z−απ)}−Γ exp{−i(k+ z−απ)}]
(15b)

For α = 0

Vf d (z) = V (z) ZIf d (z) = ZI(z)

1
and α = 2
Vf d (z) = ZI(z) ZIf d (z) = V (z)
which is original and dual to the original set of solutions for a
transmission line. For 0 < α < 1/2, solutions set (Vfd , ZIfd )
may be regarded as intermediate step between the original and dual
to the original solutions of a transmission line. Transmission line
corresponding to solutions set (Vfd , ZIfd ) may be termed as fractional
dual transmission line.

7. INPUT IMPEDANCE OF TERMINATED

FRACTIONAL NON-SYMMETRIC LINE

Input impedance of the fractional dual transmission line is defined as

Vf d (z)
Zf d = (16)
If d (z)

Using the following relations

k+ = k + kκr k− = k − kκr
210 Hussain and Naqvi

We can write
exp(ik+ z) = exp(ikz) × exp(ikκr )
exp(ik− z) = exp(ikz) × exp(−ikκr )
Using these values in equation (15a) and equation (15b), it is obtained
1
Vf d (z) = exp(iαπ){V+ (0) exp(ikκr z)+V− (0) exp(−ikκr z)}
(ZL +Z)
2 cos(kz − απ){ZL + iZ tan(kz − απ)}
1
ZIf d (z) = exp(iαπ){V+ (0) exp(ikκr z)+V− (0) exp(−ikκr z)}
(ZL +Z)
2 cos(kz − απ){iZL tan(kz − απ) + Z}
Putting these values of Vf d (z) and If d (z) in equation (16), we get

ZL + iZ tan(kz − απ)
Zf d = Z (17)
Z + iZL tan(kz − απ)
For α = 0
ZL + iZ tan(kz)
Zf d = Z (17a)
Z + iZL tan(kz)
which is the relation for input impedance at any point for a
transmission line having characteristic impedance Z. We consider it
as original transmission line. Now for α = 12 equation (17) gives

ZL + iZ cot(kz)
Zf d = Z (17b)
Z + iZL cot(kz)
which is the relation for input impedance at any point for dual to
the original transmission line. It may be noted that at a particular
point along the original transmission line input admittance becomes
the corresponding input impedance of the dual transmission line.
So, for α = 0, equation (17) represents input impedance of the
original transmission line and α = 12 represents the dual to the
original transmission line, while 0 < α < 12 , equation (17) represents
input impedance of the fractional dual transmission line. It may
also be noted that variation in the value of fractional parameter α
along the fractional transmission line corresponds to variation in the
observation point along the original transmission line. So behavior
along a transmission line may be studied by changing the fractional
parameter along the corresponding fractional transmission line while
keeping the observation point constant.
Progress In Electromagnetics Research, PIER 59, 2006 211

Setting observation point at z = 0, that is at the load, equation

(17) yields following
ZL − iZ tan(απ)
Zf d = Z
Z − iZL tan(απ)
It is obvious from above equation that
Z2
Zf d |α=0 = ZL , Zf d |α= 1 =
2 ZL
or
1
Z f d |α=0 = Z L , Z f d |α= 1 =
=YL
2 ZL
This means if original transmission line, that is α = 0, deals with
transmission line which is terminated by a load ZL then α = 1/2 deals
with transmission line which is terminated by a load having impedance
YL . For 0 < α < 12 represents new line terminated by load, which may
be regarded as intermediate step between the loads ZL and YL .
Now consider the following two special cases
If ZL = 0, then Zf d |α=0 = 0, Zf d |α= 1 = ∞
2

and
If ZL = ∞, then Zf d |α=0 = ∞, Zf d |α= 1 = 0
2

Which shows that if the original transmission line is a short-circuited

line then dual to the original line will be an open-circuited transmission
line and vice versa. For 0 < α < 12 , represents a line terminated by
an inductive type load if the original line is a short circuited line. For
0 < α < 12 , represents a line terminated by capacitive type load if the
original line is open circuited line.

8. INPUT IMPEDANCE OF MULTIPLE-SECTIONS

FRACTIONAL NON-SYMMETRIC LINE

Consider a transmission line having characteristic impedance Z1 . The

transmission line is connected to another transmission line having
characteristic impedance Z2 . The length of line having characteristic
impedance Z2 is L and the line is terminated by a load ZL . It is
assumed that load is located at z = 0. Z2in is the input impedance
just before the junction of two lines, that is at z = −L. We can
write input impedance Zin along the transmission line having intrinsic
impedance Z1 as
Z2in + iZ1 tan(k1 z)
Zin = Z1 , z < −L (18a)
Z1 + iZ2in tan(k1 z)
212 Hussain and Naqvi

where
ZL + iZ2 tan(k2 L)
Z2in = Z2 (18b)
Z2 + iZL tan(k2 L)
Using the same treatment as done in previous section we can write the
input impedance of two sections fractional dual transmission line as
Z2infd + iZ1 tan(k1 z − απ)
Zinfd = Z1 (19a)
Z1 + iZ2infd tan(k1 z − απ)
where
ZL + iZ2 tan(k2 L − απ)
Z2infd = Z2 (19b)
Z2 + iZL tan(k2 L − απ)
For α = 0, equation (19) gives
Z2infd |α=0 + iZ1 tan(k1 z)
Zinfd |α=0 = Z1 (20a)
Z1 + iZ2infd |α=0 tan(k1 z)
where
ZL + iZ2 tan(k2 L)
Z2infd |α=0 = Z2 (20b)
Z2 + iZL tan(k2 L)
1
which is given in equation (18). Similarly, for α = 2

Z2infd |α= 1 + iZ1 cot(k1 z)

Zinfd |α= 1 = Z1 2
(21a)
2 Z1 + iZ2infd |α= 1 cot(k1 z)
2

where
ZL + iZ2 cot(k2 L)
Z2infd |α= 1 = Z2 (21b)
2 Z2 + iZL cot(k2 L)
which is dual to the input impedance given in (18) or (20). For
0 < α < 1/2, input impedance Zinfd may be regarded as intermediate
step of the two input impedances given by (20) and (21).
Condition for impedance matching of transmission line network
described by (19) is
ZL + iZ2 tan(k2 L − απ)
Z2infd = Z1 = Z2 (22)
Z2 + iZL tan(k2 L − απ)
The value of fractional parameter α, in terms of the impedances,
required for the impedance matching is given below
  
1 Z2 (Z1 − ZL )
α= k2 L − tan−1   (23)
π i Z22 − Z1 ZL
Progress In Electromagnetics Research, PIER 59, 2006 213

It may be noted that for α = k2πL , equation (19) yields input impedance
as if in the circuit there is no transmission line having characteristic
impedance Z2 . From equation (23) if Z1 = ZL , then
k2 L
α=
π
is the condition for impedance matching.

REFERENCES

1. Oldham, K. B. and J. Spanier, The Fractional Calculus, Academic

Press, New York, 1974.
2. Engheta, N., “Fractional curl operator in electromagnetics,”
Microwave Opt. Tech. Lett., Vol. 17, 86–91, 1998.
3. Ozaktas, H. M., Z. Zalevsky, and M. A. Kutay, The Fractional
Fourier Transform with Applications in Optics and Signal
Processing, Wiley, New York, 2001.
4. Naqvi, Q. A., G. Murtaza, and A. A. Rizvi, “Fractional dual
solutions to Maxwell equations in homogeneous chiral medium,”
Optics Communications, Vol. 178, 27–30, 2000.
5. Lakhtakia, A., “A representation theorem involving fractional
derivatives for linear homogeneous chiral media,” Microwave Opt.
Tech. Lett., Vol. 28, 385–386, 2001.
6. Veliev, E. I. and N. Engheta, “Fractional curl operator in
reflection problems,” 10th Int. Conf. on Mathematical Methods
in Electromagnetic Theory, Ukraine, Sept. 14–17, 2004.
7. Naqvi, Q. A. and A. A. Rizvi, “Fractional dual solutions and
corresponding sources,” PIER, Vol. 25, 223–238, 2000.
8. Naqvi, Q. A. and M. Abbas, “Fractional duality and metamateri-
als with negative permittivity and permeability,” Optics Commu-
nications, Vol. 227, 143–146, 2003.
9. Naqvi, Q. A. and M. Abbas, “Complex and higher order fractional
curl operator in electromagnetics,” Optics Communications,
Vol. 241, 349–355, 2004.
10. Lindell, I. V., A. H. Sihvola, S. A. Tretyakov, and A. J. Viitanen,
Electromagnetic Waves in Chiral and Bi-isotropic Media, Artech
House, Boston, 1994.