naqvi&faryad

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High Frequency Expressions for the Field in t he CaustRegion of a Parabolic Reflector Coated with IsotropicChiral MediumM. Faryad

a& Q. A. Naqvi

b

aElectronics Department, Quaid-i-Azam University, Islamabad, Pakistan

bElectronics Department, Quaid-i-Azam University, Islamabad, Pakistan

Version of record first published: 03 Apr 2012.

To cite this art icle: M. Faryad & Q. A. Naqvi (2008): High Frequency Expressions for t he Field in t he Caust ic Region of a

Parabol ic Reflect or Coat ed w it h Isotr opic Chira l Medium, Journal of Elect romagnetic Waves and Appl icat ions, 22:7, 964-985

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J. of Electromagn. Waves and Appl., Vol. 22, 965986, 2008

HIGH FREQUENCY EXPRESSIONS FOR THE FIELDIN THE CAUSTIC REGION OF A PARABOLICREFLECTOR COATED WITH ISOTROPIC CHIRALMEDIUM

M. Faryad and Q. A. Naqvi

Electronics DepartmentQuaid-i-Azam UniversityIslamabad, Pakistan

AbstractHigh frequency field expression for a parabolic reflector

are derived using Maslovs method. Parabolic reflector has beencoated with isotropic and homogeneous chiral medium. The effectsof thickness of chiral layer, chirality parameter of the chiral mediumand permittivity of the medium are studied.

1. INTRODUCTION

Asymptotic ray theory (ART) or the geometrical optics approximationis widely used to study various kinds of problems in the areas ofelectromagnetics, acoustic waves, seismic waves, etc. [13]. It isalso well known that the geometrical optics fails in the vicinity ofcaustic. So, in order to study the field behavior near caustic, other

approach is required. Maslov proposed a method to predict the fieldin the caustic region [4]. Maslovs method combines the simplicityof asymptotic ray theory and the generality of the Fourier transformmethod. This is achieved by representing the geometrical optics fieldsin terms of mixed coordinates consisting of space coordinates and wavevector coordinates. That is by representing the field in terms of sixcoordinates. It may be noted that information of ray trajectoriesis included in both space coordinates R = (x,y,z) and wave vectorcoordinates P = (px, py, pz).

Solving the Hamiltonian equations under the prescribed initialconditions, one can construct the geometrical optics field in space R,which is valid except in the vicinity of caustic. Near the caustic, theexpression for the geometrical optics field in spatial space is rewritten


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966 Faryad and Naqvi

in mixed domain. The expression in mixed domain is related to theoriginal domain R through the asymptotic Fourier transform.

Focusing systems have been analyzed by many authors [529]. Anumber of research problems have been analyzed using the Maslovs

method [1029]. In present discussion, our interest is to studygeometrical optics (GO) fields associated with a cylindrical reflectorcoated with a layer of chiral medium. In Section 2, we have discussedgeometrical optics method and Maslovs method. In Section 3, theplane wave reflection from a chiral slab backed by perfect electricconducting (PEC) plane is considered. In Section 4, high frequencyexpression for the field of a chiral coated parabolic reflector excited byplane wave is derived. Since GO fails around caustic, the finite fieldaround caustics are derived using Maslovs method. Numerical resultsand discussion is presented in Section 5 and the paper is concluded inSection 6.

2. GEOMETRICAL OPTICS AND MASLOVS METHOD

Consider the scalar wave equation2 + k20u(r) = 0 (1)where r = (x, z), 2 = 2/x2 + 2/z2 and k0 = 00 iswavenumber of the medium. Solution of (1) may be assumed in theform of asymptotic series in powers of (jk0)

1 as

u(r) =

m=0

Am(r)

(jk0)mexp(jk0) (2)

Substituting (2) in (1) and equating coefficients we get eikonal equationas [26]

()2 1 = 0 (3)and transport equation

2A. + A2 = 0 (4)

where we have retained only A0 and denoted it with A. Since we haveassumed k0 to be large so we have neglected higher order terms.

We define wave vector p = and Hamiltonian H= (p p1)/2,so the eikonal equation becomes H = 0. Equation (3) can be solved


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Parabolic reflector coated with isotropic chiral medium 967

by method of characteristic, as follow

dx

d= px (5a)

dz

d = pz (5b)

dpxd

=H

x= 0 (5c)

dpzd

=H

z= 0 (5d)

where is parameter along the ray. Solution of Hamiltons equations(5) is

x = + px (6a)

z = + pz (6b)

px = px0 (6c)

pz = pz0 (6d)where (, ) and (px0, pz0) are initial values of (x, z) and (px, pz)respectively. The phase function is given by

= 0() +

0

d = 0() + (7)

where 0() is initial phase at (, ). The solution of transport equationis [26]

A(r) = A0()[J()]1/2 (8)

where A0() is initial field and J=D()D(0) , where D() = (x, z)/(, )

is the Jacobian of transformation from ray co-ordinates (, ) to spaceco-ordinates (x, z). Now we can write approximate solution of (1),known as GO solution, as

u(r) = u0()[J()]1/2 exp(jk0) (9)

where u0() = A0()exp{jk00()}. Equation (9) predicts infiniteamplitudes at points where J= 0, which is not realistic. This situationcan be avoided and finite field around caustic may be found as follows,using the Maslovs method [4, 15].

The solution may be assumed in the form of the Fourier transformand is given as

u(r) =

T(x, pz) exp[jk0(x, pz) + pzz]dpz (10)


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Equation (10) can be solved using stationary phase method in theregion much away from the caustic. The resulting solution is GOsolution so it should agree with GO solution given in (9). Identifyingthe unknown by comparing asymptotic solution of (10) with GO

solution given in (9), explicit expression for (x, pz) and T(x, pz) canbe obtained. Although solution agrees at stationary points ofpz butwe assume that they agree for all values of pz. The field near causticsis given as

u(r) =

k0

j2

A0()

Jpzz

1/2exp[jk0{0 + z0pz + zpz}]dpz (11)

where z0 is the value of z at stationary point of pz. The expressionJpzz can be calculated more simply as

J()pzz

= 1D(0)

(x, pz)(, )

(12)

3. REFLECTION OF PLANE WAVES FROM A CHIRALSLAB BACKED BY CONDUCTOR

Consider the problem of plane wave reflection from a chiral slab backedby PEC plane as shown in Figure 1. The region z 0 is occupied by

x

Er

z

PEC plane

Chiral Medium

Ei

d

Figure 1. Reflection from chiral slab of thickness d backed by PEC

plane.


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free space

D = 0E

B = 0H

and the region 0 z d is occupied by the chiral medium defined byconstitutive relations [31, 32].

D = (E + E)B = (H + H)

while PEC plane is lying at z = d. Define the following quantities:k0 =

00, k =

, 0 =

0/0, =

/, k1 =

k1k and

k2 =k

1+k. The incident electric field may be written as, in free space

Ei = Auy + Ak0z

k0ux +

k0x

k0uz exp(jk0zz + jk0xx) (13)

and the reflected field as

Er =

Buy + B

k0zk0

ux +k0xk0

uz

exp(jk0zz + jk0xx) (14)

and the field in the chiral layer can conveniently be written in termsof Beltrami fields as [27]

E = QL jQR (15a)H = QR jQL/ (15b)

where

QR = A1

uy+ j

k1zk1

ux+k0xk1

uz

exp(jk1zz+jk0xx)

+B1

uy+j

k1z

k1ux+

k0xk1

uz

exp(jk1zz+jk0xx) (16a)

QL = A2

uy+j

k2zk2

ux+k0xk2

uz

exp(jk2zz + jk0xx)

+B2

uy+j

k2z

k2ux+

k0xk2

uz

exp(jk2zz+jk0xx) (16b)

where k21z +k20x = k

21, k

22z +k

20x = k

22, k0z = k0 cos and k0x = k0 sin.

is the angle of incident and reflected wave in free space with z-axis. Applying the boundary conditions at z = 0 yields the following


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equations BB

= [r]

AA

+ [T]

A1A2

(17a)

B1B2

= [R]

A1A2

+ [t]

AA

(17b)

where [r], [T], [R] and [t] are 2 2 matrices. Elements of the matrices,which are Fresnel coefficients, are given as

r11 = [(20 2)(1 + 2) + 20(12 1)]/Dr22 = [(

20 2)(1 + 2) 20(12 1)]/D

r12 = 2j0(1 2)]/Dr21 = r12t11 = 2(2 + 0)/D

t22 =

2(01 + )/D

t12 = 2j(02 + )/Dt21 = 2j(1 + 0)/D

R11 = [(20 +

2)(1 2) + 20(12 1)]/DR22 = [(20 + 2)(1 2) + 20(12 1)]/DR12 = 2j2(20 2)/DR21 = 2j1(

20 2)/D

T11 = 401(2 + 0)/D

T22 = 402(01 + )/DT12 = 4j02(1 + 0)/DT21

= 4j01

(02

+ )/D

where

D = (20 + 2)(1 + 2) + 20(12 + 1)

and 1 = sec

1 (k0k1 )2 sin2 and 2 = sec

1 (k0k2 )2 sin2 .Following [28], boundary conditions at z = d yields the following

equationA1 exp(jk1zd)A2 exp(jk2zd)

=

1

k2k1z + k1k2z

k2k1z k1k2z 2jk1k2z2j k2k1z k1k2z k2k1z

B1 exp(jk1zd)B2 exp(jk2zd) (18)


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or B1B2

= [][R2][]

A1A2

(19)

where

[] =

exp(jk1zd) 00 exp(jk2zd)

and

[R2] =1

1 k0k1

2sin2 +

1

k0k2

2sin2

1 k0k1

2sin2

1

k0k2

2sin2

2j1k

0k1

2sin2

2j

1

k0k2

2sin2

1 k0k2

2sin2

1

k0k1

2sin2

From (17) and (19), we haveBB

=

[r] + [T] ([][R2][] [R])1 [t]

AA

(20)

4. GEOMETRICAL OPTICS FIELD OF PARABOLICREFLECTOR COATED WITH CHIRAL LAYER

Consider the reflection of normal incidence plane wave hitting a twodimensional cylindrical reflector, which is uniform in y axis as shownin Figure 2. We consider the case of parabolic reflector defined by

= f 2/4f (21)where f is the focal length of parabola. Consider the incident planewave making angle with normal, given as

Ei = {Ayuy + Ax(ux)} exp(jk0z) (22)


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Figure 2. Cylindrical reflector with a layer of chiral material.

where the normal is given as

n = sinux + cosuz

where

cos =2f

2 + 4f2sin =

2 + 4f2

the reflected wave is

Er = [Byuy + Bxz { cos(2)ux sin(2)uz}] exp{jk0 cos(2)z + jk0 sin(2)x} (23)

and the reflected wave vector is

p =

sin2ux

cos2uz

and the jacobian of transformation is

J() = 1 cos2 f

(24a)

and the phase function is

= + (24b)

the GO field for each component of electric field is

Erx = cos2Bxz 1 cos2

f1/2

exp{jk0(2f )cos2 jk0} (25a)


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Ery = By

1 cos2

f

1/2exp{jk0(2f )cos2 jk0} (25b)

Erz = sin2Bxz

1 cos2 f

1/2exp{jk0(2f )cos2 jk0} (25c)

Since the GO field predicts infinite amplitude at points where J = 0,so to find finite field around these points we use Maslovs method. Toevaluate field by (11), we need expression (12), which is given below

Jpzz

= sin2(2)cos2 1f

(26)

The phase function function in (12) is found as

= 2f1 cos(2)2x (27)The finite field around caustic is, using (11)

Erx =

2k0f

exp(jk0f+ j/4)

/4/4

cos2Bxz

sec exp{jk0 cos(2 )}d (28a)

Ery =

2k0f

exp(jk0fj/4)

/4/4

By

sec exp{jk0 cos(2 )}d (28b)

Erz =2k0f

exp(jk0f+ j/4)/4/4

sin2Bxz

sec exp{jk0 cos(2 )}d (28c)where x = sin and z = cos .

5. NUMERICAL RESULTS AND DISCUSSION

We have solved (28) numerically for k0 = 1, f = 100, = 0 anddifferent values of = /0, d, and . The numerical integration isperformed from = /4 to = /4. We have considered two typesof polarization for incident wave. One is (Ay = 0, Ax = 1) and theother is (Ay = 1, Ax = 0). The results are plotted in Figures 320.In all figures horizontal axis is x-axis and vertical axis corresponds to


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-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

x

|Erx|

d = 0

d = 0.2

d = 0.4

Ay= 0, A

x= 1

= 0, = 3

Figure 3.

-15 -10 -5 0 5 10 150

2

4

6

8

10

12

14

16

18

20

x

|Ery|

d = 0

d = 0.4

d = 0.8

Ay= 1, A

x= 0

= 0, = 3

Figure 4.


12/23


-15 -10 -5 0 5 10 150

2

4

6

8

10

12

x

|Erz|

d = 0

d = 0.4

d = 0.8A

y= 0, A

x= 1

= 0, = 3

Figure 5.

-15 -10 -5 0 5 10 150

0.05

0.1

0.15

0.2

0.25

0.3

0.35

x

|Erx|

d = 0

d = 0.1

d = 0.5A

y= 1, A

x= 0

= 0.5, =1

Figure 6.


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-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

8

9

x

|Erx|

d = 0

d = 0.3

d = 0.5A

y= 0, A

x= 1

= 0.5, =1

Figure 7.

-15 -10 -5 0 5 10 150

5

10

15

20

25

x

|Ery|

d = 0

d = 0.4

d = 0.8

Ay= 1, A

x= 0

= 0.5, =1

Figure 8.


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-15 -10 -5 0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

x

|Ery|

d = 0

d = 0.1

d = 0.5

Ay= 0, A

x= 1

= 0.5, =1

Figure 9.

-15 -10 -5 0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

x

|Erz|

d = 0

d = 0.4

d = 0.8

Ay= 1, A

x= 0

= 0.5, =1

Figure 10.


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-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

8

9

x

|Erz|

d = 0

d = 0.1

d = 0.5

Ay= 0, A

x= 1

= 0.5, =1

Figure 11.

-15 -10 -5 0 5 10 150

2

4

6

8

10

12

14

x

|Erx|

Ay= 0, A

x= 1

d = 0.5, =1

=0

=0.4

=0.8

Figure 12.


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-15 -10 -5 0 5 10 150

2

4

6

8

10

12

14

16

18

20

x

|Ery|

Ay= 1, A

x= 0

d = 0.5, =1

=0

=0.6

=0.8

Figure 13.

-15 -10 -5 0 5 10 150

2

4

6

8

10

12

14

x

|Erz|

Ay= 0, A

x= 1

d = 0.5, =1

=0

=0.5

=0.8

Figure 14.


17/23


-15 -10 -5 0 5 10 150

2

4

6

8

10

12

x

|Erx|

A

y= 0, A

x= 1

d = 0.5, = 0

=1

=5

=8

Figure 15.

-15 -10 -5 0 5 10 150

2

4

6

8

10

12

14

16

18

20

x

|Ery|

A

y= 1, A

x= 0

d = 0.5, = 0

=1

=4

=8

Figure 16.


18/23


-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

8

9

x

|Erz|

A

y= 0, A

x= 1

d = 0.5, = 0

=1

=3

=5

Figure 17.

-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

8

x

|Erx|

A

y= 0, A

x= 1

d = 0.1, = 0.3

=1

=5

=8

Figure 18.


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-15 -10 -5 0 5 10 150

5

10

15

20

25

x

|Ery|

Ay = 1, Ax = 0

d = 0.1, = 0.3

=1

=4

=8

Figure 19.

-15 -10 -5 0 5 10 150

1

2

3

4

5

6

7

8

x

|Erz|

A

y= 0, A

x= 1

d = 0.5, = 0.3

=1

=4

=8

Figure 20.


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the absolute value of reflected field component. All plots are taken atz = 0. We have studied the effect ofd, and on the field near focus.

Figures 35 show the effect of increase in the value of d keeping= 0 and = 3, that is ordinary dielectric case. We have shown the

|Erx| and |Erz | for (Ay = 0, Ax = 1). These figures show that increasein d results in increase in |Erx| and |Erz |. Figure 4 shows the plot of|Ery | for (Ay = 1, Ax = 0). This figure shows decrease in field strengthas d increases.

Figures 611 show the effect ofd while keeping = 0.5 and = 1for both types of polarization. Figures 7 and 11 show increase in thefield strength of |Erx| and |Erz | for (Ay = 0, Ax = 1), respectively.Figures 6 and 10 show the cross polarization effect due to chiral mediumfor polarization (Ay = 1, Ax = 0), which also increases with increasein d. For d = 0, cross polarization effect vanishes and field strengthincreases with increase in d. Figure 8 shows the plot of |Ery | for(Ay = 1, Ax = 0), which decreases with increase in d, and Figure 9shows the plot for (Ay = 0, Ax = 1). This figure shows that cross

polarization effect are zero for d = 0, but increases with increase inlayer thickness d.

Figures 1214 show the effect of chirality parameter on fieldcomponents keeping = 1 and d = 0.5. Figures 12 and 14 show theplots of |Erx| and |Erz | for (Ay = 0, Ax = 1) respectively. Thesefigures show that field strength increases with increase in chiralityparameter . Figure 13 shows that |Ery | decreases with increase in. Cross polarization effects have not been shown because they havesame trends as discussed above.

Figures 1517 show the effect of keeping d = 0.5 and = 0, thatis for ordinary dielectric case. Figures 15 and 17 show the plot of |Erx|and |Erz | for polarization (Ay = 0, Ax = 1) respectively and show theincrease in the field strength if we increase the relative permittivity of

dielectric medium coated on the surface of reflector. Figure 16, whichis for polarization (Ay = 0, Ax = 1), shows that |Ery | decreases withincrease in . Figures 1820, which are for chiral coating, show thesame trends as in Figures 1517 if we increase the relative permittivityof chiral medium.

6. CONCLUSIONS

We have studied the geometrical optics field reflected from a paraboliccylindrical reflector coated with chiral medium using Maslovs method.We have considered two types of polarization of incident plane wavewhich is normally incident on the reflector. We have studied the

reflected field component numerically and presented the plots in the


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focal plane. It is observed that increase in layer thickness, increasein chirality parameter and relative permittivity of layer, results in theincrease of |Erx| and |Erz | for (Ay = 0, Ax = 1). While |Ery | decreasesfor (Ay = 1, Ax = 0). Cross-polarized fields are present when = 0and increase with increase in d.

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characteristics of an inhomogeneous slab using Maslovs method,J. of Electromagn. Waves and Appl., Vol. 22, No. 2, 301312, 2008.

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