Zhang K., Li D. Electromagnetic Theory for Microwaves and Optoelectronics

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4.6 Electromagnetic Waves in Cylindrical Systems 201

General

9. Conical coordinates: Consists of a set of concentric spheres and

two sets of mutual orthogonal elliptic cones.

10. Ellipsoidal coordinates: Consists of a set of ellipsoids, a set of

hyperboloids of one sheet, and a set of hyperboloids of two sheets.

11. Paraboloidal coordinates: Consists of two sets of mutually or-

thogonal elliptic paraboloids, and a set of hyperbolic paraboloid.

For the details of separation of variables in the eleven coordinate systems,

please refer to [70, 71, 72, 114].

4.6 Electromagnetic Waves in

Cylindrical Systems

In an arbitrary cylindrical coordinate system, u

, u

, z, all of the functions

U, V , Π

, Π

, E

, H

satisfy the same scalar Helmholtz equation:

∇

U +

∂

∂z

+ k

U = 0. (4.109)

Applying the method of separation of variables, let

U(u

, u

, z) = U

, u

)Z(z), (4.110)

where U

denotes the transverse function and Z denotes the longitudinal

function. Substituting it into (4.109) and dividing by U yields

∇

Z/dz

= −k

. (4.111)

The ﬁrst term is a function of u

and u

only and the second term is a

function of z only. Each of them must be equal to a constant so that the sum

of them can be a constant −k

. Let

Z/dz

= −β

∇

= −T

and

+ T

= k

, β =

− T

. (4.112)

Then we have

+ β

Z = 0. (4.113)

∇

+ T

= 0. (4.114)

202 4. Time-Varying Boundary Value Problems

The ﬁrst equation (4.113) is a one-dimensional homogeneous scalar Helmholtz

equation and we have seen it in Sections 2.1 and 3.1, the latter is known as

the telegraph equation.

The solutions of (4.113) are two traveling waves propagating along +z

and −z, known as guided waves,

Z(z) = Z

−jβz

+ Z

jβz

, (4.115)

where β is the longitudinal phase coeﬃcient which is determined by k = ω

√

µ²

and T in (4.112).

The second equation (4.114) is a two-dimensional scalar Helmholtz equa-

tion which is known as the transverse wave equation and T

is the transverse

eigenvalue which is determined by the boundary conditions of the system.

The guided waves in a bounded cylindrical system are classiﬁed as follow-

ings according to the transverse eigenvalue T

(1) The TEM Mode

When T

= 0, then β = k = ω

√

µ² and v

= 1/

√

µ². This is a wave with

a velocity equal to the velocity of a plane wave in the unbounded medium,

and ∂

/∂z

= −k

. According to (4.36) and (4.39), this must be a wave with

neither electric nor magnetic ﬁeld in the direction of propagation, E

= 0,

= 0. It is known as a transverse electromagnetic wave and is denoted by

the TEM mode.

In the case of T

= 0, The equation for U

, (4.114), and the similar

equation for V

become

∇

= 0, ∇

= 0. (4.116)

So the Borgnis’ functions U and V satisfy the transverse two-dimensional

scalar Laplace equations.

Under the conditions E

= 0, H

= 0 and E = E

, H = H

, the

equations of the transverse ﬁelds (4.88) and (4.90) become

∇

× E = 0, ∇

× H = 0. (4.117)

The transverse ﬁelds are irrotational vector functions in the transverse cross-

section, and may be explained in terms of the two-dimensional gradient of

scalar potentials ϕ(u

, u

) and ψ(u

, u

) as follows:

E(u

, u

) = ∇

ϕ(u

, u

), H(u

, u

) = ∇

ψ(u

, u

). (4.118)

It has been shown in Maxwell’s equations that in source-free problems, the

ﬁelds are solenoidal vector functions, so that

∇

· E = 0, ∇

· H = 0. (4.119)

4.6 Electromagnetic Waves in Cylindrical Systems 203

Hence we have

∇

ϕ(u

, u

) = 0, ∇

ψ(u

, u

) = 0. (4.120)

It is shown that the scalar potentials of the TEM mode satisfy the two-

dimensional Laplace equations. The transverse distribution of the electric

and magnetic ﬁelds in a TEM wave are the same as those in static ﬁelds.

We come to the conclusion that the TEM mode can exist only in a system

that can support static ﬁelds. This means that the TEM mode can not be

supported by a single conductor or insulator, no matter what the conﬁgura-

tion is, and only the system composed of at least two conductors insulated

from each other can support TEM waves. This kind of system is known as

a transmission line and can be analyzed by means of the circuit approach as

well as the ﬁeld approach, refer to Chapters 3 and 5, respectively.

(2) Fast Wave modes

Here T

> 0, T is real. The ﬁelds in the system depend on the relation

between T

and k

as follows:

(a) T

< k

, β

= k

− T

> 0, β is real and β < k. Since v

= ω/β and

ω/k = 1/

√

µ², we have

√

− T

√

µ²

1 − T

√

µ²

. (4.121)

This is a traveling wave along z, and the phase velocity is larger than the

phase velocity of a plane wave in the unbounded media. So it is a fast wave

mode. In vacuum, v

> c = 1/

√

. The fact that the phase velocity

is larger than the velocity of light in vacuum does not violate the special

theory of relativity of Einstein , because the phase velocity does not bring

any matter, energy, or signal with it.

Since T

is a constant, the group velocity becomes

dω

dβ

dω/dβ

√

µ²

1 −

√

µ²

. (4.122)

So the group velocity is less then the velocity of a plane wave in the un-

bounded medium, and

µ²

, in vacuum v

= c

. (4.123)

This is the propagation state of a fast wave mode in common metallic

waveguides, refer to Chapter 5. The modes in propagation state are called

propagating modes or guided modes.

204 4. Time-Varying Boundary Value Problems

(b) T

> k

, β

= k

− T

< 0, β is imaginary. The ﬁeld is not a traveling

wave but a damping or decaying ﬁeld along z. This is the cutoﬀ state of

a waveguide mode. The modes in cutoﬀ state are called cutoﬀ modes or

evanescent modes.

= k

, β

= k

− T

= 0. This is the critical state of a waveguide

mode. Hence the transverse eigenvalue T is also known as the critical angular

wave number or cutoﬀ angular wave number,

= ω

√

µ² = T, (4.124)

where ω

denotes the cutoﬀ angular frequency of the waveguide.

(3) Slow Waves

When T

< 0, then T is imaginary, and β

= k

− T

> k

, β is real and

β > k. So

√

− T

√

µ²

1 − T

√

µ²

(4.125)

and the phase velocity along z is less then the phase velocity of a plane wave

in the unbounded medium. So it is known as a slow wave, refer to Chapters

6 and 7).

In fast-wave and TEM-wave systems, T

≥ 0, which is consistent with the

condition given in Theorem 2 for the Sturm–Liouville problems which will be

shown in Section 4.10.1. The theorem indicates that in the system with ho-

mogeneous boundary conditions, the eigenvalue of the Sturm–Liouville prob-

lem must not be negative. So fast-wave or TEM-wave systems must be

surrounded by short-circuit or open-circuit boundaries, generally conducting

walls.

On the contrary, in slow wave systems, T

< 0, which cannot be the eigen-

value of Sturm–Liouville problems with homogeneous boundary conditions.

So a system surrounded by smooth short-circuit or open-circuit boundaries

cannot support slow waves. The slow wave systems are constructed by means

of dielectric boundaries or corrugated metallic boundaries.

For a slow wave, the eigenvalue T

is no longer constant, so (4.122) and

(4.123) are no longer valid. The group velocity of a slow wave is still less then

or equal to the velocity of light in space. In some systems, two or even three

sorts of waves can be supp orted simultaneously. In most cases, the ﬁelds

related to U or the ﬁelds related to V can satisfy the boundary conditions

independently, and the waves may be classiﬁed as the following two kinds of

mode:

1. Transverse electric mode, denoted as the TE mode or H mode, with

U = 0, Π

= 0, and E

= 0. There are only a transverse electric

4.7 Solutions in Rectangular Coordinates 205

ﬁeld component and both the transverse and longitudinal magnetic

components in the TE mode.

2. Transverse magnetic mode, denoted as the TM mode or E mode, with

V = 0, Π

= 0, and H

= 0. There are only a transverse mag-

netic ﬁeld component and both the transverse and longitudinal electric

components in the TM mode.

In some cases, the ﬁelds of the TE or TM mode alone cannot satisfy the

boundary conditions, and the only possible mode is the hybrid electric and

magnetic mode denoted by HEM mode. See Chapters 6 and 7 for details.

4.7 Solution of Helmholtz’s Equations in

Rectangular Coordinates

The three axes x, y, z in a rectangular coordinate system are all Cartesian

coordinates, h

= 1, h

= 1. So we may choose any one of them as

the special coordinate u

and the vector Helmholtz equations may be solved

by means of any one of the methods given in Section 4.3.

4.7.1 Set z as u

In rectangular coordinates, if we choose x = u

, y = u

, z = u

, then

= h

= 1, and equation (4.114) becomes

∂

∂x

∂

∂y

+ T

= 0. (4.126)

Applying the method of separation of variables, let

(x, y) = X(x)Y (y), U(x, y, z) = X(x)Y (y)Z(z). (4.127)

Substituting this into (4.126) and subtracting U

yields

X/dx

Y/dy

= −T

. (4.128)

The ﬁrst term is a function of x only and the second term is a function of y

only. Each of them must be equal to a constant and the sum of them must

be equal to the constant −T

. As in the last section, we have

+ k

X = 0,

+ k

Y = 0, (4.129)

where

+ k

= T

. (4.130)

206 4. Time-Varying Boundary Value Problems

In the last section, we had the diﬀerential equation of Z (4.113):

+ β

Z = 0, and β = k

, k

+ k

+ β

= k

. (4.131)

Equations (4.129) and (4.131) are one-dimensional homogeneous scalar

Helmholtz equations or the telegraph equations, which are the same as the

equation for a plane wave and a transmission line given in Chapter 2. Their

solutions are as follows.

If k

, k

, and β are real,

X(x) = Ae

+ Be

−jk

=a sin k

x + b cos k

x =X

sin(k

x + φ

), (4.132)

Y (y) = Ce

+ De

−jk

=c sin k

y + d cos k

y =Y

sin(k

y + φ

), (4.133)

Z(z)=F e

jβz

+ Ge

−jβz

=f sin βz + g cos βz = Z

sin(βz + φ

). (4.134)

If k

, k

, and β are imaginary, let

= jK

, k

= jK

, β = jK

, (4.135)

and

+ K

= τ

, K

+ τ

= −k

. (4.136)

Then (4.129) and (4.131) become

− K

X = 0,

− K

Y = 0,

− K

Z = 0. (4.137)

The solutions become

X(x) = Ae

+ Be

−K

=a sinh K

x + b cosh K

x =X

sinh(K

x + ψ

(4.138)

Y (y) = Ce

+ De

−K

=c sinh K

y + d cosh K

y =Y

sinh(K

y + ψ

(4.139)

Z(z)=F e

+ Ge

−K

=f sinh K

z + g cosh K

z =Z

sinh(K

z + ψ

(4.140)

The above sets of functions are the eigenfunctions of the Laplace equation

as well as the Helmholtz equation in rectangular coordinates. They are called

the rectangular harmonics. They are all orthogonal and complete function

sets. k

= −K

, k

= −K

, β

= −K

are the corresponding eigenvalues,

which must be speciﬁc discrete real values for the speciﬁc boundary condi-

tions.

The exponential functions with real arguments and the hyperbolic func-

tions are monotonic increasing or decreasing functions or at most the hyper-

bolic cosine function with one minimum and the hyperbolic sine with one

zero, see Fig. 4.4(a). On the contrary, the sine and cosine functions are p eri-

odic functions with multiple zeros, see Fig. 4.4(b). The exponential functions

with imaginary arguments are generally complex.

4.7 Solutions in Rectangular Coordinates 207

-1.0

-.5

0.0

1.0

sin x

cos x

S

S

S

-3 -2 -1 0 1 2 3

-5

-4

-3

-2

-1

cosh x

sinh x

-x

S

S

S

S

(a)

(b)

sin x

cos x

-x

sinhx

coshx

Figure 4.4: Rectangular harmonics.

A single term of a harmonic function cannot certainly satisfy the speciﬁc

boundary conditions. According to the expansion theorem of the eigenvalue

problems, the solution can be expanded into series in terms of the rectangular

harmonics with initially arbitrary coeﬃcients to be chosen to satisfy the ﬁnal

boundary conditions.

The form of the solution of the function V is the same as that of the

function U . The ﬁeld components may then be obtained in terms of U and

V by (4.34)–(4.39) as follows:

∂

∂x ∂z

− jωµ

∂V

∂y

, (4.141)

∂

∂y ∂z

+ jωµ

∂V

∂x

, (4.142)

∂

∂z

+ k

U =

− k

U = T

U = −τ

U, (4.143)

∂

∂x ∂z

+ jω²

∂U

∂y

, (4.144)

∂

∂y ∂z

− jω²

∂U

∂x

, (4.145)

∂

∂z

+ k

V =

− k

V = T

V = −τ

V. (4.146)

If we are interested in the wave along the positive direction of the axis +z

only, according to (4.63)–(4.68), the ﬁeld components become:

= −jβ

∂U

∂x

− jωµ

∂V

∂y

, (4.147)

208 4. Time-Varying Boundary Value Problems

= −jβ

∂U

∂y

+ jωµ

∂V

∂x

, (4.148)

= T

U = τ

U, (4.149)

= −jβ

∂V

∂x

+ jω²

∂U

∂y

, (4.150)

= −jβ

∂V

∂y

− jω²

∂U

∂x

, (4.151)

= T

V = τ

V. (4.152)

Note that the general forms of the solutions of U and V are the same but

their boundary conditions are diﬀerent, so the ﬁnal expressions for U and V

must be diﬀerent.

4.7.2 Set x or y as u

All the three axes x, y, z in a rectangular coordinate system are Cartesian

coordinates. So we may also choose x or y as the special coordinate u

and the ﬁeld components can be explained by means of the corresponding

Borgnis’ potentials. Let U

(x)

and V

(x)

denote the Borgnis’ potentials when

x is chosen as u

, and let U

(y )

and V

(y )

denote the Borgnis’ potentials when

y is chosen as u

. All of the functions U

(x)

, V

(x)

, U

(y )

and V

(y )

satisfy the

same scalar Helmholtz equation for the functions U and V in Section 4.3.1.

So the forms of the solutions of U

(x)

, V

(x)

, U

(y )

and V

(y )

are the same as

those for U and V , (4.132)–(4.134) and (4.138)–(4.140). We must give only

the relations between the ﬁeld components and the Borgnis’ potentials U

(x)

or U

(y )

, V

(y )

(1) Set x as u

, y as u

, and z as u

The expressions of the ﬁeld components (4.34)–(4.39) become

∂

(x)

∂x

+ k

(x)

− k

(x)

, (4.153)

∂

(x)

∂y ∂x

− jωµ

∂V

(x)

∂z

, (4.154)

∂

(x)

∂z ∂x

+ jωµ

∂V

(x)

∂y

, (4.155)

∂

(x)

∂x

+ k

(x)

− k

(x)

, (4.156)

∂

(x)

∂y ∂x

+ jω²

∂U

(x)

∂z

, (4.157)

4.8 Solutions in Circular Cylindrical Coordinates 209

∂

(x)

∂z ∂x

− jω²

∂U

(x)

∂y

. (4.158)

The mode in which V

(x)

= 0 and U

(x)

6= 0 is called the TM

(x)

or the E

(x)

mode, which is also known as the LSM

(x)

mode, where all the magnetic ﬁeld

components are in a longitudinal section, the x–z plane. On the contrary,

The mode in which U

(x)

= 0 and V

(x)

6= 0 is called the TE

(x)

or the H

(x)

mode, which is also known as the LSE

(x)

mode, where all the electric ﬁeld

components are in the x–z plane.

(2) Set y as u

, z as u

, and x as u

The expressions of the ﬁeld components (4.34)–(4.39) become

∂

(y )

∂x ∂y

+ jωµ

∂V

(y )

∂z

, (4.159)

∂

(y )

∂y

+ k

(y )

− k

(y )

, (4.160)

∂

(y )

∂z ∂y

− jωµ

∂V

(y )

∂x

, (4.161)

∂

(y )

∂x ∂y

− jω²

∂U

(y )

∂z

, (4.162)

∂

(y )

∂y

+ k

(y )

− k

(y )

, (4.163)

∂

(y )

∂z ∂y

+ jω²

∂U

(y )

∂x

. (4.164)

The mode in which V

(x)

= 0 and U

(x)

6= 0 is called the TM

(x)

or LSM

(x)

mode and the mode in which U

(y )

= 0 and V

(y )

6= 0 is called the TE

(y )

LSE

(y )

mode. LSM

(x)

mode and LSE

(y )

mode are two kind of hybrid (HEM)

modes. See Chapters 5 and 6. The common TE and TM modes, where z is

set as u

, are exactly TE

(z)

and TM

(z)

modes.

4.8 Solution of Helmholtz’s Equations in

Circular Cylindrical Coordinates

For circular cylindrical coordinates,

= ρ, u

= φ, u

= z, h

= h

= 1, h

= ρ,

the equations for U (and V , the same) are scalar Helmholtz equations,

∂

∂ρ

∂U

∂ρ

∂

∂φ

∂

∂z

+ k

U = 0. (4.165)

210 4. Time-Varying Boundary Value Problems

This equation can be separated into two equations, the equation for the

longitudinal function Z is (4.113), as follows:

= −β

Z, (4.166)

where

+ β

= k

and the equation for the transverse function U

(ρ, φ), (4.114), becomes

∂

∂ρ

∂U

∂ρ

∂

∂φ

+ T

= 0. (4.167)

Let

U(ρ, φ, z) = R(ρ)Φ(φ)Z(z), (4.168)

i.e.,

(ρ, φ) = R(ρ)Φ(φ). (4.169)

Substituting this into (4.167), and multiplying by ρ

/U, we have

ρ d(ρ dR/dρ)/dρ

Φ/dφ

= −T

This equation may be separated into the following two equations:

dφ

= −ν

Φ, (4.170)

dρ

− ν

R = 0. (4.171)

The equations (4.166) and (4.170) are the same as those for rectangular

coordinates, (4.129) and (4.131). The solution of (4.166) has been given as

follows,

Z(z) = F e

jβz

+ Ge

−jβz

= f sin βz + g cos βz = sin(βz + ψ

) (4.172)

for a propagation state or

Z(z) = F e

+ Ge

−K

= f sinh K

z + g cosh K

z (4.173)

for a cutoﬀ state, where β = jK

The solution of (4.170) is given by

Φ(φ) = C

cos νφ + D

sin νφ = c

jνφ

+ d

−jνφ

, (4.174)

In (4.171), let x = T ρ, then (4.171) becomes the standard form of the

Bessel equation:

dR(x)

− ν

R(x) = 0. (4.175)