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

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5.7 Principle of Perturbation 311

The application of the principle of perturbation is to calculate the natural

frequency of a deformed cavity or a dielectric loaded cavity. For an example,

we investigate a change in the natural frequency by means of a change in

the dimensions of a circular cylindrical cavity operating at the TM

010

mode.

Any change in the length of the cavity will not change the natural frequency

because the electric and magnetic ﬁelds are perturbed simultaneously. But

squeezing the central part of the end walls will lower the natural frequency

because the electric ﬁeld is stronger at the central part, and reducing the

radius of the cavity or squeezing any part of the cylindrical wall will raise the

natural frequency because the magnetic ﬁeld is stronger there.

The other application is to measure the ﬁeld inside a cavity [32, 63].

This measurement is based on the fact that the resonant frequency change

of a cavity is proportional to the square of the ﬁeld strength at the location

where the perturbation object is placed. For this purpose, the conductor

perturbation can only be used in the region where the electric ﬁeld is superior

to the magnetic ﬁeld or vice versa. Otherwise, the dielectric perturbation

object is used to measure the electric ﬁeld and the magnetic perturbation

object is used to measure the magnetic ﬁeld.

The previous perturbation formulas are obtained for the lossless perturba-

tion object. We leave the analysis of the perturbation by a lossy perturbation

object as an exercise.

5.7.3 Cutoﬀ Frequency Perturbation of a Waveguide

Waveguide theory is a two-dimensional eigenvalue problem. The eigenvalue

of the problem, i.e., the cutoﬀ frequency of the waveguide, is just the natural

frequency of the two-dimensional resonant cavity. So the two-dimensional

forms of the perturbation formulas (5.352) and (5.367) become the pertur-

bation formulas for waveguides.

The wall perturbation formula for the waveguide cutoﬀ frequency is

∆ω

≈

∆S

µH

− ²E

²E

+ µH

, (5.370)

and the material perturbation formula for the waveguide cutoﬀ frequency is

∆ω

≈ −

∆S

∆²E

+ ∆µH

²E

+ µH

, (5.371)

where S denotes the cross section of the waveguide and ∆S denotes the cross

section of the perturbation object. Note that the perturbation object must

be a uniform cylinder in the z direction.

The wall perturbation formula (5.370) can help us to understand why the

single-mode frequency band of a ridge waveguide [22] shown in Fig. 5.42 is

broader than that of a rectangular waveguide. For the TE

mode, the ridge

is placed at the region of strong electric ﬁeld and weak magnetic ﬁeld, but for

312 5. Metallic Waveguides and Resonant Cavities

Figure 5.42: Ridge waveguides.

the TE

mode, the ridge is placed at the region of strong magnetic ﬁeld and

weak electric ﬁeld. In consequence, the cutoﬀ frequency of the TE

mode for

the rigid waveguide is lower then that for the original rectangular waveguide,

and the cutoﬀ frequency of the TE

mode for the rigid waveguide is higher

then that for the original rectangular waveguide.

The other feature of the ridge waveguide is that the characteristic

impedance of it is lower than that of the rectangular waveguide and is ad-

justable by means of adjusting the hight and the width of the ridge. So the

ridge waveguide can be used in impedance transformers or matching elements.

The ﬁeld analysis of the ridge waveguide is given as a problem, see Prob-

lem 5.16.

5.7.4 Propagation Constant Perturbation of a

Waveguide

Consider a single-mode waveguide involves a perturbation in the permittivity

of the ﬁlling medium and the perturbation is considered to be suﬃciently

weak so that the inﬂuences of the other modes can be neglected [23, 116].

The relative permittivity of the ﬁlling medium ²

in the original waveguide

is replaced by ²

+ ∆²

in the perturbed waveguide. In response of ²

∆²

, The propagation constant β of the waveguide is replaced by β + ∆β,

the transverse scalar wave function U

is replaced by U

+ ∆U

and the

transverse eigenvalue

= k

− β

= ²

− β

is replaced by

(T + ∆T )

= (²

+ ∆²

− (β + ∆β)

where k

= ω²

The equation (4.114) for the perturbed waveguide becomes

∇

+ ∆U

) +

(²

+ ∆²

− (β + ∆β)

+ ∆U

) = 0, (5.372)

5.7 Principle of Perturbation 313

where U

represents the transverse wave function for TM or TE modes, i.e.,

or V

, in Chapter 4. Multiplying this out and dropping the unperturbed

transverse wave equation, which is equal to zero, and neglecting the second-

order terms we obtain the ﬁrst-order perturbed transverse wave equation

∇

∆U

+ ²

∆U

+ ∆²

− 2β∆βU

− β

∆U

= 0. (5.373)

Multiplying the above equation by U

∗

and integrating over the cross

section S of the waveguide, we get

2β∆β

dS =

∆²

dS +

∗

∇

∆U

+ ²

∆U

∗

− β

∆U

∗

dS.

(5.374)

It can be shown that the second integral on the right-hand side vanishes.

Multiplying the complex conjugate of the unperturbed transverse wave

equation, (4.114), by ∆U

gives

∆U

∇

∗

+ ∆U

− β

∗

= 0, (5.375)

where ²

and β are assumed to be real. Thus, the last two terms in the second

integral on the right-hand side of (5.374) can be replaced by −∆U

∇

∗

and to give,

∗

∇

∆U

+ ²

∆U

∗

+ β

∆U

∗

∇

∆U

− ∆U

∇

∗

dS =

∇

·[U

∗

∇

∆U

− ∆U

∇

∗

] dS

∗

∇

∆U

− ∆U

∇

∗

]·ndl =

∗

∂∆U

∂ n

− ∆U

∂ U

∗

∂ n

dl, (5.376)

where the two dimensional Green theorem or Gauss theorem is used to con-

vert the surface integral over the cross section to a line integral around the

perimeter of the cross section, where n is the unit vector normal to the con-

tour of integration. The contour integral on the boundary is zero because

and ∆U

or ∂U

/∂ n and ∂∆U

/∂ n must vanish for guided mode in a

waveguide with short-circuit or open-circuit boundary. Then we have

∗

∇

∆U

+ ²

∆U

∗

+ β

∆U

∗

dS = 0.

Finally solving for ∆β in (5.374) we obtain the perturbation formula for

the propagation constant,

∆β =

∆²

2β

. (5.377)

In this expression, all quantities except for the perturbation of material, ∆²

are for the original unperturbed problem. One need not know the perturbed

ﬁeld to evaluate the perturbed propagation constant to the ﬁrst order in ∆²

314 5. Metallic Waveguides and Resonant Cavities

Note that, in (5.377), the perturbation of material, ∆²

, is not necessarily

uniform on the cross section although it has to be uniform along the axis z.

The perturbation formula for the propagation constant, (5.377), can also

be applied in the perturbation of dielectric waveguides, for which the surface

integral extends over the inﬁnite cross section and the contour l is at inﬁnity.

For guided modes in dielectric waveguides, U

and ∆U

vanish at inﬁnity,

refer to the next chapter.

Problems

5.1 (1) Show that the magnetic ﬁeld vector of the TE

mode in a rect-

angular waveguide is elliptically polarized, and point out the plane of

polarization.

(2) Find the position where the magnetic ﬁeld vector becomes circularly

polarized.

(3) Find the positions where the magnetic ﬁeld vector becomes linearly

polarized and point out the directions of the polarization.

5.2 Show that the conduction current on the wall of the rectangular cavity

of TE

101

mode is continuous with the displacement current in the space

inside the cavity.

5.3 The electric ﬁeld of the TEM mode satisﬁes two-dimensional Laplace’s

equation on the transverse cross section, prove that it doesn’t satisfy

the three-dimensional Laplace’s equation.

5.4 The deﬁnition of energy velocity is v

= P/W , where P denotes the

average power ﬂow through the cross section and W denotes the average

stored energy in a unit length of the waveguide.

Take the TE

mode in rectangular waveguide as an example, show that

the energy velocity is equal to the group velocity in the waveguide. Note

that the energy velocity is not always equal to the group velocity. See

Chapter 8.

5.5 Plot the ω–β diagrams of ﬁve low modes of two rectangular waveguides,

the ratios of the wide sides to the narrow sides of which are 1.5:1 and

3:1, respectively. Point out the diﬀerence of the sequences in the modes

between two waveguides.

5.6 Derive the eigenvalue equation of a circular metallic waveguide with a

thin conducting plate in it shown in Fig. 5.43. Point out the diﬀerence of

mode distributions between it and normal circular metallic waveguide.

5.7 Find the ratio of the length l to the radius a of a circular cylindrical

cavity, such that the ratio of the natural frequencies of the dominant

mode to that of the adjacent mode is 1.5:1.

Problems 315

Figure 5.43: Problem 5.6. Circular waveguide with thin conducting plate.

5.8 (1) Derive the expression for the attenuation coeﬃcient of the TE

mode in a circular waveguide due to the loss on the wall made by a

good conductor.

(2) Show that the attenuation coeﬃcient for the TE

mode is a

monotonously decreasing function with respect to the increasing fre-

quency, and the attenuation coeﬃcient for the mode with n 6= 0 has a

minimum at a certain frequency.

5.9 (1) Derive the expression for the attenuation coeﬃcient of the TM

mode in a circular waveguide due to loss on the wall made by a good

conductor.

(2) Show that the attenuation coeﬃcient for the TM

mode has a

minimum at frequency f =

√

, where f

denotes the cutoﬀ frequency

of the waveguide.

5.10 Take a lower mode as example, prove that the expressions for the ﬁelds

in the sectorial waveguide shown in Fig. 5.20(b) tend to those in the

rectangular waveguide when a − b ¿ a, a − b ¿ b, and α ¿ π.

5.11 Derive the expression for the Q factor of the TM

011

mode in a circular

cylindrical cavity due to the loss on a wall made by a go od conductor.

5.12 Derive the expression for the Q factors of the TM

101

and TE

101

modes

in a spherical cavity due to the loss on a wall made by a good conductor.

5.13 Derive the characteristic equation for the semi-spherical cavity, point

out the dominant mode, and give the expression of the natural fre-

quency of the dominant mode.

5.14 Derive the characteristic equation for the spherical-horn waveguide, and

describe the propagation characteristics of the dominant mode.

5.15

Derive the approximate characteristic equation of the dominant mo de

of a reentrant spherical cavity or capacity-loaded biconical cavity shown

in Fig. 5.44, using single-term approach in one region.

316 5. Metallic Waveguides and Resonant Cavities

Figure 5.44: Problem 5.15. Reentrant spherical cavity.

5.16 Derive the approximate cutoﬀ frequency, phase coeﬃcient and ﬁeld dis-

tribution of the dominant mo de in a symmetric ridge waveguide shown

in Fig. 5.42(b), using single-term approach in one region. Suppose the

width and the hight of the waveguide are a and b, respectively, the

width of the ridge is c and the space between two ridges is d.

5.17 A dielectric rod of radius b and permittivity ² is inserted into a circular

cylindrical cavity with radius a and length l along the axis. Derive

the expression for the change in the natural frequency of the TM

010

mode due to the insertion of the dielectric rod, by using the principle

of perturbation .

5.18 Show that the factor K of a conducting spherical perturbing object for

the electric ﬁeld is 3 and for the magnetic ﬁeld is 3/2. [32, 63]

5.19 (1) Show that the factor K of a nonmagnetic dielectric thin needle as

a perturbing object is 1 if the electric ﬁeld is tangential to the needle.

This approximation is independent of the cross-sectional shape of the

needle.

(2) Show that the factor K of a nonmagnetic dielectric thin disk as

a perturbing object is 1/²

if the electric ﬁeld is normal to the disk.

Again this approximation is independent of the shape of the disk.

(3) Show that the factor K of a nonmagnetic dielectric spherical per-

turbing object is 3/(2 + ²

Hint, in the above two problems, use a quasi-static approximation and

suppose that the unperturbed ﬁelds are uniform.

5.20 Derive the perturbation formula for a cavity, including the change in the

natural frequency and the Q factor due to introducing a lossy dielectric

perturbation object.

Chapter 6

Dielectric Waveguides and

Resonators

We have seen in the last chapter that, for single-mode operation, the dimen-

sions of waveguides and cavities must be of the same order as the operating

wavelength. The reasonable dimensions for metal work are in centimeter and

millimeter range. For this reason, metallic waveguides and cavity resonators

including coaxial lines and coaxial cavities are extensively used in microwaves,

i.e., centimeter and millimeter wave bands.

In a metallic waveguide, the wave propagation can be understood as a

plane wave being totally reﬂected between conducting boundaries and fol-

lowing a zigzag path by successive reﬂections. The same phenomenon is

observed in a dielectric slab or rod if the index of the slab or rod is larger

than that of the surrounding medium, and if the condition of total internal

reﬂection is satisﬁed. Hence, a wave may be guided without loss by a piece

of dielectric material having no metal boundaries. This kind of transmission

system is known as a dielectric waveguide. A Dielectric waveguide with a

relatively smaller cross section is much easier to fabricate. For example, a

dielectric waveguide with the thickness of a few micrometer and the width

of dozens of micrometer can easily be made by means of micro electronic

technology. As a result, dielectric waveguides are successfully used for the

millimeter to micrometer wave band, including the infrared and visible light.

For use in the millimeter wave band, a dielectric waveguide can be made

in the simple form as a thin dielectric rod or wire with rectangular or circular

cross section or a dielectric strip as for a microwave integrated circuit. For the

optical or light-wave band, two types of dielectric waveguide have been de-

veloped, they are optical waveguides and optical ﬁbers. Optical waveguides,

including the planar or slab waveguide and the strip or channel waveguide,

typically are composed of three layers of dielectric medium: a substrate, a

sheet or core, and a cover or cladding. The indices of refraction of the sub-

318 6. Dielectric Waveguides and Resonators

strate and of the cladding are slightly lower than that of the core, which

serves as the guiding layer. Optical ﬁbers, on the other hand, are made ei-

ther of fused quartz (silica) or of plastic, with their diameters ranging from

a few micrometer to about 0.5 mm. The index of refraction decreases in the

radial direction, either gradually or abruptly. The former type is known as

a graded-index ﬁber and the latter type of ﬁber is known as a step-index

ﬁber. Planar and strip waveguides are the basic components in integrated

optics or photonic integrated circuits. The optical ﬁber as a distinguished

invention has been the most important long-distance transmission medium

in communication systems.

In a dielectric waveguide, although a large majority of the power ﬂows

through the inner medium, namely the core, there is still stray power that

ﬂows through the outer one, namely the cladding, and the wave is not to-

tally conﬁned as with a metallic waveguide. The ﬁeld solution of a dielectric

waveguide is composed of a number of guided modes or conﬁned modes and

radiation modes, which form a complete set of orthogonal modes. For guided

modes, the ﬁelds in the cladding are decaying ﬁelds without transverse radi-

ation and the ﬁelds in the core are traveling waves with a small attenuation.

For radiation mo des, the ﬁelds in the cladding are traveling waves in the

transverse direction and the ﬁelds in the core become damping waves with

a large attenuation. The former corresponds to the case of total internal re-

ﬂection from the dielectric boundaries, which occurs when the incident angle

is larger than the critical angle; and the latter corresponds to the case of

transmission through the boundaries, which occurs when the incident angle

is smaller than the critical angle. The frequency limit of the guided mode

is known as the critical frequency. Usually, it is also called the cutoﬀ fre-

quency but the term cutoﬀ for a dielectric waveguide has an entirely diﬀerent

meaning than that for a metallic waveguide.

In the metallic waveguide with a uniform ﬁlling medium, the ﬁelds of a

TE or a TM mode alone can arrange themselves to satisfy the boundary con-

ditions. But in dielectric waveguides and metallic waveguides with diﬀerent

ﬁlling media, any TE or TM modes can exist by itself only in the special

case when the ﬁelds are uniform along the transverse direction of the bound-

ary. Other than this special case, no TE or TM mode alone can satisfy the

boundary conditions, thus only hybrid modes can survive there.

Dielectric resonators, which have been developed rapidly and are widely

used in microwave integrated circuits, are also discussed in this chapter.

For the generality of the theory, we assume that both permittivity and

permeability are diﬀerent for diﬀerent media. In practical use, however, most

devices have made use of media with diﬀerent permittivity, but the same

permeability µ

In order to make clear the nature and the inﬂuence of the dielectric bound-

ary on the wave modes, we start by the study of metallic waveguide ﬁlling

with diﬀerent media and form a dielectric boundary inside the waveguide.

6.1 Metallic Waveguide with Diﬀerent Filling Media 319

Figure 6.1: Metallic waveguides ﬁlled with two diﬀerent media.

6.1 Metallic Waveguide with Diﬀerent Filling

Media

The metallic waveguides discussed in the last chapter are ﬁlled with uniform

medium, in which any single TE or TM mode can exist independently. When

a waveguide is ﬁlled with diﬀerent media or partially ﬁlled with a dielectric

medium, the electric and magnetic ﬁelds must satisfy the boundary condi-

tions not only on the metallic boundaries but also on the boundary between

diﬀerent media. Rectangular metallic waveguides with two diﬀerent section-

ally uniform ﬁlling media are shown in Fig. 6.1. [81]

6.1.1 The Possible TE and TM Modes

We deal with a waveguide with two diﬀerent ﬁlling media aligned in the x

direction as shown in Fig. 6.1(a). In region 1, the constitutive parameters of

the medium are ²

, µ

and the angular wave numbers in three coordinates are

, k

y 1

, and β

; whereas in region 2, the constitutive parameters are ²

, µ

and the angular wave numbers are k

, k

y 2

, and β

. To satisfy the boundary

condition or so called phase matching condition between the two media, the

tangential angular wave numbers on both sides must be continuous:

= β

= β, k

y 1

= k

y 2

= k

. (6.1)

+ k

+ β

= k

= ω

, (6.2)

+ k

+ β

= k

= ω

. (6.3)

(1) TE Modes

For TE modes, U = 0 and E

= 0. Applying the boundary conditions on the

conducting wall, we have the functions V

and V

for the two regions

= A cos k

x cos k

−jβz

, 0 ≤ x ≤ h, (6.4)

= B cos k

(x − a) cos k

−jβz

, h ≤ x ≤ a, (6.5)

320 6. Dielectric Waveguides and Resonators

where k

= nπ/b. Substituting them into (4.147)–(4.152), we may obtain the

ﬁeld-component expressions in the two regions.

The boundary conditions at x = h are such that the tangential compo-

nents of the electric and magnetic ﬁelds E

, H

, and H

on the two sides of

the boundary must be continuous, i.e.,

y 1

(h)=E

y 2

(h) → µ

∂V

∂x

=µ

∂V

∂x

, i.e., B =

A sink

sink

(h−a)

, (6.6)

y 1

(h) = H

y 2

(h) →

∂V

∂y

∂V

∂y

, (6.7)

and

(h) = H

(h) →

+ k

. (6.8)

Substituting (6.6) into (6.7) and (6.8), we obtain two eigenvalue equations,

tan k

h = µ

tan k

(h − a) (6.9)

and

+ k

tan k

h =

+ k

tan k

(h − a). (6.10)

For guided modes, the above two equations are to be satisﬁed simultane-

ously, we ought to have

+ k

. i.e., k

= k

, µ

= µ

. (6.11)

This means that the indices of refraction or the phase velocities in the two

media must be equal. This case is trivial because the two diﬀerent media act

like uniform medium.

The other possibility is

= 0, i.e., n = 0, (6.12)

which corresponds to TE

modes. In this case,

y 1

= −jβ

∂V

∂y

= 0, and H

y 2

= −jβ

∂V

∂y

= 0.

Now (6.7) and (6.9) are no longer needed, and (6.10) alone gives the necessary

eigenvalue equation.

If we apply the condition k

= 0, the eigenvalue equation (6.10) becomes

tan k

h =

tan k

(h − a). (6.13)

Under the condition that k

= 0, from (6.2) and (6.3), we have

− k

= ω

(µ

− µ

). (6.14)