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Lecture 22: Quality Factor of Cavities and Mode Orthogonality

Lecture notes on the quality factor of cavity resonators and mode orthogonality in electromagnetic field theory. Covers Q formulas, pole relations, TM110 mode.

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Lecture 22

Quality Factor of Cavities and

Mode Orthogonality

22.1 The Quality Factor of a Cavity

22.1.1 General Concepts

The quality factor of a cavity or its Q measures how ideal or lossless a cavity resonator is.

An ideal lossless cavity resonator will sustain free oscillations forever, while most resonators

sustain free oscillations for a finite time due to losses coming from radiation, or dissipation in

the dielectric material filling the cavity, or resistive loss of the metallic part of the cavity. Be-

cause of losses, the free oscillation in a cavity has electromagnetic field with time dependence

as follows:

E ∝ e^(-αt) cos(ωt + φ1), H ∝ e^(-αt) cos(ωt + φ2) (22.1.1)

The total time-average stored energy, which is proportional to 1/4 |E|^2 + 1/4 μ |H|^2, is of the form

⟨WT⟩ = ⟨WE⟩ + ⟨WH⟩ = W0 e^(-2αt) (22.1.2)

If there is no loss, ⟨WT⟩ will remain constant. However, with loss, the average stored energy

will decrease to e^-1 of its original value at t = 1/(2α). The Q of a cavity is a measure of the

number of free oscillations the field would have before the energy stored decreases to e^-1 of

its original value (see Figure 22.1). In a time interval t = 1/(2α), the number of free oscillations

in radians is ωt or ω/(2α); hence, the Q is defined to be [31]

Q = ω/(2α) (22.1.3)

215

216 Electromagnetic Field Theory

Figure 22.1: A typical time domain response of a high Q system (courtesy of Wikipedia).

Furthermore, by energy conservation, the decrease in stored energy per unit time must be

equal to the total power dissipated in the losses of a cavity, in other words,

⟨PD⟩ = - d⟨WT⟩/dt (22.1.4)

By further assuming that WT has to be of the form in (22.1.2), then

- d⟨WT⟩/dt = 2αW0 e^(-2αt) = 2α⟨WT⟩ (22.1.5)

Hence, we can write equation (22.1.3) as

Q = ω⟨WT⟩ / ⟨PD⟩ (22.1.6)

By further letting ω = 2π/T, we lent further physical interpretation to express Q as

Q = 2π ⟨WT⟩ / (⟨PD⟩T) = 2π total energy stored / Energy dissipated/cycle (22.1.7)

In a cavity, the energy can dissipate in either the dielectric loss or the wall loss of the

cavity due to the finiteness of the conductivity.

22.1.2 Relation to the Pole Location

The resonance of a system is related to the pole of the transfer function. For instance, in our

previous examples, the reflection coefficient can be thought of as a transfer function in linear

Quality Factor of Cavities and Mode Orthogonality 217

system theory: The input is the incident wave, while the output is the reflected wave. If

we encounter the resonance of the system at a particular frequency, the reflection coefficient

becomes infinite. This infinite value can be modeled by a pole of the transfer function in the

complex ω plane. In other words, in the vicinity of the pole in the frequency domain,

R(ω) ∼ A / (ω - ωp) (22.1.8)

In principle, when ω = ωp, the reflection coefficient becomes infinite, but this does not

happen in practice due to loss in the system. In other words, ωp is not a real number and the

pole does not lie on the real ω axis; the pole is displaced slightly off the real axis to account

for loss. In fact, it is quite easy to show that the pole is at ωp = ω0 + jα.1 Then the full-width

half maximum (FWHM) bandwidth is Δω = 2α, and the Q can be written as

Q = ω0/Δω (22.1.9)

Typical plots of transfer function versus frequency for a system with different Q’s are shown

in Figure 22.2.

Figure 22.2: A typical system response versus frequency for different Q’s (courtesy of allaboutcircuits.com).

1 It is quite easy to show that the Fourier inverse transform of the above corresponds to a decaying sinusoid [45].

22.1.3 Some Formulas for Q for a Metallic Cavity

If the cavity is filled with air, then, the loss comes mainly from the metallic loss or copper-loss

from the cavity wall. In this case, the power dissipated on the wall is given by [31]

⟨PD⟩ = 1/2 Re ∮S (E × H*) · n̂ dS = 1/2 Re ∮S (n̂ × E) · H* dS (22.1.10)

where S is the surface of the cavity wall.2 Here, (n̂ × E) is the tangential component of the

electric field which would have been zero if the cavity wall is made of ideal PEC. Also, n̂

is taken to be the outward pointing normal at the surface S. However, for metallic walls,

n̂ × E = HtZm where Zm is the intrinsic impedance for the metallic conductor, viz., Zm =

μ/εm ≈ sqrt(-jμ/σ ω) = sqrt(ωμ/(2σ)) (1 + j),3 where we have assumed that εm ≈ -jσ/ω, and Ht is the

tangential magnetic field. This relation between E and H will ensure that power is flowing

into the metallic surface. Hence,

⟨PD⟩ = 1/2 Re ∮S sqrt(ωμ/(2σ)) (1 + j) |Ht|^2 dS

= 1/2 sqrt(ωμ/(2σ)) ∮S |Ht|^2 dS (22.1.11)

By further assuming that the stored electric and magnetic energies of a cavity are equal

to each other at resonance, the stored energy can be obtained by

⟨WT⟩ = 1/2 μ ∫V |H|^2 dV (22.1.12)

Written explicitly, the Q becomes

Q = sqrt(2ωμσ) ∫V |H|^2 dV / ∮S |Ht|^2 dS

= 2/δ ∫V |H|^2 dV / ∮S |Ht|^2 dS (22.1.13)

In the above, δ is the skin depth of the metallic wall. Hence, the more energy stored we can

have with respect to the power dissipated, the higher the Q of a resonating system. The lower

the metal loss, or the smaller the skin depth, the higher the Q would be.

Notice that in (22.1.13), the numerator is a volume integral and hence, is proportional to

volume, while the denominator is a surface integral and is proportional to surface. Thus, the

Q, a dimensionless quantity, is roughly proportional to

Q ∼ V / (Sδ) (22.1.14)

where V is the volume of the cavity, while S is its surface area. From the above, it is noted

that a large cavity compared to its skin depth has a larger Q than an small cavity.

2 We have used the cyclic identity that a · (b × c) = b · (c × a) = c · (a × b) in the above (see Some Useful

Mathematical Formulas).

3 When an electromagnetic wave enters a conductive region with a large β, it can be shown that the wave

is refracted to propagate normally to the surface, and hence, this formula can be applied.

22.1.4 Example: The Q of TM110 Mode

For the TM110 mode, as can be seen from the previous lecture, the only electric field is

E = ẑ Ez, where

Ez = E0 sin(πx/a) sin(πy/b) (22.1.15)

The magnetic field can be derived from the electric field using Maxwell’s equation or Faraday’s

law, and

Hx = jωε/(ω^2 με) ∂/∂y EZ = j (π/b)/(ωμ) E0 sin(πx/a) cos(πy/b) (22.1.16)

Hy = -jωε/(ω^2 με) ∂/∂xEZ = -j (π/a)/(ωμ) E0 cos(πx/a) sin(πy/b) (22.1.17)

Therefore

∫V |H|^2 dV = ∫0^-d ∫0^b ∫0^a dx dy dz [|Hx|^2 + |Hy|^2]

= |E0|^2/(ω^2 μ^2) ∫0^-d ∫0^b ∫0^a dx dy dz

[ (π/b)^2 sin^2(πx/a) cos^2(πy/b) + (π/a)^2 cos^2(πx/a) sin^2(πy/b) ]

= |E0|^2/(ω^2 μ^2) π^2/4 [a/b + b/a] d (22.1.18)

A cavity has six faces, finding the tangential exponent at each face and integrate

∮S |Ht| dS = 2 ∫0^b ∫0^a dx dy [|Hx|^2 + |Hy|^2]

+ 2 ∫0^-d ∫0^a dx dz |Hx(y = 0)|^2 + 2 ∫0^-d ∫0^b dy dz |Hy(x = 0)|^2

= 2 |E0|^2/(ω^2 μ^2) π^2ab/4 [1/a^2 + 1/b^2]

+ 2 (π/b)^2/(ω^2 μ^2) |E0|^2 ad/2

+ 2 (π/a)^2/(ω^2 μ^2) |E0|^2 bd/2

= π^2 |E0|^2/(ω^2 μ^2) [ b/(2a) + a/(2b) + ad/b^2 + bd/a^2 ] (22.1.19)

Hence the Q is

Q = 2/δ (ad/b + bd/a) / (b/2a + a/2b + ad/b^2 + bd/a^2) (22.1.20)

The result shows that the larger the cavity, the higher the Q. This is because the Q, as

mentioned before, is the ratio of the energy stored in a volume to the energy dissipated over

the surface of the cavity.

22.2 Mode Orthogonality and Matrix Eigenvalue Problem

It turns out that the modes of a waveguide or a resonator are orthogonal to each other. This

is intimately related to the orthogonality of eigenvectors of a matrix operator.4 Thus, it is

best to understand this by the homomorphism between the electromagnetic mode problem

and the matrix eigenvalue problem. Because of this similarity, electromagnetic modes are also

called eigenmodes. Thus it is prudent that we revisit the matrix eigenvalue problem (EVP).

22.2.1 Matrix Eigenvalue Problem (EVP)

It is known in matrix theory that if a matrix is hermitian, then its eigenvalues are all real.

Furthermore, their eigenvectors with distinct eigenvalues are orthogonal to each other [69].

Assume that an eigenvalue and an eigenvector exists for the hermitian matrix A. Then

A · vi = λi vi (22.2.1)

Dot multiplying the above from the left by v†i where † indicates conjugate transpose, then

the above becomes

v†i · A · vi = λi v†i · vi (22.2.2)

Since A is hermitian, then the quantity v†i ·A·vi is purely real. Moreover, the quantity v†i ·vi

is positive real. So in order for the above to be satisfied, λi has to be real.

To prove orthogonality of eigenvectors, now, assume that A has two eigenvectors with

distinct eigenvalues such that

A · vi = λi vi (22.2.3)

A · vj = λj vj (22.2.4)

Left dot multiply the first equation with v†j and do the same to the second equation with v†i,

one gets

v†j · A · vi = λi v†j · vi (22.2.5)

v†i · A · vj = λj v†i · vj (22.2.6)

Taking the conjugate transpose of (22.2.5) in the above, and since A is hermitian, their left-

hand sides (22.2.5) and (22.2.6) become the same. Subtracting the two equations, we arrive

at

0 = (λi - λj) v†j · vi (22.2.7)

For distinct eigenvalues, λi 6= λj, the only way for the above to be satisfied is that

v†j · vi = Cijδij (22.2.8)

Hence, eigenvectors of a hermitian matrix with distinct eigenvalues are orthogonal to each

other. The eigenvalues are also real.

4 This mathematical homomorphism is not discussed in any other electromagnetic textbooks.

22.2.2 Homomorphism with the Waveguide Mode Problem

We shall next show that the problem for finding the waveguide modes or eigenmodes is

analogous to the matrix eigenvalue problem. The governing equation for a waveguide mode

is BVP involving the reduced wave equation previously derived, or

∇s^2 ψi(rs) + βis^2 ψi(rs) = 0 (22.2.9)

with the pertinent homogeneous Dirichlet or Neumann boundary condition, depending on

if TE or TM modes are considered. In the above, the differential operator ∇s^2 is analogous to

the matrix operator A, the eigenfunction ψi(rs) is analogous to the eigenvector vi, and βis^2

is analogous to the eigenvalue λi.

Discussion on Functional Space

To think of a function ψ(rs) as a vector, one has to think in the discrete world.5 If one needs

to display the function ψ(rs), on a computer, one will evaluate the function ψ(rs) at discrete

N locations rls, where l = 1, 2, 3, . . . , N. For every rls or every l, there is a scalar number

ψ(rls). These scalar numbers can be stored in a column vector in a computer indexed by l.

The larger N is, the better is the discrete approximation of ψ(rs). In theory, one needs N to

be infinite to describe this function exactly.6

From the above discussion, a function is analogous to a vector and a functional space

is analogous to a vector space. However, a functional space is infinite dimensional. But in

order to compute on a computer with finite resource, such functions are approximated with

vectors. Infinite dimensional vector spaces are replaced with finite dimensional ones to make

the problem computable. Such infinite dimensional functional space is also called Hilbert

space.

It is also necessary to define the inner product between two vectors in a functional space

just as inner product between two vectors in an matrix vector space. The inner product (or

dot product) between two vectors in matrix vector space is

vti · vj = Σl=1^N vi,l vj,l (22.2.10)

The analogous inner product between two vectors in function space is7

⟨ψi, ψj⟩ = ∫S drs ψi(rs)ψj(rs) (22.2.11)

where S denotes the cross-sectional area of the waveguide over which the integration is per-

formed. The left-hand side is the shorthand notation for inner product in functional space or

the infinite dimensional Hilbert space.

5 Some of these concepts are discussed in [34, 121].

6 In mathematical parlance, the index for ψ(rs) is uncountably infinite or nondenumerable.

7 In many math books, the conjugation of the first function ψi is implied, but here, we follow the electro-

magnetic convention that the conjugation of ψi is not implied unless explicitly stated.

Another requirement for a vector in a functional Hilbert space is that it contains finite

energy or that

Ef = ∫S drs |ψi(rs)|^2 (22.2.12)

is finite. The above is analogous to that for a matrix vector v as

Em = Σl=1^N |vl|^2 (22.2.13)

The square root of the above is often used to denote the “length” or the “metric” of the

vector. Finite energy also implies that the vectors are of finite length. This length is also

called the “norm” of a vector.

22.2.3 Proof of Orthogonality of Waveguide Modes

Because of the aforementioned discussion, we see the similarity between a function Hilbert

space, and the matrix vector space. In order to use the result of the matrix EVP, one key

step is to prove that the operator ∇s^2 is hermitian. In matrix algebra, a matrix operator is

hermitian if

x†i · A · xj = (x†j · A† · xi)† = (x†j · A · xi)† = (x†j · A · xi)* (22.2.14)

The first equality follows from standard matrix algebra,8 the second equality follows if A =

A†, or that A is hermitian. The last equality follows because the quantity in the parenthesis

is a scalar, and hence, its conjugate transpose is just its conjugate.

By the same token, a functional operator ∇s^2 is hermitian if

⟨ψ*i, ∇s^2 ψj⟩ = ∫S drs ψ*i(rs) ∇s^2 ψj(rs) = (⟨ψ*j, ∇s^2 ψi⟩)* = ∫S drs ψj(rs) ∇s^2 ψ*i(rs) (22.2.15)

Using integration by parts for higher dimensions and with the appropriate boundary condition

for the function ψ(rs), the above equality can be proved.

To this end, one uses the identity that

∇s · [ψ*i(rs) ∇sψj(rs)] = ψ*i(rs) ∇s^2 ψj(rs) + ∇sψ*i(rs) · ∇sψj(rs) (22.2.16)

Integrating the above over the cross sectional area S, and invoking Gauss divergence theorem

in 2D, one gets that

∮C d l n̂ · (ψ*i(rs) ∇sψj(rs)) = ∫S drs (ψ*i(rs) ∇s^2 ψj(rs)) + ∫S drs (∇ψ*i(rs) · ∇sψj(rs)) (22.2.17)

8 (A · B · C)† = C† · B† · A† [69].

where C the the contour bounding S or the waveguide wall. By applying the boundary

condition that ψi(rs) = 0 or that n̂ · ∇sψj(rs) = 0, or a mixture thereof, then the left-hand

side of the above is zero. This will be the case be it TE or TM modes.

0 = ∫S drs (ψ*i(rs) ∇s^2 ψj(rs)) + ∫S drs (∇ψ*i(rs) · ∇sψj(rs)) (22.2.18)

Applying the same treatment to the last term (22.2.15), we get

0 = ∫S drs (ψj(rs) ∇s^2 ψ*i(rs)) + ∫S drs (∇ψ*i(rs) · ∇sψj(rs)) (22.2.19)

The above indicates that

⟨ψ*i, ∇s^2 ψj⟩ = (⟨ψ*j, ∇s^2 ψi⟩)* (22.2.20)

proving that the operator ∇s^2 is hermitian. One can then use the above property to prove the

orthogonality of the eigenmodes when they have distinct eigenvalues, the same way we have

proved the orthogonality of eigenvectors. The above proof can be extended to the case of a

resonant cavity. The orthogonality of resonant cavity modes is analogous to the orthogonality

of eigenvectors of a hermitian operator.

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