Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The travelling factor and the notation

Since the electromagnetic wave is directed along a certain direction, the analysis of the fields can be simplified. Take the $z$ axis along the guide (Figure 7.1.1). Because the guide looks the same at every $z$ , a wave travelling in the $+ z$ direction can be written with a single travelling factor, as for the plane waves of Chapter 6:

E (x, y, z) H (x, y, z) = E_{⊥} (x, y) e^{- j τ z}, = H_{⊥} (x, y) e^{- j τ z} . (7.1.1)

Here $E_{⊥}$ and $H_{⊥}$ are not transverse vectors. They still have all three components, $E_{⊥} = E_{x} a_{x} + E_{y} a_{y} + E_{z} a_{z}$ , but they depend on the cross-section coordinates $x, y$ only. All the dependence on $z$ sits in the exponential, so that for every component

\frac{\partial}{\partial z} ⟶ - j τ, \frac{\partial ^{2}}{\partial z ^{2}} ⟶ - τ^{2} . (7.1.2)

Definition · Phase constant along the guide

$τ [rad/m]$ is the phase constant of the guided wave along $z$ . It plays the role of $β$ of the plane waves of Chapter 6. With losses, or below cutoff, it becomes complex, $τ = β - j α$ , in the same way as the complex wavenumber $k = β - j α$ of Section 6.2, and $e^{- j τ z} = e^{- α z} e^{- j β z}$ . In terms of the propagation constant $γ = α + j β$ of the factor $e^{- γ z}$ , $γ = j τ$ . The wavenumber of the filling is $k = ω μ ε$ , its intrinsic impedance $η = μ / ε$ , its wave speed $v = 1/ μ ε$ and its wavelength $λ = v / f$ .

In the source-free, linear, homogeneous filling ( $ε$ , $μ$ ) the first and second Maxwell equations in time-harmonic form (Section 5.6), $\nabla \times E = - j ω μ H$ and $\nabla \times H = j ω ε E$ , can be solved for the field on the left:

H E = \frac{j}{ω μ} \nabla \times E and = - \frac{j}{ω ε} \nabla \times H . (7.1.3)

The curl equations in components

Write the curls as determinants, with the third entry of the operator row replaced by $- j τ$ according to Eq. (7.1.2):

H E = \frac{j}{ω μ} a_{x} \partial / \partial x E_{x} a_{y} \partial / \partial y E_{y} a_{z} - j τ E_{z}, = - \frac{j}{ω ε} a_{x} \partial / \partial x H_{x} a_{y} \partial / \partial y H_{y} a_{z} - j τ H_{z} . (7.1.4)

Expanding along the first row, for example $(\nabla \times E)_{x} = \partial E_{z} / \partial y - \partial E_{y} / \partial z = \partial E_{z} / \partial y + j τ E_{y}$ , gives the six scalar equations

H_{x} E_{x} = \frac{j}{ω μ} (\frac{\partial E _{z}}{\partial y} + j τ E_{y}), = - \frac{j}{ω ε} (\frac{\partial H _{z}}{\partial y} + j τ H_{y}), (7.1.5)

H_{y} E_{y} = - \frac{j}{ω μ} (\frac{\partial E _{z}}{\partial x} + j τ E_{x}), = \frac{j}{ω ε} (\frac{\partial H _{z}}{\partial x} + j τ H_{x}), (7.1.6)

H_{z} E_{z} = \frac{j}{ω μ} (\frac{\partial E _{y}}{\partial x} - \frac{\partial E _{x}}{\partial y}), = - \frac{j}{ω ε} (\frac{\partial H _{y}}{\partial x} - \frac{\partial H _{x}}{\partial y}) . (7.1.7)

The four equations (7.1.5)–(7.1.6) contain the four transverse unknowns $E_{x}, E_{y}, H_{x}, H_{y}$ together with derivatives of the two longitudinal components $E_{z}$ and $H_{z}$ . The $z$ components of the curls, Eq. (7.1.7), are not needed for what follows.

Eliminating, step by step

Take $E_{x}$ and $H_{y}$ from Eqs. (7.1.5) and (7.1.6). Multiplying out the brackets, each contains the other:

E_{x} H_{y} = - \frac{j}{ω ε} \frac{\partial H _{z}}{\partial y} + \frac{τ}{ω ε} H_{y}, = - \frac{j}{ω μ} \frac{\partial E _{z}}{\partial x} + \frac{τ}{ω μ} E_{x} . (7.1.8)

Substitute the second into the first:

E_{x} = - \frac{j}{ω ε} \frac{\partial H _{z}}{\partial y} - \frac{j τ}{ω ^{2} μ ε} \frac{\partial E _{z}}{\partial x} + \frac{τ ^{2}}{ω ^{2} μ ε} E_{x} .

With $k^{2} = ω^{2} μ ε$ the last coefficient is $τ^{2} / k^{2}$ . Collect $E_{x}$ on the left and multiply by $k^{2}$ , noting that $k^{2} / (ω ε) = ω μ$ :

(k^{2} - τ^{2}) E_{x} = - j τ \frac{\partial E _{z}}{\partial x} - j ω μ \frac{\partial H _{z}}{\partial y} . (7.1.9)

The combination on the left appears in every guide, and it gets its own name,

h^{2} h = k^{2} - τ^{2}, = k^{2} - τ^{2} [rad/m] . (7.1.10)

The same steps, done three more times ( $E_{y}$ with $H_{x}$ , then $H_{x}$ and $H_{y}$ themselves), express every transverse component through derivatives of $E_{z}$ and $H_{z}$ only.

Transverse fields from $E_z$ and $H_z$

In a uniform guide with the travelling factor $e^{- j τ z}$ and $h^{2} = k^{2} - τ^{2} \neq = 0$ ,

H_{x} E_{x} H_{y} E_{y} = - \frac{1}{h ^{2}} (j τ \frac{\partial H _{z}}{\partial x} - j ω ε \frac{\partial E _{z}}{\partial y}), = - \frac{1}{h ^{2}} (j τ \frac{\partial E _{z}}{\partial x} + j ω μ \frac{\partial H _{z}}{\partial y}), = - \frac{1}{h ^{2}} (j τ \frac{\partial H _{z}}{\partial y} + j ω ε \frac{\partial E _{z}}{\partial x}), = - \frac{1}{h ^{2}} (j τ \frac{\partial E _{z}}{\partial y} - j ω μ \frac{\partial H _{z}}{\partial x}) . (7.1.11)

In plain words

Four equations, four transverse unknowns. Solving them in pairs ( $E_{x}$ with $H_{y}$ , $E_{y}$ with $H_{x}$ ) leaves only $E_{z}$ and $H_{z}$ on the right. Know the two longitudinal components and you know the whole field. Dividing by $h^{2}$ is allowed only when $h \neq = 0$ . The case $h = 0$ is the TEM wave of the next subsection.

The generating fields Ez and Hz

Where do $E_{z}$ and $H_{z}$ come from? Every Cartesian component of the field in a source-free homogeneous region obeys the Helmholtz equation $\nabla^{2} E_{z} + k^{2} E_{z} = 0$ (Sections 5.5 and 5.6). With $\partial^{2} / \partial z^{2} \to - τ^{2}$ it loses its $z$ derivative,

\frac{\partial ^{2} E _{z}}{\partial x ^{2}} + \frac{\partial ^{2} E _{z}}{\partial y ^{2}} - τ^{2} E_{z} + k^{2} E_{z} = 0,

and since $k^{2} - τ^{2} = h^{2}$ ,

Reduced wave equation of a uniform guide

\nabla_{T}^{2} E_{z} + h^{2} E_{z} = 0, \nabla_{T}^{2} H_{z} + h^{2} H_{z} = 0, \nabla_{T}^{2} = \frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}} . (7.1.12)

In plain words

$E_{z}$ and $H_{z}$ are the generating fields. Solve the two-dimensional equation (7.1.12) over the cross-section with the boundary conditions of the walls, then differentiate with Eq. (7.1.11). The three-dimensional wave problem has become a two-dimensional one.

Quick check

Why are the $z$ derivatives in Eq. (7.1.11) gone, although the fields depend on $z$ ?

They have been replaced by $- j τ$ , Eq. (7.1.2), which is why $τ$ appears as a factor. The formulas are relations between the cross-section patterns at one $z$ . All of them carry the same $e^{- j τ z}$ , which cancels.

7.1.2 Analysis of the fields

The travelling factor and the notation

The curl equations in components

Eliminating, step by step

The generating fields Ez and Hz