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

The fields

Put $m = 1$ , $n = 0$ and $h_{10} = π / a$ into the TE formulas (7.3.23) and (7.3.21). Nothing depends on $y$ , so every $\partial / \partial y$ vanishes, and

H_{z} = C cos \frac{π x}{a} e^{- j τ z}, E_{x} = E_{z} = H_{y} = 0, E_{y} H_{x} = \frac{j ω μ}{h _{10}^{2}} \frac{\partial H _{z}}{\partial x} = - j ω μ \frac{a}{π} C sin \frac{π x}{a} e^{- j τ z}, = - \frac{j τ}{h _{10}^{2}} \frac{\partial H _{z}}{\partial x} = j τ \frac{a}{π} C sin \frac{π x}{a} e^{- j τ z} .

The constant $C$ is an amplitude of $H_{z}$ , but in practice one specifies the electric field. Call the peak electric field $E_{0} = - j ω μ a C / π$ , that is $C = j π E_{0} / (ω μ a)$ . Then $H_{x} = j τ (a / π) C sin (π x / a) = - (τ / ω μ) E_{0} sin (π x / a)$ , and with $ω μ / τ = Z_{T E}$ :

The TE$_{10}$ mode

E_{y} H_{x} H_{z} = E_{0} sin \frac{π x}{a} e^{- j τ z}, = - \frac{E _{0}}{Z _{T E}} sin \frac{π x}{a} e^{- j τ z}, = j \frac{π E _{0}}{ω μ a} cos \frac{π x}{a} e^{- j τ z}, (7.3.29)

all other components zero, with

f_{co, 10} = \frac{v}{2 a}, λ_{c, 10} = 2 a, τ = k 1 - (f_{c} / f)^{2}, Z_{T E} = \frac{η}{1 - ( f _{c} / f ) ^{2}} . (7.3.30)

In plain words

In the TE $_{10}$ mode the electric field points straight from the bottom wall to the top wall, is strongest in the middle of the guide and vanishes at the narrow walls. The magnetic field forms loops lying flat in the guide, with a transverse part $H_{x}$ in step with $E_{y}$ and a longitudinal part $H_{z}$ a quarter period out of step with it.

TE $_{10}$ is the working mode of practically every rectangular guide, for three reasons.

It has the lowest cutoff of all modes, so for a given frequency it needs the smallest guide.
It has an octave of single-mode band when $b \leq a /2$ , from $v /2 a$ to $v / a$ .
Its $E$ has one fixed direction ( $y$ ) and depends neither on $y$ nor on $b$ : the height $b$ is free to be chosen for power and loss, as the following subsections show.

The real fields

With $E_{0}$ real, the instantaneous fields are the real parts of the phasors (7.3.29) times $e^{j ω t}$ . Since $Re {j e^{j (ω t - τ z)}} = - sin (ω t - τ z)$ ,

E_{y} H_{x} H_{z} = E_{0} sin \frac{π x}{a} cos (ω t - τ z), = - \frac{E _{y}}{Z _{T E}}, = - \frac{π E _{0}}{ω μ a} cos \frac{π x}{a} sin (ω t - τ z) . (7.3.31)

Figure 7.3.5 shows them at $t = 0$ , for $f = 1.5 f_{co, 10}$ and over $1.5$ guide wavelengths. The lines of $H$ lie in planes $y =$ const and are closed. They are the contours of $sin (π x / a) sin (τ z)$ , as one checks from Eq. (7.3.31): $H_{x} \propto - \partial / \partial z$ and $H_{z} \propto \partial / \partial x$ of that function. The loops sit between the maxima of $E$ , a quarter of $λ_{g}$ away from them, and the whole pattern slides along $+ z$ at the phase velocity.

(a) Oblique view, the front wall x=0 removed. \vec E (orange) is drawn on the centre plane x=a/2, the \vec H loops (dashed) in the plane y=b/2. — (a) Oblique view, the front wall $x = 0$ removed. $E$ (orange) is drawn on the centre plane $x = a /2$ , the $H$ loops (dashed) in the plane $y = b /2$ .

(b) Top view: \odot and \otimes mark \vec E pointing out of and into the page (\pm y). — (a) Oblique view, the front wall $x = 0$ removed. $E$ (orange) is drawn on the centre plane $x = a /2$ , the $H$ loops (dashed) in the plane $y = b /2$ .

TE₁₀ as two plane waves

Split the sine of Eq. (7.3.29) into exponentials, $sin (π x / a) = (e^{j π x / a} - e^{- j π x / a}) /2 j$ :

E_{y} = \frac{E _{0}}{2 j} [e^{- j (τ z - \frac{π}{a} x)} - e^{- j (τ z + \frac{π}{a} x)}] . (7.3.32)

Each term is a uniform plane wave, with the wave vectors $τ a_{z} \pm \frac{π}{a} a_{x}$ . Their length is $τ^{2} + (π / a)^{2} = k$ by Eq. (7.3.2), the wavenumber of the filling, so these are ordinary plane waves of the medium (Chapter 6). They bounce between the narrow walls $x = 0$ and $x = a$ at the angle $θ$ to the axis given by

TE$_{10}$ as two plane waves

sin θ = \frac{π / a}{k} = \frac{λ}{2 a} = \frac{f _{co, 10}}{f}, λ_{g} = \frac{λ}{cos θ}, v_{p} = \frac{v}{cos θ}, v_{g} = v cos θ . (7.3.33)

In plain words

TE $_{10}$ is an ordinary plane wave zig-zagging between the narrow walls. The closer the frequency is to cutoff, the steeper the zig-zag, the longer the guide wavelength and the slower the energy moves down the guide.

Their $E$ ( $\pm y$ ) is parallel to the narrow walls, and at each reflection from a perfect conductor the reflection coefficient is $Γ = - 1$ (Section 6.4): incident and reflected fields cancel on the wall, so $E_{y} = 0$ at $x = 0$ and $x = a$ . Figure 7.3.6 draws the rays.

Top view of the two plane waves of TE_{10} for f=1.5f_{co,10}, where \sin\theta=1/1.5 and \theta=41.8^\circ. The waves are reflected with \Gamma=-1 by the narrow walls and progress along z with the speed v\cos\theta. — **Figure 7.3.6.** Top view of the two plane waves of TE $_{10}$ for $f = 1.5 f_{co, 10}$ , where $sin θ = 1/1.5$ and $θ = 41. 8^{\circ}$ . The waves are reflected with $Γ = - 1$ by the narrow walls and progress along $z$ with the speed $v cos θ$ .

What the picture explains

At cutoff $θ \to 9 0^{\circ}$ and $a = λ /2$ : the waves bounce straight across the guide without progress along it, and $v_{g} = v cos θ \to 0$ . Above cutoff the wavefronts meet the axis obliquely, so the crests along $z$ are spaced $λ / cos θ = λ_{g} > λ$ apart and move at $v_{p} = v / cos θ > v$ , while the energy, which travels along the rays at $v$ , advances along $z$ at $v_{g} = v cos θ < v$ . This is exactly the picture of the parallel-plate guide of Section 7.2, with the width $a$ in place of the plate spacing. The broad walls meet a normal $E$ and play no role in the bouncing: this is why $b$ is absent from $f_{co, 10}$ .

Quick check

At which angle do the plane waves of TE $_{10}$ travel at $f = 2 f_{co, 10}$ , and how long is the guide wavelength compared with $λ$ ?

$sin θ = f_{co, 10} / f = 0.5$ , so $θ = 3 0^{\circ}$ and $λ_{g} = λ / cos 3 0^{\circ} = 1.155 λ$ . The phase velocity is $1.155 v$ and the group velocity $0.866 v$ .

Surface currents and charges on the walls

On a perfect conductor the tangential $H$ is carried by a surface current and the normal $D$ ends on a surface charge (Sections 4.2.3 and 3.4.6). With $a_{n}$ the unit normal pointing into the guide,

J_{s} = a_{n} \times H, ρ_{s} = ε a_{n} \cdot E . (7.3.34)

For TE $_{10}$ , with $H = H_{x} a_{x} + H_{z} a_{z}$ and $E = E_{y} a_{y}$ :

Top wall $y = b$ , $a_{n} = - a_{y}$ : $J_{s} = - a_{y} \times (H_{x} a_{x} + H_{z} a_{z}) = H_{x} a_{z} - H_{z} a_{x}$ and $ρ_{s} = - ε E_{y}$ .
Bottom wall $y = 0$ , $a_{n} = + a_{y}$ : the negative of both, $J_{s} = - H_{x} a_{z} + H_{z} a_{x}$ , $ρ_{s} = ε E_{y}$ .
Side walls $x = 0$ ( $a_{n} = a_{x}$ ) and $x = a$ ( $a_{n} = - a_{x}$ ): only $H_{z}$ is tangential, and on both walls

$J_{s} = - j \frac{π E _{0}}{ω μ a} e^{- j τ z} a_{y}, (7.3.35)$

since $a_{x} \times H_{z} (0) a_{z} = - H_{z} (0) a_{y}$ and $- a_{x} \times H_{z} (a) a_{z} = H_{z} (a) a_{y}$ with $H_{z} (a) = - H_{z} (0)$ .

On the broad walls $J_{z} = \pm H_{x} \propto sin (π x / a)$ is largest on the centre line and $J_{x} = \mp H_{z} \propto cos (π x / a)$ is largest at the edges. The current therefore flows along $z$ near the middle of the broad walls, crosswise near their edges, and vertically on the narrow walls (Figure 7.3.7). It does not close on the walls: where it converges, charge accumulates, and the circuit closes through the displacement current $j ω ε E_{y}$ that crosses the guide from wall to wall (Exercise 7.3.4 verifies the continuity equation $\nabla_{s} \cdot J_{s} = - j ω ρ_{s}$ of Section 5.4).

TE_{10} surface current at t=0 (f=1.5f_{co,10}, 1.5\lambda_g). Upper part: the top wall seen from above. Lower part: the side wall x=0 unfolded about the common edge. The current runs along z on the centre line (dashed) of the broad wall, turns sideways towards the edges and flows vertically down or up the narrow wall. On the centre line it is purely longitudinal. — **Figure 7.3.7.** TE $_{10}$ surface current at $t = 0$ ( $f = 1.5 f_{co, 10}$ , $1.5 λ_{g}$ ). Upper part: the top wall seen from above. Lower part: the side wall $x = 0$ unfolded about the common edge. The current runs along $z$ on the centre line (dashed) of the broad wall, turns sideways towards the edges and flows vertically down or up the narrow wall. On the centre line it is purely longitudinal.

Slots

On the centre line of the broad wall $J_{s}$ is purely longitudinal. A narrow slot cut along it interrupts no current and hardly disturbs the guide: this is the slotted measuring line. A slot that cuts current lines, off the centre line or transverse to it, forces the current across a gap. The interrupted current continues as a displacement current across the slot, which then radiates like a small antenna: the slot antennas of Application 7.3.2.

The tool below shows the walls of the TE $_{10}$ guide of Problem 7.3.2 ( $a = 5 cm$ , $b = 2.5 cm$ , $f = 5 GHz$ ). Press Play: the current along the centre line is purely longitudinal while the whole pattern slides at $v_{p}$ .

InteractiveLoading simulation…

Power flow

The time-average Poynting vector is $S = \frac{1}{2} Re {E \times H^{*}}$ . With $E = E_{y} a_{y}$ and $H = H_{x} a_{x} + H_{z} a_{z}$ ,

E \times H^{*} = E_{y} H_{x}^{*} (a_{y} \times a_{x}) + E_{y} H_{z}^{*} (a_{y} \times a_{z}) = - E_{y} H_{x}^{*} a_{z} + E_{y} H_{z}^{*} a_{x} .

The longitudinal component is

S_{z} = - \frac{1}{2} Re {E_{y} H_{x}^{*}} = \frac{∣ E _{0} ∣ ^{2}}{2 Z _{T E}} sin^{2} \frac{π x}{a}, (7.3.36)

while $S_{x} = \frac{1}{2} Re {E_{y} H_{z}^{*}} = 0$ , because $E_{y} H_{z}^{*}$ is imaginary: the crosswise flow is purely reactive, energy sloshing between the narrow walls. Integrating $S_{z}$ over the cross-section, $0 < x < a$ , $0 < y < b$ , with the mean value $\frac{1}{2}$ of $sin^{2}$ over the width:

Power carried by TE$_{10}$

P_{10} = \frac{∣ E _{0} ∣ ^{2} ab}{4 Z _{T E}} = \frac{∣ E _{0} ∣ ^{2} ab}{4 η} 1 - (\frac{f _{co, 10}}{f})^{2} . (7.3.37)

In plain words

The power is the peak power density $∣ E_{0} ∣^{2} /2 Z_{T E}$ times half the cross-section. It grows with the area $ab$ and vanishes at cutoff, where $Z_{T E} \to \infty$ .

The power limit

The peak field $∣ E_{0} ∣$ must stay below the breakdown field of the filling, about $3 MV/m$ for dry air at atmospheric pressure (Section 3.4.4). For WR-90 at $10 GHz$ ( $a = 22.86 mm$ , $b = 10.16 mm$ , $Z_{T E} = 499.6 Ω$ ) this gives

P_{m a x} = \frac{( 3 \times 1 0 ^{6} ) ^{2} \times 0.02286 \times 0.01016}{4 \times 499.6} = 1.05 MW .

Power density of TE_{10} across the guide, S_z\propto\sin^2(\pi x/a), with S_{\max}=|E_0|^2/2Z_{TE}. It is the same at every height y. Its mean over the width is \frac12S_{\max}, so P=\frac12S_{\max}ab. — **Figure 7.3.8.** Power density of TE $_{10}$ across the guide, $S_{z} \propto sin^{2} (π x / a)$ , with $S_{m a x} = ∣ E_{0} ∣^{2} /2 Z_{T E}$ . It is the same at every height $y$ . Its mean over the width is $\frac{1}{2} S_{m a x}$ , so $P = \frac{1}{2} S_{m a x} ab$ .

Quick check

A TE $_{10}$ guide carries $P$ with the peak field $E_{0}$ . What happens to $P$ if (a) $E_{0}$ is doubled, (b) $b$ is halved at the same $E_{0}$ , (c) the frequency moves closer to cutoff at the same $E_{0}$ ?

(a) $P \propto ∣ E_{0} ∣^{2}$ : four times the power. (b) $P \propto b$ : half the power. (c) $Z_{T E}$ grows, so $P$ falls, and at cutoff it is zero. A guide used close to cutoff has a poor power rating.

Attenuation due to the walls

Real walls have a finite conductivity $σ_{c}$ , and the current (7.3.34) flows in a skin of depth $δ = 1/ π f μ_{c} σ_{c}$ (Section 6.2). Its loss is described by the surface resistance

R_{s} = \frac{1}{σ _{c} δ} = \frac{π f μ _{c}}{σ _{c}} [Ω] . (7.3.39)

For good conductors the fields hardly differ from those of the lossless guide, so the loss can be computed with the lossless currents (the perturbation method of Section 7.2). The power lost per unit length of guide is

P_{L}^{'} = \frac{R _{s}}{2} \oint ∣ J_{s} ∣^{2} d l, (7.3.40)

taken round the perimeter of the cross-section. With the TE $_{10}$ currents of the previous subsection:

each broad wall contributes $\frac{R _{s}}{2} \int_{0}^{a} (∣ H_{x} ∣^{2} + ∣ H_{z} ∣^{2}) d x = \frac{R _{s}}{2} \frac{a}{2} (\frac{∣ E _{0} ∣ ^{2}}{Z _{T E}^{2}} + \frac{π E _{0}}{ω μ a}^{2})$ , since the mean of $sin^{2}$ and of $cos^{2}$ over the width is $\frac{1}{2}$ ,
each narrow wall contributes $\frac{R _{s}}{2} \int_{0}^{b} ∣ H_{z} (0) ∣^{2} d y = \frac{R _{s}}{2} b \frac{π E _{0}}{ω μ a}^{2}$ .

Adding two of each (broad walls, then narrow walls):

P_{L}^{'} = R_{s} [\frac{a}{2} (\frac{∣ E _{0} ∣ ^{2}}{Z _{T E}^{2}} + \frac{π E _{0}}{ω μ a}^{2}) + b \frac{π E _{0}}{ω μ a}^{2}] . (7.3.41)

The bracket simplifies. Since $1/ Z_{T E} = τ / ω μ$ and $τ^{2} + (π / a)^{2} = k^{2}$ ,

\frac{∣ E _{0} ∣ ^{2}}{Z _{T E}^{2}} + \frac{π E _{0}}{ω μ a}^{2} = \frac{∣ E _{0} ∣ ^{2} ( τ ^{2} + π ^{2} / a ^{2} )}{ω ^{2} μ ^{2}} = \frac{∣ E _{0} ∣ ^{2} k ^{2}}{ω ^{2} μ ^{2}} = \frac{∣ E _{0} ∣ ^{2}}{η ^{2}}, \frac{π E _{0}}{ω μ a}^{2} = \frac{( π / a ) ^{2}}{k ^{2}} \frac{∣ E _{0} ∣ ^{2}}{η ^{2}} = (\frac{f _{c}}{f})^{2} \frac{∣ E _{0} ∣ ^{2}}{η ^{2}},

so $P_{L}^{'} = (R_{s} ∣ E_{0} ∣^{2} / η^{2}) [a /2 + b (f_{c} / f)^{2}]$ . The power along the guide decays as $P (z) = P (0) e^{- 2 α_{c} z}$ , so $d P / d z = - P_{L}^{'} = - 2 α_{c} P$ and $α_{c} = P_{L}^{'} /2 P_{10}$ . Dividing by twice Eq. (7.3.37):

Conductor loss of TE$_{10}$

α_{c} = \frac{R _{s}}{η b 1 - ( f _{c} / f ) ^{2}} \times [1 + \frac{2 b}{a} (\frac{f _{c}}{f})^{2}] [Np/m], (7.3.42)

with $R_{s} = π f μ_{c} / σ_{c}$ . A lossy filling adds the dielectric attenuation $α_{d} = k tan δ / (2 1 - (f_{c} / f)^{2})$ of Section 7.2. $1 Np/m = 8.686 dB/m$ .

In plain words

The wall currents heat the walls. The loss grows near cutoff, where the waves bounce almost straight across and $H_{z}$ is large, and it falls as $1/ b$ , because a taller guide spreads the same current over more wall.

Conductor loss of TE_{10} in WR-90 (22.86\times10.16\,\mathrm{mm}) from Eq. (7.3.42). Copper (5.8\times10^7\,\mathrm{S/m}): 0.140, 0.108 and 0.097\,\mathrm{dB/m} at 8.2, 10 and 12.4\,\mathrm{GHz}. Brass (1.57\times10^7\,\mathrm{S/m}) loses \sqrt{5.8/1.57}=1.92 times as much. The loss explodes near cutoff. Its minimum, 0.094\,\mathrm{dB/m} at 15.4\,\mathrm{GHz}, lies above the TE_{20} cutoff and cannot be used. — **Figure 7.3.9.** Conductor loss of TE $_{10}$ in WR-90 ( $22.86 \times 10.16 mm$ ) from Eq. (7.3.42). Copper ( $5.8 \times 1 0^{7} S/m$ ): $0.140$ , $0.108$ and $0.097 dB/m$ at $8.2$ , $10$ and $12.4 GHz$ . Brass ( $1.57 \times 1 0^{7} S/m$ ) loses $5.8/1.57 = 1.92$ times as much. The loss explodes near cutoff. Its minimum, $0.094 dB/m$ at $15.4 GHz$ , lies above the TE $_{20}$ cutoff and cannot be used.

Common mistake · Field attenuation or power attenuation

$α_{c}$ is the attenuation constant of the field, $E \propto e^{- α_{c} z}$ . The power falls twice as fast, $P \propto e^{- 2 α_{c} z}$ . In decibels the two agree, $8.686 α_{c} L dB$ for either, but a fraction of power computed as $e^{- α_{c} L}$ is wrong. Over $25 m$ of WR-137 at $7 GHz$ (Problem 7.3.4) the correct $e^{- 2 α_{c} L} = 0.732$ becomes a wrong $0.855$ .

7.3.6 The dominant TE₁₀ mode

The fields

The real fields

TE₁₀ as two plane waves

Surface currents and charges on the walls

Power flow

The power limit

Attenuation due to the walls