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

Problem 7.3.2

A TE $_{10}$ field with numbers

An air-filled rectangular waveguide with $a = 5 cm$ and $b = 2.5 cm$ carries the field

E = 2 sin (\frac{π x}{a}) cos (10 π \times 1 0^{9} t - τ z) a_{y} kV/m .

Identify the mode, find $f$ and its cutoff frequency, and check whether any other mode could propagate.
Find $τ$ , $λ_{g}$ , $v_{p}$ , $v_{g}$ and $Z_{T E}$ .
Write $H (x, z, t)$ with numbers.
Find the average power carried and the largest surface-current density on the walls.

Try it yourself first.

Strategy

Read the mode from the field: only $E_{y}$ , one half-sine across $x$ , no variation along $y$ . Then the TE $_{10}$ formulas apply with $E_{0} = 2 kV/m$ , and everything follows from $s = 1 - (f_{c} / f)^{2}$ .

Bookkeeping

$a = 0.05 m$ , $b = 0.025 m$ , $ω = 10 π \times 1 0^{9} rad/s$ , so $f = 5 GHz$ , $λ = 6 cm$ and $k = ω / c = 104.72 rad/m$ . $E_{0} = 2000 V/m$ .

Only $E_{y} \propto sin (π x / a)$ , independent of $y$ : the mode is TE $_{10}$ . $f_{co, 10} = c /2 a = 3 \times 1 0^{8} /0.1 = 3 GHz$ . The next modes are TE $_{20}$ and TE $_{01}$ at $c / a = c /2 b = 6 GHz$ and TE $_{11}$ , TM $_{11}$ at $1.5 \times 1 0^{8} 2 0^{2} + 4 0^{2} = 6.71 GHz$ , all above $5 GHz$ : only TE $_{10}$ propagates.

$f_{c} / f = 0.6$ and $s = 0.8$ :

τ λ_{g} v_{p} v_{g} Z_{T E} = 0.8 \times 104.72 = 83.78 rad/m, = \frac{6}{0.8} = 7.5 cm, = \frac{c}{0.8} = 3.75 \times 1 0^{8} m/s, = 0.8 c = 2.4 \times 1 0^{8} m/s, = \frac{120 π}{0.8} = 150 π = 471.2 Ω.

The two magnetic amplitudes are

\frac{E _{0}}{Z _{T E}} \frac{π E _{0}}{ω μ _{0} a} = \frac{2000}{471.2} = 4.244 A/m, = \frac{62.83 \times 2000}{39 478} = 3.183 A/m .

With $H_{x} = - E_{y} / Z_{T E}$ , and $H_{z}$ leading $E_{y}$ by $9 0^{\circ}$ in the phasor form, $Re {j e^{j (ω t - τ z)}} = - sin (ω t - τ z)$ gives

H_{x} H_{z} = - 4.244 sin (20 π x) cos (ω t - 83.78 z) A/m, = - 3.183 cos (20 π x) sin (ω t - 83.78 z) A/m,

with $x$ and $z$ in metres ( $π / a = 20 π rad/m$ ).

P = \frac{E _{0}^{2} ab}{4 Z _{T E}} = \frac{( 2000 ) ^{2} ( 0.05 ) ( 0.025 )}{4 \times 471.2} = 2.65 W .

On the broad walls $J_{s} = \pm (H_{x} a_{z} - H_{z} a_{x})$ : its largest value is $∣ H_{x} ∣_{m a x} = 4.24 A/m$ , along $z$ on the centre line $x = a /2$ . On the narrow walls $J_{s}$ is vertical with $∣ H_{z} (0) ∣ = 3.18 A/m$ .

Answer

TE_{10}, f = 5 GHz, f_{c} = 3 GHz τ = 83.78 rad/m, λ_{g} = 7.5 cm Z_{T E} = 471.2 Ω, P = 2.65 W ∣ J_{s} ∣_{m a x} = 4.24 A/m (broad walls), 3.18 A/m (narrow walls)

Check

$v_{p} v_{g} = 3.75 \times 2.4 \times 1 0^{16} = 9 \times 1 0^{16} = c^{2}$ . The peak power density is $E_{0}^{2} /2 Z_{T E} = 4.24 kW/ m^{2}$ at $x = a /2$ . The mean of $sin^{2}$ over the width is $\frac{1}{2}$ , so the integral across $x$ is $4244 \times 0.025 = 106 W/m$ , and times $b = 2.5 cm$ it is $2.65 W$ .

(b) Top view at t=0 (drawn for f=1.5f_c). — (a) Amplitudes across the guide.

Remark

$H_{z}$ is $9 0^{\circ}$ out of phase with $E_{y}$ , so the crosswise flow $S_{x} = \frac{1}{2} Re {E_{y} H_{z}^{*}}$ is purely reactive. The guide works at $f = 1.67 f_{c}$ , well inside its single-mode band $3$ – $6 GHz$ .

Problem 7.3.2: A TE_{10} field with numbers