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

Problem 7.2.4

Interactive: Parallel-Plate Waveguide Modes

Problem 7.2.4

TE $_{1}$ fields, power and copper loss

An air-filled parallel-plate guide with copper plates ( $σ_{c} = 5.8 \times 1 0^{7} S/m$ ) spaced $b = 2 cm$ carries the TE $_{1}$ mode at $f = 10 GHz$ . The peak electric field is $E_{0} = 1 kV/m$ .

Find $f_{co, 1}$ , $τ_{1}$ , $λ_{g}$ and $Z_{T E}$ .
Write the phasors $E_{x}$ , $H_{y}$ , $H_{z}$ with numbers (take $E_{x}$ real at $z = 0$ ).
Find the power carried per metre of width.
Find $α_{c}$ and the fraction of the power left after $10 m$ .

Try it yourself first.

Strategy

Start from $E_{x} = E_{0} sin (π y / b) e^{- j τ z}$ and the TE $_{n}$ formulas (7.2.12), $E_{x} = \frac{j ω μ}{h} C sin (h y) e^{- j τ z}$ , $H_{y} = \frac{j τ}{h} C sin (h y) e^{- j τ z}$ , $H_{z} = C cos (h y) e^{- j τ z}$ , to find the constant $C$ . The power follows from $P^{'} = b E_{0}^{2} / (4 Z_{T E})$ , the loss from the current $∣ H_{z} ∣$ on the plates.

1. Bookkeeping

$b = 0.02 m$ , $f = 10 GHz$ : $k = 2 π f / c = 209.4 rad/m$ , $h = π / b = 50 π = 157.1 rad/m$ . Copper: $R_{s} = π f μ_{0} / σ_{c} = π \times 1 0^{10} \times 4 π \times 1 0^{- 7} /5.8 \times 1 0^{7} = 26.1 m Ω$ .

2. Mode parameters (a)

$f_{co, 1} = c /2 b = 7.5 GHz$ , $f_{co, 1} / f = 0.75$ , $s = 1 - 0.5625 = 0.661$ :

τ_{1} λ_{g} Z_{T E} = 209.4 \times 0.661 = 138.5 rad/m, = \frac{2 π}{138.5} = 4.54 cm, = \frac{120 π}{0.661} = 570.0 Ω.

3. Phasors (b)

Matching $E_{x}$ gives

C = \frac{h E _{0}}{j ω μ _{0}} = \frac{157.1 \times 1000}{j 2 π \times 1 0 ^{10} \times 4 π \times 1 0 ^{- 7}} = - j 1.989 A/m,

and then $H_{y} = \frac{j τ}{h} C sin (h y) e^{- j τ z} = \frac{138.5}{157.1} \times 1.989 sin (h y) e^{- j τ z}$ :

E_{x} H_{y} H_{z} = 1000 sin (50 π y) e^{- j 138.5 z} V/m, = 1.755 sin (50 π y) e^{- j 138.5 z} A/m, = - j 1.989 cos (50 π y) e^{- j 138.5 z} A/m .

4. Power (c)

$S_{z} = \frac{1}{2} Re {E_{x} H_{y}^{*}} = \frac{E _{0}^{2}}{2 Z _{T E}} sin^{2} (π y / b)$ , with the peak $877 W/ m^{2}$ , and the average of $sin^{2}$ over the gap is $\frac{1}{2}$ :

P^{'} = \int_{0}^{b} S_{z} d y = \frac{b E _{0}^{2}}{4 Z _{T E}} = \frac{0.02 \times 1 0 ^{6}}{4 \times 570.0} = 8.77 W per metre of width .

5. Loss (d)

α_{c} = \frac{2 R _{s} ( f _{co} / f ) ^{2}}{η _{0} b s} = \frac{2 ( 0.0261 ) ( 0.5625 )}{377 \times 0.02 \times 0.661} = 5.89 \times 1 0^{- 3} Np/m = 0.051 dB/m .

After $10 m$ the power is multiplied by $e^{- 2 α_{c} \times 10} = e^{- 0.118} = 0.889$ .

Answer

Z_{T E} = 570 Ω, P^{'} = 8.77 W/m, α_{c} = 5.89 \times 1 0^{- 3} Np/m = 0.051 dB/m, 88.9 % left after 10 m

Compare $P^{'} = 8.77 W/m$ and $α_{c} = 5.89 \times 1 0^{- 3} Np/m$ below with steps 4 and 5, then switch to TM $_{1}$ to see its larger copper loss.

InteractiveLoading simulation…

Check

$E_{x} / H_{y} = 1000/1.755 = 570 Ω = Z_{T E}$ , and $H_{z}$ is $9 0^{\circ}$ out of phase with $E_{x}$ . The loss from first principles: on each plate the tangential field is $H_{z}$ only, $∣ H_{z} ∣ = 1.989 A/m$ , so both plates together dissipate $P_{L}^{'} = 2 \times \frac{1}{2} R_{s} ∣ C ∣^{2} = 0.0261 \times 1.98 9^{2} = 0.103 W$ per square metre of guide, and $P_{L}^{'} / (2 P^{'}) = 0.103/17.5 = 5.89 \times 1 0^{- 3} Np/m$ , the same value.

(b) Power density across the gap. — (a) The field lines of the solution.

Read the graphs

The power flows mainly in the middle of the gap, away from the plates, and the plates see only $H_{z}$ , here $1.99 A/m$ . That is why the TE $_{1}$ loss is small: a TM $_{1}$ mode in the same guide would lose $α_{c} = 2 R_{s} / (η_{0} b s) = 0.091 dB/m$ , almost twice as much, because its full transverse $H$ is tangential to the plates.

Problem 7.2.4: TE_1 fields, power and copper loss