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

About the further exercises of Section 7.3

Ten further exercises in the spirit of Problems 7.3.1–7.3.5, the home exercises and the worked examples, graded from easy to harder. Exercises 7.3.1–7.3.2 are about cutoffs and dimensions, Exercises 7.3.3–7.3.5 about power, wall currents and loss, Exercises 7.3.6–7.3.8 about dispersion and pairs of modes, and Exercises 7.3.9–7.3.10 about many modes and degenerate modes. They use only the results listed at the head of this subsection, together with the dielectric attenuation $α_{d} = k tan δ / (2 s)$ .

Exercise 7.3.4

Wall currents, charges and slots

An air-filled WR-90 guide ( $a = 22.86 mm$ , $b = 10.16 mm$ ) carries $100 W$ in the TE $_{10}$ mode at $10 GHz$ .

Find $E_{0}$ , $∣ H_{x} ∣_{m a x}$ , $∣ H_{z} ∣_{m a x}$ and the largest surface charge density.
Write $J_{s}$ and $ρ_{s}$ on the top wall and verify $\nabla_{s} \cdot J_{s} = - j ω ρ_{s}$ .
Five narrow slots are cut: (1) along $z$ on the centre line of the broad wall, (2) along $z$ at $x = a /4$ , (3) across the broad wall, centred on $x = a /2$ , (4) vertical in a narrow wall, (5) in a narrow wall, tilted $1 5^{\circ}$ from the vertical. Which radiate? Give the peak current density each one interrupts.

Wall current at one instant and slots 1–4.

Try it yourself first.

Strategy

On a perfect conductor $J_{s} = a_{n} \times H$ and $ρ_{s} = ε_{0} a_{n} \cdot E$ . A narrow slot is harmless when it runs along the current lines and radiates when it cuts them: the interrupted current must jump the gap as displacement current, and the slot becomes a source.

$f_{co, 10} = 6.562 GHz$ , $s = 0.755$ , $Z_{T E} = 499.6 Ω$ , $ω = 2 π \times 1 0^{10} rad/s$ :

E_{0} ∣ H_{x} ∣_{m a x} ∣ H_{z} ∣_{m a x} ∣ ρ_{s} ∣_{m a x} = \frac{4 \times 499.6 \times 100}{0.02286 \times 0.01016} = 29.3 kV/m, = \frac{E _{0}}{Z _{T E}} = 58.7 A/m, = \frac{π E _{0}}{ω μ _{0} a} = 51.1 A/m, = ε_{0} E_{0} = 0.259 μ C/ m^{2} (x = a /2) .

On the top wall $a_{n} = - a_{y}$ :

J_{s} ρ_{s} = H_{x} a_{z} - H_{z} a_{x}, = - ε_{0} E_{y} = - ε_{0} E_{0} sin \frac{π x}{a} e^{- j τ z} .

Then, using $H_{x} = - (τ / ω μ_{0}) E_{y}$ ,

\nabla_{s} \cdot J_{s} = - \frac{\partial H _{z}}{\partial x} - j τ H_{x} = j \frac{E _{0}}{ω μ _{0}} [(\frac{π}{a})^{2} + τ^{2}] sin \frac{π x}{a} e^{- j τ z} = j \frac{k ^{2} E _{0}}{ω μ _{0}} sin \frac{π x}{a} e^{- j τ z} .

Since $k^{2} / (ω μ_{0}) = ω ε_{0}$ , this is $j ω ε_{0} E_{0} sin \frac{π x}{a} e^{- j τ z} = - j ω ρ_{s}$ . Numerically $(π / a)^{2} + τ^{2} = 137. 4^{2} + 158. 0^{2} = 43865 m^{- 2} = k^{2}$ .

On the broad wall the current crossing a longitudinal slot is $J_{x} = ∣ H_{z} ∣_{m a x} cos (π x / a)$ , the current crossing a transverse slot $J_{z} = ∣ H_{x} ∣_{m a x} sin (π x / a)$ . On the narrow walls the current is vertical, $∣ J_{y} ∣ = ∣ H_{z} ∣_{m a x}$ .

(1) x = a /2 : J_{x} = 0, no radiation, (2) x = a /4 : J_{x} = 51.1 cos 4 5^{\circ} = 36.1 A/m, radiates, (3) cuts J_{z} = 58.7 A/m, radiates strongly, (4) parallel to J_{y}, no radiation, (5) 51.1 sin 1 5^{\circ} = 13.2 A/m, radiates weakly .

Answer

E_{0} = 29.3 kV/m, ∣ H_{x} ∣_{m a x} = 58.7 A/m ∣ H_{z} ∣_{m a x} = 51.1 A/m, ρ_{s} = 0.259 μ C/ m^{2} slots 2, 3, 5 radiate (36.1, 58.7, 13.2 A/m) slots 1, 4 do not

Check

$∣ H_{z} ∣_{m a x} /∣ H_{x} ∣_{m a x} = 51.1/58.7 = 0.870 = (f_{c} / f) / s = 0.656/0.755$ .

Remark

Slot 1 is the slotted measuring line. Slot 2 is the element of the broad-wall array of Application 7.3.2: the offset sets how much power each slot takes. Slot 3 is a series slot, slot 5 the element of narrow-wall arrays, whose tilt sets the coupling and whose alternating sign gives the extra $18 0^{\circ}$ . The continuity check of (b) shows why the broad-wall current must turn sideways near the edges: it closes through the charge on which the field $E_{y}$ ends.

cf. Home exercise 7.3.3

Exercise 7.3.4: Wall currents, charges and slots