Worked example 7.3.2: A slotted-waveguide array for a marine radar

About the worked examples of Section 7.3

Three short calculations tied to the applications, with $f_{c}$ , $λ_{g}$ , $Z_{T E}$ , the power limit and the copper loss of the TE $_{10}$ mode, and no new physics. Constants: $c = 3 \times 1 0^{8} m/s$ , $η_{0} = 120 π Ω$ , copper $σ_{c} = 5.8 \times 1 0^{7} S/m$ , breakdown of dry air $3 MV/m$ . A good way to use them: read the given quantities, produce the formula from the corresponding application yourself, and only then put the numbers in.

Worked example 7.3.2

A slotted-waveguide array for a marine radar

Given. A $1.2 m$ long WR-90 guide ( $a = 22.86 mm$ ) with longitudinal slots on alternate sides of the centre line, spaced $d = λ_{g} /2$ at the design frequency $f_{0} = 9.4 GHz$ , shorted at the far end.

Find. (a) $λ$ , $λ_{g}$ and $d$ , (b) the number of slots and the beam width (uniform excitation), (c) the phase step $Δ φ = τ d - π$ and the beam direction at $9.2$ and $9.6 GHz$ , (d) the position of the short.

Try it yourself first.

Strategy

Adjacent slots are $λ_{g} /2$ apart, so the guide wave reaches them $18 0^{\circ}$ apart, and alternating the side of the centre line adds another $18 0^{\circ}$ : at $f_{0}$ all slots radiate in phase. A uniform array of length $N d$ has $θ_{3 dB} \approx 0.886 λ / (N d)$ , and a phase step $Δ φ$ between elements tilts the beam to $sin θ_{0} = Δ φ / (k d)$ .

$λ = c / f_{0} = 3.19 cm$ , $f_{co, 10} = c /2 a = 6.562 GHz$ and $s = 1 - (6.562/9.4)^{2} = 0.716$ :

λ_{g} = \frac{3.19}{0.716} = 4.457 cm, d = \frac{λ _{g}}{2} = 2.229 cm .

$1.2/0.02229 = 53.8$ , so $N = 53$ slots and $N d = 1.181 m$ :

θ_{3 dB} \approx 0.886 \times \frac{0.0319}{1.181} = 0.0239 rad = 1.3 7^{\circ} .

The spacing stays $2.229 cm$ while $τ = k^{2} - (π / a)^{2}$ changes:

9.2 GHz : τ = 135.1 rad/m, τ d = 172. 5^{\circ}, Δ φ = - 7. 5^{\circ}, θ_{0} = arcsin \frac{- 0.1317}{4.295} = - 1.7 6^{\circ}, 9.6 GHz : τ = 146.8 rad/m, τ d = 187. 4^{\circ}, Δ φ = + 7. 4^{\circ}, θ_{0} = arcsin \frac{0.1291}{4.481} = + 1.6 5^{\circ} .

The short must place a maximum of the standing wave at the last slot: $λ_{g} /4 = 1.11 cm$ beyond the centre of the last slot.

Answer

λ_{g} = 4.457 cm, d = 2.229 cm, N = 53 θ_{3 dB} = 1.3 7^{\circ} θ_{0} = - 1.7 6^{\circ} (9.2 GHz) θ_{0} = + 1.6 5^{\circ} (9.6 GHz) short 1.11 cm beyond the last slot

Check

At $9.4 GHz$ , $τ d = (2 π / λ_{g}) (λ_{g} /2) = π$ exactly, so $Δ φ = 0$ and the beam is broadside. The sign is right: above $f_{0}$ the guide wavelength shrinks, the phase per slot exceeds $π$ and each slot lags its predecessor more, which tilts the beam towards the load end.

Worked example 7.3.2. Array factor of the 53 slots, |\sin(N\psi/2)/(N\sin(\psi/2))| with \psi=kd\sin\theta-\Delta\varphi, in dB. At 9.4\,\mathrm{GHz} the beam is broadside and 1.37^\circ wide at -3\,\mathrm{dB}, with the first side lobes of a uniform array at -13.3\,\mathrm{dB}. A shift of \pm200\,\mathrm{MHz} moves the beam by -1.76^\circ and +1.65^\circ, more than its own width. — **Figure 7.3.20.** Worked example 7.3.2. Array factor of the $53$ slots, $∣ sin (N ψ /2) / (N sin (ψ /2)) ∣$ with $ψ = k d sin θ - Δ φ$ , in dB. At $9.4 GHz$ the beam is broadside and $1.3 7^{\circ}$ wide at $- 3 dB$ , with the first side lobes of a uniform array at $- 13.3 dB$ . A shift of $\pm 200 MHz$ moves the beam by $- 1.7 6^{\circ}$ and $+ 1.6 5^{\circ}$ , more than its own width.

Remark

A $\pm 200 MHz$ change tilts the beam by more than its own width (Figure 7.3.20): the array is a narrow-band antenna. Used on purpose, the same effect gives frequency-scanned radars that steer the beam without moving parts. Real arrays taper the slot offsets for low side lobes, which widens the beam a little.

cf. Application 7.3.2