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

Problem 6.3.5

Half-wave matching plate

Find the minimum thickness of a fibreglass ( $ε_{r} = 4.9$ , $μ_{r} = 1$ , $σ \approx 0$ ) radome for an antenna on an airplane (the antenna needs to be protected from weather) such that (1) the radome appears transparent to the antenna radiation at $f = 10$ GHz and (2) the mechanical requirement that the radome be at least $2$ cm thick is met. Assume that the electromagnetic waves radiated (or received) by the antenna are normally incident and planar at the radome surface.

(b) Over which band around $10$ GHz does this radome reflect less than $1 %$ of the power? Compare with the thinnest transparent radome.

Adapted from: B. M. Notaroš, Electromagnetics, Example 10.29.

Try it yourself first.

Strategy

Air is on both sides of the wall, $η_{1} = η_{3} = η_{0}$ and $η_{2} = η_{0} / 4.9$ . Referred to the front face, $Γ_{d}$ of the previous problem simplifies to

Γ_{d} = \frac{j ( \frac{η _{2}}{η _{1}} - \frac{η _{1}}{η _{2}} ) tan β _{2} d}{2 + j ( \frac{η _{2}}{η _{1}} + \frac{η _{1}}{η _{2}} ) tan β _{2} d},

which vanishes when $tan β_{2} d = 0$ , that is for $d = m λ_{2} /2$ , whatever the impedance step. Pick the smallest half-wave multiple above $2$ cm, then follow $∣ Γ_{d} ∣^{2}$ with frequency.

In the fibreglass

v_{2} λ_{2} = \frac{c}{ε _{r}} = 1.355 \times 1 0^{8} m/s, = \frac{v _{2}}{f} = 1.3553 cm .

The transparent thicknesses are $d = m λ_{2} /2 = 0.678, 1.355, 2.033, 2.711, \dots$ cm for $m = 1, 2, 3, 4, \dots$ . The smallest one above $2$ cm is $m = 3$ .

Answer

d_{m i n} = \frac{3 λ _{2}}{2} = 2.03 cm

With $η_{2} / η_{1} = 1/ ε_{r}$ , $∣ Γ_{d} ∣^{2}$ is Eq. (6.3.36),

∣ Γ_{d} ∣^{2} β_{2} d = \frac{( ε _{r} - 1 ) ^{2} sin ^{2} β _{2} d}{4 ε _{r} + ( ε _{r} - 1 ) ^{2} sin ^{2} β _{2} d}, = mπ \frac{f}{f _{0}} (f_{0} = 10 GHz) .

$∣ Γ_{d} ∣^{2} = 0.01$ gives

sin^{2} β_{2} d = \frac{0.04 ε _{r}}{0.99 ( ε _{r} - 1 ) ^{2}} = \frac{0.196}{15.06} = 0.01302,

$∣ sin β_{2} d ∣ = 0.1141$ , so $∣ β_{2} d - mπ ∣ < 0.1143$ rad and

\frac{f}{f _{0}} - 1 < \frac{0.1143}{mπ} = {0.0121, 0.0364, m = 3 (d = 2.03 cm), m = 1 (d = 6.78 mm) .

Answer

m = 3 : 9.88 - 10.12 GHz (240 MHz), m = 1 : 9.64 - 10.36 GHz (730 MHz)

Check

At $9$ GHz the formula gives $33.7 %$ for the thick wall ( $β_{2} d = 2.7 π$ ) and $6.9 %$ for the thin one ( $β_{2} d = 0.9 π$ ). Midway between windows the wall is an odd number of quarter waves, $η_{in} = η_{2}^{2} / η_{0}$ and $∣ Γ_{d} ∣ = \frac{ε _{r} - 1}{ε _{r} + 1} = 0.661$ , the largest possible reflection, $43.7 %$ , which is also what the formula gives for $sin^{2} β_{2} d = 1$ $✓$ .

Problem 6.3.5. Fraction of the power reflected by the two half-wave walls against frequency. Both are transparent at 10 GHz, the dots mark 9 GHz. Neither ever reflects more than \big(\frac{\varepsilon_r-1}{\varepsilon_r+1}\big)^2=43.7\,\%. — **Figure 6.3.13.** Problem 6.3.5. Fraction of the power reflected by the two half-wave walls against frequency. Both are transparent at $10$ GHz, the dots mark $9$ GHz. Neither ever reflects more than $(\frac{ε _{r} - 1}{ε _{r} + 1})^{2} = 43.7 %$ .

Remark

The thick wall is three half waves thick: its phase $β_{2} d = 3 π f / f_{0}$ runs three times faster with frequency, so its window is three times narrower (Figure 6.3.13). At $9$ GHz it already reflects a third of the power. Real radomes reconcile strength and bandwidth with sandwich walls, two thin skins on a low-permittivity foam core.

Problem 6.3.5: Half-wave matching plate