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

About the further exercises of Section 6.3

Ten further exercises at normal incidence, in the spirit of the problems, the home exercises and the worked examples of this section, each group of five graded from easy to harder. They use only the toolbox at the head of this subsection.

Exercise 6.3.10

Shielding with aluminium foil

A uniform plane wave falls normally from air on a household aluminium foil ( $σ = 3.5 \times 1 0^{7}$ S/m, $t = 16$ μm) with air behind it. The shielding effectiveness is $SE = 20 lo g_{10} ∣ E^{i} / E^{t} ∣$ . (a) Find the skin depth at $1$ MHz and at $1$ GHz, and the frequency at which $t = δ$ . (b) At $1$ GHz the multiple reflections inside the foil are negligible. Estimate SE as the loss at the two faces plus the absorption in the metal. (c) At $1$ MHz treat the foil as a resistive sheet (Exercise 6.3.9) and find SE. What would the estimate of (b) give? (d) Sketch SE from $10$ kHz to $2$ GHz.

Try it yourself first.

Strategy

Aluminium is a good conductor at all these frequencies, $η_{2} = (1 + j) R_{s}$ with $∣ η_{2} ∣ ≪ η_{0}$ . The exact three-media result for a slab with air on both sides follows from Eq. (6.3.33) with $η_{1} = η_{3} = η_{0}$ and $j β_{2} \to γ$ :

\frac{E ^{t}}{E ^{i}} = \frac{4 η _{0} η _{2} e ^{- γ t}}{( η _{0} + η _{2} ) ^{2} - ( η _{0} - η _{2} ) ^{2} e ^{- 2 γ t}} .

For $t ≫ δ$ the second term of the denominator (the multiple reflections) is negligible. For $t ≪ δ$ expand $e^{- γ t}$ instead.

$δ = 1/ π f μ_{0} σ$ : $85.1$ μm at $1$ MHz and $2.69$ μm at $1$ GHz. $t = δ$ at $f = 1/ (π μ_{0} σ t^{2}) = 28.3$ MHz.

At $1$ GHz, $R_{s} = π f μ_{0} / σ = 0.0106 Ω$ , $∣ η_{2} ∣ = 0.0150 Ω$ , $t / δ = 5.95$ . Dropping the multiple reflections,

SE \approx 20 lo g_{10} \frac{( η _{0} + η _{2} ) ^{2}}{4 η _{0} η _{2}} + 8.686 \frac{t}{δ} \approx 20 lo g_{10} \frac{η _{0}}{4∣ η _{2} ∣} + 8.686 \times 5.95 = 76.0 + 51.7 = 127.6 dB .

For $γ t ≪ 1$ the exact formula becomes $E^{t} / E^{i} \approx 1/ (1 + γ t η_{0} /2 η_{2})$ , and in a good conductor $γ / η_{2} = σ$ : the foil acts as a sheet with $R_{s} = 1/ (σ t) = 1/ (3.5 \times 1 0^{7} \times 16 \times 1 0^{- 6}) = 1.79$ m $Ω$ , and

SE = 20 lo g_{10} (1 + \frac{η _{0}}{2 R _{s}}) = 20 lo g_{10} (1.06 \times 1 0^{5}) = 100.5 dB .

The estimate of (b) would give $106.0 + 1.6 = 107.6$ dB: $7.1$ dB too much, because in a foil thinner than $δ$ the waves reflected back and forth inside add to the first transmitted wave.

SE stays at $100.5$ dB from the lowest frequencies up to a few MHz (the sheet regime, independent of $f$ ), rises once $t > δ$ , reaches $102.5$ dB at $100$ MHz and $127.6$ dB at $1$ GHz, and then grows with the absorption term $\propto f$ (Figure 6.3.20, from the exact formula).

Answer

δ = 85.1 μ m (1 MHz), 2.69 μ m (1 GHz), t = δ at 28.3 MHz, SE = 100.5 dB (1 MHz), 127.6 dB (1 GHz)

Check

The exact formula gives $100.47$ dB at $1$ MHz and $127.61$ dB at $1$ GHz $✓$ . The face term of (b) equals $∣1/ τ_{12} τ_{21} ∣$ in dB, with $τ_{12} \approx 2 η_{2} / η_{0}$ entering the metal and $τ_{21} \approx 2$ leaving it.

Exercise 6.3.10. Plane-wave shielding effectiveness of a 16 μm aluminium foil. Below about 10 MHz the foil is thinner than a skin depth and acts as a resistive sheet (100.5 dB), where the estimate without multiple reflections (dashed) is too optimistic. Above about 100 MHz the two agree. — **Figure 6.3.20.** Exercise 6.3.10. Plane-wave shielding effectiveness of a $16$ μm aluminium foil. Below about $10$ MHz the foil is thinner than a skin depth and acts as a resistive sheet ( $100.5$ dB), where the estimate without multiple reflections (dashed) is too optimistic. Above about $100$ MHz the two agree.

Remark

A plane-wave SE of $100$ dB means that a foil without holes is an excellent shield. Real enclosures are limited by seams, cable entries and ventilation holes, and at low frequencies by magnetic near fields, whose wave impedance is far below $η_{0}$ : the face term then shrinks, and a thin nonmagnetic foil shields a $50$ Hz magnetic field hardly at all.

cf. Problem 6.3.6

Exercise 6.3.10: Shielding with aluminium foil