Exercise 6.3.5: A resistive film in front of a metal plate

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 6.3.5

A resistive film in front of a metal plate

A thin resistive film of sheet resistance $R_{s} = 377 Ω$ (per square) is held $d = 7.5$ mm in front of a flat metal plate (a PEC), with air between them, and a plane wave falls normally on the film. The film carries the surface current $J_{S} = E_{t a n} / R_{s}$ , so $H_{t a n}$ jumps by $J_{S}$ across it while $E_{t a n}$ is continuous. (a) Show that the air gap backed by the metal has, at the film, the impedance $η_{g} = j η_{0} tan β_{0} d$ . (b) Show that the incident wave sees $R_{s}$ and $η_{g}$ in parallel, and that $Γ = - 1/ (1 + 2 j tan β_{0} d)$ . (c) At which frequencies does the screen reflect nothing, and at which everything? (d) Find the band around $10$ GHz where less than $1 %$ of the power is reflected. (e) What is reflected at $10$ GHz if the film comes out with $R_{s} = 250 Ω$ ? Where does the power that is not reflected go?

Try it yourself first.

Strategy

Use the transmission-line picture of Figure 6.3.2a: the gap is a line of length $d$ with $η_{0}$ , terminated by $η_{3} = 0$ . The film adds a current $E / R_{s}$ to the magnetic field without changing $E$ , which is what a resistor connected across the line does.

With $η_{2} = η_{0}$ and $η_{3} = 0$ the layer formula Eq. (6.3.39) gives $η_{g} = η_{0} \frac{0 + j η _{0} tan β _{0} d}{η _{0} + 0} = j η_{0} tan β_{0} d$ , a pure reactance.

Let $E_{f}$ be the tangential field at the film. Behind it $H_{b} = E_{f} / η_{g}$ , and in front $H_{f} = H_{b} + J_{S} = E_{f} / η_{g} + E_{f} / R_{s}$ . The impedance seen from the front is therefore

η_{in} = \frac{E _{f}}{H _{f}} = (\frac{1}{R _{s}} + \frac{1}{η _{g}})^{- 1} = \frac{j η _{0} R _{s} t}{R _{s} + j η _{0} t}, t = tan β_{0} d .

With $R_{s} = η_{0}$ , $η_{in} = η_{0} \frac{j t}{1 + j t}$ and

Γ ∣Γ ∣^{2} = \frac{η _{in} - η _{0}}{η _{in} + η _{0}} = \frac{j t - ( 1 + j t )}{j t + ( 1 + j t )} = - \frac{1}{1 + 2 j tan β _{0} d}, = \frac{1}{1 + 4 tan ^{2} β _{0} d} .

$Γ = 0$ when $tan β_{0} d \to \infty$ , $d = (2 m + 1) λ_{0} /4$ : with $d = 7.5$ mm at $f = 10, 30, 50, \dots$ GHz. There the shorted quarter-wave gap is an open circuit and the incident wave sees only $R_{s} = η_{0}$ . $∣Γ∣ = 1$ when $tan β_{0} d = 0$ , $d = m λ_{0} /2$ : at $20, 40, \dots$ GHz the film sits on a zero of the standing-wave $E$ and carries no current.

$∣Γ ∣^{2} < 0.01$ requires $tan^{2} β_{0} d > 99/4$ , $∣ tan β_{0} d ∣ > 4.975$ , that is $∣ β_{0} d - π /2∣ < arctan (1/4.975) = 0.198$ rad. Since $β_{0} d = \frac{π}{2} f / f_{0}$ ,

\frac{f}{f _{0}} - 1 < \frac{0.198}{π /2} = 0.126 ⟹ 8.74 < f < 11.26 GHz .

At $8$ and $12$ GHz, $tan β_{0} d = \pm 3.08$ and $∣Γ ∣^{2} = 2.6 %$ ( $- 15.9$ dB).

At $10$ GHz still $η_{in} = R_{s}$ : $Γ = \frac{250 - 377}{250 + 377} = - 0.203$ , $R = 4.1 %$ ( $- 13.9$ dB). The metal transmits nothing, so the fraction $1 - ∣Γ ∣^{2}$ that is not reflected is dissipated as Joule heat in the film, $\frac{1}{2} ∣ J_{S} ∣^{2} R_{s}$ per unit area.

Answer

Γ = - \frac{1}{1 + 2 j tan β _{0} d}, Γ = 0 at 10, 30, \dots GHz, ∣Γ∣ = 1 at 20, 40, \dots GHz, R < 1 % for 8.74 - 11.26 GHz

Check

Energy: $∣Γ∣ \leq 1$ for every $t$ , with equality only when the film carries no current ( $t = 0$ ) $✓$ . At very low frequency the gap is a short circuit and $Γ \to - 1$ , as for the bare metal.

Exercise 6.3.5. Reflected power of the Salisbury screen (R_s=377\ \Omega, d=7.5 mm). The reflection vanishes at the odd multiples of 10 GHz and is total at the even ones. — **Figure 6.3.18.** Exercise 6.3.5. Reflected power of the Salisbury screen ( $R_{s} = 377 Ω$ , $d = 7.5$ mm). The reflection vanishes at the odd multiples of $10$ GHz and is total at the even ones.

Remark

This is the oldest radar-absorbing screen (a Salisbury screen): a resistive sheet of $377 Ω$ per square a quarter wavelength in front of the surface to be hidden. It is thin and light but works over a band of only about $\pm 13 %$ for $R < 1 %$ , and the resonances repeat at odd multiples of the design frequency (Figure 6.3.18). A metal film of conductivity $σ$ and thickness $t_{f} ≪ δ$ has $R_{s} = 1/ (σ t_{f})$ , so $377 Ω$ needs $σ t_{f} = 2.65$ mS.

cf. Problem 6.3.2