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

Problem 6.3.3

Interactive: Normal Incidence and Layers

Problem 6.3.3

A dielectric slab in air

A lossless nonmagnetic dielectric slab with $ε_{r} = 4$ and thickness $d = 2.5$ cm stands in air. A uniform plane wave of frequency $f = 1.5$ GHz is incident normally on it (the three-part problem with media 1 and 3 both air).

Find the reflection coefficient $Γ_{d}$ at the front face of the slab and the fractions of the incident power that are reflected and transmitted.
Find the standing-wave ratio in front of the slab.
For which thicknesses is the slab transparent at $1.5$ GHz, and at which frequencies up to $6$ GHz is the $2.5$ cm slab transparent?

Try it yourself first.

Strategy

Seen from the front face, the slab followed by air is a line of length $d$ terminated by $η_{3} = η_{0}$ , with the input impedance Eq. (6.3.39). Everything depends on the electrical thickness $β_{2} d$ , so compute it first. The special-case formula Eq. (6.3.34) serves as a second route.

$β_{1} = 2 π f / c = 10 π$ rad/m ( $λ_{1} = 20$ cm) and $β_{2} = 2 β_{1} = 20 π$ rad/m ( $λ_{2} = 10$ cm), so

β_{2} d = 20 π \times 0.025 = \frac{π}{2} ⟹ d = \frac{λ _{2}}{4}, tan β_{2} d \to \infty.

The layer is a quarter-wave transformer. With $η_{2} = 60 π Ω$ ,

η_{in} Γ_{d} = \frac{η _{2}^{2}}{η _{3}} = \frac{( 60 π ) ^{2}}{120 π} = 30 π Ω, = \frac{30 π - 120 π}{30 π + 120 π} = - 0.6.

The slab is lossless and the air behind it has the impedance of the air in front, so $R = ∣ Γ_{d} ∣^{2} = 36 %$ and $T = 1 - R = 64 %$ , that is $∣ τ ∣ = 0.8$ .

$S = \frac{1 + 0.6}{1 - 0.6} = 4$ .

$Γ_{d} = 0$ requires $η_{in} = η_{1}$ , which for $η_{3} = η_{1}$ happens when $tan β_{2} d = 0$ , $β_{2} d = mπ$ , $d = m λ_{2} /2$ . At $1.5$ GHz: $d = 5, 10, 15, \dots$ cm. For $d = 2.5$ cm,

f_{m} = \frac{m c}{2 d ε _{r}} = m \times \frac{3 \times 1 0 ^{8}}{2 \times 0.025 \times 2} = m \times 3 GHz : 3 and 6 GHz .

Answer

Γ_{d} = - 0.6, R = 36 %, T = 64 %, S = 4 transparent for d = m \times 5 cm, and at 3 and 6 GHz

Read $Γ_{d} = - 0.6$ and $S = 4$ below, then move $f$ to $3 GHz$ , where the slab disappears.

InteractiveLoading simulation…

Check

The special-case formula Eq. (6.3.34) with $η_{1} = 1$ , $η_{2} = \frac{1}{2}$ (in units of $η_{0}$ ) and $tan β_{2} d \to \infty$ gives

Γ = - e^{2 j β_{1} d} \frac{1 - 1/4}{1 + 1/4} = - e^{j π /2} \times 0.6 = - j 0.6,

referred to $y = 0$ . Moved to the front face with Eq. (6.3.38), $Γ_{d} = Γ e^{- 2 j β_{1} d} = - j 0.6 e^{- j π /2} = - 0.6$ $✓$ . The transmission coefficient Eq. (6.3.33), with $D = (1.5) (1.5) j - (0.5) (0.5) (- j) = 2.5 j$ , has $∣ τ ∣ = 4 \times \frac{1}{2} /2.5 = 0.8$ $✓$ .

(a) |E(y)| at 1.5 GHz, wave incident from y>d. — (a) $∣ E (y) ∣$ at $1.5$ GHz, wave incident from $y > d$ .

(b) |\Gamma_d|^2 of the 2.5 cm slab, 0 to 6.5 GHz. — (a) $∣ E (y) ∣$ at $1.5$ GHz, wave incident from $y > d$ .

Remark

The reflected power is periodic in frequency with period $3$ GHz: zero at multiples of $3$ GHz (half-wave thickness), maximum, $36 %$ , at $1.5, 4.5, \dots$ GHz (odd quarter-wave thickness), Figure 6.3.11b. A quarter-wave slab is the worst choice for a window, the reason radome walls are cut to $λ_{2} /2$ .

Problem 6.3.3: A dielectric slab in air