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

Coatings, windows, radome walls and layered ground all have two interfaces: a layer of medium 2, of thickness $d$ , between medium 1, from which the wave comes, and medium 3, into which it leaves (Figure 6.3.6). For this problem we use coordinates in which the layer occupies $0 < y < d$ , medium 1 the region $y > d$ and medium 3 the region $y < 0$ . The incident wave travels along $- a_{y}$ with its electric field along $a_{x}$ . In each outer medium there is one wave per direction allowed by the physics (medium 3 extends to $- \infty$ , so it carries only a wave leaving the layer), and inside the layer there are two, one going down ( $a$ ) and one coming back up ( $b$ ), produced by the reflection at $y = 0$ :

incident: E^{i} = E_{0}^{i} e^{j β_{1} y} a_{x}, H^{i} = \frac{E _{0}^{i}}{η _{1}} e^{j β_{1} y} a_{z}, reflected: E^{r} = E_{0}^{r} e^{- j β_{1} y} a_{x}, H^{r} = - \frac{E _{0}^{r}}{η _{1}} e^{- j β_{1} y} a_{z}, in the layer: E^{a} = E_{0}^{a} e^{j β_{2} y} a_{x}, H^{a} = \frac{E _{0}^{a}}{η _{2}} e^{j β_{2} y} a_{z}, E^{b} = E_{0}^{b} e^{- j β_{2} y} a_{x}, H^{b} = - \frac{E _{0}^{b}}{η _{2}} e^{- j β_{2} y} a_{z}, transmitted: E^{t} = E_{0}^{t} e^{j β_{3} y} a_{x}, H^{t} = \frac{E _{0}^{t}}{η _{3}} e^{j β_{3} y} a_{z} . (6.3.26)

A wave travelling along $- a_{y}$ varies as $e^{+ j β y}$ , and $a_{x} \times a_{z} = - a_{y}$ fixes the sign of its magnetic field by Eq. (6.3.5). The waves travelling up, along $+ a_{y}$ , have the opposite sign of $H$ , exactly as the reflected wave of a single interface.

(a) The five waves of the three-part problem.

(b) The two special cases. — (a) The five waves of the three-part problem.

The four boundary conditions

The tangential components $E_{x}$ and $H_{z}$ are continuous on both faces. At $y = d$ :

E_{0}^{i} e^{j β_{1} d} + E_{0}^{r} e^{- j β_{1} d} = E_{0}^{a} e^{j β_{2} d} + E_{0}^{b} e^{- j β_{2} d}, (6.3.27)

\frac{E _{0}^{i} e ^{j β_{1} d} - E _{0}^{r} e ^{- j β_{1} d}}{η _{1}} = \frac{E _{0}^{a} e ^{j β_{2} d} - E _{0}^{b} e ^{- j β_{2} d}}{η _{2}}, (6.3.28)

and at $y = 0$ :

E_{0}^{a} + E_{0}^{b} = E_{0}^{t}, \frac{E _{0}^{a} - E _{0}^{b}}{η _{2}} = \frac{E _{0}^{t}}{η _{3}} . (6.3.29)

These are four linear equations for the four unknowns $E_{0}^{r}$ , $E_{0}^{a}$ , $E_{0}^{b}$ and $E_{0}^{t}$ . They are solved from the far side inwards.

Step 1. The pair Eq. (6.3.29) at $y = 0$ , added and subtracted after multiplying the second by $η_{2}$ , gives the two waves in the layer in terms of the transmitted one:

E_{0}^{a} E_{0}^{b} = \frac{η _{3} + η _{2}}{2 η _{3}} E_{0}^{t}, = \frac{η _{3} - η _{2}}{2 η _{3}} E_{0}^{t} . (6.3.30)

Step 2. At $y = d$ , form Eq. (6.3.27) $+ η_{1} \times$ Eq. (6.3.28) and Eq. (6.3.27) $- η_{1} \times$ Eq. (6.3.28). The first combination contains only $E_{0}^{i}$ on the left, the second only $E_{0}^{r}$ :

2 E_{0}^{i} e^{j β_{1} d} 2 E_{0}^{r} e^{- j β_{1} d} = \frac{η _{2} + η _{1}}{η _{2}} E_{0}^{a} e^{j β_{2} d} + \frac{η _{2} - η _{1}}{η _{2}} E_{0}^{b} e^{- j β_{2} d}, = \frac{η _{2} - η _{1}}{η _{2}} E_{0}^{a} e^{j β_{2} d} + \frac{η _{2} + η _{1}}{η _{2}} E_{0}^{b} e^{- j β_{2} d} .

Step 3. Insert Eq. (6.3.30). The first equation becomes $2 E_{0}^{i} e^{j β_{1} d} = E_{0}^{t} D / (2 η_{2} η_{3})$ with the common denominator

D = (η_{3} + η_{2}) (η_{1} + η_{2}) e^{j β_{2} d} - (η_{3} - η_{2}) (η_{1} - η_{2}) e^{- j β_{2} d}, (6.3.31)

and the second gives $E_{0}^{r}$ in the same way. Dividing by $E_{0}^{i}$ :

The three-part problem

For a layer of medium 2 and thickness $d$ between media 1 and 3, with the amplitudes $E_{0}^{r}$ and $E_{0}^{i}$ referred to $y = 0$ ,

Γ = \frac{E _{0}^{r}}{E _{0}^{i}} = - \frac{e ^{2 j β_{1} d}}{D} \times [(η_{3} + η_{2}) (η_{1} - η_{2}) e^{j β_{2} d} - (η_{3} - η_{2}) (η_{1} + η_{2}) e^{- j β_{2} d}], (6.3.32)

τ = \frac{E _{0}^{t}}{E _{0}^{i}} = \frac{4 η _{2} η _{3} e ^{j β_{1} d}}{D}, (6.3.33)

with $D$ from Eq. (6.3.31).

In plain words

The layer turns the two-wave bookkeeping of one interface into a five-wave one, but the answer is still a ratio of impedance products. Everything about the layer enters through its impedance $η_{2}$ and its electrical thickness $β_{2} d$ .

Checks

No layer: with $η_{2} = η_{1}$ and $β_{2} = β_{1}$ the second term of $D$ vanishes, $D = 2 η_{1} (η_{3} + η_{1}) e^{j β_{1} d}$ , and

Γ = - e^{2 j β_{1} d} \cdot \frac{- ( η _{3} - η _{1} ) 2 η _{1} e ^{- j β_{1} d}}{D} = \frac{η _{3} - η _{1}}{η _{3} + η _{1}},

the single interface of Eq. (6.3.10). Units: $τ$ is a ratio of two products of impedances. The factor $e^{2 j β_{1} d}$ in $Γ$ appears because $E_{0}^{r}$ and $E_{0}^{i}$ are the amplitudes referred to $y = 0$ , a distance $d$ beyond the front face where the reflection actually happens. At the front face the incident phasor is $E_{0}^{i} e^{j β_{1} d}$ and the reflected one $E_{0}^{r} e^{- j β_{1} d}$ , and their ratio, $Γ e^{- 2 j β_{1} d}$ , is free of this factor (Section 6.3.6).

Special cases

When the first and the third media are the same ( $η_{3} = η_{1}$ , a slab in a single medium), the numerator of Eq. (6.3.32) becomes

(η_{1}^{2} - η_{2}^{2}) (e^{j β_{2} d} - e^{- j β_{2} d}) = 2 j (η_{1}^{2} - η_{2}^{2}) sin β_{2} d

and the denominator

D = (η_{1} + η_{2})^{2} e^{j β_{2} d} - (η_{1} - η_{2})^{2} e^{- j β_{2} d} = 4 η_{1} η_{2} cos β_{2} d + 2 j (η_{1}^{2} + η_{2}^{2}) sin β_{2} d .

Dividing both by $2 cos β_{2} d$ ,

Γ = - e^{2 j β_{1} d} \times \frac{j ( η _{1}^{2} - η _{2}^{2} ) tan β _{2} d}{2 η _{1} η _{2} + j ( η _{1}^{2} + η _{2}^{2} ) tan β _{2} d}, (6.3.34)

∣Γ ∣^{2} = \frac{( η _{1}^{2} - η _{2}^{2} ) ^{2} sin ^{2} β _{2} d}{4 η _{1}^{2} η _{2}^{2} cos ^{2} β _{2} d + ( η _{1}^{2} + η _{2}^{2} ) ^{2} sin ^{2} β _{2} d} . (6.3.35)

For nonmagnetic media in air, $η_{2} / η_{1} = 1/ ε_{r}$ , and after multiplying through by $ε_{r} cos^{2} β_{2} d$ and using $(ε_{r} + 1)^{2} - 4 ε_{r} = (ε_{r} - 1)^{2}$ this becomes the convenient form

∣Γ ∣^{2} = \frac{( ε _{r} - 1 ) ^{2} sin ^{2} β _{2} d}{4 ε _{r} + ( ε _{r} - 1 ) ^{2} sin ^{2} β _{2} d} . (6.3.36)

When the third medium is a perfect conductor ( $η_{3} = 0$ ),

Γ = - e^{2 j β_{1} d} \frac{η _{1} - j η _{2} tan β _{2} d}{η _{1} + j η _{2} tan β _{2} d} . (6.3.37)

For lossless media the numerator and the denominator of Eq. (6.3.37) are complex conjugates and have equal moduli, so $∣Γ∣ = 1$ : all the power comes back, and only its phase depends on $d$ .

Read the special cases

(i) $Γ = 0$ whenever $tan β_{2} d = 0$ , that is $d = m λ_{2} /2$ : a half-wave slab is invisible, whatever its permittivity. (ii) A lossless layer on a metal cannot absorb anything. With a lossy $η_{2}$ the same layer becomes an absorber: the Salisbury screen of Exercise 6.3.5 and radar-absorbing paint work this way.

6.3.5 Two interfaces: the three-part problem

The four boundary conditions

Checks

Special cases