# 8.1 Wave Reflection and Transmission at Normal Incidence

We know from Chapter 2 that, when a guided wave encounters a junction between two transmission lines with different characteristic impedances, the incident wave is partly reflected back toward the source and partly transmitted across the junction onto the other line. The same happens to a uniform plane wave when it encounters a boundary between two material half-spaces with different characteristic impedances. In fact, the situation depicted in Fig. 8-2(b) has an exact analogue in the transmission-line configuration of Fig. 8-2(a). The boundary conditions governing the relationships between the electric and magnetic fields in Fig. 8-2(b) map one-to-one onto those we developed in Chapter 2 for the voltages and currents on the transmission line.

# Figure 8-2 Discontinuity between two different transmission lines is analogous to that between two dissimilar media.

(a) Boundary between transmission lines

Figure 8 2-a Full Alternative Text

(b) Boundary between different media

Figure 8 2-b Full Alternative Text

For convenience, we divide our treatment of wave reflection by and transmission through planar boundaries into two parts. In this section, we confine our discussion to the normal-incidence case depicted in Fig. 8-3(a), and in Sections 8-2 to 8-4, we examine the more general oblique-incidence case depicted in Fig. 8-3(b). We will show the basis for the analogy between the transmission-line and plane-wave configurations so that we may use transmission-line equivalent models, tools (e.g., Smith chart), and techniques (e.g., quarter-wavelength matching) to expeditiously solve plane-wave problems.

# Figure 8-3 Ray representation of wave reflection and transmission at (a) normal incidence and (b) oblique incidence. (c) Wavefront representation of oblique incidence.

(a) Normal incidence

Figure 8 3-a Full Alternative Text

(b) Ray representation of oblique incidence

Figure 8 3-b Full Alternative Text

(c) Wavefront representation of oblique incidence

Figure 8 3-c Full Alternative Text

Before proceeding, however, we should explain the notion of rays and wavefronts and the relationship between them, as both are used throughout this chapter to represent electromagnetic waves. A ray is a line representing the direction of flow of electromagnetic energy carried by a wave; therefore, it is parallel to the propagation unit vector k ˆ. A wavefront is a surface across which the phase of a wave is constant; it is perpendicular to the wavevector k ˆ. Hence, rays are perpendicular to wavefronts. The ray representation of wave incidence, reflection, and transmission shown in Fig. 8-3(b) is equivalent to the wavefront representation depicted in Fig. 8-3(c). The two representations are complementary; the ray representation is easier to use in graphical illustrations, whereas the wavefront representation provides greater physical insight into what happens to a wave when it encounters a discontinuous boundary.

# 8.1.1 Boundary between Lossless Media

A planar boundary located at z=0 (Fig. 8-4(a)) separates two lossless, homogeneous, dielectric media. Medium 1 has permittivity ε 1 and permeability μ 1 and fills the half-space z≤0. Medium 2 has permittivity ε 2 and permeability μ 2 and fills the half-space z≥0. An x-polarized plane wave, with electric and magnetic fields (E i,H i) propagates in medium 1 along direction k ˆ i=z ˆ toward medium 2. Reflection and transmission at the boundary at z=0 result in a reflected wave, with electric and magnetic fields (E r,H r), traveling along direction k ˆ r=−z ˆ in medium 1, and a transmitted wave, with electric and magnetic fields (E t,H t), traveling along direction k ˆ t=z ˆ in medium 2. On the basis of the formulations developed in Sections 7-2 and 7-3 for plane waves, the three waves are described in phasor form by:

Incident Wave

E˜i(z)=x ˆ E 0 i e−j k 1 z, (8.1a)
H˜i(z)=z ˆ×E˜i(z)η 1=y ˆ E 0 i η 1 e−j k 1 z. (8.1b)

Reflected Wave

E˜r(z)=x ˆ E 0 r e j k 1 z, (8.2a)
H˜r(z)=(−z ˆ)×E˜r(z)η 1=−y ˆ E 0 r η 1 e j k 1 z. (8.2b)

Transmitted Wave

E˜t(z)=x ˆ E 0 t e−j k 2 z, (8.3a)
H˜t(z)=z ˆ×E˜t(z)η 2=y ˆ E 0 t η 2 e−j k 2 z. (8.3b)

# Figure 8-4 The two dielectric media separated by the x–y plane in (a) can be represented by the transmission-line analogue in (b).

(a) Boundary between dielectric media

Figure 8 7-a Full Alternative Text

(b) Transmission-line analogue

Figure 8 7-b Full Alternative Text

The quantities E 0 i, E 0 r, and E 0 t are, respectively, the amplitudes of the incident, reflected, and transmitted electric fields at z=0 (the boundary between the two media). The wavenumber and intrinsic impedance of medium 1 are k 1=ω μ 1 ε 1 and η 1=μ 1/ε 1, and those for medium 2 are k 2=ω μ 2 ε 2 and η 2=μ 2/ε 2.

The amplitude E 0 i is imposed by the source responsible for generating the incident wave; therefore, it is assumed known. Our goal is to relate E 0 r and E 0 t to E 0 i. We do so by applying boundary conditions for the total electric and magnetic fields at z=0. According to Table 6-2, the tangential component of the total electric field is always continuous across a boundary between two contiguous media, and in the absence of current sources at the boundary, the same is true for the total magnetic field. In the present case, the electric and magnetic fields of the incident, reflected, and transmitted waves are all tangential to the boundary.

The total electric field E˜1(z) in medium 1 is the sum of the electric fields of the incident and reflected waves, and a similar statement applies to the magnetic field H˜1(z). Hence,

Medium 1

E˜1(z)=E˜ i(z)+E˜ r(z)=x ˆ(E 0 i e−j k 1 z+E 0 r e j k 1 z), (8.4a)
H˜1(z)=H˜ i(z)+H˜ r(z)=y ˆ 1 η 1(E 0 i e−j k 1 z−E 0 r e j k 1 z). (8.4b)

With only the transmitted wave present in medium 2, the total fields are

Medium 2

E˜2(z)=E˜ t(z)=x ˆ E 0 t e−j k 2 z, (8.5a)
H˜2(z)=H˜ t(z)=y ˆ E 0 t η 2 e−j k 2 z. (8.5b)

At the boundary (z=0), the tangential components of the electric and magnetic fields are continuous. Hence,

E˜1(0)=E˜2(0)or E 0 i+E 0 r=E 0 t, (8.6a)
H˜1(0)=H˜2(0)or E 0 i η 1−E 0 r η 1=E 0 t η 2. (8.6b)

Solving these equations for E 0 r and E 0 t in terms of E 0 i gives

E 0 r=η 2−η 1 η 2+η 1 E 0 i=Γ E 0 i, (8.7a)
E 0 t=2 η 2 η 2+η 1 E 0 i=τ E 0 i, (8.7b)

where

The quantities Γ and τ are called the reflection and transmission coefficients. For lossless dielectric media, η 1 and η 2 are real; consequently, both Γ and τ are real also. As we see in Section 8-1.4, the expressions given by Eqs. (8.8a) and (8.8b) are equally applicable when the media are conductive, even though in that case η 1 and η 2 may be complex. Hence, Γ and τ may be complex as well. From Eqs. (8.8a) and (8.8b), it is easily shown that Γ and τ are interrelated as

For nonmagnetic media,

η 1=η 0 ε r 1,and η 2=η 0 ε r 2,

where η 0 is the intrinsic impedance of free space. In this case, Eq. (8.8a) may be expressed as

# 8.1.2 Transmission-Line Analogue

The transmission-line configuration shown in Fig. 8-4(b) consists of a lossless transmission line with characteristic impedance Z 01 connected at z=0 to an infinitely long lossless transmission line with characteristic impedance Z 02. The input impedance of an infinitely long line is equal to its characteristic impedance. Hence, at z=0, the voltage reflection coefficient (looking toward the boundary from the vantage point of the first line) is

Γ=Z 02−Z 01 Z 02+Z 01,

which is identical in form to Eq. (8.8a). The analogy between plane waves and waves on transmission lines does not end here. To demonstrate the analogy further, equations pertinent to the two cases are summarized in Table 8-1. Comparison of the two columns shows that there is a one-to-one correspondence between the transmission-line quantities (V˜,I˜,β,Z 0) and the plane-wave quantities (E˜,H˜,k,η).

The simultaneous presence of incident and reflected waves in medium 1 (Fig. 8-4(a)) gives rise to a standing-wave pattern. By analogy with the transmission-line case, the standing-wave ratio in medium 1 is defined as

The distance from the boundary to where the magnitude of the electric field intensity in medium 1 is a maximum (denoted l max) is described by the same expression as that given by Eq. (2.70) for the voltage maxima on a transmission line, namely

where λ 1=2 π/k 1 and θ r is the phase angle of Γ (i.e., Γ=|Γ|e j θ r, and θ r is bounded in the range −π<θ r≤π). The expression for l max is valid not only when the two media are lossless dielectrics, but also when medium 1 is a low-loss dielectric. Moreover, medium 2 may be either a dielectric or a conductor. When both media are lossless dielectrics, θ r=0 if η 2>η 1, and θ r=π if η 2<η 1.

The spacing between adjacent maxima is λ 1/2, and the spacing between a maximum and the nearest minimum is λ 1/4. The electric-field minima occur at

# 8.1.3 Power Flow in Lossless Media

Medium 1 in Fig. 8-4(a) is host to the incident and reflected waves, which together comprise the total electric and magnetic fields E˜1(z) and H˜1(z) given by Eqs. (8.11a) and (8.12a) of Table 8-1. Using Eq. (7.100), the net average power density flowing in medium 1 is

S av 1(z)=1 2 R e[E˜1(z)×H˜1∗(z)]=1 2 R e x ˆ E 0 i(e−j k 1 z+Γ e j k 1 z)y ˆ E 0 i∗η 1(e j k 1 z−Γ∗e−j k 1 z)=z ˆ|E 0 i|2 2 η 1(1−|Γ|2), (8.18)

which is analogous to Eq. (2.106) for the lossless transmission-line case. The first and second terms inside the bracket in Eq. (8.18) represent the average power density of the incident and reflected waves, respectively. Thus,

S av 1=S av i+S av r (8.19a)

with

S av i=z ˆ|E 0 i|2 2 η 1, (8.19b)
S av r=−z ˆ|Γ|2|E 0 i|2 2 η 1=−|Γ|2 S av i. (8.19c)

# Table 8-1 Analogy between plane-wave equations for normal incidence and transmission-line equations with both under lossless conditions.

Table 8 1 Full Alternative Text

Even though Γ is purely real when both media are lossless dielectrics, we chose to treat it as complex, thereby providing in Eq. (8.19c) an expression that is also valid when medium 2 is conducting.

The average power density of the transmitted wave in medium 2 is

S av 2(z)=1 2 R e[E˜2(z)×H˜2∗(z)]=1 2 R e x ˆ τ E 0 i e−j k 2 z×y ˆ τ∗E 0 i∗η 2 e j k 2 z=z ˆ|τ|2|E 0 i|2 2 η 2. (8.20)

Through the use of Eqs. (8.8a and b), it can be easily shown that for lossless media

which leads to

S av 1=S av 2.

This result is expected from considerations of power conservation.

# Example 8-1: Radar Radome Design

A 10 GHz aircraft radar uses a narrow-beam scanning antenna mounted on a gimbal behind a dielectric radome, as shown in Fig. 8-5. Even though the radome shape is far from planar, it is approximately planar over the narrow extent of the radar beam. If the radome material is a lossless dielectric with ε r=9 and μ r=1, choose its thickness d such that the radome appears transparent to the radar beam. Structural integrity requires d to be greater than 2.3 cm.

# Figure 8-5 Antenna beam “looking” through an aircraft radome of thickness d (Example 8-1).

Solution: Figure 8-6(a) shows a small section of the radome on an expanded scale. The incident wave can be approximated as a plane wave propagating in medium 1 (air) with intrinsic impedance η 0. Medium 2 (the radome) is of thickness d and intrinsic impedance η r, and medium 3 (air) is semi-infinite with intrinsic impedance η 0. Figure 8-6(b) shows an equivalent transmission-line model with z=0 selected to coincide with the outside surface of the radome, and the load impedance Z L=η 0 represents the input impedance of the semi-infinite air medium to the right of the radome.

# Figure 8-6 (a) Planar section of the radome of Fig. 8-5 at an expanded scale and (b) its transmission-line equivalent model (Example 8-1).

Figure 8 6 Full Alternative Text

For the radome to “appear” transparent to the incident wave, the reflection coefficient must be zero at z=−d, thereby guaranteeing total transmission of the incident power into medium 3. Since Z L=η 0 in Fig. 8-6(b), no reflection takes place at z=−d if Z in=η 0, which can be realized by choosing d=n λ 2/2 [see Section 2-8.4], where λ 2 is the wavelength in medium 2 and n is a positive integer. At 10 GHz, the wavelength in air is λ 0=c/f=3 cm, while in the radome material it is

λ 2=λ 0 ε r=3 cm 3=1 cm.

Hence, by choosing d=5 λ 2/2=2.5 cm, the radome is both nonreflecting and structurally stable.

# Example 8-2: Yellow Light Incident upon a Glass Surface

A beam of yellow light with wavelength 0.6 μ m is normally incident in air upon a glass surface. Assume the glass is sufficiently thick as to ignore its back surface. If the surface is situated in the plane z=0 and the relative permittivity of glass is 2.25, determine:

1. the locations of the electric field maxima in medium 1 (air),

2. the standing-wave ratio, and

3. the fraction of the incident power transmitted into the glass medium.

Solution: (a) We begin by determining the values of η 1, η 2, and Γ:

η 1=μ 1 ε 1=μ 0 ε 0≈120 π (Ω),η 2=μ 2 ε 2=μ 0 ε 0⋅1 ε r≈120 π 2.25=80 π (Ω),Γ=η 2−η 1 η 2+η 1=80 π−120 π 80 π+120 π=−0.2.

Hence, |Γ|=0.2 and θ r=π. From Eq. (8.16), the electric-field magnitude is maximum at

l max=θ r λ 1 4 π+n λ 1 2=λ 1 4+n λ 1 2(n=0,1,2,…)

with λ 1=0.6 μ m.

(b)

S=1+|Γ|1−|Γ|=1+0.2 1−0.2=1.5.

(c) The fraction of the incident power transmitted into the glass medium is equal to the ratio of the transmitted power density given by Eq. (8.20) to the incident power density, S av i=|E 0 i|2/2 η 1:

S av 2 S av i=τ 2|E 0 i|2 2 η 2/|E 0 i|2 2 η 1=τ 2 η 1 η 2.

In view of Eq. (8.21),

S av 2 S av i=1−|Γ|2=1−(0.2)2=0.96, or 96%.

# 8.1.4 Boundary between Lossy Media

In Section 8-1.1, we considered a plane wave in a lossless medium incident normally on a planar boundary of another lossless medium. We now generalize our expressions to lossy media. In a medium with constitutive parameters (ε,μ,σ), the propagation constant γ=α+j β and the intrinsic impedance η c are both complex. General expressions for α, β, and η c are given by Eqs. (7.66a), (7.66b), and (7.70), respectively, and approximate expressions are given in Table 7-2 for the special cases of low-loss media and good conductors. If media 1 and 2 have constitutive parameters (ε 1,μ 1,σ 1) and (ε 2,μ 2,σ 2) (Fig. 8-7), then expressions for the electric and magnetic fields in media 1 and 2 can be obtained from Eqs. (8.11) through (8.14) of Table 8-1 by replacing j k with γ and η with η c. Thus,

Medium 1

E˜1(z)=x ˆ E 0 i(e−γ 1 z+Γ e γ 1 z), (8.22a)
H˜1(z)=y ˆ E 0 i η c 1(e−γ 1 z−Γ e γ 1 z), (8.22b)

Medium 2

E˜2(z)=x ˆ τ E 0 i e−γ 2 z, (8.23a)
H˜2(z)=y ˆ τ E 0 i η c 2 e−γ 2 z. (8.23b)

Here, γ 1=α 1+j β 1, γ 2=α 2+j β 2, and

Because η c 1 and η c 2 are, in general, complex, Γ and τ may be complex as well.

# Figure 8-7 Normal incidence at a planar boundary between two lossy media.

(a) Boundary between dielectric media

Figure 8 7-a Full Alternative Text

(b) Transmission-line analogue

Figure 8 7-b Full Alternative Text

# Example 8-3: Normal Incidence on a Metal Surface

A 1 GHz x-polarized plane wave traveling in the +z direction is incident from air upon a copper surface. The air-to-copper interface is at z=0, and copper has ε r=1, μ r=1, and σ=5.8×10 7 S/m. If the amplitude of the electric field of the incident wave is 12 (mV/m), obtain expressions for the instantaneous electric and magnetic fields in the air medium. Assume the metal surface to be several skin depths deep.

Solution: In medium 1 (air), α=0,

β=k 1=ω c=2 π×10 9 3×10 8=20 π 3(rad/m),η 1=η 0=377 (Ω),λ=2 π k 1=0.3 m.

At f=1 GHz, copper is an excellent conductor because

ε′′ε′=σ ω ε r ε 0=5.8×10 7 2 π×10 9×(10−9/36 π)=1×10 9≫1.

Use of Eq. (7.77c) gives

η c 2=(1+j)π f μ σ=(1+j)π×10 9×4 π×10−7 5.8×10 7 1/2=8.25(1+j)(m Ω).

Since η c 2 is so small compared to η 0=377 (Ω) for air, the copper surface acts, in effect, like a short circuit. Hence,

Γ=η c 2−η 0 η c 2+η 0≈−1.

Upon setting Γ=−1 in Eqs. (8.11) and (8.12) of Table 8-1, we obtain

E˜1(z)=x ˆ E 0 i(e−j k 1 z−e j k 1 z)=−x ˆ j 2 E 0 i sin k 1 z, (8.25a)
H˜1(z)=y ˆ E 0 i η 1(e−j k 1 z+e j k 1 z)=y ˆ 2 E 0 i η 1 cos k 1 z. (8.25b)

With E 0 i=12 (mV/m), the instantaneous fields associated with these phasors are

E 1(z,t)=R e[E˜1(z) e j ω t]=x ˆ 2 E 0 i sin k 1 z sin ω t=x ˆ 24 sin(20 π z/3)sin(2 π×10 9 t) (mV/m),H 1(z,t)=R e[H˜1(z) e j ω t]=y ˆ 2 E 0 i η 1 cos k 1 z cos ω t=y ˆ 64 cos(20 π z/3)cos(2 π×10 9 t) (μ A/m).

Plots of the magnitude of E 1(z,t) and H 1(z,t) are shown in Fig. 8-8 as a function of negative z for various values of ω t. The wave patterns exhibit a repetition period of λ/2, and E and H are in phase quadrature (90° phase shift) in both space and time. This behavior is identical with that for voltage and current waves on a shorted transmission line.

# Figure 8-8 Wave patterns for fields E 1(z,t) and H 1(z,t) of Example 8-3.

Figure 8 8 Full Alternative Text

Concept Question 8-1: What boundary conditions were used in the derivations of the expressions for Γ and τ?

Concept Question 8-2: In the radar radome design of Example 8-1, all the incident energy in medium 1 ends up getting transmitted into medium 3, and vice versa. Does this imply that no reflections take place within medium 2? Explain.

Concept Question 8-3: Explain on the basis of boundary conditions why it is necessary that Γ=−1 at the boundary between a dielectric and a perfect conductor.