# 7.4 Plane-Wave Propagation in Lossy Media

To examine wave propagation in a lossy (conducting) medium, we return to the wave equation given by Eq. (7.15),

∇2 E˜−γ 2 E˜=0(7.61)

with

γ 2=−ω 2 μ ε c=−ω 2 μ(ε′−j ε′′),(7.62)

where ε′=ε and ε′′=σ/ω. Since γ is complex, we express it as

γ=α+j β,(7.63)

where α is the medium’s attenuation constant and β its phase constant. By replacing γ with (α+j β) in Eq. (7.62), we obtain

(α+j β)2=(α 2−β 2)+j 2 α β=−ω 2 μ ε′+j ω 2 μ ε′′.(7.64)

The rules of complex algebra require the real and imaginary parts on one side of an equation to equal, respectively, the real and imaginary parts on the other side. Hence,

α 2−β 2=−ω 2 μ ε′,(7.65a)
2 α β=ω 2 μ ε′′.(7.65b)

Solving these two equations for α and β gives

For a uniform plane wave with electric field E˜=x ˆ E˜x(z) traveling along the z direction, the wave equation given by Eq. (7.61) reduces to

d 2 E˜x(z)d z 2−γ 2 E˜x(z)=0.(7.67)

The general solution of the wave equation given by Eq. (7.67) comprises two waves: one traveling in the +z direction and another traveling in the −z direction. Assuming only the former is present, the solution of the wave equation leads to

E˜(z)=x ˆ E˜x(z)=x ˆ E x 0 e−γ z=x ˆ E x 0 e−α z e−j β z.(7.68)

The associated magnetic field H˜ can be determined by applying Eq. (7.2b): ∇×E˜=−j ω μ H˜, or using Eq. (7.39a): H˜=(k ˆ×E˜)/η c, where η c is the intrinsic impedance of the lossy medium. Both approaches give

H˜(z)=y ˆ H˜y(z)=y ˆ E˜x(z)η c=y ˆ E x 0 η c e−α z e−j β z,(7.69)

where

We noted earlier that in a lossless medium, E(z,t) is in phase with H(z,t). This property no longer holds true in a lossy medium because η c is complex. This fact is demonstrated later in Example 7-4.

From Eq. (7.68), the magnitude of E˜x(z) is given by

|E˜x(z)|=|E x 0 e−α z e−j β z|=|E x 0|e−α z,(7.71)

which decreases exponentially with z at a rate dictated by the attenuation constant α. Since H˜y=E˜x/η c, the magnitude of H˜y also decreases as e−α z. As the field attenuates, part of the energy carried by the electromagnetic wave is converted into heat due to conduction in the medium. As the wave travels through a distance z=δ s with

the wave magnitude decreases by a factor of e−1≈0.37 (Fig. 7-13). At depth z=3 δ s, the field magnitude is less than 5% of its initial value, and at z=5 δ s, it is less than 1%.

# Figure 7-13 Attenuation of the magnitude of E˜x(z) with distance z. The skin depth δ s is the value of z at which |E˜x(z)|/|E x 0|=e−1 or z=δ s=1/α.

Figure 7 13 Full Alternative Text

In a perfect dielectric, σ=0 and ε′′=0. Use of Eq. (7.66a) yields α=0; therefore, δ s=∞. Thus, in free space, a plane wave can propagate indefinitely with no loss in magnitude. On the other extreme, in a perfect conductor, σ=∞ and use of Eq. (7.66a) leads to α=∞. Hence δ s=0. If the outer conductor of a coaxial cable is designed to be several skin depths thick, it prevents energy inside the cable from leaking outward and shields against penetration of electromagnetic energy from external sources into the cable.

The expressions given by Eqs. (7.66a), (7.66b), and (7.70) for α, β, and η c are valid for any linear, isotropic, and homogeneous medium. For a perfect dielectric (σ=0), these expressions reduce to those for the lossless case (Section 7-2), wherein α=0, β=k=ω μ ε, and η c=η. For a lossy medium, the ratio ε′′/ε′=σ/ω ε, which appears in all these expressions, plays an important role in classifying how lossy the medium is. When ε′′/ε′≪1, the medium is considered a low-loss dielectric, and when ε′′/ε′≫1, it is considered a good conductor. In practice, the medium may be regarded as a low-loss dielectric if ε′′/ε′< 10−2, as a good conductor if ε′′/ε′>10 2, and as a quasi-conductor if 10−2≤ε′′/ε′≤10 2. For low-loss dielectrics and good conductors, the expressions given by Eq. (7.66) can be significantly simplified, as shown next.

# 7.4.1 Low-Loss Dielectric

From Eq. (7.62), the general expression for γ is

γ=j ω μ ε′1−j ε′′ε′1/2.(7.73)

For |x|≪1, the function (1−x)1/2 can be approximated by the first two terms of its binomial series; that is, (1−x)1/2≈1−x/2. By applying this approximation to Eq. (7.73) for a low-loss dielectric with x=j ε′′/ε′ and ε′′/ε′≪1, we obtain

γ≈j ω μ ε′1−j ε′′2 ε′.(7.74)

The real and imaginary parts of Eq. (7.74) are

We note that the expression for β is the same as that for the wavenumber k of a lossless medium. Applying the binomial approximation (1−x)−1/2≈(1+x/2) to Eq. (7.70) leads to

η c≈μ ε′1+j ε′′2 ε′=μ ε 1+j σ 2 ω ε.(7.76a)

In practice, because ε′′/ε′=σ/ω ε<10−2, the second term in Eq. (7.76a) often is ignored. Thus,

which is the same as Eq. (7.31) for the lossless case.

# 7.4.2 Good Conductor

When ε′′/ε′>100, Eqs. (7.66a), (7.66b), and (7.70) can be approximated as

In Eq. (7.77c), we used the relation given by Eq. (1.53): j=(1+j)/2. For a perfect conductor with σ=∞, these expressions yield α=β=∞ and η c=0. A perfect conductor is equivalent to a short circuit in a transmission line equivalent.

Expressions for the propagation parameters in various types of media are summarized in Table 7-1.

# Example 7-4: Plane Wave in Seawater

A uniform plane wave is traveling in seawater. Assume that the x−y plane resides just below the sea surface and the wave travels in the +z direction into the water. The constitutive parameters of seawater are ε r=80, μ r=1, and σ=4 S/m. If the magnetic field at z=0 is

H(0,t)=y ˆ 100 cos(2 π×10 3 t+15∘)(mA/m),

1. obtain expressions for E(z,t) and H(z,t), and

2. determine the depth at which the magnitude of E is 1% of its value at z=0.

Solution: (a) Since H is along y ˆ and the propagation direction is z ˆ, E must be along x ˆ. Hence, the general expressions for the phasor fields are

E˜(z)=x ˆ E x 0 e−α z e−j β z,(7.78a)
H˜(z)=y ˆ E x 0 η c e−α z e−j β z.(7.78b)

To determine α, β, and η c for seawater, we begin by evaluating the ratio ε′′/ε′. From the argument of the cosine function of H(0,t), we deduce that ω=2 π×10 3 (rad/s). Therefore, f=1 kHz. Hence,

ε′′ε′=σ ω ε=σ ω ε r ε 0=4 2 π×10 3×80×(10−9/36 π)=9×10 5.

This qualifies seawater as a good conductor at 1 kHz and allows us to use the good-conductor expressions given in Table 7-1:

α=π f μ σ=π×10 3×4 π×10−7×4=0.126 (Np/m),(7.79a)
β=α=0.126 (rad/m),(7.79b)
η c=(1+j)α σ=(2 e j π/4)0.126 4=0.044 e j π/4 (Ω).(7.79c)

# Table 7-1 Expressions for α, β, η c, u p, and λ for various types of media.

Table 7 1 Full Alternative Text

As no explicit information has been given about the electric field amplitude E x 0, we should assume it to be complex; that is, E x 0=|E x 0|e j ϕ 0. The wave’s instantaneous electric and magnetic fields are given by

E(z,t)=ℜ e[ x ˆ|E x 0|e j ϕ 0 e−α z e−j β z e j ω t ]=x ˆ|E x 0|e−0.126 z cos(2 π×10 3 t−0.126 z+ϕ 0) (V/m),(7.80a)
H(z,t)=ℜ e[ y ˆ |E x 0|e j ϕ 0 0.044 e j π/4 e−α z e−j β z e j ω t ]=y ˆ 22.5|E x 0|e−0.126 z cos(2 π×10 3 t−0.126 z+ϕ 0−45∘) (A/m).(7.80b)

At z=0,

H(0,t)=y ˆ 22.5|E x 0|cos(2 π×10 3 t+ϕ 0−45∘)(A/m).(7.81)

By comparing Eq. (7.81) with the expression given in the problem statement,

H(0,t)=y ˆ 100 cos(2 π×10 3 t+15∘)(m A/m),

we deduce that

22.5|E x 0|=100×10−3

or

|E x 0|=4.44(m V/m)

and

ϕ 0−45∘=15∘or ϕ 0=60∘.

Hence, the final expressions for E(z,t) and H(z,t) are

E(z,t)=x ˆ 4.44 e−0.126 z cos(2 π×10 3 t−0.126 z+60∘)(m V/m),(7.82a)
H(z,t)=y ˆ 100 e−0.126 z cos(2 π×10 3 t−0.126 z+15∘)(m A/m). (7.82b)

(b) The depth at which the amplitude of E has decreased to 1% of its initial value at z=0 is obtained from

0.01=e−0.126 z

or

z=ln(0.01)−0.126=36.55 m≈37 m.