# 7.2 Plane-Wave Propagation in Lossless Media

The properties of an electromagnetic wave, such as its phase velocity u p and wavelength λ, depend on the angular frequency ω and the medium’s three constitutive parameters: ε, μ, and σ. If the medium is nonconducting (σ=0), the wave does not suffer any attenuation as it travels; hence, the medium is said to be lossless. Because in a lossless medium ε c=ε, Eq. (7.14) becomes

γ 2=−ω 2 μ ε.(7.17)

For lossless media, it is customary to define the wavenumber k as

In view of Eq. (7.17), γ 2=−k 2 and Eq. (7.15) becomes

# 7.2.1 Uniform Plane Waves

For an electric field phasor defined in Cartesian coordinates as

E˜=x ˆ E˜x+y ˆ E˜y+z ˆ E˜z,(7.20)

substitution of Eq. (7.12) into Eq. (7.19) gives

(∂2∂x 2+∂2∂y 2+∂2∂z 2)(x ˆ E˜x+y ˆ E˜y+z ˆ E˜z)+k 2(x ˆ E˜x+y ˆ E˜y+z ˆ E˜z)=0.(7.21)

To satisfy Eq. (7.21), each vector component on the left-hand side of the equation must vanish. Hence,

∂2∂x 2+∂2∂y 2+∂2∂z 2+k 2 E˜x=0,(7.22)

and similar expressions apply to E˜y and E˜z.

If this happens to be the x−y plane, then E and H do not vary with x or y. Hence, ∂E˜x/∂x=0 and ∂E˜x/∂y=0, so Eq. (7.22) reduces to

d 2 E˜x d z 2+k 2 E˜x=0.(7.23)

Similar expressions apply to E˜y, H˜x, and H˜y. The remaining components of E˜ and H˜ are zero; that is, E˜z=H˜z=0. To show that E˜z=0, let us consider the z component of Eq. (7.6d),

z ˆ∂H˜y∂x−∂H˜x∂y=z ˆ j ω ε E˜z.(7.24)

Since ∂H˜y/∂x=∂H˜x/∂y=0, it follows that E˜z=0. A similar examination involving Eq. (7.6b) reveals that H˜z=0.

For the phasor quantity E˜x, the general solution of the ordinary differential equation given by Eq. (7.23) is

E˜x(z)=E˜x+(z)+E˜x−(z)=E x 0+e−j k z+E x 0−e j k z,(7.25)

where E x 0+ and E x 0− are constants to be determined from boundary conditions. The solution given by Eq. (7.25) is similar in form to the solution for the phasor voltage V˜(z) given by Eq. (2.54a) for the lossless transmission line. The first term in Eq. (7.25), containing the negative exponential e−j k z, represents a wave with amplitude E x 0+ traveling in the +z direction. Likewise, the second term (with e j k z) represents a wave with amplitude E x 0− traveling in the −z direction. Assume for the time being that E˜ only has a component along x (i.e., E˜y=0) and that E˜x is associated with a wave traveling in the +z direction only (i.e., E x 0−=0). Under these conditions,

E˜(z)=x ˆ E˜x+(z)=x ˆ E x 0+e−j k z.(7.26)

To find the magnetic field H˜ associated with this wave, we apply Eq. (7.6b) with E˜y=E˜z=0:

∇×E˜=x ˆ y ˆ z ˆ∂∂x∂∂y∂∂z E˜x+(z)0 0=−j ω μ(x ˆ H˜x+y ˆ H˜y+z ˆ H˜z).(7.27)

For a uniform plane wave traveling in the +z direction,

∂E x+(z)/∂x=∂E x+(z)/∂y=0.

Hence, Eq. (7.27) gives

H˜x=0,(7.28a)
H˜y=1−j ω μ∂E˜x+(z)∂z,(7.28b)
H˜z=1−j ω μ∂E x+(z)∂y=0. (7.28c)

Use of Eq. (7.26) in Eq. (7.28b) gives

H˜y(z)=k ω μ E x 0+e−j k z=H y 0+e−j k z,(7.29)

where H y 0+ is the amplitude of H˜y(z) and is given by

H y 0+=k ω μ E x 0+.(7.30)

For a wave traveling from the source toward the load on a transmission line, the amplitudes of its voltage and current phasors, V 0+ and I 0+, are related by the characteristic impedance of the line, Z 0. A similar connection exists between the electric and magnetic fields of an electromagnetic wave. The intrinsic impedance of a lossless medium is defined as

where we used the expression for k given by Eq. (7.18).

In view of Eq. (7.31), the electric and magnetic fields of a +z-propagating plane wave with E field along x ˆ are

E˜(z)=x ˆ E˜x+(z)=x ˆ E x 0+e−j k z,(7.32a)
H˜(z)=y ˆ E˜x+(z)η=y ˆ E x 0+η e−j k z.(7.32b)

# Figure 7-4 A transverse electromagnetic (TEM) wave propagating in the direction k ˆ=z ˆ. For all TEM waves, k ˆ is parallel to E×H.

Figure 7 4 Full Alternative Text

Other examples of TEM waves include waves traveling on coaxial transmission lines (E is along r ˆ, H is along ϕ^, and the direction of travel is along z ˆ) and spherical waves radiated by antennas.

In the general case, E x 0+ is a complex quantity with magnitude |E x 0+| and phase angle ϕ+. That is,

E x 0+=|E x 0+|e j ϕ+.(7.33)

The instantaneous electric and magnetic fields therefore are

E(z,t)=R e E˜(z)e j ω t=x ˆ|E x 0+|cos(ω t−k z+ϕ+)(V/m),(7.34a)

and

H(z,t)=R e H˜(z)e j ω t=y ˆ|E x 0+|η cos(ω t−k z+ϕ+)(A/m).(7.34b)

Because E(z,t) and H(z,t) exhibit the same functional dependence on z and t, they are said to be in phase; when the amplitude of one of them reaches a maximum, the amplitude of the other does so too. The fact that E˜ and H˜ are in phase is characteristic of waves propagating in lossless media.

From the material on wave motion presented in Section 1.4, we deduce that the phase velocity of the wave is

and its wavelength is

In a vacuum, ε=ε 0 and μ=μ 0, and the phase velocity u p and the intrinsic impedance η given by Eq. (7.31) are

u p=c=1 μ 0 ε 0=3×10 8 (m/s),(7.37)
η=η 0=μ 0 ε 0=377 (Ω)≈120 π (Ω),(7.38)

where c is the velocity of light and η 0 is called the intrinsic impedance of free space.

# Example 7-1: EM Plane Wave in Air

This example is analogous to the “Sound Wave in Water” problem given by Example 1-1.

The electric field of a 1 MHz plane wave traveling in the +z direction in air points along the x direction. If this field reaches a peak value of 1.2 π (mV/m) at t=0 and z=50 m, obtain expressions for E(z,t) and H(z,t). Then plot them as a function of z at t=0.

Solution: At f=1 MHz, the wavelength in air is

λ=c f=3×10 8 1×10 6=300 m,

and the corresponding wavenumber is k=(2 π/300) (rad/m). The general expression for an x-directed electric field traveling in the +z direction is given by Eq. (7.34a) as

E(z, t)=x^| E x 0+ |cos(ω t−k z+ϕ+)=x^ 1.2 π cos(2 π×10 6 t−2 π z 300+ϕ+) (mV/m).

The field E(z,t) is maximum when the argument of the cosine function equals zero or a multiple of 2 π. At t=0 and z=50 m, this condition yields

−2 π×50 300+ϕ+=0 or ϕ+=π 3.

Hence,

E(z,t)=x ˆ 1.2 π cos 2 π×10 6 t−2 π z 300+π 3 (m V/m),

and from Eq. (7.34b), we have

H(z,t)=y ˆ E(z,t)η 0=y ˆ 10 cos 2 π×10 6 t−2 π z 300+π 3(μ A/m),

where we have used the approximation η 0≈120 π (Ω).

At t=0,

E(z,0)=x ˆ 1.2 π cos 2 π z 300−π 3(m V/m),H(z,0)=y ˆ 10 cos 2 π z 300−π 3(μ A/m).

Plots of E(z,0) and H(z,0) as a function of z are shown in Fig. 7-5.

# Figure 7-5 Spatial variations of E and H at t=0 for the plane wave of Example 7-1.

Figure 7 5 Full Alternative Text

# 7.2.2 General Relation between E and H

It can be shown that, for any uniform plane wave traveling in an arbitrary direction denoted by the unit vector k ˆ, the electric and magnetic field phasors E˜ and H˜ are related as

The relations given by Eqs. (7.39a and b) are valid not only for lossless media but for lossy ones as well. As we see later in Section 7-4, the expression for η of a lossy medium is different from that given by Eq. (7.31). As long as the expression used for η is appropriate for the medium in which the wave is traveling, the relations given by Eqs. (7.39a and b) always hold.

# Wave Propagating Along +z with E Along x^

Let us apply Eq. (7.39a) to the wave given by Eq. (7.32a). The direction of propagation k ˆ=z ˆ and E˜=x ˆ E˜x+(z). Hence,

H˜=1 η k ˆ×E˜=1 η(z ˆ×x ˆ)E˜x+(z)=y ˆ E˜x+(z)η,(7.40)

which is the same as the result given by Eq. (7.32b).

# Wave Propagating Along −z with E Along x ˆ

For a wave traveling in the −z direction with electric field given by

E˜=x ˆ E˜x−(z)=x ˆ E x 0−e j k z,(7.41)

application of Eq. (7.39a) gives

H˜=1 η(−z ˆ×x ˆ)E˜x−(z)=−y ˆ E˜x−(z)η=−y ˆ E x 0−η e j k z.(7.42)

Hence, in this case, H˜ points in the negative y direction.

# Wave Propagating Along +z with E Along x ˆ and y ˆ

In general, a uniform plane wave traveling in the +z direction may have both x and y components, where E˜ is given by

E˜=x ˆ E˜x+(z)+y ˆ E˜y+(z),(7.43a)

and the associated magnetic field is

H˜=x ˆ H˜x+(z)+y ˆ H˜y+(z).(7.43b)

Application of Eq. (7.39a) gives

H˜=1 η z ˆ×E˜=−x ˆ E˜y+(z)η+y ˆ E˜x+(z)η.(7.44)

By equating Eq. (7.43b) to Eq. (7.44), we have

H˜x+(z)=−E˜y+(z)η,H˜y+(z)=E˜x+(z)η.(7.45)

These results are illustrated in Fig. 7-6. The wave may be considered the sum of two waves: one with electric and magnetic components (E x+,H y+) and another with components (E y+,H x+). In general, a TEM wave may have an electric field in any direction in the plane orthogonal to the direction of wave travel, the associated magnetic field is also in the same plane, and its direction is dictated by Eq. (7.39a).

# Figure 7-6 The wave (E,H) is equivalent to the sum of two waves: one with fields (E x+,H y+) and another with (E y+,H x+) with both traveling in the +z direction.

Figure 7 6 Full Alternative Text

Concept Question 7-1: What is a uniform plane wave? Describe its properties, both physically and mathematically. Under what conditions is it appropriate to treat a spherical wave as a plane wave?

Concept Question 7-2: >Since E˜ and H˜ are governed by wave equations of the same form [Eqs. (7.15) and (7.16)], does it follow that E˜=H˜? Explain.

Concept Question 7-3: If a TEM wave is traveling in the y ˆ direction, can its electric field have components along x ˆ, y ˆ, and z ˆ? Explain.