# 7.3 Wave Polarization

In the most general case, the locus of the tip of E is an ellipse, and the wave is said to be elliptically polarized. Under certain conditions, the ellipse may degenerate into a circle or a straight line, in which case the polarization state is called circular or linear, respectively.

It was shown in Section 7-2 that the z components of the electric and magnetic fields of a z-propagating plane wave are both zero. Hence, in the most general case, the electric field phasor E˜(z) of a +z-propagating plane wave may consist of an x component, x ˆ E˜x(z), and a y component, y ˆ E˜y(z), or

E˜(z)=x ˆ E˜x(z)+y ˆ E˜y(z)(7.46)

with

E˜x(z)=E x 0 e−j k z,(7.47a)
E˜y(z)=E y 0 e−j k z,(7.47b)

where E x 0 and E y 0 are the amplitudes of E˜x(z) and E˜y(z), respectively. For the sake of simplicity, the plus sign superscript has been suppressed; the negative sign in e−j k z is sufficient to remind us that the wave is traveling in the positive z direction.

The two amplitudes E x 0 and E y 0 are, in general, complex quantities with each characterized by a magnitude and a phase angle. The phase of a wave is defined relative to a reference state, such as z=0 and t=0 or any other combination of z and t. As will become clear from the discussion that follows, the polarization of the wave described by Eqs. (7.46) and (7.47) depends on the phase of E y 0 relative to that of E x 0 but not on the absolute phases of E x 0 and E y 0. Hence, for convenience, we assign E x 0 a phase of zero and denote the phase of E y 0, relative to that of E x 0, as δ. Thus, δ is the phase difference between the y and x components of E˜. Accordingly, we define E x 0 and E y 0 as

E x 0=a x,(7.48a)
E y 0=a y e j δ,(7.48b)

where a x=|E x 0|≥0 and a y=|E y 0|≥0 are the magnitudes of E x 0 and E y 0, respectively. Thus, by definition, a x and a y may not assume negative values. Using Eqs. (7.48a and b) in Eqs. (7.47a and b), the total electric field phasor is

E˜(z)=(x ˆ a x+y ˆ a y e j δ)e−j k z,(7.49)

and the corresponding instantaneous field is

E(z,t)=R e E˜(z)e j ω t=x ˆ a x cos(ω t−k z)+y ˆ a y cos(ω t−k z+δ).(7.50)

When characterizing an electric field at a given point in space, two of its attributes that are of particular interest are its magnitude and direction. The magnitude of E(z,t) is

|E(z,t)|=[E x 2(z,t)+E y 2(z,t)]1/2=[a x 2 cos 2(ω t−k z)+a y 2 cos 2(ω t−k z+δ)]1/2.(7.51)

The electric field E(z,t) has components along the x and y directions. At a specific position z, the direction of E(z,t) is characterized by its inclination angle ψ, defined with respect to the x axis and given by

ψ(z,t)=tan−1 E y(z,t)E x(z,t).(7.52)

In the general case, both the intensity of E(z,t) and its direction are functions of z and t. Next, we examine some special cases.

# 7.3.1 Linear Polarization

Under these conditions, Eq. (7.50) simplifies to

E(0,t)=(x ˆ a x+y ˆ a y)cos(ω t−k z), (in phase)(7.53a)
E(0,t)=(x ˆ a x−y ˆ a y)cos(ω t−k z). (out of phase)(7.53b)

Let us examine the out-of-phase case. The field’s magnitude is

|E(z,t)|=[a x 2+a y 2]1/2|cos(ω t−k z)|,(7.54a)

and the inclination angle is

ψ=tan−1−a y a x.(out of phase)(7.54b)

We note that ψ is independent of both z and t. Figure 7-7 displays the line segment traced by the tip of E at z=0 over a half of a cycle. The trace would be the same at any other value of z as well. At z=0 and t=0, |E(0,0)|=[a x 2+a y 2]1/2. The length of the vector representing E(0,t) decreases to zero at ω t=π/2. The vector then reverses direction and increases in magnitude to [a x 2+a y 2]1/2 in the second quadrant of the x–y plane at ω t=π. Since ψ is independent of both z and t, E(z,t) maintains a direction along the line making an angle ψ with the x axis while oscillating back and forth across the origin.

# Figure 7-7 Linearly polarized wave traveling in the +z direction (out of the page).

Figure 7 7 Full Alternative Text

If a y=0, then ψ=0∘ or 180∘, and the wave is x-polarized; conversely, if a x=0, then ψ=90∘ or −90∘, and the wave is y-polarized.

# 7.3.2 Circular Polarization

We now consider the special case when the magnitudes of the x and y components of E˜(z) are equal, and the phase difference δ=±π/2. For reasons that become evident shortly, the wave polarization is called left-hand circular when δ=π/2, and right-hand circular when δ=−π/2.

# Left-Hand Circular (LHC) Polarization

For a x=a y=a and δ=π/2, Eqs. (7.49) and (7.50) become

E˜(z)=(x ˆ a+y ˆ a e j π/2)e−j k z=a(x ˆ+j y ˆ)e−j k z,(7.55a)

E(z,t)=ℜ e[ E˜(z) e j ω t ]=x ˆ a cos(ω t−k z)+y ˆ a cos(ω t−k z+π/2)=x ˆ a cos(ω t−k z)−y ˆ a sin(ω t−k z).(7.55b)

The corresponding field magnitude and inclination angle are

|E(z,t)|=E x 2(z,t)+E y 2(z,t)1/2=[a 2 cos 2(ω t−k z)+a 2 sin 2(ω t−k z)]1/2=a (7.56a)

and

ψ(z,t)=tan−1 E y(z,t)E x(z,t)=tan−1−a sin(ω t−k z)a cos(ω t−k z)=−(ω t−k z). (7.56b)

We observe that the magnitude of E is independent of both z and t, whereas ψ depends on both variables. These functional dependencies are the converse of those for the linear polarization case.

At z=0, Eq. (7.56b) gives ψ=−ω t; the negative sign implies that the inclination angle decreases as time increases. As illustrated in Fig. 7-8(a), the tip of E(t) traces a circle in the x−y plane and rotates in a clockwise direction as a function of time (when viewing the wave approaching). Such a wave is called left-hand circularly polarized because when the thumb of the left hand points along the direction of propagation (the z direction in this case) the other four fingers point in the direction of rotation of E.

# Figure 7-8 Circularly polarized plane waves propagating in the +z direction (out of the page).

(a) LHC polarization

Figure 7 8-a Full Alternative Text

(b) RHC polarization

Figure 7 8-b Full Alternative Text

# Right-Hand Circular (RHC) Polarization

For a x=a y=a and δ=−π/2, we have

|E(z,t)|=a, ψ=(ω t−k z).(7.57)

The trace of E(0,t) as a function of t is shown in Fig. 7-8(b). For RHC polarization, the fingers of the right hand point in the direction of rotation of E when the thumb is along the propagation direction. Figure 7-9 depicts a right-hand circularly polarized wave radiated by a helical antenna.

# Figure 7-9 Right-hand circularly polarized wave radiated by a helical antenna.

Figure 7 9 Full Alternative Text

# Example 7-2: RHC-Polarized Wave

An RHC-polarized plane wave with electric field magnitude of 3 (mV/m) is traveling in the +y direction in a dielectric medium with ε=4 ε 0, μ=μ 0, and σ=0. If the frequency is 100 MHz, obtain expressions for E(y,t) and H(y,t).

Solution: Since the wave is traveling in the +y direction, its field must have components along the x and z directions. The rotation of E(y,t) is depicted in Fig. 7-10, where y ˆ is out of the page. By comparison with the RHC-polarized wave shown in Fig. 7-8(b), we assign the z component of E˜(y) a phase angle of zero and the x component a phase shift of δ=−π/2. Both components have magnitudes of a=3 (mV/m). Hence,

E˜(y)=x ˆ E˜x+z ˆ E˜z=x ˆ a e−j π/2 e−j k y+z ˆ a e−j k y=(−x ˆ j+z ˆ)3 e−j k y(m V/m),

and application of Eq. (7.39a) gives

H˜(y)=1 η y ˆ​×E˜(y)=1 η y ˆ​×(−x ˆ j+z ˆ)3 e−j k y=3 η(z ˆ j+x ˆ)e−j k y (mA/m).

# Figure 7-10 Right-hand circularly polarized wave of Example 7-2.

Figure 7 10 Full Alternative Text

With ω=2 π f=2 π×10 8 (rad/s), the wavenumber k is

k=ω ε r c=2 π×10 8 4 3×10 8=4 3 π(r a d/m),

and the intrinsic impedance η is

η=η 0 ε r≈120 π 4=60 π(Ω).

The instantaneous fields E(y,t) and H(y,t) are

E(y,t)=R e E˜(y)e j ω t=R e(−x ˆ j+z ˆ)3 e−j k y e j ω t=3[x ˆ sin(ω t−k y)+z ˆ cos(ω t−k y)] (mV/m)

and

H(y,t)=ℜ e[ H˜(y) e j ω t ]=ℜ e[ 3 η(z ˆ j+x ˆ)e−j k y e j ω t ]=1 20 π[x ˆ cos(ω t−k y)−z ˆ sin(ω t−k y)] (mA/m).

# 7.3.3 Elliptical Polarization

Plane waves that are not linearly or circularly polarized are elliptically polarized. That is, the tip of E(z,t) traces an ellipse in the plane perpendicular to the direction of propagation. The shape of the ellipse and the field’s handedness (left-hand or right-hand) are determined by the values of the ratio (a y/a x) and the phase difference δ.

The polarization ellipse shown in Fig. 7-11 has its major axis with length a ξ along the ξ direction and its minor axis with length a η along the η direction. The rotation angle γ is defined as the angle between the major axis of the ellipse and a reference direction, which is chosen here to be the x axis, with γ being bounded within the range −π/2≤γ≤π/2. The shape of the ellipse and its handedness are characterized by the ellipticity angle χ, which is defined as

tan χ=±a η a ξ=±1 R,(7.58)

with the plus sign corresponding to left-handed rotation and the minus sign corresponding to right-handed rotation. The limits for χ are −π/4≤χ≤π/4. The quantity R=a ξ/a η is called the axial ratio of the polarization ellipse, and it varies between 1 for circular polarization and ∞ for linear polarization. The polarization angles γ and χ are related to the wave parameters a x, a y, and δ by*

tan 2 γ=(tan 2 ψ 0)cos δ (−π/2≤γ≤π/2),(7.59a)
sin 2 χ=(sin 2 ψ 0)sin δ (−π/4≤χ≤π/4),(7.59b)

where ψ 0 is an auxiliary angle defined by

tan ψ 0=a y a x 0≤ψ 0≤π 2.(7.60)

# Figure 7-11 Polarization ellipse in the x−y planewith the wave traveling in the z direction (out of the page).

Figure 7 11 Full Alternative Text

Sketches of the polarization ellipse are shown in Fig. 7-12 for various combinations of the angles (γ,χ). The ellipse reduces to a circle for χ=±45∘ and to a line for χ=0.

# Figure 7-12 Polarization states for various combinations of the polarization angles (γ,χ) for a wave traveling out of the page.

Figure 7 12 Full Alternative Text

Since the magnitudes a x and a y are, by definition, nonnegative numbers, the ratio a y/a x may vary between zero for an x-polarized linear polarization and ∞ for a y-polarized linear polarization. Consequently, the angle ψ 0 is limited to the range 0≤ψ 0≤90∘. Application of Eq. (7.59a) leads to two possible solutions for the value of γ, both of which fall within the defined range from −π/2 to π/2. The correct choice is governed by the following rule:

γ>0 if cos δ>0,γ<0 if cos δ<0.

# Example 7-3: Polarization State

Determine the polarization state of a plane wave with electric field

E(z,t)=x ˆ 3 cos(ω t−k z+30∘)−y ˆ 4 sin(ω t−k z+45∘) (mV/m).

Solution: We begin by converting the second term to a cosine reference:

E=x ˆ 3 cos(ω t−k z+30∘)−y ˆ 4 cos(ω t−k z+45∘−90∘)=x ˆ 3 cos(ω t−k z+30∘)−y ˆ 4 cos(ω t−k z−45∘).

The corresponding field phasor E˜(z) is

E˜(z)=x ˆ 3 e−j k z e j 30∘−y ˆ 4 e−j k z e−j 45∘=x ˆ 3 e−j k z e j 30∘+y ˆ 4 e−j k z e−j 45∘e j 180∘=x ˆ 3 e−j k z e j 30∘+y ˆ 4 e−j k z e j 135∘,

where we have replaced the negative sign of the second term with e j 180∘ in order to have positive amplitudes for both terms, thereby allowing us to use the definitions given earlier. According to the expression for E˜(z), the phase angles of the x and y components are δ x=30∘ and δ y=135∘, giving a phase difference of δ=δ y−δ x=135∘−30∘=105∘. The auxiliary angle ψ 0 is obtained from

ψ 0=tan−1 a y a x=tan−1 4 3=53.1∘.

From Eq. (7.59a),

tan 2 γ=(tan 2 ψ 0)cos δ=tan 106.2∘cos 105∘=0.89,

which gives two solutions for γ, namely γ=20.8∘ and γ=−69.2∘. Since cos δ<0, the correct value of γ is −69.2∘. From Eq. (7.59b),

sin 2 χ=(sin 2 ψ 0)sin δ=sin 106.2∘sin 105∘=0.93 or χ=34.0∘.

The magnitude of χ indicates that the wave is elliptically polarized, and its positive polarity specifies its rotation as left handed.

Concept Question 7-4: An elliptically polarized wave is characterized by amplitudes a x and a y and by the phase difference δ. If a x and a y are both nonzero, what should δ be in order for the polarization state to reduce to linear polarization?

Concept Question 7-5: Which of the following two descriptions defines an RHC-polarized wave: A wave incident upon an observer is RHC-polarized if its electric field appears to the observer to rotate in a counterclockwise direction (a) as a function of time in a fixed plane perpendicular to the direction of wave travel or (b) as a function of travel distance at a fixed time t?