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Chapter 6 Plane-Wave Propagation • We have seen that a time-varying electric field E(t)

produces a time-varying magnetic field H(t) and, conversely, a time-varying magnetic field produces a time-varying electric field.

This cyclic pattern generates electromagnetic waves (EM). • When the propagation is guided by a material structure,

the EM wave is said to be traveling in a guided medium.

Example of guided medium:

- A coaxial transmission line,

z

0

x

E

E

1.2π (mV/m)

λ

y

H H

10 (µA/m)

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- Earth’s surface and ionosphere constitute parallel boundaries for radio transmissions in the HF band (3 to 30 MHz).

• EM wave can also travel in unbounded media; light

waves emitted by the sun, radio transmission by antennas.

• We can model wave propagation on a transmission line

either in terms of voltages across the line and currents through its conductors, OR, in terms of the electric and magnetic fields in the dielectric medium between the conductors.

- to the voltage is associated an electric field

- to the current is associated a magnetic field

Transmitter

Earth's Surface

Ionosphere

g

Rg

RL+

-

E E

E E

H H

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Uniform plane wave

Aperture

Observer

• When energy (electromagnetic wave) is emitted by a source, such as an antenna, it expands outwardly from the source in the form of spherical waves.

The wave travels at the same speed in all directions and therefore expands at the same rate; (yet, the antenna may radiate more energy along some preferred directions).

• The wavefront of a spherical wave appears

approximately plane to an observer very far away from the source, as if it were a part of a uniform plane wave.

This approximation of plane-wave is used in this chapter to develop a physical understanding of wave propagation.

Radiatingantenna

Sphericalwavefront

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6.1 Time-Harmonic Fields The Maxwell’s equations in the phasor domain are given by

∇.E = ρv / ε (1-a) ∇×E = - jωµH (1-b) ∇.H = 0 (1-c) ∇×H = J + jωεE (1-d)

where the phasor vector E(x,y,z) corresponds to the

instantaneous field E(x,y,z,t) = Re [E(x,y,z) e jωt]

6.1.1 Complex Permittivity In a medium with conductivity σ, the current density J is related to E by J = σE. Then equation (1-d) can be written as

∇×H = (σ + jωε)E = jω (ε - j σ/ω)E = jωεcE

εc is called the complex permittivity • For a lossless medium, σ = 0. So the current density J is

also null (non-conducting medium) and it follows that εc = ε

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6.1.2 Wave Equations for a Charge-Free Medium

In a charge-free medium ρv=0, and the Maxwell’s equations become

∇.E = 0 (2-a) ∇×E = - jωµH (2-b) ∇.H = 0 (2-c) ∇×H = jωεcE (2-d)

• To describe the propagation of the EM wave we need to

obtain the expressions of E and H as a function of the spatial variables (x, y, z).

Taking the curl of equation (2-b), gives

∇×(∇×E) = - jωµ(∇×H)

Using equation (2-d), we obtain

∇×(∇×E) = ω2µεcE (j2=-1)

But ∇×(∇×E) = ∇(∇.E) - ∇2E, and equation (2-a) states that ∇.E = 0,

Thus we obtain the homogeneous wave equation for E:

∇2E + ω2µεcE = 0

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Introducing the propagation constant γ2 = - ω2µεc we obtain

∇2E - γ2 E = 0 We obtain the same equation with the magnetic field H: ∇2H - γ2 H = 0

6.2 Plane-Wave Propagation in Lossless Media

If the medium is non-conducting (σ=0), it is said to be lossless, and the wave that travels through it is not attenuated. In that case, we have εc = ε. For a lossless medium, we introduce the wavenumber k defined by k = ω(µε)½ The wave equation for E becomes: ∇2E + k2 E = 0

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6.2.1 Uniform Plane Waves In Cartesian coordinates, the electric field phasor is

E = Exx + Eyy + Ezz

Substituting into the wave equation, we have

( ∂2/∂x2+∂2/∂y2+∂2/∂z2 ) (Exx+Eyy+Ezz) + k2(Exx+Eyy+Ezz)=0 It's equivalent to say that each phasor component must satisfy this equation:

( ∂2/∂x2+∂2/∂y2+∂2/∂z2 + k2 ) Ex = 0 To obtain the above equation you have to take the scalar product of the vector equation with x. Similar equations apply to Ey and Ez. • A uniform plane wave is characterized by electric and

magnetic fields that have uniform properties at all points across an infinite plane, meaning that the properties do not depend of the position on the plane.

Assuming that the infinite plane is the x-y plane, then E and H do not vary with x and y. Hence, ∂Ex/∂x = 0 and ∂Ex/∂y = 0. The direction of propagation is then the z axis. The wave equation for Ex is reduced to d2Ex/dz2 + k2Ex =0 The same relation is true for Ey: d2Ey/dz2 + k2Ey =0 www.mywbut.com

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Concerning the z-component of E, we can prove that Ez =0 Using the equation (2-d): z(∂Hy/∂x - ∂Hx/∂y) = z jωεEz

Since ∂Hy/∂x = ∂Hx/∂y = 0, it follows that Ez =0. The same set of equations is obtained for the magnetic field H.

d2Hx/dz2 + k2Hx =0, d2Hy/dz2 + k2Hy =0, and Hz =0 This means that a plane wave has no electric- or magnetic-field components along its direction of propagation. • The general solution of this equation: d2Ex/dz2 + k2Ex =0

is Ex(z) = Ex+(z) + Ex

-(z) = Exo+ e-jkz + Exo

- ejkz where Exo

+ and Exo- are constant to be determined.

The first term containing the exponential e-jkz, represents a wave of amplitude Exo

+, traveling in the +z-direction. The form of the solution is similar to the one obtained for the voltage of a transmission line.

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______EXAMPLE_____________________________ • Assuming for the time being that E has only one

component along x (i.e., Ey =0), and that Ex consists of a wave traveling in the +z-direction only (i.e., Exo

- =0).

Hence, E(z) = Ex+(z) x = Exo

+ e-jkz x • We apply equation (2-b) to find the magnetic field H(z): | x y z | ∇∇×E = | ∂/∂x ∂/∂y ∂/∂z | = -jωµ(Hxx + Hyy + Hzz) | Ex

+(z) 0 0 | For a uniform plane wave traveling in the z-direction, ∂Ex(z)/∂x = ∂Ex(z)/∂y = 0 ∂Ex

+(z)/∂y = ∂Ex+(z)/∂y = 0

Hence, Hx = 0

Hy = (-1/jωµ) ∂Ex+(z)/∂z = (k/ωµ) Exo

+ e-jkz Hz = 0 The intrinsic impedance of a lossless medium is defined by η = ωµ / k = (µ/ε)½ (Ω) Finally, the expression of the electric and magnetic fields

E(z) = Exo+ e-jkz x

H(z) = (Exo+/η) e-jkz y

The fields are perpendicular to each other, and both perpendicular to the direction of wave propagation. ______________________________________ END____

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Characteristics of a transverse electromagnetic wave. Instantaneous fields: Exo

+ = |Exo+| e jφ+ E(z,t) = Re [E(z) e jωt]

= |Exo+| cos(ωt – kz +φ+) x

H(z,t) = (|Exo

+|/η) cos(ωt – kz +φ+) y Phase velocity: up = ω / k = 1/(µε)½ (m/s) Wavelength: λ = 2π / k = up / f (m)

6.2.2 General Relation between E and H For a uniform plane wave traveling in an arbitrary direction denoted by k, it can be shown that

H = (1/η) k×E E = -η k×H

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6.3 Wave Polarization The polarization of a uniform plane wave describes the locus of the tip of the electric field vector, at a given point in space, as a function of time. In the general case, the locus is an ellipse, and the wave is called elliptically polarized. Consider the electric field of a +z-propagating plane wave,

E(z) = Exo e-jkz x + Eyo e-jkz y

The amplitudes Exo and Eyo are in general complex quantities, with a magnitude and a phase angle. • Wave polarization depends on the phase of Eyo relative to

that of Exo, but not on the absolute phases of Exo and Eyo. • Hence, for convenience, we choose the phase of Exo as

our reference, and δ the phase difference between Exo and Eyo.

We define, Exo = |Exo| e j0 = ax Eyo = |Eyo| e jδ = ay e jδ ax (= |Exo|) and ay (= |Eyo|) are positive by definition.

Locus Tip of the vector Fixed point P x E

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The total electric field phasor is then given by

E(z) = (ax x + ay e jδ y) e-jkz and the corresponding instantaneous field is

E(z,t) = Re [ E(z) e jωt ] = ax cos(ωt – kz) x + ay cos(ωt – kz + δ) y

The instantaneous field is necessary to find the wave polarization. • Intensity of the electric field:

If E(z,t) = Ex(z,t) x + Ey(z,t) y

Then |E(z,t)| = |Ex2(z,t) + Ey

2(z,t)|½ • Direction of the electric field:

Defined by the inclination angle Ψ, in the x-y plane, between E(z,t) and the reference axis, which is the x-axis in this example:

Ψ(z,t) = tan-1 (Ey(z,t) / Ex(z,t))

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ay

Ey

Exax-ax

-ay

t =

t = 0

y

xz

E

6.3.1 Linear Polarization • A wave is said to be linearly polarized if Ex(z,t) and

Ey(z,t) are in phase (i.e. δ=0) or out of phase ( δ=π) To trace the tip of E(z,t) we fixed z and let the time vary. We usually choose the plane at z=0. Then we have,

E(0,t) = ( ax x + ay y ) cosωt (in-phase) E(0,t) = ( ax x – ay y ) cosωt (out-of-phase)

Let us examine the case when δ=π :

The modulus of E(0,t) varies as cosωt :

|E(0,t)| = |ax

2 + ay2|½ |cosωt|

and the inclination angle is constant:

Ψ = tan-1 (ay / ax)

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6.3.2 Circular Polarization In this special case, the magnitudes of the x- and y-components of E(z,t) are equal, and the phase difference δ = ±π/2 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 ax = ay = a and δ=π/2,

E(z) = (a x + a e jπ/2 y) e-jkz = a (x + j y) e-jkz

and E(z,t) = Re [ E(z) e-jωt ]

= a cos(ωt–kz)x + a cos(ωt–kz+π/2)y = a cos(ωt–kz)x + a sin(ωt–kz)y

The corresponding modulus and inclination angle are given by

|E(z,t)| = |Ex

2(z,t) + Ey2(z,t)|½ = a (constant)

Ψ(z,t) = tan-1 (Ey(z,t) / Ex(z,t)) = - (ωt – kz) (varying)

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In this case, the modulus of E is constant, whereas Ψ depends of z and t. At a fixed position on the z-axis, say z = 0, we have Ψ = -ωt The negative sign means that Ψ decreases when time increase. This wave is left-hand circularly polarized because when the thumb of the left hand points along z, the other four fingers point in the direction of rotation of E.

(a) LHC polarization

a

y z

z xa

E

(b) RHC polarization

z

a

y

z xa

E

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6.4 Electromagnetic Power Density For any wave with an electric field E and a magnetic field H, the Poynting vector S is defined as S = E × H The unit of S is (V/m) × (A/m) = (W/m2) The direction of S is along the direction of propagation, k. S represents the power per unit area carried by the wave. The total power that flows through an aperture (or opening) is

P = S.n dA (W) Aperture

where n is the surface unit vector.

uniform plane wave

Since E and H are both functions of time, so is S. That's why we defined a constant quantity called the average power density of the wave Sav. Sav = (1/2) Re [ E × H* ] (W/m2)

S

A

n

k

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In a lossless medium, for a uniform plane wave, traveling in the +z-direction,

E(z) = Ex(z) x + Ey(z) y

= Exo e-jkz x + Eyo e-jkz y H(z) = (1/η) z x E = So H*(z) = And E × H* =

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