Plane Waves

1 Review dielectrics

2 Plane waves in the time domain

3 Plane waves in the frequency domain

4 Plane waves in lossy and dispersive media

5 Phase and group velocity

6 Wave polarization

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I. Review dielectrics

Simple medium: D = εE

Dispersive medium: D = ε(ω)E

Anisotropic medium: Permittivity as a tensor εxx εxy εxzεyx εyy εyzεzx εzy εzz

(1)

Conducting medium:

εe = εR − j (εI + σ/ω) (2)

We will be assuming that B = µH throughout the course

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II. Plane waves in the time domain

Assuming that we are in a linear, isotropic, homogeneous, lossless, andtime-invariant medium, Maxwell’s source-free equations become

∇× H = ε∂E

∂t(3)

∇× E = −µ∂H

∂t(4)

∇ · E = 0 (5)

∇ · H = 0 (6)

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Taking the curl of both sides of the first equation and using a vectoridentity results in

∇2H = εµ∂2

∂t2H (7)

We can also obtain

∇2E = εµ∂2

∂t2E (8)

Both are vector wave equations

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These involve a lot of terms, so let’s simplify by assuming fields vary onlyalong x

∂2Ex

∂x2= µε

∂2Ex

∂t2(9)

∂2Ey

∂x2= µε

∂2Ey

∂t2(10)

∂2Ez

∂x2= µε

∂2Ez

∂t2(11)

Requiring ∇ · E = 0 shows that Ex could be at most a constant, set it tozero since it is not interesting. Remaining equations are of the form

∂2f

∂x2=

1

v2

∂2f

∂t2(12)

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It is easy to show that these equations admit solutions of the form

f = f1(x − vt) + f2(x + vt) (13)

where f1 and f2 are arbitrary twice-differentiable functions (D’Alambertsolutions). Accordingly, we find

E = y

[f1(x − t

√µε

) + f2(x +t√µε

)

]+ z

[f3(x − t

√µε

) + f4(x +t√µε

)

]and

H =

√ε

µ[−y (f3 − f4) + z (f1 − f2)] (14)

follows from Maxwell’s equations

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These solutions represent traveling waves because a disturbance will movethrough space undistorted as time progresses. Functions of x − vtpropagate in the +x direction, while functions of x + vt propagate in the−x direction. Furthermore,

v =1√µε

(15)

turns out to be the velocity of these waves. Given the direction ofpropagation of a pure traveling wave, k,

H =(k × E

)/η (16)

E = −η(k × H

)(17)

where the wave impedance η is given by

η =k × E

H=

õ

ε(18)

This is a TEM (transverse electromagnetic) field!Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 31, 2018 7 / 25

0 1 2 3 40

0.5

1

x

f 1(x

,t=

0)

(a)

0 1 2 3 40

0.5

1

u

f 1(u

)

(b)

0 1 2 3 40

0.5

1

x

f 1(x

,t=

2 (

µ ε

)1/2

) (c)

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H

k^

E

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Electromagnetic plane waves are ideal for information transmission inthe simplest medium because time-domain signals propagateundistorted from one location to another

D’Alembert’s solution, however, can be shown to be invalid if themedium is not simple - signals distort when propagated

The “bandwidth” of a medium (i.e. the frequency spread of a signalwhich can be reliably transmitted) is thus related to the complexity ofthe propagation medium

More simple media have larger bandwidths - free space has infinitebandwidth!

Trying to remove medium effects (which are often unpredictable)usually is very difficult

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III. Plane waves in the frequency domain

For sinusoidal time-dependence, we need only specify the amplitude andphase of the fields as functions of space, this leads to phasor description

Ey (x , t) = Re[E y (x)e jωt

](19)

and we can work in terms of E y (x) (the phasor) only. A monochromaticplane wave traveling in the +x direction can be written as

E = (yA + zB) e−jkx (20)

H =1

η(−yB + zA) e−jkx (21)

with k = ω√µε = ω

v = 2πλ . Sinusoidal in time and space!

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A plane wave traveling in an arbirtrary direction can be derived similarly as

E = E 0e−j[kxx+kyy+kzz] = E 0e

−jk·r (22)

H =1

ωµk × E =

1

ηk × E (23)

where k = xkx + y ky + zkz = kk = kω√µε and

E 0 · k = 0 (24)

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Plane wave examples Consider the following phasor plane waves in free

space

E = xe−j2πy

H = ze−jπz

E = xe j8π(y+z)

E = (2x + y − 3z) e−j2π(x+y+z)/√

3

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IV. Plane waves in lossy and dispersive media

Since lossy media are dispersive in general, signals clearly cannotpropagate undistorted in a lossy medium since different frequencycomponents propagate at different velocities! It is easiest to work in thefrequency domain since things are simple for individual frequencies.

∇2H = −ω2µεH (25)

is now the vector wave equation in for phasors, and assuming variationsonly along x for simplicity gives

∂2Hy

∂x2= −ω2µεHy (26)

∂2Hz

∂x2= −ω2µεHz (27)

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These linear differential equations have exponentials as their solution

H = (yM + zN) e jkx + (yP + zR) e−jkx (28)

The corresponding electric field follows

E =k

ωε

[(−yN + zM) e jkx + (yR − zP) e−jkx

](29)

where

k = ω√µε (30)

is now complex! Careful in choice of sign for square root! Complex kindicates we have both phase variation (real part of k) and amplitudedecay (imaginary part of k)

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A plane wave propagating in any direction can be written as

E = E 0e−jk(x cosα+y cos ζ+z cos δ) = E 0e

−jk·r (31)

H = k × E/ωµ = k × E/η (32)

where k = kk = kω√µε if all components of k are in phase with one

another. Actually all that is required to solve the differential equations is

k · k = ω2µε (33)

which allows for more general behaviors, as we will see in Chapter 6.

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V. Phase and group velocity

Although plane waves in lossy media attenuate in space, we can stillidentify a phase associated with each point and thus a wavelength

λ = 2π/kR (34)

The phase velocity, vp of a wave determines how rapidly a point ofconstant phase moves, and is given by

vp =dx

dt|ωt−kRx=const =

ω

kR(35)

However, in a dispersive medium, different signal frequencies will travelwith different velocities, so how can we define an “effective signalvelocity”?

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0 1 2 3 40

0.2

0.4

0.6

0.8

1

x

Ey(x

,t=

0)

(a)

0 1 2 3 40

0.2

0.4

0.6

0.8

1

x

Ey(x

,t=

t 1)

(b)

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A convenient quantity is the group velocity, given by

vg =dx

dt|t− dkR

dω(ωc )x=const

=1

dkRdω (ωc)

(36)

which is an effective signal velocity derived through a linear assumption fork versus ω, as can be shown through the modulated sine wave example ofthe notes. However, when signal distortion is large, the concept of groupvelocity becomes unclear - from which points of the signal should velocitybe measured? - so caution should be used in describing signal velocities indispersive media

vg = vp (37)

in a non-dispersive medium!

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VI. Wave Polarization

Our example phasor field was

E (x) = (yA + zB) e−jkx (38)

What does this look like in the time domain?

E (x , t) = Re[E (x)e jωt

]= Re

{[yA + zB]e jωt−jkx

}(39)

Letting

A = Ae ja

B = Be jb (40)

we get

E (x , t) = yAe−kI x cos(ωt − kRx + a) + zBe−kI x cos(ωt − kRx + b)

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Setting

r = Ae−kI x (41)

s = Be−kI x (42)

and

u = ωt − kRx (43)

this simplifies to

Ey (x , t) = r cos(u + a) (44)

Ez(x , t) = s cos(u + b) (45)

which are the equations of an ellipse! Linear and circular polarizations arespecial cases.

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z

y

Linear polarization

t= /2t=3 /2

t=0

z

y

Elliptical polarization

x

x

x

(a) Elliptical Polarization (b) Linear Polarization

z

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Polarization examples

Consider the following phasor plane waves in free space

E = xe−j2πy (linear)

H = (x + j y)e−jπz (circular)

E = (x + 2(y − z)/√

2)e j8π(y+z) (linear)

E = (2x + j y − (2 + j)z) e−j2π(x+y+z)/√

3 (eliptical)

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Poynting Vector

We can also derive relations from Maxwell’s equations that decribethe transfer of energy due to electromagnetic fields.

The relevant quantity is the Poynting Vector which for phasor fields iscomputed as

S =1

2Re[E × H

∗]

The amplitude of the Poynting vector has units of Watts per squaremeter, and is the power per unit area “carried” by the electromageticfield.

The direction of the Poynting vector indicates the direction of powerflow.

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Poynting Vector for a plane wave

Consider a phasor plane wave with k = kR − jk I = kk :

E = E 0e−jk·r

The Poynting vector then has the form

S = kRe{

1

2η∗

} ∣∣E 0

∣∣2 e−2k I ·r

if all components of k have the same phase.

In a lossless medium this reduces to

S = k1

2η

∣∣E 0

∣∣2In both cases the direction of power flow is the direction ofpropagation of the plane wave.

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