Notes 15 Plane Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE3317/SectionJackson/Class...

Notes 15 Plane Waves

1

ECE 3317Applied Electromagnetic Waves

Prof. David R. JacksonFall 2020

ocean

Ex

z

Introduction to Plane Waves

A plane wave is the simplest solution to Maxwell’s equations for a wave that travels through free space.

The wave does not requires any conductors – it exists in free space.

A plane wave is a good model for radiation from an antenna, if we are far enough away from the antenna.

2

*12

S E H= ×

E

Hz

S

x

The Electromagnetic Spectrum

3

http://en.wikipedia.org/wiki/Electromagnetic_spectrum

0 /c fλ =

Source Frequency Wavelength

U.S. AC Power 60 Hz 5000 km

ELF Submarine Communications 500 Hz 600 km

AM radio (KTRH) 740 kHz 405 m

TV ch. 2 (VHF) 60 MHz 5 m

FM radio (Sunny 99.1) 99.1 MHz 3 m

TV PBS ch. 8 (VHF ch 8) 183 MHz 1.6 m

TV KPRC ch. 2 (UHF ch. 35) 599 MHz 50 cm

Cell Phone (4G) 1.9 GHz 16 cm

μ-wave oven 2.45 GHz 12 cm

Police radar (X-band) 10.5 GHz 2.85 cm

Cell Phone (5G mm wave) 28 GHz 1.1 cm

THz 1000 GHz 0.3 mm

Light 5×1014 [Hz] 0.60 µm

X-ray 1018 [Hz] 3 Å

The Electromagnetic Spectrum (cont.)

4

0 /c fλ =

TV and Radio Spectrum

VHF TV: 55-216 MHz (channels 2-13)

Band I : 55-88 MHz (channels 2-6)Band III: 175-216 MHz (channels 7-13)

FM Radio: (Band II) 88-108 MHz

UHF TV: 470-806 MHz (channels 14-69)

AM Radio: 520-1610 kHz

Digital TV broadcast takes place primarily in UHF and VHF Band III.

5

Note: Monopole antenna are ideally about one-quarter of a wavelength in length.

Dipole antennas are ideally about one-half of a wavelength in length.

Vector Wave Equation

Start with Maxwell’s equations in the phasor domain:

E j HH J j E

ωµωε

∇× = −∇× = +

Faraday’s law

Ampere’s law

Assume free space:

0J Eσ= =

0

0

E j HH j E

ωµωε

∇× = −∇× =

0 0,ε ε µ µ= =

We then have:

Ohm’s law:

6

Vector Wave Equation (cont.)

Take the curl of the first equation and then substitute the second equation into the first one:

( ) ( )( )

0

0 0

E j H

j j E

ωµ

ωµ ωε

∇× ∇× = − ∇×

= −

0 0 0k ω µ ε≡

( ) 20 0E k E∇× ∇× − = Vector wave equation

Define:

Then

0

0

E j HH j E

ωµωε

∇× = −∇× =

Wavenumber of free space [rad/m]

7

Vector Helmholtz Equation

( ) 20 0E k E∇× ∇× − =

Recall the vector Laplacian identity:

( ) ( )2V V V∇ ≡∇ ∇⋅ −∇× ∇×

Hence we have

( ) 2 20 0E E k E∇ ∇⋅ −∇ − =

Also, from the divergence of the vector wave equation, we have:

( ) 20 0E k E∇⋅ ∇× ∇× − ∇⋅ =

0E∇⋅ =8

( ) 20 0E k E∇× ∇× − =

Hence we have:

Vector Helmholtz equation2 20 0E k E∇ + =

Recall the property of the vector Laplacian in rectangular coordinates:

2 2 2 2ˆ ˆ ˆx y zV x V y V z V∇ = ∇ + ∇ + ∇

Taking the x component of the vector Helmholtz equation, we have

2 20 0x xE k E∇ + = Scalar Helmholtz equation

9

Reminder:This identity only holds in rectangular coordinates.

Vector Helmholtz Equation (cont.)

Plane Wave Field

Assume

Then

or

ˆ ( )xE x E z=

2 20

2 2 2202 2 2

2202

0

0

0

x x

x x xx

xx

E k EE E E k Ex y z

d E k Edz

∇ + =

∂ ∂ ∂+ + + =

∂ ∂ ∂

+ =

Solution: 00( ) jk z

xE z E e−= 0 0 0( )k ω µ ε=

The electric field is polarized in the x direction, and the wave is propagating (traveling) in the z direction.

00( ) jk z

xE z E e+=

For wave traveling in the negative z direction:

10

Ex

y

z

where

0( ) jkzxE z E e−=

k ω µε=

For a plane wave traveling in a lossless dielectric medium (does not have to be free space):

0

0

r

r

ε ε εµ µ µ==

(wavenumber of dielectric medium)

Plane Wave Field (cont.)

11


The electric field of a plane wave in a lossless medium propagates in zexactly as does the voltage on a lossless transmission line:

kβ →TL ⇒ PW :


12

All of the formulas that we had for a lossless transmission line now hold for a plane wave, using this notation change.

TLpv ω

β= PW

pvkω

=Example:

The H field is found from:

so

E j Hωµ∇× = −

( )( )

( )

1 ˆ

1 ˆ

1 ˆ ( )

x

x

x

H x E zj

d Eyj dz

y jk Ej

ωµ

ωµ

ωµ

= − ∇×

= −

= − −

ˆ ˆ ˆ

x y z

x y z

Ex y z

E E E

∂ ∂ ∂∇× =

∂ ∂ ∂


13


Intrinsic Impedance

We then have

Hence

ˆ xkH y Eωµ

=

x

y

EH k

ωµ ωµ µ ηεω µε

= = = =

where

µηε

≡ Intrinsic impedance of the medium

y xkH Eωµ

=

00

0

376.730313 [ ]µη ηε

= = Ω

Free-space:

14

70

120 2

08

4 10 [H/m]1 8.854 10 [F/m]

2.99792458 10 [m/s] ( )c

c

µ π

εµ

−

−

×

= ×

≡ ×

exact

so

0 0

1cµ ε

=

Poynting Vector

The complex Poynting vector is given by

0

0

ˆ1ˆ

jkz

jkz

E x E e

H y E eη

−

−

=

=

( )*12

S E H= ×

Hence we have:

( )

( ) ( )

*

00

*0 0

20

1 ˆ21 ˆ

21 ˆ

2

jkz jkz

jkz jkz

ES z E e e

z E e E e

z E

η

η

η

− −

− +

=

=

=

220ˆ [VA/m ]

2E

S zη

=

( ) ( )2

0 2ˆRe [W/m ]2E

t S zη

= =S

15

(no VARS)

For a wave traveling in a lossless dielectric medium:

Phase Velocity

pvkω

=

1p dv c

µε= =

From our previous discussion on phase velocity for transmission lines, we know that

Hence we have

(speed of light in the dielectric material)

pv ωω µε

=

so

Notes:

All plane waves travel at the same speed in a lossless medium, regardless of the frequency. This implies that there is no dispersion, which in turn implies that there is no distortion of the signal. The phase velocity of a plane wave in a lossless medium is the same as that of a wave on a lossless

transmission line that is filled with the same material.16

Wavelength

2kπλ =

2 2 2 12

dck ff fπ π πλ

ω µε π µε µε= = = = =

dcf

λ =

From our previous discussion on wavelength for transmission lines, we know that

Hence

Also, we can write

00

2kπλ =For free space:

Free space: 0cf

λ =

17

0

r r

λλµ ε

=

2πλβ

=

Hence, for a plane wave (letting β → k):

0

dcc

λλ

=

Note:

Summary (Lossless Case)

0

0

20

1

2

jkzx

jkzy

z

E E e

H E e

ES

η

η

−

−

=

=

=

1p d

r r

cv ckω

µε µ ε= = = = 02 d

r r

ck f

λπλε µ

= = =

18

k ω µε=

0r

r

µµη ηε ε

= = 00

0

376.730313 [ ]µηε

= Ω

EH

x

y

z

S

Lossy Medium

E j HH J j E

ωµωε

∇× = −∇× = +

J Eσ=

Return to Maxwell’s equations:

Assume Ohm’s law:

( )H E j E

j Eσ ωεσ ωε

∇× = +

= +

Ampere’s law

We define an effective (complex) permittivity εc that accounts for conductivity:

cj jωε σ ωε= + c j σε εω ≡ −

19

Ocean

Ex

z

Set

c

E j HH j E

ωµωε

∇× = −∇× =

Maxwell’s equations then become:

The form is exactly the same as we have for the lossless case, with

cε ε→

Hence we have:

0

01

jkzx

jkzy

E E e

H E eη

−

−

=

=

ck ω µε=

c

µηε

=

(complex)

(complex)

20

Lossy Medium (cont.)

E j HH j E

ωµωε

∇× = −∇× =

Lossy Lossless

Examine the wavenumber:

k k jk′ ′′= −

0 0jkz jk z k z

xE E e E e e′ ′′− − −= =

cε k

ck ω µε= c j σε εω = −

00

kk′ ≥′′ ≥

Denote:

kk

βα

′↔′′↔

Compare with lossy TL:

21

/2j jz z e z eθ θ

π θ π

= =

− < ≤

Principal branch:


( ) 0jk z k z

xE z E e e′ ′′− −=

λ

z

0k zE e ′′−

2kπλ =′

22

( ),0x zE

( ) ( )0 0, cos k zx z t E t k z eω φ ′′−′= − +E

0t =

00 0

jE E e φ=


1/pd k′′≡

The “depth of penetration” dp is defined.

k ze ′′−

z

( )xE z 1

1 0.37e−

pd

23

( ) 0jk z k z

xE z E e e′ ′′− −=

(choose E0 = 1)


1pk d′′ =

( )2 2

* * 2 20 0*

1 1ˆ ˆ ˆ2 2 2 2

k z j k zx y

E ES E H z E H z e z e eφ

η η′′ ′′− −= × = = =

0

01

jk z k zx

jk z k zy

E E e e

H E e eη

′ ′′− −

′ ′′− −

=

=

The complex Poynting vector is

j

c

e φµη ηε

= =

( )2

20Re cos2

k zz z

ES t S eφ

η′′−= = 2k ze ′′−

z

( )zS z1

1 /pd k′′=

2 0.14e− =

24


Note: The angle between the Ex and Hy phasors is φ.

Example

Ocean water:

0

814 [S/m]

rεσµ µ

===

Assume f = 2.0 GHz:

00

c rj jσ σε ε ε εω ωε

= − = −

( )( )0 81 35.95 [F/m]c jε ε= −

0 0 0 0c rc rck kω µ ε ω µ ε ε ε= = = ( )386.022 81.816 [1/m]k j= −

1/pd k′′= 0.013 [m]pd =

2 / kλ π ′= 0.016 [m]λ =

(These values are fairly constant up through microwave frequencies.)

25

( )81 35.95rc jε = −

386.022 [rad/m]k′ =

81.816 [nepers/m]k′′ =

Example (cont.)

f dp [m]

1 [Hz] 251.6

10 [Hz] 79.6

100 [Hz] 25.2

1 [kHz] 7.96

10 [kHz] 2.52

100 [kHz] 0.796

1 [MHz] 0.262

10 [MHz] 0.080

100 [MHz] 0.0262

1.0 [GHz] 0.013

10.0 [GHz] 0.012

100 [GHz] 0.012

The depth of penetration into ocean water is shown for various frequencies.

26

1/pd k′′=

Note: The relative permittivity

of water starts changing at very high frequencies (above

about 2 GHz), but this is ignored here.

0

814 [S/m]

rεσµ µ

===

Loss Tangent

c j σε εω = −

Denote: c c cjε ε ε′ ′′= −

The loss tangent is defined as:

tan c

c

εδε

′′≡

′

We then have: tan σδωε

=

27

Note:The loss tangent

characterizes how lossy the material is.

Recall:

( )

( )( )

0 0

0

0

0

1

1 tan

1 tan / 2

ck j

j

j

j

σω µ ε ω µ εω

σω µ εωε

ω µ ε δ

ω µ ε δ

= = −

= −

= −

≈ −

Low-Loss Limit: tanδ << 1

1 1 / 2, 1( )z z z+ ≈ + <<Using

We approximate the wavenumber for small loss tangent:

012

k σω µ εωε

′′ ≈

0

2k µ σ

ε ′′ ≈ 0

2pd ε

µ σ ≈

(independent of frequency)

or

28

f dp [m] tanδ

1 [Hz] 251.6 8.88×108

10 [Hz] 79.6 8.88×107

100 [Hz] 25.2 8.88×106

1 [kHz] 7.96 8.88×105

10 [kHz] 2.52 8.88×104

100 [kHz] 0.796 8.88×103

1 [MHz] 0.262 888

10 [MHz] 0.080 88.8

100 [MHz] 0.0262 8.88

1.0 [GHz] 0.013 0.888

10.0 [GHz] 0.012 0.0888

100 [GHz] 0.012 0.00888

Ocean water

29

“Low-loss” region

0

tanr

σδωε ε

=

Low-Loss Limit: tanδ << 1 (cont.)

0

814 [S/m]

rεσµ µ

===

Dielectric Polarization Loss

The permittivity ε can also be complex, due to molecular and atomic polarization loss (friction at the molecular and atomic levels).

c j σε εω = −

Example: distilled water: σ ≈ 0 (but it heats up well in a microwave oven!).

30

jε ε ε′ ′′= − (complex permittivity)

Notation:

Polarization Loss (cont.)

31

Complex relative permittivity for pure (distilled) water

0σ ≈

rε ′

rε ′′

Frequency [GHz]

0r rjε ε ε

ε′ ′′= −


c j σε εω ≡ −

( )c j j σε ε εω ′ ′′= − −

c j σε ε εω

′ ′′= − +

tan c

c

ε ε σδε ε ωε′′ ′′

≡ = +′ ′ ′

32

,c cσε ε ε εω

′ ′ ′′ ′′= = +

Effective permittivity with polarization loss accounted for:

The loss tangent now accounts for conductivity

and polarization loss.


( )0 1 tanc r jε ε ε δ= −

33

( )( )( )

0

0

1

1 tan

1 tan

1 tan

c c c

cc

c

c

rc

r

j

j

j

j

j

ε ε ε

εεε

ε δ

ε ε δ

ε ε δ

′ ′′= −

′′′= − ′ ′= −

′= −

′= −

Effective permittivity is now related to loss tangent:

r rε ε′Note : is usually just called in most books.

Example:

Teflon

2.1rε =

tan 0.001δ =

( ) ( )0 2.1 1 0.001c jε ε= −( )cε ε′ ′=


Practical notes:

34

For some materials (most good conductors), it is the conductivity that is approximately constant with frequency.

Ocean water: σ ≈ 4 [S/m]

For other materials (most good insulators), it is the loss tangent that is approximately constant with frequency.

Teflon: tanδ ≈ 0.001

For materials with significant conductivity (like earth or ocean) the conductivity effect usually dominates the loss tangent (and hence ε can be assumed to be approximately real). For good insulators (like Teflon), the conductivity is almost zero and the polarization loss usually dominates the loss tangent.

Appendix: Summary of Formulas

35


x

y

EH

η=

0r

r

µ µη ηε ε

= =

00

0

376.730313 [ ]µηε

= Ω

dr r

ccµ ε

=

02 d

r r

ck f

λπλε µ

= = =

k ω µε=

0cf

λ =

Lossless

p dv ckω

= =

Appendix: Summary of Formulas

36

tan c

c

εδε

′′≡

′

( )0

tan 0r

σδ εωε ε

′′= =

( )0 1 tanc r jε ε ε δ= −c j σε εω = −

c c cjε ε ε′ ′′= −

tan ε σδε ωε′′

= +′ ′


x

y

EH

η=

c

µηε

=

2kπλ =′

k k jk′ ′′= −

1 /pd k′′≡

ck ω µε=

Lossy

0 rε ε ε′ =

Notes 15 Plane Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE3317/SectionJackson/Class...

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