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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
0( ) jkzxE z E e−=
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 :
Plane Wave Field (cont.)
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
∂ ∂ ∂∇× =
∂ ∂ ∂
0( ) jkzxE z E e−=
13
Plane Wave Field (cont.)
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:
Lossy Medium (cont.)
( ) 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=
Lossy Medium (cont.)
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)
Lossy Medium (cont.)
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
Lossy Medium (cont.)
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′ ′′= −
Polarization Loss (cont.)
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.
Polarization Loss (cont.)
( )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′ ′=
Polarization Loss (cont.)
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
0( ) jkzxE z E e−=
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′′
= +′ ′
0( ) jkzxE z E e−=
x
y
EH
η=
c
µηε
=
2kπλ =′
k k jk′ ′′= −
1 /pd k′′≡
ck ω µε=
Lossy
0 rε ε ε′ =