Download Electromagnetic Intro

Electromagnetism
INEL 4152
Sandra Cruz-Pol, Ph. D.
ECE UPRM
Mayagüez, PR
Electricity => Magnetism

In 1820 Oersted discovered that a steady
current produces a magnetic field while
teaching a physics class.
Cruz-Pol, Electromagnetics
UPRM
Would magnetism would produce
electricity?

Eleven years later,
and at the same time,
Mike Faraday in
London and Joe
Henry in New York
discovered that a
time-varying magnetic
field would produce
an electric current!
Vemf
d
 N
dt

L E  dl   t s B  dS
Cruz-Pol, Electromagnetics
UPRM
Electromagnetics was born!

This is the principle of
motors, hydro-electric
generators and
transformers operation.
This is what Oersted discovered
accidentally:
D 

L H  dl  s  J  t   dS
*Mention some examples of em waves
Cruz-Pol, Electromagnetics
UPRM
Cruz-Pol, Electromagnetics
UPRM
Electromagnetic Spectrum
Cruz-Pol, Electromagnetics
UPRM
Some terms
E
= electric field intensity [V/m]
 D = electric field density
 H = magnetic field intensity, [A/m]
 B = magnetic field density, [Teslas]
Cruz-Pol, Electromagnetics
UPRM
Maxwell Equations
in General Form
Differential form Integral Form
  D  v
 D  dS   v dv
s
 B  0
v
 B  dS  0
s
B
 E  
t
D
 H  J 
t
Gauss’s Law for E
field.
Gauss’s Law for H
field. Nonexistence
of monopole

Faraday’s Law
L E  dl   t s B  dS
D 

H

dl

J


L
s  t   dS
Cruz-Pol, Electromagnetics
UPRM
Ampere’s Circuit
Law
Maxwell’s Eqs.
 v
J  
t
 Also
the equation of continuity
D
 Maxwell added the term  t to Ampere’s
Law so that it not only works for static
conditions but also for time-varying
situations.
 This added term is called the displacement
current density, while J is the conduction
current.
Cruz-Pol, Electromagnetics
UPRM
Maxwell put them together

And added Jd, the
displacement current
 H  dl   J  dS  I
L
enc
I
S1
 H  dl   J  dS  0
L
S1
I
L
S2
d
dQ
L H  dl  S J d  dS  dt S D  dS  dt  I
2
2
S2
At low frequencies J>>Jd, but at radio frequencies both
Cruz-Pol, Electromagnetics
terms are comparable in magnitude.
UPRM
Moving loop in static field

When a conducting loop is moving inside a
magnet (static B field, in Teslas), the force on a
charge is

 
F  Qu  B
 

F  Il B
Cruz-Pol, Electromagnetics
UPRM
Encarta®
Who was NikolaTesla?
 Find
out what inventions he made
 His relation to Thomas Edison
 Why is he not well know?
Cruz-Pol, Electromagnetics
UPRM
Special case

Consider the case of a lossless medium

with no charges, i.e. .  v  0
 0
The wave equation can be derived from Maxwell
equations as
 E    c E  0
2
2
What is the solution for this differential equation?
 The equation of a wave!
Cruz-Pol, Electromagnetics
UPRM
Phasors & complex #’s
Working with harmonic fields is easier, but
requires knowledge of phasor, let’s review
 complex numbers and
 phasors
Cruz-Pol, Electromagnetics
UPRM
COMPLEX NUMBERS:
 Given
a complex number z
z  x  jy  re
j
 r  r cos   jr sin 
where r | z |
x  y is the magnitude
2
  tan
1
2
y
is the angle
x
Cruz-Pol, Electromagnetics
UPRM
Review:
1/ j =
 Addition,
 Subtraction,
j=
1/
 Multiplication,
 Division,
 Square
example :
Root,
 Complex Conjugate
e
j 45o
/ j=
e
j 45o
- j=
3e
j 90 o
2e
Cruz-Pol, Electromagnetics
UPRM
2e
+j=
+ j 45o
+ j 45o
=
+10e
+ j 90 o
=
For a time varying phase
 Real
and imaginary parts are:
  t  
j
Re{re }  r cos(t   )
j
Im{re }  r sin( t   )
Cruz-Pol, Electromagnetics
UPRM
PHASORS
a sinusoidal current I (t )  I o cos(t   )
j jt
I
e
equals the real part of o e
 For
j
I
e
 The complex term o
which results from
jt
dropping the time factor e is called the
phasor current, denoted by I s (s comes
from sinusoidal)
Cruz-Pol, Electromagnetics
UPRM
To change back to time domain
 The
phasor is multiplied by the time factor,
ejt, and taken the real part.
A  Re{ As e
jt
}
Cruz-Pol, Electromagnetics
UPRM
Advantages of phasors
 Time
derivative is equivalent to
multiplying its phasor by j
A
 jAs
t
 Time
integral is equivalent to dividing by
the same term.
As
 At  j
Cruz-Pol, Electromagnetics
UPRM
Time-Harmonic fields
(sines and cosines)
 The
wave equation can be derived from
Maxwell equations, indicating that the
changes in the fields behave as a wave,
called an electromagnetic field.
 Since any periodic wave can be
represented as a sum of sines and
cosines (using Fourier), then we can deal
only with harmonic fields to simplify the
equations.
Cruz-Pol, Electromagnetics
UPRM
Maxwell Equations
for Harmonic fields
Differential form*

 DE
 v  v
Gauss’s Law for E field.
 BH
0

Gauss’s Law for H field.
No monopole
0
  E   jH
 E  
B
t
  H  J  jE
D
 H  J 
t
* (substituting
Faraday’s Law
Ampere’s Circuit Law
D   E andCruz-Pol,
H  Electromagnetics
B)
UPRM
A wave

Start taking the curl of Faraday’s law
    Es   j  H s

Then apply the vectorial identity
    A  (  A)   2 A

And you’re left with
(  Es )   Es   j (  j ) Es
2
  Es
2
Cruz-Pol, Electromagnetics
UPRM
A Wave
 E  E  0
2
2
Let’s look at a special case for simplicity
without loosing generality:
•The electric field has only an x-component
•The field travels in z direction
Then we have
E ( z, t )
whose general solution is
E(z)  Eo e z  Eo' e z
Cruz-Pol, Electromagnetics
UPRM
To change back to time domain

From phasor
E xs ( z )  Eo e

 z
 Eo e
 z (  j )
…to time domain
E ( z , t )  Eo e
 z

cos(t  z ) x
Cruz-Pol, Electromagnetics
UPRM
Ejemplo 9.23
 In
free space,
 50

8
E  cos(10 t  kz)  V / m

 Find
k, Jd and H using phasors and
maxwells eqs.
Cruz-Pol, Electromagnetics
UPRM
Several Cases of Media
1.
2.
3.
4.
Free space (  0,    o ,   o )
Lossless dielectric (  0,    r  o ,    r o or    )
Lossy dielectric (  0,    r  o ,    r o )
Good Conductor (  ,    o ,    r o or    )
Recall: Permittivity
o=8.854 x 10-12[ F/m]
Permeability
o= 4p x 10-7 [H/m]
Cruz-Pol, Electromagnetics
UPRM
1. Free space
There are no losses, e.g.

E ( z, t )  A sin( t  z ) x
Let’s define
 The phase of the wave
 The angular frequency
 Phase constant
 The phase velocity of the wave
 The period and wavelength
 How does it moves?
Cruz-Pol, Electromagnetics
UPRM
3. General Case
(Lossy Dielectrics)

In general, we had

E ( z , t )  Eo e z cos(t  z ) x
 2  j (  j )
    j
 Re  2   2   2   2 
2 

 2   2    2   2 2
From this we obtain
2



 

 1 
 
  1
2 

  



2



 

 1 
 
  1
2 
 




So , for a known material and frequency, we can find j
Cruz-Pol, Electromagnetics
UPRM
Intrinsic Impedance, h

If we divide E by H, we get units of ohms and
the definition of the intrinsic impedance of a
medium at a given frequency.
given E = Eoe-g z x
Find
H
|E|
j
h

|H |
  j
E(z, t) = Eoe-a z cos(w t - b z)x
Eo - a z
H (z, t) = e cos(w t - b z - qh ) ŷ
h
Cruz-Pol, Electromagnetics
UPRM
 h h
[]
*Not in-phase
for a lossy
medium
Note…
E(z, t) = Eoe-a z cos(w t - b z)x
Eo - a z
H (z, t) = e cos(w t - b z - qh ) ŷ
h
E(z) = Eoe-a z e- j ( b z ) x
Eo -a z - j ( b z-qh )
H (z) =
e e
ŷ

h
E and H are perpendicular to one another
 Travel is perpendicular to the direction of
propagation
 The amplitude is related to the impedance
 And so is the phase
Cruz-Pol, Electromagnetics
UPRM
Loss Tangent

If we divide the conduction current by the
displacement current
J cs
J ds
 Es



 tan   loss tangent
j Es 
http://fipsgold.physik.uni-kl.de/software/java/polarisation
Cruz-Pol, Electromagnetics
UPRM
Relation between tan and c
 

  H  E  j E  j 1  j
E

 

 j c E
The complex permittivi ty is
 

 c   1  j    ' j ' '
 

" 
The loss tangent can be defined also as tan   
 ' 
Cruz-Pol, Electromagnetics
UPRM
2. Lossless dielectric
(  0,    r  o ,   r o or    )
 Substituting
in the general equations:
  0,    

1
2p
u 




 o
h
0

Cruz-Pol, Electromagnetics
UPRM
Review: 1. Free Space
(  0,    o ,   o )

Substituting in the general equations:
  0,       / c

1
2p
u 
c



 o o
h
o o
0  120p   377 
o

E ( z , t )  Eo cos(t  z ) x V / m
H ( z, t ) 
Eo
ho
cos(t  z ) yˆ
A/ m
Cruz-Pol, Electromagnetics
UPRM
4. Good Conductors
(  ,    o ,   r o )

Substituting in the general equations:
  

2

2
u 



2p
Is water a good
conductor???


h
45o


E ( z , t )  Eo e z cos(t  z ) x [V / m]
H ( z, t ) 
Eo
e z cos(t  z  45o ) yˆ [ A / m]

Cruz-Pol, Electromagnetics
 o UPRM
Summary
Lossless
medium
(=0)
Any medium


h
uc

 r r 
c
2

 

 1 
  1
2 

  



 
2

 

 1 
  1
2 
 




Low-loss medium
(”/’<.01)
0



2
 
 
Good conductor
(”/’>100)
pf
pf

(1  j )

j
  j


/
1
1
4pf



2p/up/f
**In free space;


up
up
up
f
f
f
Cruz-Pol,xElectromagnetics
o =8.85
10-12 F/m
UPRM
o=4p x 10-7 H/m
Units
[Np/m]
[rad/m]
[ohm]
[m/s]
[m]
Skin depth, d
E ( z , t )  Eo e
 z

cos(t  z ) x [V / m]
We know that a wave attenuates in a lossy medium
until it vanishes, but how deep does it go?

Is defined as the
depth at which the
electric amplitude is
decreased to 37%
1
e  0.37  (37%)
e
z
e
1
at z  1 /   d
d  1 /  [m]
Cruz-Pol, Electromagnetics
UPRM
Short Cut …

You can use Maxwell’s or use
 1

H  kˆ  E
h


E  h kˆ  H
where k is the direction of propagation of the wave,
i.e., the direction in which the EM wave is
traveling (a unitary vector).
Cruz-Pol, Electromagnetics
UPRM
Waves

Static charges > static electric field, E
 Steady current > static magnetic field, H
 Static magnet > static magnetic field, H

Time-varying current > time varying E(t) & H(t) that are
interdependent > electromagnetic wave

Time-varying magnet > time varying E(t) & H(t) that are
interdependent > electromagnetic wave
Cruz-Pol, Electromagnetics
UPRM
EM waves don’t need a medium
to propagate



Sound waves need a
medium like air or water
to propagate
EM wave don’t. They
can travel in free space in
the complete absence of
matter.
Look at a “wind wave”;
the energy moves, the
plants stay at the same
place.
Cruz-Pol, Electromagnetics
UPRM
Exercises: Wave Propagation in
Lossless materials

A wave in a nonmagnetic material is given by

H  zˆ50 cos(109 t  5 y ) [mA/m]
Find:
(a) direction of wave propagation,
(b) wavelength in the material
(c) phase velocity
(d) Relative permittivity of material
(e) Electric field phasor

 j5 y
[V/m]
Answer: +y, up= 2x108 m/s, 1.26m, 2.25, E   xˆ12.57e
Cruz-Pol, Electromagnetics
UPRM
Power in a wave

A wave carries power and transmits it
wherever it goes
The power density per
area carried by a wave
is given by the
Poynting vector.
See Applet by Daniel RothCruz-Pol,
at
Electromagnetics
UPRM
http://fipsgold.physik.uni-kl.de/software/java/oszillator/index.html
Poynting Vector Derivation

Start with E dot Ampere’s
E 

E    H  E  
t 

E    H   E  E  E  

Apply vector identity
   A  B   B    A  A    B  or in this case :
  H  E   E    H   H    E 

And end up with:
1 E
H    E     H  E   E  
2 t
2
Cruz-Pol, Electromagnetics
UPRM
2
E
t
Poynting Vector Derivation…

Substitute Faraday in 1rst term
H 
1 E 2

2
H  
    H  E   E  

t
2 t


H   H  H 

As in derivative of square function : H    

t  2
t

and if invert the order, it' s (-)
  H  E     E  H 

 H 2
2 t
   E  H   E 2 
 E 2
2 t
Rearrange
  E 2  H 2 
  E 2
  E  H   

t
2Electromagnetics
t
 2 Cruz-Pol,
UPRM 
Poynting Vector Derivation…

Taking the integral wrt volume
   E  H dv  
v

  2  2
2
E

H
dv


E
dv




t v  2
2

v
Applying Theorem of Divergence
  2  2
2


E

H

dS


E

H
dv


E
dv


S


t v  2
2

v
Total power across
surface of volume

Rate of change of
stored energy in E or H
Ohmic losses due to
conduction current
Which means that the total power coming out of a
volume is either due to the electric or magnetic field
energy variations or is lost in ohmic losses.
Cruz-Pol, Electromagnetics
UPRM
Power: Poynting Vector

Waves carry energy and information
 Poynting says that the net power flowing out of a
given volume is = to the decrease in time in
energy stored minus the conduction losses.
  
P  EH
2
[W/m ]
Cruz-Pol, Electromagnetics
UPRM
Represents the
instantaneous
power density
vector associated
to the
electromagnetic
wave.
Time Average Power

The Poynting vector averaged in time is



T
T 

 *


1
1
1
Pave   P dt   E  H dt  Re Es  H s
T0
T0
2


For the general case wave:
Es  Eo e z e  jz xˆ [V / m]
Hs 
Eo
h
e z e  jz yˆ [ A / m]
Pave
Eo2  2z

e
cos h zˆ
2h
Cruz-Pol, Electromagnetics
UPRM
[W/m 2 ]
Total Power in W
The total power through a surface S is

Pave   Pave  dS [W ]
S



Note that the units now are in Watts
Note that power nomenclature, P is not cursive.
Note that the dot product indicates that the surface
area needs to be perpendicular to the Poynting
vector so that all the power will go thru. (give example
of receiver antenna)
Cruz-Pol, Electromagnetics
UPRM
Exercises: Power
1. At microwave frequencies, the power density considered
safe for human exposure is 1 mW/cm2. A radar radiates
a wave with an electric field amplitude E that decays
with distance as |E(R)|=3000/R [V/m], where R is the
distance in meters. What is the radius of the unsafe
region?
 Answer: 34.64 m
2. A 5GHz wave traveling In a nonmagnetic medium with
r=9 is characterized by E  yˆ 3 cos(t  x)  zˆ 2 cos(t  x)[V/m]
Determine the direction of wave travel and the average
power density carried by the wave
2

E
 Answer: P  o e  2 cos  aˆ   xˆ 0.05 [W/m 2 ]
ave
2h1
h kElectromagnetics
Cruz-Pol,
UPRM
TEM wave
x
x
z
z
y
Transverse ElectroMagnetic = plane wave
 There are no fields parallel to the direction
of propagation,
 only perpendicular (=transverse).
 If have an electric field Ex(z)

…then must have a corresponding magnetic
field Hx(z)
 The

direction of propagation is
aE x aH = a k
Cruz-Pol, Electromagnetics
UPRM
PE 10.7
In free space, H=0.2 cos (t-x) z A/m. Find
the total power passing through a
 square plate of side 10cm on plane x+z=1
x
Answer; Ptot = 53mW
 square
Hz
plate at z=3
Ey
Answer; Ptot = 0mW!
Cruz-Pol, Electromagnetics
UPRM
Polarization:
Why do we care??
 Antenna applications –


Remote Sensing and Radar Applications –


Antenna can only TX or RX a polarization it is designed to support.
Straight wires, square waveguides, and similar rectangular systems
support linear waves (polarized in one direction) Round waveguides,
helical or flat spiral antennas produce circular or elliptical waves.
Many targets will reflect or absorb EM waves differently for different
polarizations. Using multiple polarizations can give more
information and improve results.
Absorption applications –

Human body, for instance, will absorb waves with E oriented from
head to toe better than side-to-side, esp. in grounded cases. Also,
the frequency at which maximum absorption occurs is different for
these two polarizations. This has ramifications in safety guidelines
and studies.
Cruz-Pol, Electromagnetics
UPRM
Polarization of a wave
IEEE Definition:
The trace of the tip of the
E-field vector as a
function of time seen from
behind.
Simple cases
 Vertical, Ex
x
x
y
y
x
Ex ( z )  Eo cos(t  z ) xˆ
Exs ( z )  Eo e  jz

Horizontal, Ey
yy
http://fipsgold.physik.uniCruz-Pol, Electromagnetics
kl.de/software/java/polarisation/
UPRM
x
Polarization

In general, plane wave has 2 components; in x & y
E ( z )  xˆE x  yˆE y

And y-component might be out of phase wrt to xcomponent, d is the phase difference between x and y.
E x  Eoxe  j z
x
E y  Eoy e  j z d
Ex
x
y
y
Cruz-Pol, Electromagnetics
UPRM
Ey
Front View
Several Cases
polarization: ddy-dx =0o or ±180on
 Circular polarization: dy-dx =±90o &
Eox=Eoy
 Elliptical polarization: dy-dx=±90o &
Eox≠Eoy, or d≠0o or ≠180on even if Eox=Eoy
 Unpolarized- natural radiation
 Linear
Cruz-Pol, Electromagnetics
UPRM
Linear polarization

Front View
d =0
x
E x  E o e  j z
Ex
E y  E o e  j z

y
Ey
@z=0 in time domain
E x  E xo cos(t)
E y  E yo cos(t)
Back View:
x
y
Cruz-Pol, Electromagnetics
UPRM
Circular polarization

Both components have
same amplitude Eox=Eoy,
 d =d y-d x= -90o = Right
circular polarized (RCP)
 d =+90o = LCP
E x  E xo cos(t)
E y  E yo cos(t  90 o )
in phasor :
E  xˆE xo  yˆ E yo e  j 90  xˆE xo  jE yo yˆ
Cruz-Pol, Electromagnetics
UPRM
Elliptical polarization

X and Y components have different amplitudes Eox≠Eoy, and d
=±90o or d ≠±90o and Eox=Eoy

Or d ≠0,180o,
Or any other phase difference, for example d =56o

Cruz-Pol, Electromagnetics
UPRM
Polarization example
All light
comes out
Unpolarized
radiation
enters
Nothing comes
out this time.
Polarizing glasses
Cruz-Pol, Electromagnetics
UPRM
sin(   90 )  cos( )
sin(   180 )  sin(  )
cos(  90 )  sin ( )
cos(  180 )   cos( )
o
Example
o

o
o
Determine the polarization state of a plane wave
with electric field:
a. E ( z, t )  xˆ3cos(t - z  30o ) - ŷ4sin( t - z  45o )
o
o
b. E ( z, t )  xˆ5cos(t - z  45 )  ŷ10sin( t - z  45 )
c. E ( z, t )  xˆ 4cos(t - z  45 ) - ŷ4sin( t - z  45 )
o
-jz
ˆ
ˆ
d. Es ( z )  14( x-jy )e
Cruz-Pol, Electromagnetics
UPRM
o
.
d=105, Elliptic
.
d=0, linear a 30o
c.
+180, LP a 45o
d.
-90, RHCP
Cell phone & brain

Computer model for Cell phone
Radiation inside the Human Brain
 SAR Specific Absorption Rate
[W/Kg] FCC limit 1.6W/kg,
~.2mW/cm2 for 30mins
http://www.ewg.org/cellphoneradiation/Get-a-SaferPhone/Samsung/Impression+%28SGH-a877%29/
Cruz-Pol, Electromagnetics
UPRM
Human absorption

 30-300
MHz is
where the human

body absorbs RF
energy most
efficiently


* The FCC limit in the US for public exposure
from cellular telephones at the ear level is a
SAR level of 1.6 watts per kilogram (1.6 W/kg)
as averaged over one gram of tissue.
**The ICNIRP limit in Europe for public
exposure from cellular telephones at the ear
level is a SAR level of 2.0 watts per kilogram
(2.0 W/kg) as averaged over ten grams of
tissue.
http://handheld-safety.com/SAR.aspx
http://www.fcc.gov/Bureaus/Engineering_Technology/Docume
nts/bulletins/oet56/oet56e4.pdf
Cruz-Pol, Electromagnetics
UPRM
Radar bands
Band Name
Nominal Freq
Range
HF, VHF, UHF
3-30 MHz0, 30-300 MHz, 3001000MHz
Specific Bands
138-144 MHz
216-225, 420-450 MHz
890-942
Application
TV, Radio,
Clear air, soil moist
L
1-2 GHz (15-30 cm)
1.215-1.4 GHz
S
2-4 GHz (8-15 cm)
2.3-2.5 GHz
2.7-3.7>
C
4-8 GHz (4-8 cm)
5.25-5.925 GHz
TV stations, short range
Weather
X
8-12 GHz (2.5–4 cm)
8.5-10.68 GHz
Cloud, light rain, airplane
weather. Police radar.
Ku
12-18 GHz
13.4-14.0 GHz, 15.7-17.7
K
18-27 GHz
24.05-24.25 GHz
Ka
27-40 GHz
33.4-36.0 GHz
V
40-75 GHz
59-64 GHz
W
75-110 GHz
millimeter
110-300 GHz
76-81 GH, 92-100 GHz
Cruz-Pol, Electromagnetics
UPRM
Weather observations
Cellular phones
Weather studies
Water vapor content
Cloud, rain
Intra-building comm.
Rain, tornadoes
Tornado chasers
Microwave Oven
Most food is lossy media at
microwave frequencies,
therefore EM power is lost
in the food as heat.

Find depth of penetration if
meat which at 2.45 GHz has
the complex permittivity
given.
The power reaches the inside
as soon as the oven in
turned on!
 c   o (30  j1)

j o j c
2pf

( j  30)
c
 4.7  j 281 [/m]
d  1/   21.3 cm
Cruz-Pol, Electromagnetics
UPRM
Decibel Scale

In many applications need comparison of two
powers, a power ratio, e.g. reflected power,
attenuated power, gain,…
 The decibel (dB) scale is logarithmic
G
P1
P2
 V12 /R 
 P1 
V 
G[dB]  10 log    10 log  2   20 log  1 
 P2 
 V2 
 V2 /R 

Note that for voltages, fields, and electric
currents, the log is multiplied by 20 instead of 10.
Cruz-Pol, Electromagnetics
UPRM
Attenuation rate, A

Represents the rate of decrease of the magnitude
of Pave(z) as a function of propagation distance
 Pave(z) 
  10 log e 2z
A  10 log 
 Pave( 0 ) 
 20z log e  - 8.68z  - dB z [dB]

where
 dB[dB/m ]  8.68 [ Np/m]
Cruz-Pol, Electromagnetics
UPRM

Submarine antenna
A submarine at a depth of 200m uses a wire
antenna to receive signal transmissions at
1kHz.
 Determine the power density incident upon
the submarine antenna due to the EM wave
with |Eo|= 10V/m.

[At 1kHz, sea water has r=81, =4].
Pave

Eo2  2z

e
cos h zˆ
2h
[W/m 2 ]
At what depth the amplitude of E has
decreased to 1% its initial value at z=0 (sea
surface)?
Cruz-Pol, Electromagnetics
UPRM
Exercise: Lossy media
propagation
For each of the following determine if the material is low-loss dielectric,
good conductor, etc.
(a) Glass with r=1, r=5 and =10-12 S/m at 10 GHZ
(b) Animal tissue with r=1, r=12 and =0.3 S/m at 100 MHZ
(c) Wood with r=1, r=3 and =10-4 S/m at 1 kHZ
Answer:
(a)
(b)
(c)
low-loss,  8.4x1011 Np/m,  468 r/m,  1.34 cm, up1.34x108, hc168 
general,  9.75, 12, 52 cm, up0.5x108 m/s, hc39.5j31.7 
Good conductor,  6.3x104,  6.3x104,  10km, up0.1x108, hc6.281j 
Cruz-Pol, Electromagnetics
UPRM