# Electromagnetic Intro by ewghwehws

VIEWS: 0 PAGES: 67

• pg 1
```									Electromagnetism
INEL 4152

Sandra Cruz-Pol, Ph. D.
ECE UPRM
Mayagüez, PR
2005
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,                                   d
Vemf    N
London and Joe
Henry in New York
discovered that a
time-varying magnetic
field would produce

an electric current!
 E  dl   t  B  dS
L               s

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 
 H  dl    J  t   dS
L          s         
*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
Gauss’s Law for E
  D  v          D  dS   v dv
s            v
field.
Gauss’s Law for H
 B  0
 B  dS  0              field. Nonexistence
s                         of monopole
 E  
t         E  dl   t  B  dS
L               s

D                        D         Ampere’s Circuit
 H  J 
t      
L
H  dl    J 
s
  dS
t         Law
Cruz-Pol, Electromagnetics
UPRM
Maxwell’s Eqs.
v
 J  
 Also the equation of continuity                      t
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
displacement current

 H  dl   J  dS  I
L         S1
enc   I

I

 H  dl   J  dS  0
S1

L              S2                                      L
S2
d              dQ
 H  dl  S J d  dS  dt S D  dS  dt  I
L            2                2

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
 0
   with no charges, i.e. .    v  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
j
z  x  jy  re         r  r cos   jr sin 

where r | z |        x  y is the magnitude
2          2

y
  tan    1
is the angle
x
Cruz-Pol, Electromagnetics
UPRM
Review:
 Subtraction,
 Multiplication,
 Division,
 Square Root,
 Complex Conjugate

Cruz-Pol, Electromagnetics
UPRM
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
 Fora sinusoidal current I (t )  I o cos(t   )
j  j t
equals the real part of I o e e

 The complex term   I o e j which results from
dropping the time factor e jt 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.

j t
A  Re{ As e                   }

Cruz-Pol, Electromagnetics
UPRM
 Time derivative  is equivalent to
multiplying its phasor by j
A
 jAs
t
is equivalent to dividing by
 Time integral
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*

Gauss’s Law for E field.
  v
D E   v
Gauss’s Law for H field.
BH
 0           0                      No monopole

  E   jH
 E  
B
t

Ampere’s Circuit Law
  H  J  jE
 H  J 
D
t

* (substituting   D   E and H  B )
Cruz-Pol, Electromagnetics
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 )
whosegeneral solution is
E(z)  Eo e z  Eo e z
'

Cruz-Pol, Electromagnetics
UPRM
To change back to time domain
   From phasor

 z                  z (  j  )
E xs ( z )  Eo e           Eo e
   …to time domain
 z                  
E ( z , t )  Eo e          cos(t   z ) x

Cruz-Pol, Electromagnetics
UPRM
Several Cases of Media
1.   Free space (  0,    o ,    o )
2.   Lossless dielectric (  0,    r  o ,    r o or    )
3.   Lossy dielectric (  0,    r  o ,    r o )
4.   Good Conductor (  ,    o ,    r o or    )

o=8.854 x 10-12[ F/m]
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. Lossy Dielectrics                                                                  
E ( z , t )  Eo e z cos(t   z ) x
(General Case)
   In general, we had      j (  j )
2                                     j

 Re  2   2   2   2 
 2   2   2    2   2 2

   From this we obtain

      
2                   
2  
      1       1 and        1       1
2 
          

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.

j  h 
h             h                                       []
  j

 z                 
E ( z, t )  Eo e          cos(t  z ) x
*Not in-phase
Eo
H ( z, t )         e z cos(t  z  h ) y
ˆ                      for a lossy
h                                                    medium
Cruz-Pol, Electromagnetics
UPRM
Note…
 z                 
E ( z, t )  Eo e          cos(t  z ) x
Eo
H ( z, t )         e z cos(t  z  h ) y
ˆ
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       Es   
           tan   loss tangent
J ds     j Es 

Cruz-Pol, Electromagnetics
UPRM
Relation between tan and c
       
  H  E  j E  j 1  j    E
       
 j c E

The complex permittivity 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            
o o
h           0  120p   377 
o


E ( z, t )  Eo cos(t  z ) x V / m
Eo
H ( z, t )         cos(t  z ) y
ˆ      A/ m
ho
Cruz-Pol, Electromagnetics
UPRM
4. Good Conductors
(  ,    o ,    r o )
   Substituting in the general equations:

  
2
  2                   2p                       Is water a good
u                                              conductor???
                     

h    45o


E ( z , t )  Eo e z cos(t  z ) x [V / m]
Eo
H ( z, t )          e z cos(t  z  45o ) y [ A / m]
ˆ

 o UPRM
Cruz-Pol, Electromagnetics
Skin depth, d
   Is defined as the
1
depth at which the               e  0.37  (37%)
electric amplitude is                z
decreased to 37%                 e         e   1
at z  1 /   d

d  1 /  [m]

Cruz-Pol, Electromagnetics
UPRM
Short Cut …
   You can use Maxwell’s or use

 1   
ˆ
H  kE
h
          
ˆ
E  h k  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  z 50 cos(10 9 t  5 y ) [mA/m]
ˆ
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
Answer: +y, up= 2x108 m/s, 1.26m, 2.25, E   x12 .57 e
ˆ                [V/m]

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.

at
See Applet by Daniel RothCruz-Pol, Electromagnetics
UPRM
Poynting Vector Derivation
               E 
 Start with E dot Ampere E    H  E  
               t 

E
E    H   E  E  E  
t
   Apply vectorial identity
   A  B   B    A  A    B  or in this case :
  H  E   E    H   H    E 

   And end up with
1 E                           2
H    E     H  E   E                 2

2 t
Cruz-Pol, Electromagnetics
UPRM
Poynting Vector Derivation…
   Substitute Faraday in 1rst term
    H                        1 E 2
H           H  E   E  
2

     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                              E 2
               E  H   E 2 
2 t                               2 t
Rearrange
  E 2  H 2 
  E  H           
 2 Cruz-Pol,2 t     E 2
 t                 
Electromagnetics
UPRM
Poynting Vector Derivation…
   Taking the integral wrt volume
  2  2
   E  H dv           E  H dv   E dv
t   2
2

v                         v
2       v

   Applying theory of divergence
  2  2 
 E  H  dS     E  H dv   E 2dv
S
t v  2 2      v

Rate of change of         Ohmic losses due to
Total power across
stored energy in E or H   conduction current
surface of volume
   Which simply 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.

                                      Represents the
P  EH          [W/m ]       2
instantaneous
power vector
associated to the
electromagnetic
wave.
Cruz-Pol, Electromagnetics
UPRM
Time Average Power
   The Poynting vector averaged in time is
1  
                         
     1 
T        T
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 x [V / m]
ˆ                                    2
Eo 2z
Pave      e   cosh z
ˆ   [W/m2 ]
Hs 
Eo
e z e  jz y [ A / m]
ˆ                              2h
h

Cruz-Pol, Electromagnetics
UPRM
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

Cruz-Pol, Electromagnetics
UPRM
Exercises: Power
1. At microwave frequencies, the power density considered
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?
a
2. A 5GHz wave traveling In  nonmagnetic medium with
r=9 is characterized by E  y3 cos(t   x)  z 2 cos(t  x)[V/m]
ˆ               ˆ
Determine the direction of wave travel and the average
power density carried by the wave

 Answer: Pave   x 0.05 [W/m ]
ˆ           2

Cruz-Pol, Electromagnetics
UPRM
x

TEM wave
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 = ak
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
Hz
 square plate at   z=1, 0
Ey

Cruz-Pol, Electromagnetics
UPRM
Polarization of a wave
IEEE Definition:
The trace of the tip of the                                x
x
E-field vector as a
function of time seen from
behind.

Simple cases                                          y
y
 Vertical, Ex                                                   x
x
 jz
Exs ( z )  Eo e
Ex ( z )  Eo cos(t  z ) x
ˆ

   Horizontal, Ey
y   y
Cruz-Pol, Electromagnetics
UPRM
Polarization:
Why do we care??
 Antenna 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, often) Round
waveguides, helical or flat spiral antennas produce circular or
elliptical waves.
   Remote Sensing and Radar Applications –
   Many targets will reflect or absorb EM waves differently for different
polarizations. Using multiple polarizations can give different
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
   In general, plane wave has 2 components; in x & y
E ( z )  xE x  yE y
ˆ      ˆ

   And y-component might be out of phase wrt to x-
component, d is the phase difference between x and y.
 j z
E x  Eo e                                                   x

E y  Eo e j z d                                     Ex

x
y
y
Ey

Cruz-Pol, Electromagnetics       Front View
UPRM
Several Cases
 Linear  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

Cruz-Pol, Electromagnetics
UPRM
Linear polarization
Front View
   d =0                                                            x

E x  Eo e  j z                                       Ex

E y  Eo e j z                                 y
Ey
   @z=0 in time domain
Ex  Exocos(t)
E y  E yocos(t)
Back View:
x

y
Cruz-Pol, Electromagnetics
UPRM
Circular polarization
 Both components have
same amplitude Eox=Eoy,           E x  E xocos(t)
 d =d y-d x= -90o = Right          E y  E yocos(t  90o )
circular polarized (RCP)          in phasor :

 d =+90o = LCP                     E x  xE xo  y Eyoe  j 90  xE xo  jE yo y
ˆ       ˆ               ˆ             ˆ

Cruz-Pol, Electromagnetics
UPRM
Elliptical polarization

   X and Y components have different amplitudes
Eox≠Eoy, and d =±90o

   Or d ≠±90o and Eox=Eoy,
Cruz-Pol, Electromagnetics
UPRM
Polarization example

All light
comes out

Unpolarized
enters
Nothing comes
out this time.

Polarizing glasses

Cruz-Pol, Electromagnetics
UPRM
sin(  90  sin(  
sin(  180 ))  cos( ))
oo

Example
cos(  180 ))  sin(())
(  90   cos  oo

   Determine the polarization state of a plane wave
with electric field:
a. E ( z , t )  x3cos(t - z  30 o ) - y4sin(t - z  45 o )
ˆ                        ˆ

b.E ( z, t )  x3cos(t - z  45 )  y8sin(t - z  45 )
ˆ                      ˆ
o                      o

c. E ( z, t )  x 4cos(t - z  45 o ) - y4sin(t - z  45 o )
ˆ                         ˆ
a.   Elliptic
ˆ-j z )e -jy
d. Es ( z )  14 ( x ˆ                                  b.   -90, RHEP
c.   +90, LHCP
Cruz-Pol, Electromagnetics
UPRM               d.   -90, RHCP
Cell phone & brain
   Computer model for
inside the Human
Brain

Cruz-Pol, Electromagnetics
UPRM
Nominal Freq
Band Name
Range
Specific Bands                   Application
HF, VHF, UHF                                    138-144 MHz
3-30 MHz0, 30-300 MHz, 300-
1000MHz
890-942

L            1-2 GHz (15-30 cm)                    1.215-1.4 GHz         Clear air, soil moist
2.3-2.5 GHz          Weather observations
S            2-4 GHz (8-15 cm)                          2.7-3.7>        Cellular phones
TV stations, short range
C             4-8 GHz (4-8 cm)                    5.25-5.925 GHz
Weather
Cloud, light rain, airplane
X           8-12 GHz (2.5–4 cm)                    8.5-10.68 GHz
Ku                 12-18 GHz                  13.4-14.0 GHz, 15.7-17.7   Weather studies
K                  18-27 GHz                      24.05-24.25 GHz        Water vapor content
Ka                 27-40 GHz                       33.4-36.0 GHz         Cloud, rain
V                  40-75 GHz                        59-64 GHz            Intra-building comm.

W                 75-110 GHz                  76-81 GH, 92-100 GHz
Cruz-Pol, Electromagnetics
UPRM
Microwave Oven
Most food is lossy media at
microwave frequencies,
therefore EM power is lost
in the food as heat.
 c   o (30  j1)
   Find depth of penetration if
meat which at 2.45 GHz has                          j o j c
the complex permittivity
2pf
given.                                                 ( j  30)
c
 4.7  j 281 [/m]
The power reaches the inside
as soon as the oven in                      d  1/   21.3 cm
turned on!
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
P
G    1
P2
P             V12 /R      V 
G[dB]  10 log  1   10 log  2   20 log  1 
P             V /R        V 
 2            2            2

   Note that for voltages, the log is multiplied by 20
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) 
A  10 log 
 P (0)               
  10 log e  2z   
 ave     
 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].

2
Eo 2z
Pave      e   cosh z
ˆ             [W/m2 ]
2h
   At what depth the amplitude of E has
decreased to 1% its initial value at z=0
(sea surface)?          Cruz-Pol, Electromagnetics
UPRM
Summary
Lossless      Low-loss medium   Good conductor   Units
Any medium                     medium           (”/’<.01)      (”/’>100)
(=0)

     
      
 1 
2  
  1
0
       
pf
[Np/m]
2 
          
                          2       

                                                                  pf

h           j                                                     (1  j )
     [ohm]

  j                                                            

uc           /                           1                   1            4pf
[m/s]
                             
up               up              up
         2p/up/f
f                                f
[m]
f
**In free space;              Cruz-Pol, Electromagnetics  =4p 10-7 H/m
o =8.85 10-12 F/m             o
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