Electromagnetic Intro by ewghwehws

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									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
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Would magnetism would produce
         electricity?
   Eleven years later,
    and at the same time,                                   d
                                                Vemf    N
    Mike Faraday in                                         dt
    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
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    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
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Cruz-Pol, Electromagnetics
          UPRM
Electromagnetic Spectrum




        Cruz-Pol, Electromagnetics
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              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
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                 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
          B                                      Faraday’s Law
  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
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             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
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        Maxwell put them together
   And added Jd, the
    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.
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      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
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                                                  Encarta®
      Who was NikolaTesla?
 Find out  what inventions he made
 His relation to Thomas Edison
 Why is he not well know?




                Cruz-Pol, Electromagnetics
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                  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
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      Phasors & complex #’s
Working with harmonic fields is easier, but
  requires knowledge of phasor, let’s review
 complex numbers and
 phasors




                Cruz-Pol, Electromagnetics
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      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
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                     Review:
 Addition,
 Subtraction,
 Multiplication,
 Division,
 Square Root,
 Complex Conjugate



                    Cruz-Pol, Electromagnetics
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    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
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                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
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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
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     Advantages of phasors
 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
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       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
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                   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
                                              Faraday’s Law

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

* (substituting   D   E and H  B )
                             Cruz-Pol, Electromagnetics
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                          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
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                         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
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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
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        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
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                 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
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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
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         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
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                              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
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                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
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   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
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         2. Lossless dielectric
       (  0,    r  o ,    r o or    )

 Substituting in   the general equations:

            0,    
                 1         2p
         u             
                          
             o
         h   0
            


                    Cruz-Pol, Electromagnetics
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                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
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           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
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                 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
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                            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
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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
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    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
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                   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
http://www.netzmedien.de/software/download/java/oszillator/
     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
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      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
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               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
  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
                               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
        Answer; Ptot = 53mW
                                                  Hz
 square plate at   z=1, 0
                                         Ey
     Answer; Ptot = 0mW!

                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
 Unpolarized- natural radiation




                 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
radiation
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
    Cell phone Radiation
    inside the Human
    Brain




                   Cruz-Pol, Electromagnetics
                             UPRM
                                   Radar bands
                     Nominal Freq
 Band Name
                         Range
                                                     Specific Bands                   Application
HF, VHF, UHF                                    138-144 MHz
                3-30 MHz0, 30-300 MHz, 300-
                     1000MHz
                                                216-225, 420-450 MHz           TV, Radio,
                                                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
                                                                                  weather. Police radar.
      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
                                                                               Rain, tornadoes
                                                  UPRM
   millimeter          110-300 GHz                                             Tornado chasers
               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
    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) 
          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
                                                                                     [rad/m]



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
Answer:
(a)   low-loss,  8.4x1011 Np/m,  468 r/m,  1.34 cm, up1.34x108, hc168 
(b)   general,  9.75, 12, 52 cm, up0.5x108 m/s, hc39.5j31.7 
(c)   Good conductor,  6.3x104,  6.3x104,  10km, up0.1x108, hc6.281j 



                               Cruz-Pol, Electromagnetics
                                         UPRM

								
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