waves

Document Sample

```					ELECTROMAGNETIC WAVES

Uniform plane waves (TEM) in lossless media
Solutions – time-domain
E r,t  = aE E cos t -  r  = aE E cos t - a  r 
E r,t  = aE E cos t -   x cos x + y cos y + z cosz 
                                        

Solutions – frequency-domain
E r,t  = Re E e-jt
E(r, ) = aEE e-j r

The electric and magnetic vectors perpendicular and both are perpendicular to
the direction of propagation.

aE  aH = a 

1
H =       a  E or use Faraday’s law

E =  H  a or use Ampere’s law

                2
=                =
                

v = f =     (general)

1
v=          (lossless only)


all kinds of applets on transmission lines, waves etc.
http://www.educypedia.be/electronics/javatransmissinlines.htm

another excellent site
http://www.amanogawa.com
2

Example: TEM wave propagation             ☼
   x       3y  
An E-field is given by E = a z 50cos 109 t - 5
 +
2            V/m . ( = o) Find

           2 

i) wave number, wave vector, direction of travel
ii) velocity
iii) wavelength
iv) wave (or intrinsic) impedance
v)   H (time-domain vector), H (frequency-domain vector)

EM waves
3

Energy, conduction current, displacement current         ☼
From the point form of Ohm’s law, J = E, two points can be appreciated:
1.   The conduction current and the electric field vary alike. Changes in the
electric field produce like variations in the conduction current.
2.   The conduction current and the electric field are in the same direction.

The result is that plots of E and J versus time are scaled versions of one another
as shown below. From Joule’s law,
dP
=E J
dv

Given that E and J are in phase and parallel, the power dissipated per volume in
conductive material is simply the product of the magnitudes of E and J.
dP
=E J
dv

EM waves
4

Displacement current density is the time rate-of-change of the electric flux
density vector. The flux density can be broken into two pieces, the first being
 oE which is present even in the absence of material being present (it’s there
when material is present too) and the second being P which is the polarization
vector associated with the separation of bound change.

The displacement current is associated with the time rate of change of these two components.
D =  E =  r  o E =  oE + ( r - 1) oE    D =  oE + P
D          E     P
JD =     = o         +
t          t     t
Note that both components of the displacement current have the same phase
and direction (true at least for ―low‖ frequencies—more on this on the next page).

This allows us to frame the discussion of power absorbed by a dielectric in terms
of the behavior of bound change with respect to changes in electric field.

Below, a sinusoidal variation in E is assumed and the corresponding bound charge
separation is tracked with respect to time and correlated to the associated
displacement current. The power per unit volume absorbed by a dielectric is E JD

EM waves
5

If the material under consideration have both displacement and conduction
currents present, there will be both energy storage and average power
dissipation. The net current will be between 0° and 90° out-of-phase with the
electric field.

This situation can also occur in dielectrics at high frequencies in which the time
constants associated with the dynamics of the bound charge mass can no longer
be neglected with respect to the time variation of the field.

Displacement current at high-frequencies

The result in either case, whether the material has both conduction and
displacement current or whether the frequency is sufficiently high so that the
displacement current is not in phase with the electric field, the result is that there
is an average power lost from the field in the material.

Conduction loss and dielectric loss both lead to the same result, which is that the
propagation constant becomes complex,  =  + j . Conduction loss is
accounted for by a non-zero conductivity, , in Ampere's law, and dielectric loss
is accounted for by including an imaginary component in the permittivity ( - jp).

EM waves
6

Waves in lossy material
Consider Ampere and Faraday (with no charge or current sources). There are two
basic source for loss – conductive loss and dielectric loss. The dielectric loss is readily
incorporated by including an imaginary component in the permittivity [( - jp) used
here—other notations are used, e.g. (' - j")] . It is important to note that this
imaginary component describes how the bound charge separation lags the electric field
excitation and leads to loss. Conductive loss is due to a non-zero conductivity .
Note: no current sources DOES NOT mean there are no currents. In conductive
materials (0), currents exist in response to an electric field in conductive material.
E                      E
 H = J +        =  E +  - j p 
t                     t
H
  E = -
t
Combining these equations
                                 E
    E =    g E  -  2E = -          H = -  E +  - j p  

t              t                t 

E      2E
2E -       -  2 = 0
t     t

In the frequency domain,
                                            
 2E - j E +  2    - j p  E =  2E +  2    - j p - j  E =  2E +  2    - j p - j  E = 0
                                            
                                                             
 2E +  2   E = 0              where  =  - j p - j  is the complex permittitivty 
                                                              

In the frequency domain, solutions are

E = aE E e -    r

                              +  p
where  =  a = j   a = j    - j p - j  = j   1 + j
                               
 =  + j   a 

Terminology
        complex propagation vector
        complex propagation constant
         attenuation constant (equal to zero for lossless materials)
         phase constant


IMPORTANT NOTE: From this point, p will be neglected for simplicity so that  =  - j                   .


EM waves
7

Details of solution
The solution must satisfy the wave equation, 2E +  2  E = 0.

E = aE E e -     r

E = aE E e j  e
-      + j  a r          

E = aE E e j  e
-                   
+ j  x cos x + y cos y + z cosz    

 2       2       2                     - + j  x cos x + y cos y + z cosz  
 2 +           +     2  E
a E e j e                                                   
 x       y       z                                                                 
2

- + j  x cos x + y cos y + z cosz  
+  2   aE E e j e
                                                           = 0.
                                                          

Each derivative pulls down -  + j  cos  i from the exponential. The derivatives
hear are second derivatives.

-  + j  cos                  x
2
+ -  + j  cos  y  + -  + j  cos  z 
                    
2
                    
2
 E +   E = 0
2

-  + j  cos 
2                    2
x                          
+ cos2 y + cos2 z  E +  2  E = 0

The sum of the squares of the directional cosines must equal zero.

-  + j  cos 
2                        2
x                          
+ cos2 y + cos2 z  E +  2  E = 0

 2 E +  2  E = 0                                            2 = - 2             = j 

The result is the propagation constant is complex  =  + j . The propagation
vector also indicates direction of travel,  =  a =  + j  a .

Propagation along a 

E = aE E e -     r

E = aE E e

- a r                    cos  t -   a r  +  + 
                       

EM waves
8

Plane waves in lossy materials
E = aE E e -    r

E = a E E e j e
-           
+ j  x cos x + y cos y + z cos z   

E = aE E e

- a r         e-  a r  e j

In the time domain

E = aE E e

 a r           cos  t -   a r  +  
                     

- x cosx + y cosy + z cosz                 cos  t -  x cos  + y cos  + z cos  +  
E = aE E e                                                                     x         y         z   
EM waves in lossy material are exponentially decaying sinusoids                                     ☼
Consider an electromagnetic wave, pointing in ax direction, traveling in a lossy
material in the az direction (assume  = 0).

E = a xEe- z


- 2 =  2  ( - j
)


 = j  ( - j ) =  + j


In the time domain,

E = a xEe- zcos( t -  z)

The intrinsic impedance is complex (E and H are NOT in phase for lossy materials)

                                                              j
=       =                                 =                = e
                                            
 1 - j                      -j
              
               

1
H = a              E or use Faraday’s law

E =  H  a or use Ampere’s law

2                                                                          1
 =                                     v = f =       (general)              v=        (lossless only)
                                                                          

EM waves
9

Example: Wave propagation in lossy material                    ☼
- z
An electric field is given as E = a x 100e          V/m is traveling through material with
=0.1/m, r=1, r=4. The frequency is 2.45 GHz. Find  and  and the dB/m
attenuation in the material.

EM waves
10

Waves and power flow in lossy material ☼
What is the relation between E and H
in the diagram shown?

Take E as pointing along ax, traveling in az direction and take phase of E z = 0 = 0° .
t=0

S = E  H = a xEe- z cos( t -  z)  a yHe- zcos( t -  z -  )

cos   + cos(2 t - 2 z -  )
EH -2 z
= az      e                                       
2

time-average Poynting vector
1
2
         1
S = Re E  H = Re a xEe- z e-j z  a yH e  e- z e j z
% %
2
                  j

1                    1
= a z EH cos   e-2 z
j -2 z
= a zRe EHe e
2                    2

EM waves
11

Special Cases ☼
High loss (good conductors) ~  ? 

               
 = j  (1 - j      )  j  -j
              

j2                j
=          =  j =  e 2
j
1+j       
=         =              1 + j  =    + j
2             2

Low loss (good dielectric) ~  = 

       
 = j   1 - j 
       
          1   
2

 j  1 - j      +       

     2   8    
                 1   
2

         + j  1 +         
2              
   8     
=  + j

For very low loss, the second order
term is often neglected.
 
        + j 
2 

EM waves
12

Good dielectrics (low loss) and good conductors (high loss)
Note that, in the expression for complex permittivity, the imaginary term within the
parentheses gives the ratio between conduction current and displacement current.

        
 =  1 - j
        


Jc =  E  Jc =  E
D      E
Jd =      =        Jd = jD = j E
t      t
Jc     E              
=        =      = -j
Jd    j E   j       

―good dielectrics‖ are defined by        Jd >> Jc

―good conductors‖ are defined by         Jc >> Jd

Example: Good conductor? Good dielectric?
Find the propagation constant for f = 100Hz,  = 10/m, r=1, r=10. Is the
material a good conductor or a good dielectric? Neither?

Example: Copper
At what frequency does Cu cease to be a good conductor?
=5.8(107)/m, r=1, r=1.

EM waves
13

High loss - skin depth            ☼
The distance a wave must travel into a good conductor for its amplitude to be
reduced by a factor of e-1 is called skin depth, .
Recall that, for a good conductor, conduction current is much greater than
displacement current ( >> ).         Jc = E,      Jd = dD/dt =  dE/dt

- 2 =  2  (1 - j      )


             
   + j =                     +j
2                2

 is the attenuation constant and is the reciprocal of skin depth.
Consider an electric wave pointing in the
x-direction and traveling in the z-direction:

This implies that the magnetic field
must lag the electric field
High loss - wave impedance
(E =  H )
                                              
=       =                                         =      45°
                                          
 1 - j                 - j 
                       

EM waves
14

Example: skin effect         ☼
The distance in a good conductor that is required to reduce the amplitude of the
field vectors by a factor of e-1 is called the skin depth, .

Assuming copper:
z
-
E(z) = E(z) = Eo e- z = Eo e  = Eo e-1 for z =  =  -1

2        1    1     0.0661
 =  1 =           =             
         2   5.8f      f

copper, non-magnetic
 = o = 4(10 ) H/m      = 5.8(10 ) S/m
-7                   7

frequency (Hz)      60 Hz         1 KHz   1 MHz     1 GHz
skin depth        8.5 mm     2.1 mm     66.1 m   2.1 m

EM waves
15

Skin effect in circuits        ☼
Consider the important special case in which the skin depth is
much smaller than the cross sectional dimensions of the
conductor.
In this case, the field are confined to a thin layer around the
surface. Even though the fields actually decay exponentially, it is
sometimes convenient to suppose the current density is
uniformly distributed over a thin layer—the thickness of which
turns out to be a skin depth (=).

V
R =
I

The AC resistance of conductors is a function of frequency since
the skin depth is a function of frequency.
i)    circular conductors

ii)   square conductors

iii) stranded wire

EM waves
16

Example: Skin effect       ☼
A voltage, vs(t) = 10 cos (2ft) V is applied to a 100  load and a solid copper
conductor as shown. The radius of the conductors is 2 mm and their length is 5 cm.
a)    What power is dissipated in the copper conductor at 60 Hz?

b)   What is the resistance of the conductors at 4 GHz? What power is
dissipated in the copper conductors at 4 GHz?

A voltage, vs(t) = 10 cos (2ft) V is applied to a similar conductor-load, this time
the conductors consist of a 10 m copper coating with the interior of the
conductors being a good insulator like Teflon.

a)   What is the resistance of the Cu-coated conductors at 60 Hz?
What power is dissipated by the Cu-coated conductors at 60 Hz?

b)   What is the resistance of the Cu-coated conductors at 4 GHz?
What power is dissipated by the Cu-coated conductors at 4 GHz?

EM waves
17

Poynting’s theorem          ☼

 B 
H      E    =H   -   
 t 
John Henry Poynting
    D 
E       H   =E   J +     
     t 

Using this and the vector identity
    E  H   =H     E   -E      H

EM waves
18

Poynting’s theorem ☼
Looking at the integral form of Poynting’s theorem, it is readily seen that the
theorem is an expression of conservation of energy IF Poynting’s vector is power
flux density in W/m2.

Conservation of energy is the justification for the interpretation of S = E  H as
power flux density.

Example: units
What are the units of a) E  D b) H  B c) S ?

Example: Calculations of power flow             ☼
An electric vector, pointing in the y-direction, travels in the z-direction. At z=0, the
electric vector has an amplitude of 2 V/m. f=10 MHz with material properties
r=1, r=4, =10-5/m.
Calculate the terms in the integral form of Poynting’s theorem for the cube
0 < x < 2 m, 0 < y < 2 m, 0 < z < 2 m.

EM waves
19

Instantaneous and time-averaged power density          ☼
Calculate Poynting’s vector and the time-averaged Poynting’s vector for the
plane wave below, traveling in air.
-j y
E = a x 2e     2     V/m

Repeat, this time for the wave traveling in a lossy material.
- + j  y
E = ax E e                   V/m = a x E e y e j y V/m

EM waves
20

EM waves incident on boundaries (normal incidence)         ☼
How are EM waves reflected and transmitted when a plane dielectric boundary is
encountered? Consider the situation below:

Define a transmission coefficient, T, and reflection coefficient, .
Et              Er
T=              =
Ei              Ei
-j z
Let Ei = a xEie 1
Boundary conditions:     E1tan   z=0
= E2tan   z=0
H1tan   z=0
= H2tan   z=0

EM waves
21

Transmission and reflection ☼
Find the reflection and transmission coefficients for a wave traveling in a printed
circuit board made from FR-4 (most common PCB material—fiberglass epoxy
base) into air.

FR-4     r=4.4,   r=1
air      r=1,     r=1

EM waves
22

Polarization       ☼
Polarization refers to the path traced by the electric vector in a traveling
electromagnetic wave. Assume a wave in traveling in the +az direction and take
the point-of-view seen in front of the oncoming EM wave. For a given value of z,
what path is traced by the electric vector in the x-y plan as time advances?

In general, a wave traveling in the az direction can have an x-component and a y-
component. The resultant vector traces an ellipse—referred to as elliptical polarization—
in the x-y plan as it travels, the shape of which depends on Ex, Ey, x, and y.

E = axExcos t -  z + x  + ayEycos t -  z + y 

Linear polarization is a special case
E = a xEx cos  t -  z + x    E = ayEycos t -  z + y     E = a xE x cos  t -  z  + a yE y cos  t -  z 

When Ex = Ey and x and y differ by ±90°, the result
is circular polarization.

For circular and elliptical polarization, the sense of the rotation is referred to as
left-handed or right-handed, depending on which hand, if the fingers follow the
rotation of E, the thumb is in the direction of propagation.

Polarization examples ☼
What is the polarization for E = a x 2 cos  t -  z + 45°  - a y 10 cos t -  z - 90°  ?       (LHEP)

What is the polarization for E = a x 2 cos  t -  z + 45°  + a y 2 cos  t -  z - 45°  ?       (RHCP)

What is the polarization for E = a x 3 cos  t -  z + 45°  + a y cos  t -  z + 45°  ?         (LP,  = 18.43°)

EM waves
23

EM waves incident on boundaries (oblique incidence)
Now consider an EM wave, traveling in region 1, which encounters a second
region at an angle other than normal. In this case, we must consider two types of
polarization.
1.   Perpendicular polarization: E  to plane of incidence (POI).
2.   Parallel polarization: E P to POI

E  POI                                       E P to POI

Definitions
POI – plane formed by propagation vector, , and the vector normal to the
boundary between regions 1 and 2.

Polarization: in general, any non-random orientation of an electric and magnetic
field. In particular, polarization more usually describes the path the electric
vector takes, in planes of constant phase as the wave travels.

Note: We’re considering only linear polarization, where the electric vector
points in one direction. It reaches its maximum positive, grows smaller,
becomes zero then negative, becomes maximum negative, grows smaller,
becomes zero, etc.

That is the vector remains on a straight line—thus, linear polarization.

Any TEM wave can be expressed in terms of linearly polarized waves.

EM waves
24

Refection and transmission at dielectric boundaries ( E  POI)           ☼

1st, give expressions for the electric and magnetic vectors.
Then, to find reflection and transmission coefficients, require tangential H and E
to be continuous at the boundary (z = 0).

EM waves
25

Refection and transmission at dielectric boundaries ( E P POI)

 2cos t - 1cos i                  2 2cos i
 =                               =
 2cos t + 1cos i              2cos t + 1cos i

EM waves
26

Example: Reflection and transmission                ☼
A uniform plane wave, in air, impinges on a dielectric material at i = 45°. The
transmitted wave propagates with t = 30°. The frequency is 300 MHz and the
electric vector is perpendicular to the plane-of-incidence.
i)    Find 2 in terms of 0 (assume 1=2=o)

ii)   Find the reflection coefficient, .
iii) Find the transmission coefficient, T.

iv) Give expressions for the incident E and H -fields, the
reflected E and H fields, and the transmitted E and H -fields.

EM waves
27

EM waves normally incident on good conductors          ☼
Consider EM waves encountering a conducting material

2 - 1                 22
=                    T=
2 + 1               2 + 1

EM waves
28

Reflection and transmission for an air/metal interface            ☼
A 1-MHz plane wave, in air is incident on a copper sheet. The amplitude of the
incident E-field is 100 V/m. Find , T, , and the |E| and |H| fields at a distance of
 into the copper.

EM waves
29

Transmission through a conducting sheet – an air/metal/air barrier         ☼
To shield from potential EM interference, a product manufacturer is
considering using a copper sheet with a thickness, t, of several skin
depths (say 4 or 5). As a measure of the shielding available, find an
expression for the ratio of the magnitudes of the electric fields,
Eo/Ei, at f = 106 Hz.

In developing the expression, incorporate the good-conductor
approximations for the copper shield.
copper = o,   copper = o )
7
(copper = 5.810 S/m,

EM waves
30

Standing Waves         ☼
Consider normal incidence. In steady-state, the incident and reflected waves interact
to produce, in region 1, a traveling wave and a standing wave.


-j
E1 = Ei + Er                        choose reference phase Ei = Eie                                               2    e-j z
                                         
-j            -j z          j           -j
E1 = Eie        2    e           + e          Ei e        2    e j z
                                                                                  
-j                              j                             -j                        -j
E1 = Eie        2    e-j z + Eie             2    e j z + Eie             2    e-j z - Eie        2    e-j z

EM waves
31

Standing Wave Ratio (SWR)         ☼
E     1+
SWR = max =
Emin   1-

=
SWR - 1
SWR + 1
  = 

The SWR ratio can readily be measured—so too  =  .

For no reflection, SWR=1. As more of the incident wave is reflected, the SWR
ratio grows.

EM waves

```
DOCUMENT INFO
Categories:
Tags:
Stats:
 views: 7 posted: 12/9/2011 language: pages: 31