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PPT No 4 Sound waves; Wave impedance; Reflection - PDF

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PPT No 4 Sound waves; Wave impedance; Reflection - PDF Powered By Docstoc
					            PPT No. 5 E&M Waves

Mechancial waves require media capable of storing
kinetic and potential energies.
Electromagentic waves require media capable of
storing electric (capacitive) and magnetic (inductive)
energies.

LC transmission line
Consider a transmission line having an inductance per
unit length L / l (H/m) and capacitance per unit length
C / l (F/m).
Kirchhoff's voltage theorem
                          L ∂i               ∂V L ∂i
V ( x ) = V ( x + Δx ) + Δ x = V ( x ) + Δ x   + Δx
                          l    ∂t            ∂x l   ∂t
∂V           L ∂i
       =−
 ∂x          l ∂t
Current theorem
                        C    ∂V
i ( x ) = i ( x + Δx ) + Δx
                        l     ∂x
 ∂i        C ∂V
     =−
∂x          l ∂t
∂ 2V          L ∂ 2i C L ∂ 2V
        =−            =
 ∂x 2         l ∂t∂x l l ∂t 2
∂ 2V         1 ∂ 2V
        =
 ∂t 2 C L ∂x 2
           l l
                           1
Wave velocity c =
                          CL
                          l l
In vacuum, L / l = μ0 , C / l = ε 0 . Then
      1
c=            = 3.0 × 108 m/sec
     ε 0 μ0
Impedance
Let V ( x, t ) =V0sin ( kx − ωt ) , i ( x, t ) =i0sin ( kx − ωt )
  ∂V     L ∂i           L
in    =−      . kV0 = − ( −ω )i0
   ∂x    l ∂t           l
V0 L ω     L/l       μ0
  =    =         =      = 377 Ohms in vacuum.
i0 l k    C /l       ε0
Examples:
                          μ0 1 ⎛ b ⎞
Coaxial cable Z =                    (Ω )
                          ε 2π ⎜ a ⎟
                              ln
                                 ⎝ ⎠
                                 μ0 1 ⎛ d ⎞
Parallel wire lines Z =                   ⎟ (Ω )
                                 ε0 π ⎜ a ⎠
                                     ln
                                        ⎝

                                  b
                                         ε,μ

                            2a

                           wire radius a

                            d
           L/l     1C 2 1L 2
V = Zi =        i→     V =        i
           C /l    2 l       2 l
In electromagnetic wave, electric energy = magnetic energy
In mechanical wave, potential energy = kinetic energy


                        C 2
Total energy density = V
                        l
Energy transfer rate = wave velocity × energy density
         C 2 V2
Power = c V =   (W)
         l    Z
Example: A dc volatge source 6V is suddenly connected
to a 50 Ω coaxial cable filled with insulator of permittivity ε = 2ε 0 .
Assuming no reflection, find the current wave.


                 V     6V
The current is i =  =       = 0.12 A. The wave velocity
                 Z 50 Ω
is 1/ εμ0 = c / 2 = 0.71c. The power delivered by the
                   V2
source is P = Vi =    = 0.72 W.
                   Z
Example Discuss how the voltage and current waves develop after the switch is closed.


                                                   50
After the switch is closed, a voltage pulse of         × 10 = 5 V propagtes toward the 20Ω load.
                                                 50+50
                                      20-50
After τ = l / c, reflected voltage is        × 5 = −2.14 V and the line voltage is 5-2.14=2.86 V.
                                      20+50
There is no further reflection at the source, Therefore, the final voltage is 2.86 V and final current is
2.86 V/20 Ω = 0.143 A.

                      10 V
DC theory gives i =        = 0.143 A.
                      70 Ω
Quarter wavelength transformer for impedance matching


To achieve impedance matching between two impedances Z1 and Z 2 , insert Z 3 = Z1 Z 2
quarter wavelength long (or thick). The reflection at the first boundary is
     Z − Z1
Γ1 = 3
     Z 3 + Z1
and that at the second boundary is
      Z − Z3
Γ2 = 2
      Z3 + Z2
                                                             λ          λ
The second reflected wave travels additional distance of         ×2 =       and the total reflection is
                                                             4          2
Z 3 − Z1 Z 2 − Z 3
        −
Z 3 + Z1 Z 3 + Z 2
                                     Z 3 − Z1 Z 2 − Z 3
The condition for no refelction is           −          = 0 or Z 3 = Z1 Z 2
                                     Z 3 + Z1 Z 3 + Z 2
Anti-reflection coating: Coating material should have impedance of 377 × 251 = 307Ω
                                                                1         1 λair
or index of refaction of n = 1.5 = 1.22. Coating thickness is     λcoat =        . If λair = 550 nm
                                                                4         4 1.22
                                       1 550
(green), coating thickness should be    ×     = 112 nm. Coating material like MgO_2 has
                                       4 1.22
the desired permittivity.

    http://physics.usask.ca/%7Ehirose/ep225/emref.htm
Example: For impedance matching between 50 Ω coaxial cable and 300 Ω coaxial cable
can be achieved if a third cable having an impedance of 50 × 300 = 122 Ω and λ /4 long
is inserted.

                                                                         1
Example: A resistive plate with thickness d having a conductiviy σ =        (S/m) is placed at
                                                                        Z0d
a distance λ /4 in front of a conducting plate. Explain how reflection can be avoided.

Wave reflection by a conductor plate is almost complete. At λ /4, the voltage is maximum and
current is minimum → open circuit. Then the incient wave sees an admittance of
      1         1
Y =       ×d =    or impedance of Z 0 and impedance matching is achieved. Such antireflection
    Z0d        Z0
technique has important military application.

                 resistive
                 plate             metal




                             λ/4
Brewster's angle
                                           sin θ
Oblique incidence at angle θ . Snell's law         = ng
                                           sin θ '
Electric field in the incidence plane and magnetic field out of the page.
Continuity of H-field H i + H r = H t
Continuity of tangential components of E field
Ei cos θ − Er cos θ = Et cos θ '
                                     Z0
Ei = Z 0 H i , Er = Z 0 H r , Et =      Ht
                                     ng
                        1                1
( H i − H r ) cos θ =      H t cos θ ' =    ( H i + H r ) cos θ '
                        ng               ng
       tan (θ − θ ' )
Hr =                  Hi
       tan (θ + θ ' )
If θ + θ ' = π / 2, tan (θ + θ ' ) = ∞ and reflection becomes zero.
Brewster's angle θ B = tan −1 ( ng ) . If ng = 1.5, θ =56.3° .
Example: Reflection at air-glass boundary. nglass = 1.5
                              μ0                      μ0   1
Impedance of air Z a =           = 377 Ω. Glass Z g =    =   × 377 = 251 Ω
                              ε0                      ε 1.5
Electric field reflection is
251 − 377
            = −0.2
251 + 377
Energy (power) reflection is ( −0.2 ) = 4%.
                                         2




Example: Brewster's angle
               π
If θ + θ ' =       in refraction and if the electric field is in the incidence plane, reflection can be avoided.
               2




                                                  θ θ
                                                          π/2
                                                     θ'
                  Ei
                                       Er



                         θ    θ
          n=1
           ng
                             θ'

                                       Et




Electric field in the incident plane
Magnetic field parallel to the boundary
                                                 π
In this case, reflection vanishes if θ + θ ' =       or
                                                 2
θ = arctan ( ng ) : Brewster's angle
If ng = 1.5, θ = 56.3
y
    0.20



    0.15



    0.10



    0.05



    0.00
           0      10     20     30      40     50     60     70     80
                                                                   x

           If the electric field is in the incident plane,
              tan (θ − θ ' )
           r=                In the figure r 2 is plotted.
              tan (θ + θ ' )
           The Brewster's angle is at θ = tan −1 ( n ) = 56.3 if n = 1.5.
         http://physics.usask.ca/%7Ehirose/ep225/emwave.htm

Polarization
In plane waves propagating in z direction, E x and H y (e.g., laser light)
Sun and most light sources emit randomly polarized light.
Scattered and reflected light is polarized to some dgree.
Reflection at Brewster's angle → linearly polarized normal to the incident
plane.

Example: Find the intensity of unpolarized light when it goes through
3 polarizer plates at angles 0 , θ , and 90 .
                                                 1
The first plate cuts y component. → intensity      I0
                                                 2
                           1
After the second plate,      I 0 cos2 θ
                           2
                        1
After the third plate,    I 0 cos2 θ sin 2 θ
                        2
                                      1
If θ = 45 , the intensity becomes I 0
                                      8
Doppler shift


Doppler effect in E&M wave is fundamentally different from
that in sound waves.
                       cs − Vo
Sound wave      f '=           f0
                       cs − Vs
                      1− β2                 V
E&M wave f ' =                 f 0 where β = r
                   1 − β cos θ               c
Vr = relative velocity between the source and observer


Example: A distant nebula is receding from the Earth at a velocity 0.4c.
                                       1 − ( 0.4 )
                                                     2

The frequency is Doppler shifted by                      = 0.655 and the wavelength
                                       1 + 0.4
       1
by         = 1.523.
     0.655
Example: Police radar emits 4 GHz microwave and detects reflection
from a moving car. If the car is moving at 150 km/hr toward the radar,
what is the frequency change?

                        1                  V 150000 / 3600
The car detects f ' =       f 0 where β = =                    = 1.4 × 10−7
                      1− β                  c          3 × 108
                             1
The radar detects f '' =            f = (1 + 2 β ) f 0
                         (1 − β )
                                  2 0


Δf = 2 β f 0 = 1120 Hz
  Radiation of E&M waves




Electric field profile of accelerated charge


The charge is accelerated for a short duration after which undergoes
drift motion. The kink in the field line has a transverse field component
which is the radiation field. For radiation of electromagnetic waves,
charge must be accelerated (or decelerated).
Radiation electric field
Charge q is accelerated for a duration Δt after which
it travels at a constant velocity v = aΔt.
Coulomb field
          q             q
EC =             =
       4πε 0 r     4πε 0 ( ct )
               2                2


     ER vt sin θ
From    =
     EC   c Δt
we find the radiation field
        qa
ER =          sin θ
     4πε 0c r
           2
Radiation power
                                                   2
              ER 2   1 ⎛ qa                  ⎞
Poynting flux      =     ⎜             sin θ ⎟
              Z0     Z 0 ⎝ 4πε 0 c 2 r       ⎠
Radiation power is
                   π   ER 2                1 2 q2 a 2
P=r            ∫            2π sin θ dθ =
           2                      3
                                                      (W)
               0       Z0                 4πε 0 3 c 3
where
    π                      4
∫0
        sin 3 θ dθ =
                           3
Example: An electron having an energy of 1 keV undergoes cyclotron motion in
a magnetic field of 2 T. Find the acceleration and radiation power.
Lorentz force = evB
                      2E   2000 × 1.6 × 10−19
The velocity is v =      =             −31
                                              =1.87 × 107 m/sec
                      me      9.1 × 10
Force is evB = 1.6 × 10−19 × 1.87 × 107 × 2=5.98 × 10-12 N
                     F
Acceleration is a =     =6.6 × 1018 m/sec 2
                     me
                           1 2 e2 a 2
Radiation power is P =                = 2.46 W=1.54 × 103 eV/sec
                       4πε 0 3 c  3
Cyclotron
Radiation
(Non-relativistic)



Bending magnet              e-beam    B into page
Synchrotron Radiation                               Radaition beam
(relativistic)
Radiation power increases
as γ4, frequency as γ2
                                     R
                                      orbit curvature raidus




Wiggler
Free electron laser
Wavelength
Spatial period/γ2
Radiation by Antenna
Radiation by antenna is caused by oscillatory acceleration of electrons. If electron
is under oscillatory motion at frequency ω , the acceleration is
a = ωv
The radiation power per one electron is
      1  2 e 2ω 2 v 2
P=
   4πε 0 3 c 3
If there are many electrons in an antenna l (m) long, Il = qv and
      1  2 I 2 l 2ω 2 Z 0
P=                   =    ( kl ) I 2 (W)
                                2

   4πε 0 3 c 3         6π
    Z0
R=     ( kl ) is called radiation resistance.
             2

    6π
The radiation resistance of λ /2 antenna is R = 73.1 Ω.
Other radiation mechanisms

Laser and maser (QM mechanism)
Ionized gas = plasma (QM)
Cherenkov radiation by energetic particles in matter
Bremsstrahlung (braking radiation): collision between unlike charges
Transition radiation due to passage of charges at a boundary
of dielectrics

				
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