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

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```					            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

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.

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.

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).
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
2
ER 2   1 ⎛ qa                  ⎞
Poynting flux      =     ⎜             sin θ ⎟
Z0     Z 0 ⎝ 4πε 0 c 2 r       ⎠
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
(Non-relativistic)

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

Wiggler
Free electron laser
Wavelength
Spatial period/γ2
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.

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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