EE3321 ELECTROMAGNETIC FIELD THEORY

Report
Week 10
Intrinsic Impedance
Plane Wave Reflection
Reflection, Transmission, Refraction
Polarization



Any medium through which
plane waves propagate have a
property called intrinsic
impedance or electromagnetic
impedance, denoted by η.
The intrinsic impedance is
analogous to the characteristic
impedance of a cable in
transmission line theory.
For a given frequency, the
intrinsic impedance of a
homogeneous material is
given by the ratio of the E field
phasor to the H field phasor.


Consider a plane wave with fields
E = Ei e
jωt
e
–jkz
ax
B = Bi e
jωt
e
–jkz
ay
Recall that these fields are related by
– jk Ei e
jωt
e
–jkz
ay = – jω Bi e
jωt
e
–jkz
ay

Since H = B / μ it follows that
– jk Ei e
jωt
e
–jkz
ay = – jωμ Hi e
kEi = ωμ Hi
η = Ei/Hi = ωμ/k
η = (μ/є)1/2
jωt
e
–jkz
ay


In free space μr = 1 and εr = 1.
Therefore the intrinsic impedance of free
space ηo is given by
ηo = 120 π
which is approximately 377 Ω.

For a region with slight electrical conductivity
(σ > 0, e.g. seawater), the equation becomes

This means that the E and H fields will be out
of phase.

Exercise: What is the intrinsic impedance of a
perfect conductor?


When a plane wave passes
from one dielectric medium (A)
to another medium (B) with
different intrinsic impedance, a
fraction of its energy will be
reflected and the remainder
will be transmitted through the
interface.
In this case, a reflection
coefficient is used to describe
the amplitude of the reflected
part of the wave relative to the
amplitude of the incident wave.
The arrows indicate the
direction of propagation


In particular, the reflection coefficient is the
complex ratio of the electric field phasor of
the reflected wave (E − ) to that of the incident
wave (E + ).
This is typically represented with a Γ (capital
gamma) and can be written as:
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

Let the intrinsic impedances of the first and
second media be η1 and η2, respectively.
At normal incidence the reflection coefficient of
medium 2 relative to 1 is given by
Exercise: A plane wave hits a mirror at normal
incidence. Determine the reflection coefficient.

Incident wave
E i = Ei e
jωt
Hi = Hi e
e
jωt

–jkz
e
ax
–jkz
and η1 = Ei / Hi
ay
Reflected wave
E r = Er e
jωt
Hr = Hr e
e
jωt
jkz
e
ax
jkz
and η1 = - Er /Hr
ay

At the interface, the components of the electric field and
magnetic field parallel to the surface must be continuous.

Consequently the following boundary condition must be
satisfied:

Ei + Er = E t
Ei + ΓEi = Et
Hi + Hr = Ht
Ei /η1 – ΓEi/η1 = Et/η2
This yields
(1 – Γ)/η1 = (1 + Γ)/ η2
Γ = (η2 – η1) / (η2 + η1)

When the electric field Ei is incident upon a perfectly conducting plate
located a z = 0, the reflected field is
Er = ΓEi e
+jωt
e
+jkz
x.
The total field in region 1
E 1 = E i + E r = Ei e

(e
–jkz
+ Γe
+jkz)
x
Since Γ = –1, the expression simplifies to
E 1 = E i + E r = Ei e
E1 = – 2j Ei e

+jωt
+jωt
+jωt
(e
–jkz
–e
sin (kz) x
Taking of the real part of E1 yields
E1 = 2Ei sin (ωt) sin (kz) x
+jkz)
x

Exercise: Derive an expression for the total
electric field if the reflection coefficient is Γ =
1.

Standing Wave in a Microwave Oven

Consider the stationary electric field in a
microwave oven:
E = Eo e


jωt
sin (kz) ax
V/ m
The oven operates at a typical frequency of 2.45
GHz. The width of the oven W is 2.5 λ.
The electric field is zero on the side walls of the
oven (i.e. at z = 0 and z = W).
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What is the width of the oven?
Is this a traveling wave or a standing wave?
What are the boundary conditions in the
oven?
How many hot spots are there?
Where is |E| the strongest? In other words,
where do you put the turkey?

The transmission coefficient τ is used when a plane
wave encounters a change in the intrinsic impedance.

The transmission coefficient describes the amplitude
of the transmitted wave relative to the incident wave.


The transmission coefficient is a measure of how
much of an electromagnetic wave passes through a
surface.
The transmission coefficient τ is defined as the ratio
of the transmitted electric field to the ratio of the
incident electric field.

When the electric field Ei is incident upon a surface located a z =
0, the reflected field is
E t = τ Eo e
+jωt
e
–jkz
x.

At the interface, the component of the electric field parallel to
the surface must be continuous.

Consequently the following boundary condition must be
satisfied:
Ei(0) + Er(0) = Et(0)
Ei + ΓEi = τ Ei
or
1+Γ=τ



When an electromagnetic field moves
from a medium of a given refractive
index n1 into a second medium with
refractive index n2, both reflection and
refraction of the plane wave may
occur.
An incident plane wave PO strikes at
point O the interface between two
media of refractive indexes n1 and n2.
Part of the wave is reflected as wave
OQ and part refracted as ray OS. The
angles that the incident, reflected and
refracted waves make to the normal of
the interface are given as θi, θr and θt,
respectively.



The law of reflection states that the
direction of the incident plane wave
and the direction of the reflected
plane wave make the same angle with
respect to the surface normal.
That is, the angle of incidence equals
the angle of reflection:
θi = θr
Exercise: An electromagnetic plane
wave is reflected by a perfect
conductor. The angle of incidence is
45o. What percentage of the wave
will be reflected at 30o?

Snell's law (also known as
Descartes' law or the law of
refraction) describes the
relationship between the
angles of incidence and
refraction, when referring
an electromagnetic wave,
passing through a
boundary between two
different isotropic media,
such as water and glass.


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
The law says that the ratio of the sines of
the angles of incidence and of refraction
is a constant that depends on the media.
The refraction of an electromagnetic
wave occurs at the interface between two
media of different refractive indices,
with n2 > n1.
Since the velocity is lower in the second
medium (v2 < v1), the angle of refraction
θ2 is less than the angle of incidence θ1.
That is, the ray in the higher-index
medium is closer to the normal.



When light moves from a dense to
a less dense medium, such as
from water to air, Snell's law
cannot be used to calculate the
refracted angle when the resolved
sine value is higher than 1.
At this point, light is reflected in
the incident medium, known as
internal reflection.
Before the ray totally internally
reflects, the light refracts at the
critical angle; it travels directly
along the surface between the
two refractive media, without a
change in phases like in other
forms of optical phenomena.


In order to calculate the critical angle, let θ2 =
90o and solve for θcrit:
When θ1 > θcrit, no refracted ray appears, and
the incident ray undergoes total internal
reflection from the interface medium.

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The index of refraction of water at optical
wavelengths is approximately 1.33.
Suppose that a swimmer is underwater
looking up at the water-air interface.
Determine the minimum angle at which she
starts seeing the bottom of the pool reflected
on the interface.
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Polarization is a property of
waves that describes the
orientation of their oscillations.
A polarizing filter, such as a
pair of polarizing sunglasses,
can be used to observe this
effect by rotating the filter
while looking through it at the
reflection off of a distant
horizontal surface.
At certain rotation angles, the
reflected light will be reduced
or eliminated.

For a simple harmonic wave, where the amplitude of the electric
vector varies in a sinusoidal manner in time, the two components
have exactly the same frequency:
Ex = Ex exp [ jθx ] exp [ j(ωt – kzz) ] ax
Ey = Ey exp [ jθy ] exp [ j(ωt – kzz) ] ay

The two components may not have the same amplitude:
Ex ≠ Ey

The two components may not have the same phase, that is they
may not reach their maxima and minima at the same time:
θx ≠ θy



In the figure below, the E field
is propagating in the + z
direction.
The field has two components
Ex and Ey.
The components are both in
phase and orthogonal
(perpendicular) to each other.


In the case of circular
polarization, the two
orthogonal components Ex
and Ey have exactly the same
amplitude and are exactly
ninety degrees out of phase.
In this case one component is
zero when the other
component is at maximum or
minimum amplitude.

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The direction the field rotates in,
depends on which of the two
phase relationships exists.
Depending on which way the
electric vector rotates these cases
are called
◦ right-hand circular polarization and
◦ left-hand circular polarization.
In electrical engineering, the
rotation is defined as seen from
the source, such as from a
transmitting antenna.


More generally, the two
components are not in phase
and do not have the same
amplitude.
In this case the polarization is
called elliptical polarization
because the electric vector
traces out an ellipse in the
plane (the polarization
ellipse).


Read book sections 7-1, 7-2, 7-3, 8-1, 8-2
Solve end-of-chapter problems 7.1, 7.5,
7.10, 8.1, 8.17

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