Download Wave Reflection and Transmission – Oblique Incidence

ECE 3300 Wave Reflection and Transmission – Oblique Incidence
Wave Reflection and Transmission – Oblique Incidence
A plane wave (shown by the solid lines) is propagating so that it hits a boundary at
oblique incidence θ (anything other than normal incidence, when θ would be zero). The
direction of propagation
of the plane wave is
shown by the light arrow.
The Plane of incidence is
defined as the plane that
includes the DOP and the
vector normal to surface.
The plane of incidence is
the paper.
There are two possible polarizations for the electric field, as shown above.
Parallel Polarization: E field is parallel to Plane of incidence (the paper)
Also called Transverse Magnetic (magnetic field is perpendicular
(transverse) to plane of incidence)
Perpendicular Polarization: E field is perpendicular to Plane of incidence (the paper)
Also called Transverse Electric (electric field is perpendicular (transverse)
to plane of incidence)
This lecture is about the perpendicular polarization case (E is y-polarized).
We have three waves of interest (incident, reflected, transmitted)
The angles can be found this way:
θi = θr
k1 sinr = k2 sint (Snell’s law, derived later)
For each wave, we have to define three values:
The polarization vector represents the orientation of the E vector. In the case of
perpendicular polarization, all of the waves have y polarization (out of the paper).
The propagation constants ki = kr and kt are found from…
The direction of propagation for all three waves (this is just geometry):
xi ,r ,t  Direction of propagation
xi  x sin  i  z cos  i
xr  x sin  r  z cos  r
xt  x sin  t  z cos  t
Then the waves are specified as:
Ei = |Ei| e – j ki xi y
Er = |Er| e – j kr xr y
Et = |Et| e – j kt xt y
Now lets worry about themagnetic fields
The magnetic fields are always perpendicular to both E and the direction of propagation.
Hi = |Hi| e – j ki xi yi
Hr = |Hr| e – j kr xr yr
Ht = |Ht| e – j kt xt yt
Their polarization vectors are given by (geometry, see figure above):
yˆ i   xˆ cos  i  zˆ sin  i
yˆ r  xˆ cos  r  zˆ sin  r
yˆ t   xˆ cos  t  zˆ sin  t
Their magnitudes are given by: |H| = |E| / η (where ηi = ηr and ηt are defined in your text)
Apply Boundary Conditions:
We are now going to apply the boundary conditions to the TOTAL fields in region 1 and
region 2.
Total field in region 1: sum of incident and reflected fields
Total field in region 2: transmitted field only.
From boundary conditions, we know that
Tangential E fields = across boundary
For the perpendicular polarization case we are doing, all E fields (y-components) are
tangential to the boundary
~i
~r
~t
E y  E y  E y
at z  0
Ei y e  jk1 x sin  i  Er y e  jk1 x sin  r  Et y e  jk2 x sin  t
From boundary conditions, we know that the
Tangential H fields = across boundary (ONLY if there is no current distribution, which is
the case here)
The x-component of H fields is tangential to this boundary for perpendicular polarization.
~i
~r
~t
H x  H x  H x

E i y
1
cos i e
 jk1 x sin i
at z  0

E r y
1
cos r e
 jk1 x sin r

E t y
2
cos t e  jk2 x sin t
The only way both the E and H boundary conditions can be met is if
k1 sini = k1 sinr = k2 sint
This is called PHASE MATCHING
This gives us a derivation of Snell’s Law:
From the first set of equalities: k1 sini = k1 sinr we get i =r
From the second set of equalities: k1 sinr = k2 sint we can calculate t
We can also calculate the Reflection and Transmission Coefficients. These are used to
find the reflected and transmitted electric field magnitudes (and from there, the reflected
and transmitted magnetic field magnitudes).
E r 0  2 cos  i  1 cos  t
  i 
E  0  2 cos  i  1 cos  t
E t 0
2 2 cos  i
  i 
E  0  2 cos  i  1 cos  t
   1  
Now, how do we use this:
Typical question:
Given a 10 V/m electric field incident at 30 degrees to an interface between air (region 1)
and dionized water (er = 80, sig = 0 S/m), find the electric and magnetic fields in each
region.
Steps
1)
2)
3)
4)
Find η and k for each region.
Find θ for each region from Snell’s law
Find the reflection and transmission coefficients.
Find the electric field magnitudes from the reflection and transmission
coefficients.
5) Find the magnetic field magnitudes from the electric field magnitudes and
characteristic impedances.
6) Write the equations for incident, reflected, and transmitted fields.