wave plane reflections
Maxwell, time harmonic, transverse-to-z
collecting all the terms,
assuming time harmonic solutions using Ohms law assuming there is no component of either E or H in the z direction
Maxwells equations reduce to
E y z
= j H x
H y z
= ( + j ) Ex
E y x
Ex =0 y
Ex = j H y z
things to notice
H x = ( + j ) E y z
H y x
i D = v i B = 0
H x =0 y
J =E D= E
Ey is connected to Hx via d/dz and Ex is connected to Hy via d/dz and Hy is connected to Ex via d/dz and Hx is connected to Ey via d/dz and Ey and Ex are connected via d/dx and d/dy Hy and Hx are connected via d/dx and d/dy
1
r o
B= H
r o
EE 325, Dept. of ECE, Univ. of Texas at Austin
Dean P. Neikirk 2004, last update April 27, 2004
plane wave reflections
Uniform plane wave solution to Maxwells equations
the complete, time harmonic solution is
E plane = Exo e jt z x
wave
H plane = ( Exo ) e jt z y
wave
=
j ( + j ) = + j
2
= j ( )
d d = vg = d d
Ps ,ave =
1
is called the complex propagation constant
=
vp =
direction of propagation
Eo H o
k = j EH EH
k = j ( + j )
2
1 Re ES H S 2
(
)
x
E plane = Exo e
wave
j ( t k z )
Dean P. Neikirk 2004, last update April 27, 2004
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Plane waves and boundaries
we have found the TEM traveling wave solution to Maxwells equations that would work in a world that is made on one and only one material what would happen if there were a simple dielectric interface, i.e., half the world is filled with 1, and half is 2 ? we must still satisfy boundary conditions at the interface between the materials medium 1 in anticipation of what we x need to solve this problem, r1 lets assume now that as r1 a result of the incident wave there will be a transmitted 1 medium 2 wave and a reflected wave r2 r2 incident = 1 = j1 ( 1 + j1 ) Eincident 2 kin
kincident = j 1 z
Eincident = Ex+1o e 1z x
H incident
y z
H incident = Ex+1o 1 e 1z y
Dean P. Neikirk 2004, last update April 27, 2004
(
)
1 = j (1 1 )
3
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Plane waves and boundaries
as a result of the incident wave there will be a transmitted wave and a reflected wave
incident = 1 =
j1 ( 1 + j1 )
transmit = 2 =
j2 ( 2 + j 2 )
Etransmit = Ex+2 o e 2 z x
+ H tramsit = Ex 2o 2 e 2 z y
Eincident = Ex+1o e 1z x H incident = Ex+1o 1 e 1z y
(
)
(
)
Eincident
kin
H incident
x
Etransmit
ktransmit
reflect = 1 =
j1 ( 1 + j1 )
medium 1 r1 r1 1 y
H transmit
Ereflect = Ex1o e + 1z x H reflect = Ex1o 1 e+ 1z y
Dean P. Neikirk 2004, last update April 27, 2004
H reflect
kreflect
4
(
)
Ereflect
medium 2 r2 r2 2
z
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Fields at the interface
the total fields at the interface between the two materials (i.e., at z=0) are
infinitesimally to the left of the interface
+ Eleft = Ex1o + Ex1o
H left = Ex+1o 1 + Ex1o 1 H right = Ex+2 o 2
(
) ( )
)
infinitesimally to the right of the interface
Eright = Ex+2 o
(
since the fields are tangential to the interface, and were assuming there is no surface current in this problem, the fields must be CONTINUOUS across the interface
Eleft = Eright
+ Ex+1o + Ex1o = Ex 2 o
H left = H right
(E
+ x1o
1 ) + ( Ex1o 1 ) = ( Ex+2o 2 )
so we have two unknowns, Ereflect (Ex1o-) and Etransmit (Ex2o+)
Dean P. Neikirk 2004, last update April 27, 2004 5 EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Transmitted and reflected fields
two unknowns, two equations
+ Ex+1o + Ex1o = Ex 2 o
medium 1
medium 2
Ex+1o Ex1o
2 x1o
(E
+ x1o
+ x1o
1 ) + ( Ex1o 1 ) = ( Ex+2o 2 )
x1o 1 + x1o x1o
Ex+2 o
( E ) + ( E ) = ( E + E ) (E ) (E ) = (E ) + (E
1 + x1o 1 + x1o 2 x1o 2
1 )
+ (1 1 ) (1 2 ) Ex1o = (1 2 ) + (1 1 ) Ex1o
E
x1o
(1 1 ) (1 2 ) E + = 2 1 E + = x1o 1 2 ) + (1 1 ) 2 + 1 x1o (
j ( + j ) =
6
= j = j
Dean P. Neikirk 2004, last update April 27, 2004
j ( + j )
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Reflection coefficient
E
x1o
(1 1 ) (1 2 ) E + = 2 1 E + = x1o 1 2 ) + (1 1 ) 2 + 1 x1o (
medium 1
medium 2
we now have simple relation that gives the ratio of the reflected electric field to the incident electric field the reflection coefficient is
Ex+1o Ex1o
Ex+2 o
E
x1o
= E E
+ x1o
Ex1o 2 1 E = + = Ex1o 2 + 1
=
j ( + j )
for our assumed coordinate system the sign of will tells us which way the reflected electric field points
Eincident pointed in the +x^ direction if is positive, then Ereflect also points in the +x^ direction if is negative, then Ereflect points in the -x^ direction
Dean P. Neikirk 2004, last update April 27, 2004 7 EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Transmission coefficient
recall we had the equation from continuity of total tangential electric field at the interface, and we also have the reflection coefficient, so
+ Ex+1o + Ex1o = Ex 2 o
Ex1o = E Ex+1o
E
+ x 2o
medium 1
medium 2
E
+ x1o
+ (E E
+ x1o
)=E
+ x 2o
= (1 + E ) E
Ex+1o
+ x1o
Ex1o
Ex+2 o
we define the transmission coefficient to be the ratio of the transmitted electric field to the incident electric field
Ex+2 o 2 2 = E = 1+ E = 1+ 2 1 = + 2 + 1 2 + 1 Ex1o
=
Dean P. Neikirk 2004, update last April 27, 2004 8
j ( + j )
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Magnetic fields
recall we had the simple relation between the electric and magnetic fields
+ H o ,incident = H y1o y = ( Ex+1o 1 ) y
H o ,reflect = H y1o y = Ex1o 1 y
(
)
+ + H tramsit = H y 2 o y = ( H y1o + H y1o ) y
so we can also define a magnetic field reflection coefficient
H =
H y1o
H
+ y1o
( = (E
Ex1o 1 + x1o 1
)=E E )
))
x1o + x1o
= E
and the transmitted magnetic field is
+ + + + H y 2 o = H y1o + H y1o = H y1o 1 + H y1o H y1o
( (
H
+ + = H y1o (1 + H ) = H y1o (1 E )
H =
+ H y 2o + y1o
= 1+ H = 1 E
9 EE 325, Dept. of ECE, Univ. of Texas at Austin
Dean P. Neikirk 2004, last update April 27, 2004
plane wave reflections
Power flow: reflected power
Poynting vectors at the interface (z = 0)
2 1 * 1 1 1 + + Ex1o 1 y = Ex+1o Re Pin ,ave = Re Ein H in = Re Ex1o x z 2 2 2 1 * * 1 1 Ex1o 1 y Preflect ,ave = Re Erefl H refl = Re Ex1o x 2 2
(
)
(
)
(
)
(
) (
)
1 1 2 1 1 + 2 = Ex1o Re ( z ) = E Ex1o Re (z) 2 2 1 * 1 * 1 1 2 + 2 = E Ex1o Re (z ) 2 1 *
so the ratio of the incident power to the reflected power is
Preflect ,ave Pin ,ave
Dean P. Neikirk 2004, last update April 27, 2004
= E
10
2
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Power flow: transmitted power
Poynting vectors at the interface (z = 0) Pin ,ave =
1 + 2 1 Ex1o Re z 2 1 *
Ptrans ,ave =
* 1 1 + Re Etrans H trans = Re Ex 2 o x Ex+2 o 2 y 2 2 1 1 + 2 1 1 + 2 = Ex 2 o Re ( z ) = E Ex1o Re ( z) 2 2 2 * 2 *
(
)
(
) (
)
1 1 2 + 2 = E Ex1o Re (z) 2 2 *
so the ratio of transmitted power to incident power is
Ptrans ,ave Pin ,ave
Dean P. Neikirk 2004, last update April 27, 2004
= E
2
Re (1 2 *) Re (1 1 *)
11
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Example: two lossless dielectrics
assume the two materials are lossless, = 0, = o so here medium 1
medium 2 =r2o
=
j ( j ) = j o r o
=r1o
Ex+1o Ex1o
= o r o
Ex1o E = + = Ex1o
limits:
r1 << r2 r1 >> r2 ~ -1, ~ 0 ~ +1, ~ 2
Ex+2 o
o r 2 o o r1 o 1 r 2 1 r1 r1 r 2 = = o r 2 o + o r1 o r1 + r 2 1 r 2 + 1 r1
r1 r 2 2 r1 E = 1+ = r1 + r 2 r1 + r 2
1 < E < 1
0 <E < 2
12 EE 325, Dept. of ECE, Univ. of Texas at Austin
Dean P. Neikirk 2004, last update April 27, 2004
plane wave reflections
Example: two lossless dielectrics
assume the two materials are lossless, = 0, = o lets assume that r1 < r2
1 = o r1 o > o r 2 o = 1
E =
1 r 2 1 r1
1 r 2 + 1 r1
r1 r 2 = <0 r1 + r 2
E =
Ex+1o H
+ y1o
2 r1
medium 1 =r1o
medium 2 =r2o
r1 + r 2
H y1o
>0
H y1o
Ex+2 o = Ex+1o H
+ y 2o
H =
H
+ y1o
= E > 0
+ + Ex1o = Ex1o = Ex1o
H =
+ H y 2o
H
+ y1o
= 1+ H > 0
Dean P. Neikirk 2004, last update April 27, 2004
13
EE 325, Dept. of ECE, Univ. of Texas at Austin
plane wave reflections
Example: two lossless dielectrics, power flow
assume two lossless materials, = 0, r1 , r2 , = o
E =
r1 r 2 r1 + r 2
E =
2 r1
r1 + r 2
>0
1 = o r1 o
2
2 ...