Biomedical Engineering Reference
In-Depth Information
By combining (3.20) and (3.23), we find the total transmission and
reflection coe
cients forTE excitation:
T
ee
=
T
ee
+
T
ee
=
1
−
jk
0
sec
θα
e
yy
mm
zz
+
sin
θ
tan
θα
2
d
x
d
y
T
me
=
T
me
+
T
me
=−
jk
0
α
e
xy
e
xz
+
sin
θα
2
d
x
d
y
(3.24)
R
ee
=
R
ee
+
R
ee
=−
jk
0
sec
e
yy
mm
zz
θα
+
sin
θ
tan
θα
2
d
x
d
y
jk
0
α
.
e
xy
+
θα
e
xz
sin
R
e
me
+
R
me
=−
R
me
=
2
d
x
d
y
Also, in this case,
T
ee(mm)
=
1
+
R
ee(mm)
,
T
em(me)
=
R
em(me)
,as
expected.
3.2.3
Circularly Polarized Plane Wave Incidence
As we are interested in the overall chiral response of the metasur-
face, we can now transform the reflection and transmission tensors
obtained in the previous subsections from a linear into a circular
basis, assuming circularly polarized inputs. This is easily obtained
by considering the following transformation:
1
2
(
T
ee
−
jT
em
+
jT
me
+
T
mm
)
T
LL
=
1
2
(
T
LR
=
−
T
ee
−
jT
em
−
jT
me
+
T
mm
)
(3.25)
1
2
(
T
RL
=
−
T
ee
+
jT
em
+
jT
me
+
T
mm
)
1
2
(
T
ee
+
jT
em
−
jT
me
+
T
mm
).
which constructs the transmission matrix for circularly polarized
inputs
T
RR
=
T
LL
T
LR
T
RL
T
RR
,
T
cp
=
(3.26)
where the second letter in the subscript of each element refers to
the polarization of excitation (LCP or RCP) and the first one to the
transmitted polarization. We can express the elements of (3.26) in
