Lecture 5: Polarization and Related Antenna Parameters
(Polarization of EM fields – revision. Polarization vector. Antenna
polarization. Polarization loss factor and polarization efficiency.)
1. Introduction and Definition
The polarization of the EM field describes the orientation of its vectors at a
given point and how it varies with time. In other words, it describes the way the
direction and magnitude of the field vectors (usually E) change in time.
Polarization is associated with TEM time-harmonic waves where the H vector
relates to the E vector simply by H= rˆ × E / η .
In antenna theory, we are concerned with the polarization of the field in the
plane orthogonal to the direction of propagation—this is the plane defined by
the vectors of the far field. Remember that the far field is a quasi-TEM field.
The polarization is the locus traced by the extremity of the time-varying field
vector at a fixed observation point.
According to the shape of the trace, three types of polarization exist for
harmonic fields: linear, circular and elliptical. Any polarization can be
represented by two orthogonal linear polarizations, ( Ex , E y ) or ( EH , EV ), the
fields of which may, in general, have different magnitudes and may be out of
phase by an angle δ L .
y
y
E
ω
z
x
(a) linear polarization
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z
y
E
x
(b) circular polarization
z
E
x
(c) elliptical polarization
1
• If δ L = 0 or nπ , then linear polarization results.
ωt = 0
ωt = 0
eH (t ) only
eH (t )
eV (t )
ωt = π
ωt = π
eV (t )
eH (t ) only
eH (t )
Animation: Linear Polarization, δ L = 0 , Ex = E y
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• If δ L = π / 2 (90 ) and | Ex |=| E y | , then circular polarization results.
ω=
t2 ωt1 + π / 2
ωt1
Animation: Clockwise Circular Rotation
• In the most general case, elliptical polarization is defined.
ωt = 0
ωt = π / 2
Animation: Counter-clockwise Elliptical Rotation
It is also true that any type of polarization can be represented by a righthand circular and a left-hand circular polarizations ( EL , ER ). [Animation]
Next, we review the above statements and definitions, and introduce the
new concept of polarization vector.
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2. Field Polarization in Terms of Two Orthogonal Linearly Polarized
Components
The polarization of any field can be represented by a set of two orthogonal
linearly polarized fields. Assume that locally a far-field wave propagates along
the z-axis. The far-zone field vectors have only transverse components. Then,
the set of two orthogonal linearly polarized fields along the x-axis and along
the y-axis, is sufficient to represent any TEMz field. We use this arrangement to
introduce the concept of polarization vector.
The field (time-dependent or phasor vector) is decomposed into two
orthogonal components:
(5.1)
e =e x + e y ⇒ E =E x + E y ,
E x = Ex xˆ
e x Ex cos (ωt − β z ) xˆ
=
(5.2)
⇒
E y = E y e jδ L yˆ.
e y E y cos (ωt − β z + δ L ) yˆ
=
At a fixed position (assume z = 0 ), equation (5.1) can be written as
e(t ) = xˆ ⋅ Ex cos ωt + yˆ ⋅ E y cos(ωt + δ L )
(5.3)
j
δ
L
ˆ
ˆ
⇒ E = x ⋅ Ex + y ⋅ E y e
Case 1: Linear polarization:
=
δ L n=
π , n 0,1, 2,
e(t ) = xˆ ⋅ Ex cos(ωt ) + yˆ ⋅ E y cos(ωt ± nπ )
(5.4)
⇒ E = xˆ ⋅ Ex ± yˆ ⋅ E y
y
Ey
δ L = 2 kπ
⇒t > 0
t
E
y δ= (2k + 1)π
L
E
Ex x
⇒t < 0
Ey
Ex
t
x
 Ey 
t = ± arctan  
 Ex 
(a)
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(b)
4
Case 2: Circular polarization:
π

0,1, 2,
Ex =
Ey =
Em and d L =
±  + nπ  , n =
2


=
e(t ) xˆ Ex cos(ωt ) + yˆ E y cos[ωt ± (π / 2 + nπ )]
(5.5)
⇒ E= Em (xˆ ± jyˆ )
=
E Em (xˆ + jyˆ )
=
E Em (xˆ − jyˆ )
π
δL =
+ + 2nπ
π
δL =
− − 2nπ
2
ωt =
3π
ωt =
y
2
5π
4
z
ωt = π
ωt =
2
ωt =
7π
4
ωt =
ωt = 0
x
3π
4
ωt =
π
ωt =
π
4
2
If +zˆ is the direction of
propagation: counterclockwise
(CCW) or left-hand polarization
3π
4
ωt =
π
2
ωt =
ωt = π
5π
ωt =
4
y
z
3π
2
4
ωt = 0
x
ωt =
ωt =
π
7π
4
If +zˆ is the direction of
propagation: clockwise (CW) or
right-hand polarization
Note that the sense of rotation changes if the direction of propagation
changes. In the example above, if the wave propagates along −zˆ , the plot to the
left, where
=
E Em (xˆ + jyˆ ) , corresponds to a right-hand (RH) wave, while the
plot to the right, where
=
E Em (xˆ − jyˆ ) , corresponds to a left-hand (LH) wave.
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A snapshot of the field vector (at a particular moment of time) along the axis
of propagation is given below for a right-hand circularly polarized (RHCP)
wave. Pick an observing position along the axis of propagation (see the plane
defined by the x and y axes in the plot below) and imagine that the whole
helical trajectory of the tip of the field vector moves along the wave vector .
Are you going to see the vector rotating clockwise or counter-clockwise as you
look along ? (Ans.: Clockwise, which is equivalent to RH sense of rotation.)
[Hayt, Buck, Engineering Electromagnetics, 8th ed., p. 399]
Case 3: Elliptic polarization
The field vector at a given point traces an ellipse as a function of time. This
is the most general type of polarization, obtained for any phase difference
. Mathematically, the linear and the circular
and any ratio
polarizations are special cases of the elliptical polarization. In practice,
however, the term elliptical polarization is used to indicate polarizations other
than linear or circular.
(5.6)
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Show that the trace of the time-dependent vector is an ellipse:
e y (=
t ) E y (cos ωt ⋅ cos δ L − sin ωt ⋅ sin δ L )
 e (t ) 
1−  x 
 Ex 
e (t )
t
and sin ω=
cos ωt = x
Ex
2
2
 e y (t ) 
 ex (t ) 
 ex (t )   e y (t ) 
sin 2 δ L =
2
cos
δ
−
+
L


 E 
 E  E 
 x 
 x  y 
 Ey 
or (dividing both sides by sin 2 δ L ),
(5.7)
1=
x 2 (t ) − 2 x(t ) y (t )cos δ L + y 2 (t ) ,
where
ex (t )
cos ωt
,
=
x(t ) =
Ex sin δ L sin δ L
e y (t )
cos(ωt + δ L )
.
y (t ) =
=
sin δ L
E y sin δ L
Equation (5.7) is the equation of an ellipse centered in the xy plane. It
describes the trajectory of a point of coordinates x(t) and y(t), i.e., normalized
ex (t ) and e y (t ) values, along an ellipse where the point moves with an angular
frequency ω .
As the circular polarization, the elliptical polarization can be right-handed
or left-handed, depending on the relation between the direction of propagation
and the sense of rotation.
2
ey (t )
Ey
mi
t
Ex ex (t )
)
OB
ma
(
(2
xis
a
jor
xis
ra
no
)
A
2O
E
ω
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The parameters of the polarization ellipse are given below. Their derivation
is given in Appendix I.
a) major axis ( 2 × OA )
1 2
Ex + E y2 + Ex4 + E y4 + 2 Ex2 E y2 cos(2δ L ) 
OA =


2
b) minor axis ( 2 × OB )
1 2
OB =
Ex + E y2 − Ex4 + E y4 + 2 Ex2 E y2 cos(2δ L ) 


2
(5.8)
(5.9)
c) tilt angle τ
1
2
 2 Ex E y

δ
cos
L

2 − E2
E
x
y

t = arctan 
(5.10)
A
Note: Eq. (5.10) produces an infinite number of angles, τ = (arctanA)/2
± nπ / 2 , n = 1,2,….Thus, it gives not only the angle which the major
axis of the ellipse forms with the x axis but also the angle of the minor
axis with the x axis. In spherical coordinates, τ is usually specified
with respect to the θ̂ direction
d) axial ratio
=
AR
major axis OA
=
minor axis OB
(5.11)
Note: The linear and circular polarizations as special cases of the elliptical
polarization:
π

• If δ L =
±  + 2nπ  and Ex = E y , then OA
= OB
= E=
E y ; the ellipse
x
2

becomes a circle.
• If δ L = nπ , then OB = 0 and t = ± arctan( E y / Ex ) ; the ellipse collapses
into a line.
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3. Field Polarization in Terms of Two Circularly Polarized Components
The representation of a complex vector field in terms of circularly polarized
components is somewhat less easy to perceive but it is actually more useful in
the calculation of the polarization ellipse parameters. This time, the total field
phasor is represented as the superposition of two circularly polarized waves,
one right-handed and the other left-handed. For the case of a wave propagating
along − z [see Case 2 and Eq. (5.5)],
(5.12)
E= E R ( xˆ + jyˆ ) + E L ( xˆ − jyˆ ) .
Here, ER and EL are in general complex phasors. Assuming a relative phase
difference of δ=
ϕ L − ϕ R , one can write (5.12) as
C
(5.13)
E = eR (xˆ + jyˆ ) + eL e jδC (xˆ − jyˆ ) ,
where eR and eL are real numbers.
The relations between the linear-component and the circular-component
representations of the field polarization are easily found as
(5.14)
E = xˆ ( E R + E L ) + yˆ j ( E R − E L )
((

((
Ex
Ey
E= E R + E L
⇒ x
=
E y j( ER − EL )
(5.15)
=
ER 0.5( E x − jE y )
⇒
=
EL 0.5( E x + jE y ).
(5.16)
4. Polarization Vector and Polarization ratio
The polarization vector is the normalized phasor of the electric field vector.
It is a complex-valued vector of unit magnitude, i.e., ρˆ L ⋅ ρˆ ∗L =
1.
E y jδ L
E
E
ρˆ L =
=
xˆ x + yˆ
e , Em =
Ex2 + E y2
Em
Em
Em
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(5.17)
9
The polarization vector takes the following specific forms:
Case 1: Linear polarization (the polarization vector is real-valued)
Ey
E
, Em =+
Ex2 E y2
ρˆ =±
xˆ x yˆ
Em
Em
where Ex and E y are real numbers.
(5.18)
Case 2: Circular polarization
1
ρˆ L = ( xˆ ± jyˆ ) , Em =2 E x =2 E y
2
(5.19)
The polarization ratio is the ratio of the phasors of the two orthogonal
polarization components. In general, it is a complex number:
rL rL
=
δL
e=
E y E y e jδ L
EV
or =
rL
=
E x
Ex
E H
(5.20)
Point of interest: In the case of circular-component representation, the
polarization ratio is defined as
E R
.
(5.21)
=
rC rC=
e jδC
E L
The circular polarization ratio rC is of particular interest since the axial ratio of
the polarization ellipse AR can be expressed as
r +1
.
(5.22)
AR = C
rC − 1
Besides, its tilt angle with respect to the y (vertical) axis is simply
(5.23)
τV = δ C / 2 .
Comparing (5.10) and (5.23) readily shows the relation between the phase
difference δC of the circular-component representation and the linear
polarization ratio rL = rL e jδ L :
 2r

(5.24)
δ C = arctan  L 2 cos δ L  .
1
−
r


L
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We can calculate rC from the linear polarization ratio rL making use of (5.11)
and (5.22):
=
AR
rC + 1
=
rC − 1
1 + rL2 + 1 + rL4 + 2rL2 cos(2δ L )
.
(5.25)
δ
1+ − 1+
Using (5.24) and (5.25) allows for the switching between the representation of
the wave polarization in terms of linear and circular components.
rL2
rL4
+ 2 rL2 cos(2 L )
5. Antenna Polarization
The polarization of a transmitting antenna is the polarization of its
radiated wave in the far zone.
The polarization of a receiving antenna is the polarization of a plane
wave, incident from a given direction, which, for a given power flux density,
results in maximum available power at the antenna terminals.
By convention, the antenna polarization is defined by the polarization
vector of the wave it transmits. Therefore, the antenna polarization vector is
determined according to the definition of antenna polarization in a transmitting
mode. Notice that the polarization vector of a wave in the coordinate system of
the transmitting antenna is represented by its complex conjugate in the
coordinate system of the receiving antenna:
(5.26)
ρˆ rw = (ρˆ tw )∗ .
The conjugation is without importance for a linearly polarized wave since its
polarization vector is real. It is, however, important in the cases of circularly
and elliptically polarized waves.
This is illustrated in the figure below with a right-hand CP wave. Let the
coordinate triplet ( x1t , x2t , x3t ) represent the coordinate system of the transmitting
antenna while ( x1r , x2r , x3r ) represents that of the receiving antenna. Note that in
antenna theory the plane of polarization is usually given in spherical
coordinates by (xˆ 1, xˆ 2 ) ≡ (θˆ , φˆ ) and the third axis obeys xˆ 1 × xˆ 2 =
xˆ 3 , i.e.,
xˆ 3 = rˆ . Since the transmitting and receiving antennas face each other, their
coordinate systems are oriented so that xˆ t3 = −xˆ 3r (i.e., rˆ r = −rˆ t ). If we align the
axes xˆ 1t and xˆ 1r , then xˆ t2 = −xˆ r2 must hold. This changes the sign in the
imaginary part of the wave polarization vector.
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RHCP wave
x2t
=
ρˆ tw
x1t
x1r
wt = 0
wt = 0
x3t
kˆ = xˆ t3
wt = π / 2
1 t
(xˆ 1 − jxˆ t2 )
2
x2r
x3r
wt = π / 2
=
ρˆ rw
kˆ = −xˆ 3r
1 r
(xˆ 1 + jxˆ r2 )
2
Bearing in mind the definitions of antenna polarization in transmitting and
receiving modes, we conclude that the transmitting-mode polarization vector
of an antenna is the conjugate of its receiving-mode polarization vector.
6. Polarization Loss Factor (Polarization Efficiency)
Generally, the polarization of the receiving antenna is not the same as the
polarization of the incident wave. This is called polarization mismatch.
The polarization loss factor (PLF) characterizes the loss of EM power due
to the polarization mismatch:
(5.27)
PLF
= | ρˆ i ⋅ ρˆ a |2 .
The above definition is based on the representation of the incident field and the
antenna polarization by their polarization vectors in a common coordinate
system, e.g., that of the receiving antenna. If the incident field is
Ei = Emi ρˆ i ,
then the field of the same magnitude that would produce maximum received
power at the antenna terminals is
Ea = Emi ρˆ a .
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[Balanis]
If the antenna is polarization matched, then PLF = 1, and there is no
polarization power loss. If PLF = 0 , then the antenna is incapable of receiving
any signal. Note that
(5.28)
0 ≤ PLF ≤ 1.
The polarization efficiency has the same meaning as that of the PLF.
In a communication link, the PLF has to be expressed by the polarization
vectors of the transmitting and receiving antennas, ρ̂Tx and ρ̂ Rx , respectively.
Both of these are defined in the respective antenna’s coordinate system.
However, these two coordinate systems have their radial unit vectors pointing
in opposite directions, i.e., rˆRx = −rˆTx as illustrated in the figure below.
Therefore, either ρ̂Tx or ρ̂ Rx has to be conjugated when calculating the PLF (it
does not matter which one). For example, if the reference coordinate system is
that of the receiving antenna, then
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= ρˆ ∗Tx ⋅ ρˆ Rx
PLF
The expression PLF
= ρˆ Tx ⋅ ρˆ ∗Rx
2
.
(5.29)
is also correct.
θ̂ Rx
θ̂Tx
r̂Tx
φ̂Tx
2
ρ̂Tx
ρˆ ∗Rx
φ̂Rx
r̂Rx
ρˆ ∗Tx
ρ̂ Rx
Examples
Example 5.1. The electric field of a linearly polarized EM wave is
Ei= xˆ ⋅ Em ( x, y )e − j β z .
It is incident upon a linearly polarized antenna whose polarization is
Ea = (xˆ + yˆ ) ⋅ E (r ,θ ,ϕ ) .
Find the PLF.
2
1
1
PLF =⋅
xˆ
(xˆ + yˆ ) =
2
2
PLF[dB] = 10log10 0.5 = −3 dB
Example 5.2. A transmitting antenna produces a far-zone field, which is
RH circularly polarized. This field impinges upon a receiving antenna,
whose polarization (in transmitting mode) is also RH circular. Determine
the PLF.
Both antennas (the transmitting one and the receiving one) are RH
circularly polarized in transmitting mode. Assume that a transmitting
antenna is located at the center of a spherical coordinate system. The farzone field it would produce is described as
far
Em θˆ ⋅ cos ωt + φˆ ⋅ cos(ωt − π / 2)  .
E=
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This is a RHCP field with respect to the outward radial direction. Its
polarization vector is
θˆ − jφˆ
.
ρˆ =
2
This is exactly the polarization vector of the transmitting antenna.
z
Eϕ
θ
ϕ
r
Eθ
y
x
This same field Ε far is incident upon a receiving antenna, which has the
ˆ a (θˆ a − jφˆ a ) / 2 in its own coordinate system
polarization vector ρ=
(ra ,θ a ,ϕa ) . However, Ε far propagates along −rˆa in the (ra ,θ a ,ϕa )
coordinate system, and, therefore, its polarization vector is
θˆ a + jφˆ a
.
ρˆ i =
2
The PLF is calculated as per (5.27):
| (θˆ a + jφˆ a )(θˆ a − jφˆ a ) |2
2
=
PLF =
| ρˆ i ⋅ ρˆ a | =
1,
4
=
PLF[dB] 10log
=
10 1 0 .
There is no polarization loss.
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Exercise: Show that an antenna of RH circular polarization (in transmitting
mode) cannot receive LH circularly polarized incident wave (or a wave
emitted by a left-circularly polarized antenna).
Appendix I
Find the tilt angle τ , the length of the major axis OA, and the length of the
minor axis OB of the ellipse described by the equation:
2
 e y (t ) 
 ex (t ) 
 ex (t )   e y (t ) 
−
+
δ
sin 2 δ =
2
cos

 .
 E 
 E  E 
E
 x 
 x  y 
 y 
2
(A-1)
ey (t )
Ey
mi
t
(2
Ex ex (t )
)
OB
xi
ra
jo
ma
xis
ra
no
A)
O
s (2
E
ω
Equation (A-1) can be written as
a ⋅ x 2 − b ⋅ xy + c ⋅ y 2 =
1,
(A-2)
where
x = ex (t ) and y = e y (t ) are the coordinates of a point of the ellipse
centered in the xy plane;
1
a= 2 2 ;
Ex sin δ
2cos δ
;
b=
Ex E y sin 2 δ
1
c= 2 2 .
E y sin δ
After dividing both sides of (A-2) by ( xy ) , one obtains
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x
y 1
a −b+c = .
y
x xy
Introducing x=
(A-3)
y e y (t )
, one obtains that
=
x ex (t )
1
x2 = 2
cx − bx + a
1+ x 2
.
x = 2
⇒ ρ x) = + =
(A-4)
cx − bx + a
Here, ρ is the distance from the center of the coordinate system to the point on
the ellipse. We want to know at what values of ξ the maximum and the
minimum of ρ occur ( ξ min , x max ). This will produce the tilt angle τ . We also
want to know the values of ρ max (major axis) and ρ min (minor axis). Then, we
have to solve
d (ρ 2 )
= 0 , or
dξ
2( a − c )
2
(A-5)
ξξ
0 , where x m ≡ x min , x max .
m −
m −1 =
b
(A-5) is solved for the angle τ, which relates to ξmax as
tan
x=
=
t ( y / x )max .
max
(A-6)
Substituting (A-6) in (A-5) yields:
2
 sin ττ

 sin 
−
2
C
0
(A-7)



 −1 =
 cosττ

 cos 
where
E y2 − Ex2
a−c
.
=
C =
2 Ex E y cos δ
b
Multiplying both sides of (A-7) by cos2 τ and re-arranging results in
2
2
cos
0.
ττττ
− sin 
+ 2((
C sin((
⋅ cos =
((
((
2(
x2
y2
x 2 (1 +
2)
cos(2ττ
)
C sin(2 )
Thus, the solution of (A-7) is
tan(2t ) = −1 / C
or
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 2 Ex E y cos δ 
1
π
(A-8)
.
arctan 
;
t
t
=
+
 2 1
2 − E2
2
2
E
x
y


The angles τ1 and τ2 are the angles between the major and minor axes with the x
axis. Substituting τ1 and τ 2 back in ρ (see A-4) yields the expressions for OA
and OB.
t1=
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