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 /  . This is why it suffices to know
the polarization of the E vector only.
In antenna theory, we are concerned with the polarization of the field in the
plane orthogonal to the direction of propagation (the polarization plane)—this
is the plane defined by the far-zone vectors E and H. 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 linearly polarized fields, ( xˆ E x , yˆ E y ) or ( E H , EV ),
which, in general, may have different magnitudes and phases. What matters is
their phase difference  L =  y −  x .
y
y
E

z
x
(a) linear polarization
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z
y
E
x
(b) circular polarization
z
E
x
(c) elliptical polarization
1
• If  L = n (n is integer), the field is linearly polarized, i.e., the locus
traced by the tip of the E vector as time flows is a line since
E(t ) = (xˆ mx  yˆ m y )cos(t ) .
(5.1)
The magnitude of the total field is Em = mx2 + m2y .
t = 0
t = 0
Ex (t )
Ex (t ) only
E y (t )
t = 
t = 
E y (t )
Ex (t ) only
Ex (t )
Animation: Linear Polarization,  L = 0 , mx = m y
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• If  L =  / 2 (90 ) and mx = m y , then the field is circularly polarized.
This is evident from the fact that the magnitude of the E-field vector
below remains the same at all times:
E(t ) = m  xˆ cos(t ) + yˆ cos(t   / 2)  = m  xˆ cos(t )  yˆ sin(t )  (5.2)
indicating that the tip of the E-field vector traces a circle as time flows.
t2 = t1 +  / 2
t1
Animation: Clockwise Circular Rotation
• In the most general case, the polarization is elliptical.
t = 0
t =  / 2
Animation: Counter-clockwise Elliptical Rotation
Alternatively, any type of polarization can be represented by a right-hand
circular and a left-hand circular polarizations ( E LH , E RH ). [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 the superposition 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:
e(t ) = e x (t ) + e y (t )  E = E x + E y ,
(5.3)
e x (t ) = Ex cos (t −  z ) xˆ
E x = Ex e − j  z xˆ
(5.4)

e y (t ) = E y cos (t −  z +  L ) yˆ
E y = E y e − j  z e j L yˆ .
At a fixed position (assume z = 0 ), equation (5.3) can be written as
e(t ) = xˆ  Ex cos t + yˆ  E y cos(t +  L )
(5.5)
j

L
ˆ
ˆ
 E = x  Ex + y  E y e
The x-component is always assumed to have a zero phase, setting the reference
for the phase of the y-component. As a result, the x-component is always real
positive whereas the y-component may be complex.
Case 1: Linear polarization:  L = n , n = 0,1,2,
e(t ) = xˆ  Ex cos(t ) + yˆ  E y cos(t  n )
(5.6)
 E = xˆ  Ex  yˆ  E y
Note that the phasor vector of a linearly polarized wave is real-valued.
y
Ey
 L = 2 k
  0

E
y  = (2k + 1)
L
E
  0
Ey
Ex

Ex x
x
 Ey 

 Ex 
 =  arctan 
(a)
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(b)
4
Case 2: Circular polarization:


Ex = E y = Em and  L =   + n  , n = 0,1,2,
2

e(t ) = xˆ Ex cos(t ) + yˆ E y cos[t  ( / 2 + n )]
(5.7)
 E = Em (xˆ  jyˆ )
Note that the phasor of a circularly polarized wave is complex with the xcomponent being real-positive whereas the y-component is purely imaginary.
E = Em (xˆ + jyˆ )
E = Em (xˆ − jyˆ )


 L = − − 2n
L = +
t =
2
3
t =
y
2
5
4
z
t = 
t =
+ 2n
2
t =
3
t =
4
7
4
t = 0
x
3
4
t =

t =

4
2
If +zˆ is the direction of propagation:
counterclockwise (CCW) or lefthand (LH) polarization
t =

2
t =
t = 
t =
y
z
5
4

4
t = 0
x
t =
7
4
3
2
If +zˆ is the direction of propagation:
clockwise (CW) or right-hand (RH)
polarization
t =
The sense of rotation, clockwise or counterclockwise, depends on the
direction into which the observer looks. To avoid this ambiguity, in antenna
engineering, the direction into which the observer looks must coincide with the
direction of wave propagation. In the plot on the left, where E( z ) = Em (xˆ + jyˆ ) ,
if the wave propagates along −zˆ (into the plot), we have a right-hand (RH or
CW) wave. In the plot on the right, where E( z ) = Em (xˆ − jyˆ ) , if the wave
propagates along −zˆ (into the plot), we have a left-hand (LH or CCW) wave.
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Vice versa, if the wave propagates along +zˆ , then the left plot shows a LH (or
CCW) wave, whereas the right plot shows a RH (or CW) wave.
The terms RH and LH are associated with the use of the right hand or the
left hand to relate the direction of propagation to the direction of rotation of the
E-field vector. The thumb of the hand has to be aligned with the direction of
propagation. With this alignment, one has to determine whether it is the fingers
of the right hand or those of the left hand that show properly the direction of
rotation of the E-field.
There is a simple rule of thumb that can be used to tell whether a wave has
RH or LH sense of rotation. If the sign in the propagation factor of the wave
matches the sign of the y-component in the field pahsor, then the wave is
RHCP. If the signs do not match, the wave is LHCP. Assume, for example, a
wave travelling along + z . The propagation factor is then e − j  z . The phasor
E( z ) = Em (xˆ + jyˆ )  e− j  z
(5.8)
represents an LHCP wave. This is the case in the left plot on the previous page.
On the contrary, the phasor
E( z ) = Em (xˆ − jyˆ )  e − j  z
(5.9)
represents an RHCP wave. This is the case in the right plot. For a wave
travelling along − z , the propagation factor is e + j  z and so
E( z ) = Em (xˆ + jyˆ )  e+ j  z
(5.10)
is an RHCP wave whereas
E( z ) = Em (xˆ − jyˆ )  e + j  z
(5.11)
is LHCP.
A snapshot of the field vector along the axis of propagation is given below
for an 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 k . Are you going to see the vector rotating clockwise or counterclockwise as you look along k ? (Ans.: Clockwise, which is equivalent to RH
sense of rotation.)
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[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 
and any ratio ( Ex / E y ) . Mathematically, the linear and the circular
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.
e(t ) = xˆ Ex cos(t ) + yˆ E y cos(t +  L )
(5.12)
j

L
ˆ
ˆ
 E = xEx + yE y e
Show that the trace of the time-dependent vector is an ellipse:
e y (t ) = E y (cos t  cos  L − sin t  sin  L )
e (t )
 e (t ) 
cos t = x
and sin t = 1 −  x 
Ex
 Ex 
2
2
 e (t ) 
 e (t ) 
 e (t )   e (t ) 
sin 2  L =  x  − 2  x   y  cos  L +  y 
 Ex 
 Ex   E y 
 Ey 
or (dividing both sides by sin 2  L ),
(5.13)
1 = x 2 (t ) − 2 x(t ) y(t )cos  L + y 2 (t ) ,
where
e y (t )
ex (t )
cos(t )
cos(t +  L )
=
and y (t ) =
.
x(t ) =
=
E y sin  L
sin  L
Ex sin  L
sin  L
2
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Equation (5.13) 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  .
Similarly to the circular polarization, the elliptical polarization can be righthanded or left-handed, depending on the relation between the direction of
propagation and the sense of rotation.
ey (t )
Ey
no
mi
)
Ex ex (t )
O
(2
(

xis
ra
B)
xis
ra
o
j
ma
A
2O
E

The parameters of the polarization ellipse are given below. Their derivation is
given in Appendix I.
a) major axis ( 2  OA )
1 2
OA =
Ex + E y2 + Ex4 + E y4 + 2 Ex2 E y2 cos(2 L ) 


2
b) minor axis ( 2  OB )
1 2
OB =
Ex + E y2 − Ex4 + E y4 + 2 Ex2 E y2 cos(2 L ) 


2
(5.14)
(5.15)
c) tilt angle 
 
 2E E
1
 = arctan  2 x y2 cos  L  
2
 Ex − E y
 2
(5.16)
A
Note: Eq. (5.16) produces an infinite number of angles, τ = (arctanA)/2
 n / 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
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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
(5.17)
Note: The linear and circular polarizations as special cases of the elliptical
polarization:
• If  L =  ( ( / 2) + 2n ) and E x = E y , then OA = OB = Ex = E y ; the
ellipse becomes a circle.
• If  L = n , then OB = 0 and  =  arctan( E y / E x ) ; the ellipse collapses
into a line.
3. Field Polarization in Terms of Two Circularly Polarized Components
The representation of a complex vector field as a superposition of circularly
polarized components is somewhat less intuitive but it is actually more
convenient 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. (5.7)],
E = ERH (xˆ + jyˆ ) + ELH (xˆ − jyˆ ) .
(5.18)
Here, ERH and ELH are, in general, complex phasors. Assuming a relative phase
difference of  C =  R −  L , one can write (5.18) as
E = eR e jC ( xˆ + jyˆ ) + eL ( xˆ − jyˆ ) ,
(5.19)
where eR and eL are real numbers.
The relations between the linear-component and the circular-component
decompositions are easily found by re-arranging (5.18) as
E = xˆ ( ERH + ELH ) + yˆ j ( ERH − ELH )
(5.20)
Ex


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Ex = ERH + ELH
E y = j ( ERH − ELH )
ERH = 0.5( Ex − jE y )
ELH = 0.5( Ex + jE y ).
Ey
(5.21)
(5.22)
9
4. Polarization Vector and Polarization Ratio of a Plane Wave
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 j L
E
E
(5.23)
= xˆ x + yˆ
e , Em = E  E
Em
Em
Em
The expression in (5.23) assumes a wave decomposition into linearly polarized
(x and y) components, thereby the subscript L. Polarization vector in terms of
RHCP and LHCP components is also used. The polarization vector defined in
(5.23) takes the following specific forms in the cases of linearly, circularly and
elliptically polarized waves.
ρˆ L =
Case 1: Linear polarization (the polarization vector is real-valued)
Ey
E
ρˆ = xˆ x  yˆ
,
Em
Em
where E x and E y are real.
(5.24)
Case 2: Circular polarization (the polarization vector is complex-valued)
1
ρˆ L =
(5.25)
( xˆ  jyˆ ) , Em = 2E x = 2E y
2
The polarization ratio is the ratio of the phasors of the two orthogonal
polarization components. In general, it is a complex number:
rL = rL
e L
E y E y e j L
=
=
Ex
Ex
(5.26)
Point of interest: In the case of circular-component representation, the
polarization ratio is defined as
E
rC = rC e jC = RH .
(5.27)
ELH
The circular polarization ratio rC is of particular interest since the axial ratio of
the polarization ellipse AR can be expressed very simply as
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AR =
rC + 1
.
rC − 1
Besides, its tilt angle with respect to the y (vertical) axis is simply
V =  C / 2 + n , n = 0, 1,... .
(5.28)
(5.29)
Comparing (5.16) and (5.29) readily shows the relation between the phase
difference δC of the circular-component representation and the linear
polarization ratio rL = rLe j L :
 2r

(5.30)
 C = arctan  L 2  cos  L  .
1
−
r


L
We can calculate rC from the linear polarization ratio rL = rL e j L by making
use of (5.14), (5.15), (5.17) and (5.28):
1 + rL2 + 1 + rL4 + 2rL2 cos(2 L )
rC + 1
.
(5.31)
AR =
=
rC − 1
1 + rL2 − 1 + rL4 + 2rL2 cos(2 L )
Using (5.30) and (5.31) allows for the switching between the representation of
the wave polarization in terms of linear and circular components.
5. Antenna Polarization
In general terms, 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 an incident plane wave, which, for a
given power flux density, results in maximum available power at the antenna
terminals.
Mathematically, the antenna’s polarization is defined by the polarization
vector of the wave it transmits and in the coordinate system where the
antenna is at the origin. Here, it is important to appreciate the fact that the
polarization vector ρˆ tw of a wave in the coordinate system of the transmitting
antenna (the wave moves away from the origin along +rˆ t ) is the conjugate of
this wave’s polarization vector ρˆ rw in the coordinate system of receiving
antenna, in the coordinate system of which the wave moves toward the origin
along −rˆ r ):
ρˆ rw = (ρˆ tw ) .
(5.32)
The conjugation is without importance for a linearly polarized wave since its
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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 RHCP 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. In antenna analysis,
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 , and so it is xˆ 3  rˆ . Since
the transmitting and receiving antennas face each other, their coordinate
systems are oriented so that xˆ t3 = −xˆ 3r ( rˆ r = −rˆ t ). If we align the axes xˆ 1t and xˆ 1r
( θˆ t = θˆ r ), then xˆ t2 = − xˆ r2 ( φˆ t = −φˆ r ) must hold in order to have xˆ t3 = −xˆ 3r
( rˆ r = −rˆ t ). This changes the sign of the x̂2 ( θ̂ ) field vector component. This
leads to a change of sign in the imaginary part of the wave’s polarization
vector.
RHCP wave
x2t
x1t
x1r
t = 0
t = 0
kˆ =
t =  / 2
ρˆ tw =
x3t
xˆ t3
1 t
(xˆ 1 − jxˆ t2 )
2
x2r
x3r
t =  / 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 in a common coordinate system 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. The PLF is defined so that it attains a value of 1 (or
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100%, or 0 dB) if there is no polarization mismatch, i.e., the antenna receives
the maximum possible power for the given incident power density of the wave.
A PLF equal to 0 ( − dB) indicates complete polarization mismatch and
inability to capture power from the incident wave. Thus,
(5.33)
0  PLF  1.
Note that the polarization loss has nothing to do with dissipation. It can be
viewed as a “missed opportunity” to capture as much power from the incident
wave as possible. The polarization efficiency has the same meaning as PLF.
Let us denote the polarization vector of a wave incident upon a receiving
antenna as ρˆ w . In the coordinate system where the receiving antenna is at the
origin, this vector describes a wave propagating along −rˆ . Let the polarization
vector of the wave that the receiving antenna would produce if it were to
operate in transmitting mode is ρ̂ A . Remember that, in the coordinate system
where the antenna is at the origin, ρ̂ A describes a wave propagating along +rˆ .
With these polarization vectors, the PLF is defined as
(5.34)
PLF =| ρˆ w  ρˆ A |2 .
Note that if ρˆ w = ρˆ A , then PLF = ρˆ A  ρˆ A = 1, and we obtain the maximum
possible received power at the antenna terminals. Remember that the
transmitting-mode and receiving-mode polarization vectors of a wave are
mutually conjugate? This means that ρˆ A is nothing but the wave the receiving
antenna would generate if it were to transmit in the direction of the incidentwave propagation. Thus, the optimal polarization of the incident wave is the
one that matches the polarization of the wave produced by the receiving
antenna if it were to operate in transmitting mode.
Here are some simple examples:
1) if ρˆ w = ρˆ A = xˆ , then PLF=1;
2) if ρˆ w = xˆ and ρˆ A = yˆ , then PLF=0;
3) if ρˆ w = ρˆ A = xˆ + jyˆ , then PLF=0;
4) if ρˆ w = xˆ + jyˆ and ρˆ A = xˆ − jyˆ ( ρˆ w = ρˆ A ), then PLF=1.
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[Balanis, 2nd ed.]
In a communication link, the PLF has to be expressed by the polarization
vectors of the transmitting and receiving antennas, ρ̂Tx and ρ̂Rx , respectively.
Both of these are defined in the coordinate system of the respective antenna as
the polarization of its transmitted wave. 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 ρ̂Rx 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
2
(5.35)
PLF = ρˆ Tx  ρˆ Rx .
The expression PLF = ρˆ Tx  ρˆ Rx
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is also correct.
14
θ̂ Rx
θ̂Tx
r̂Tx
φ̂Tx
ρ̂Tx
ρˆ Rx
φ̂Rx
r̂Rx
ρˆ Tx
ρ̂ Rx
Examples
Example 5.1. The electric field of a linearly polarized EM wave is
Ei ( x, y, z ) = xˆ  Em ( x, y )e − j  z .
It is incident upon a linearly polarized receiving antenna at the origin,
which, if in transmitting mode, would generate the field
EA ( x, y, z ) = (xˆ + yˆ )  e j  z , z < 0.
Find the PLF.
The incident wave Ei travels along + z toward the antenna at the origin.
This means that it arrives from the left (on the z axis) at the Rx antenna.
Since the antenna is at the origin, when it transmits, its field E A propagates
along − z in accordance with the propagation factor e j  z for z < 0 as
indicated in the problem statement above.
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 (transmitting and receiving) are RH-CP circularly polarized
in transmitting mode. Therefore, their polarization vectors in their
respective coordinate systems are:
ρˆ Tx = (θˆ t − jφˆ t ) / 2 ,
ρˆ Rx = (θˆ r − jφˆ r ) / 2 ,
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where θˆ r = θˆ t = θˆ and φˆ r = −φˆ t = φˆ .
z
E

r
E

y
x
The PLF is calculated as per (5.35) i.e., using the polarization vectors as
defined in the coordinate systems of the Tx and Rx antennas:
| (θˆ + jφˆ )  (θˆ − jφˆ ) |2
PLF =| ρˆ *Tx  ρˆ Rx |2 =
= 1,
4
Notice that the conjugation of ρ̂Tx simply casts it in the coordinate system
of the Rx antenna.
PLF[dB] = 10log10 1 = 0 .
There is no polarization loss. When transmitting with an RHCP antenna in
a communication link, it is best to receive with RHCP antenna.
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  , the length of the major axis OA, and the length of the
minor axis OB of the ellipse described by the equation:
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2
 e (t ) 
 e (t ) 
 e (t )   e (t ) 
sin 2  =  x  − 2  x   y  cos  +  y  .
 Ex 
 Ex   E y 
 Ey 
2
(A-1)
ey (t )
Ey
no
mi

)
O
(2
ax
OA
xis
ra
Ex ex (t )
B)
jor
ma
2
is (
E

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 
2cos 
b=
;
Ex E y sin 2 
1
c= 2 2 .
E y sin 
After dividing both sides of (A-2) by ( xy ) , one obtains
x
y 1
a −b+c = .
(A-3)
y
x xy
y e (t )
Introducing  = = y , one obtains that
x ex (t )
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x2 =
1
c 2 − b + a
1+  2
  ) = + =
 = 2
.
(A-4)
c − b + a
Here,  is the distance from the center of the coordinate system to the point on
the ellipse. We want to know at what values of  the maximum and the
minimum of  occur (  min ,  max ). This will produce the tilt angle  . We also
want to know the values of  max (major axis) and  min (minor axis). Then, we
have to solve
d ( 2 )
= 0 , or
d
2( a − c )
(A-5)
m2 −
m − 1 = 0 , where  m   min ,  max .
b
(A-5) is solved for the angle τ, which relates to ξmax as
 max = tan  = ( y / x )max .
(A-6)
Substituting (A-6) in (A-5) yields:
2
 sin  
 sin  
(A-7)

 − 2C 
 −1 = 0
cos

cos





where
E y2 − Ex2
a−c
.
C=
=
b
2 Ex E y cos 
Multiplying both sides of (A-7) by cos2  and re-arranging results in
cos2  − sin 2  + 2C sin   cos = 0 .
2(
x2
y2
cos(2 )
Thus, the solution of (A-7) is
x 2 (1 +
2)
C sin(2 )
tan(2 ) = −1 / C
or
 2 Ex E y cos  
1

(A-8)
.
;

=

+
2
1

2 − E2
2
E
2
x
y


The angles τ1 and τ2 are the angles between the major and minor axes with the x
axis. Substituting 1 and  2 back in  (see A-4) yields the expressions for OA
and OB.
1 = arctan 
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