Yzowledge - Antenna

Polarizations Part 4: Axial Ratio Measurement

2026-03-01T15:32:00-05:00

Measuring the axial ratio (AR) of a circularly polarized antenna is important because it quantifies how close the antenna’s polarization is to ideal circular polarization, which directly affects signal quality, polarization matching, and system performance.

Measurement Method

A linearly polarized antenna is usually used to measure the axial ratio of a circularly polarized antenna. The method relies on the fact that a linearly polarized antenna responds differently to the major and minor axes of the polarization ellipse.

A circularly (or elliptically) polarized wave can be represented as two orthogonal linear components. When a linearly polarized probe is rotated about its boresight axis:

The received power varies sinusoidally.
The maximum received power corresponds to alignment with the major axis of the polarization ellipse.
The minimum received power corresponds to alignment with the minor axis.

The axial ratio is then obtained from the ratio of these two field magnitudes.

Because received power is proportional to the square of the electric field:

$$\begin{equation}AR=\frac{E_{max}}{E_{min}}=\sqrt{\frac{P_{max}}{P_{min}}} \end{equation} $$

In dB:

$$\begin{equation}AR_{dB}=P_{max,dB}-P_{min,dB} \end{equation} $$

Polarization Purity Impact

When the linear polarization reference antenna has limited cross-polarization isolation (also called cross-pol discrimination or XPD), it will affect the accuracy of axial ratio measurements. The reference antenna's cross-polarization component introduces measurement errors that can make the circular polarization antenna appear better or worse than its actual performance.

Understanding the Error Mechanism

An ideal linear polarization antenna transmits only in its main polarization (e.g., vertical) with zero energy in the orthogonal polarization (e.g., horizontal). However, real antennas have finite cross-polarization isolation, meaning they transmit some unwanted energy in the orthogonal polarization.

When measuring a circular polarization antenna's axial ratio, the linear antenna's cross-polarization component adds to the measurement, corrupting the true polarization pattern. This is particularly problematic because circular polarization requires precise balance between orthogonal components.

Quantifying the Measurement Error

The polarization match factor between practical linear polarization and circular polarization antennas is (presented in Polarizations: Part3):

$$\begin{equation}\begin{split}\rho(\phi)&=\frac{\cos^2\beta+\gamma_{CP}^2\sin^2\beta+\gamma_{LP}^2\left(\sin^2\beta+\gamma_{CP}^2\cos^2\beta\right)+2\gamma_{LP}\left[\left(1-\gamma_{CP}^2\right)\sin\beta\cos\beta\cos\phi+s\gamma_{CP}\sin\phi\right]}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \\ \\ &= \frac{1}{2}+\frac{2s\gamma_{LP}\gamma_{CP}\sin\phi}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}+\frac{(1-\gamma_{LP}^2)(1-\gamma_{CP}^2)\cos2\beta}{2(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}+\frac{\gamma_{LP}(1-\gamma_{CP}^2)\cos\phi\sin 2\beta}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \\ \\ &=A+B\cos 2\beta+C\sin 2\beta\end{split}\end{equation}$$

where

$$\begin{equation}\begin{cases}A=\frac{1}{2}+\frac{2s\gamma_{LP}\gamma_{CP}\sin\phi}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \\ \\ B=\frac{(1-\gamma_{LP}^2)(1-\gamma_{CP}^2)}{2(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \\ \\ C=\frac{\gamma_{LP}(1-\gamma_{CP}^2)\cos\phi}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \end{cases}\end{equation}$$

$\rho(\phi)$ is a sinusoid of the form $A+\sqrt{B^2+C^2}\cos (2\beta-2\beta_0)$, where $\tan 2\beta_0=\tan 2\delta \cos\phi$, $\gamma_{CP}=\tan \delta$. Therefore,

$$\begin{equation}\begin{cases}\rho_{max}=A+\sqrt{B^2+C^2} \\ \\ \rho_{min}=A-\sqrt{B^2+C^2} \end{cases}\end{equation}$$

and,

$$\begin{equation}\sqrt{B^2+C^2}=\frac{(1-\gamma_{CP}^2)\sqrt{1+2\gamma_{LP}^2\cos\phi+\gamma_{LP}^4}}{2(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \end{equation}$$

The measured axial ratio is:

$$\begin{equation}AR_{meas}=\sqrt{\frac{\rho_{max}}{\rho_{min}}}=\sqrt{\frac{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)+4\gamma_{LP}\gamma_{CP}\sin\phi+(1-\gamma_{CP}^2)\sqrt{1+2\gamma_{LP}^2\cos 2\phi+\gamma_{LP}^4}}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)+4\gamma_{LP}\gamma_{CP}\sin\phi-(1-\gamma_{CP}^2)\sqrt{1+2\gamma_{LP}^2\cos 2\phi+\gamma_{LP}^4}}} \end{equation}$$

$s$ is taken as $+1$ with loss of generality.

For perfect linear polarization, $\gamma_{LP}=0$, then $AR_{meas}=\frac{1}{\gamma_{CP}}$, as expected.

For perfect circular polarization, $\gamma_{CP}=1$, then $AR_{meas}=1$, as expected too.

The true axial ratio is $AR_{true}=\frac{1}{\gamma_{CP}}$. The measurement error in dB is:

$$\begin{equation}Err_{dB}=AR_{meas,dB}-AR_{true,dB}=20\log_{10}\left(\frac{AR_{meas}}{AR_{true}}\right)=20\log_{10}\left(\gamma_{CP}AR_{meas}\right) \end{equation}$$

The measurement error or uncertainty depends on the cross polarization phase difference ($\phi$) than the colinear polarization of the linearly polarized antenna. $\phi$ is typically unknown and random.

Perfect Probe Case

If $\phi=0^\circ, 180^\circ$, the formula reduces to $\mathbf{AR}_{meas}=\frac{1}{\gamma_{CP}}=\mathbf{AR}_{true}$. For this case, both co-pol and cross-pol voltages scale together, and the measured AR is unaffected.

Worst-Case Error

The error is maximized when the $\sin\phi$ and $\cos 2\phi$ terms interact to minimize the denominator. This typically occurs near $\phi = \pm 90^\circ$ (quadrature leakage). This gives $\beta=0$ which means the major axis of the circular polarization is aligned with the linear polarization. The linear polarization is actually an elliptical polarization with major axis same as the colinear direction. $\phi = \pm 90^\circ$ corresponds to the two rotation senses of the elliptical polarization that defines the two worst-case errors.

$$\begin{equation}AR_{meas,worst}=\begin{cases}\frac{1+\gamma_{LP}\gamma_{CP}}{\gamma_{LP}+\gamma_{CP}}=\frac{AR_{true}\sqrt{XPD_{LP}}+1}{AR_{true}+\sqrt{XPD_{LP}}},&if\space \phi=\frac{\pi}{2} \\ \\ \left| \frac{1-\gamma_{LP}\gamma_{CP}}{\gamma_{LP}-\gamma_{CP}}\right|=\left| \frac{AR_{true}\sqrt{XPD_{LP}}-1}{AR_{true}-\sqrt{XPD_{LP}}}\right|,&if\space \phi=-\frac{\pi}{2} \end{cases}\end{equation}$$

The measured axial ratio will be within the estimations of the two worst cases, as illustrated in the figure below.

Polarizations Part 3: Polarization Match Special Cases

2026-02-27T14:43:00-05:00

Special cases of polarization match in a transmit-receive system are studied herein.

Polarization-Matched Antennas

If we have two polarization-matched antennas in a transmit-receive system, the polarization match factor is equal to 1. Thus

$$\begin{equation} \rho=\frac{(1+p_1p_2)(1+p_1^*p_2^*)}{(1+p_1p_1^*)(1+p_2p_2^*)}=1 \end{equation}$$

It gives:

$$\begin{equation}(p_1-p_2^*)(p_2-p_1^*)=0 \end{equation}$$

which means

$$\begin{equation}p_1=p_2^* \end{equation}$$

and

$$\begin{equation}q_1=q_2^* \end{equation}$$

Therefore,

$$\begin{equation}\begin{cases}AR_1=AR_2 \\ \\ \tau_1=-\tau_2 \end{cases}\end{equation}$$

$\tau_1$ and $\tau_2$ are defined in different coordinate systems. $\tau_1=-\tau_2$ means the semi-major axes of the two polarizations are coincide. For polarization-match antennas, the rotation senses of the two polarization ellipses are identical when described in the appropriate coordinate systems. If we think of both antennas transmitting a right elliptic wave, for example, the two waves will appear to rotate in opposite directions at a point in space at which they "meet."

Cross-Polarized Antennas

Two antennas in a transmit-receive configuration that are so polarized that no signal is received are said to be cross-polarized. Thus, $\rho=0$. Then,

$$\begin{equation}p_1=-\frac{1}{p_2} \end{equation}$$

and

$$\begin{equation}q_1=-\frac{1}{q_2} \end{equation}$$

Therefore,

$$\begin{equation}\begin{cases}AR_1=AR_2 \\ \\ \tau_1+\tau_2=\pm\frac{\pi}{2} \end{cases}\end{equation}$$

For cross-polarized antennas, the major axis of one polarization ellipse coincides with the minor axis of the other. The rotation senses of the two polarization ellipses are different when described in the appropriate coordinate systems.

Ideal Linear Polarization to Elliptical Polarization

The polarization match factor between an ideal linear polarization and an elliptical polarization is:

$$\begin{equation}\rho=\frac{1}{2}+\frac{(AR^2-1)\cos2\beta}{2(AR^2+1)} \end{equation}$$

where $AR$ is the axial ratio of the elliptical polarization, $\beta$ is the angle between the linear polarization and semi-major axis of the elliptical polarization.

For linear polarization to circular polarization case, $AR=1$, then $\rho=\frac{1}{2}$ which is the well known 3dB power mismatch loss.

Practical Linear to Practical Circular

In practical linear polarization antenna, there is always power leakage from colinear polarization to cross polarization. The power leakage is represented by cross polarization discrimination (XPD). The linear antenna field can be written as:

$$\begin{equation}\mathbf{E_{LP}}=E_{0,LP}\left(\hat{x}+\frac{1}{\sqrt{XPD_{LP}}}e^{j\phi}\hat{y} \right)=\frac{E_{0,LP}}{\sqrt{1+\gamma_{LP}^2}}\left(\hat{x}+\gamma_{LP}e^{j\phi}\hat{y} \right) \end{equation}$$

where $XPD_{LP}=\frac{|E_x|^2}{|E_y|^2}$, $\gamma_{LP}=\frac{1}{\sqrt{XPD_{LP}}}$, $\phi$ is the phase difference between $E_x$ and $E_y$, $E_x$ is considered as colinear direction. In practice, the linear polarization cross-pol phase is unknown and probably random.

The practical linear polarization with limited XPD is actually an elliptical polarization whose semi-major axis direction depends on $XPD_{LP}$ and $\phi$ and is not necessarily aligned with the colinear direction.

The practical circular polarization field is give as:

$$\begin{equation}\mathbf{E_{CP}}=\frac{E_{0,CP}}{\sqrt{\gamma_{CP}^2+1}}\left[\left(\cos\beta-js\gamma_{CP}\sin\beta \right)\hat{x}+\left(\sin\beta+js\gamma_{CP}\cos\beta\right)\hat{y}\right] \end{equation}$$

where $\gamma_{CP}=\frac{1}{AP_{CP}}$, $AP_{CP}$ is the axial ratio, $\beta$ is the angle between colinear direction of the linear polarization and semi-major axis of the circular polarization, and $s=\pm 1$ to indicate the rotation sense.

The polarization match factor can be derived as:

$$\begin{equation}\begin{split}\rho(\phi)&=\frac{|\mathbf{E_{LP}^*}\cdot\mathbf{E_{CP}}|^2}{|\mathbf{E_{LP}^*}|^2|\mathbf{E_{CP}}|^2} \\ \\ &=\frac{1}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}\left|(\cos\beta-js\gamma_{CP}\sin\beta)+\gamma_{LP}e^{-j\phi}(\sin\beta+js\gamma_{CP}\cos\beta)\right|^2 \\ \\ &=\frac{\cos^2\beta+\gamma_{CP}^2\sin^2\beta+\gamma_{LP}^2\left(\sin^2\beta+\gamma_{CP}^2\cos^2\beta\right)+2\gamma_{LP}\left[\left(1-\gamma_{CP}^2\right)\sin\beta\cos\beta\cos\phi+s\gamma_{CP}\sin\phi\right]}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)} \\ \\ &= \frac{1}{2}+\frac{(1-\gamma_{LP}^2)(1-\gamma_{CP}^2)\cos2\beta}{2(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}+\frac{\gamma_{LP}\sqrt{4\gamma_{CP}^2+(1-\gamma_{CP}^2)^2\sin^2 2\beta}\cos(\phi-\phi_0)}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}\end{split}\end{equation}$$

where $\tan\phi_0=\frac{2s\gamma_{CP}}{(1-\gamma_{CP}^2)\sin2\beta}=\frac{s\tan 2\delta}{\sin 2\beta}$, $\gamma_{CP}=\tan\delta$.

In practical antenna, cross-pol phase $\phi$ is typically unknown and often varies with frequency or mechanical alignment. The extrema can be calculated:

$$\begin{equation}\begin{cases}\rho_{max}=\bar{\rho}+ \frac{\gamma_{LP}\sqrt{4\gamma_{CP}^2+(1-\gamma_{CP}^2)^2\sin^2 2\beta}}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}, &if\space \phi=\phi_0 \\ \\ \rho_{min}=\bar{\rho}- \frac{\gamma_{LP}\sqrt{4\gamma_{CP}^2+(1-\gamma_{CP}^2)^2\sin^2 2\beta}}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}, &if\space \phi=\pi+\phi_0 \end{cases} \end{equation}$$

where $\bar{\rho}$ is the average term:

$$\begin{equation}\rho=\bar{\rho}=\frac{1}{2}+\frac{(1-\gamma_{LP}^2)(1-\gamma_{CP}^2)\cos2\beta}{2(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)}, \space if \space \phi=\pm\frac{\pi}{2}+\phi_0\end{equation}$$

If $\phi$ is assumed to be random with a uniform distribution in $[0,2\pi]$, then

$$\begin{equation}\begin{cases}\left<\cos\phi \right>=0 \\ \\ \left<\sin\phi \right>=0 \\ \\ \left<e^{-j\phi} \right>=0 \end{cases}\end{equation}$$

The cross terms vanish. Thus, the averaged polarization match factor is:

$$\begin{equation}\left<\rho \right>=\bar{\rho}\end{equation}$$

In practice, $\beta$ is also unknown. The global extrema of $\rho$ are:

$$\begin{equation}\begin{cases}\rho_{gmax}=\frac{(1+\gamma_{LP}\gamma_{CP})^2}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)},&if \space \beta=0,\phi=\pm\frac{\pi}{2} \\ \\ \rho_{gmin}=\frac{(1-\gamma_{LP}\gamma_{CP})^2}{(1+\gamma_{LP}^2)(1+\gamma_{CP}^2)},&if \space \beta=0,\phi=\mp\frac{\pi}{2}\end{cases}\end{equation}$$

Polarizations Part 2: Polarization Match

2026-02-22T10:00:00-05:00

It is obvious that when two antennas are used in a communication system, they should be matched in polarization so that the available power at the receiving antenna can be fully utilized. A polarization match factor is developed and can be given in terms of the standard polarization parameters.

Effective Length of An Antenna

The transmitted field of any antenna can be written as:

$$\begin{equation}\mathbf{E}^t(r,\theta,\phi)=\frac{jZ_0I}{2\lambda r}e^{-jkr}\mathbf{h}(\theta,\phi)\end{equation}$$

where $Z_0$ is the intrinsic impedance of free space, $k$ is the free space propagation constant, $\lambda$ is the wavelength, and $I$ is an input current at an arbitrary pair of terminals. $\mathbf{h}(\theta,\phi)$ is effective length of a general antenna. The effective length does not necessarily correspond to a physical length of the antenna. $\mathbf{h}$ is a complex vector to describe an elliptical polarized field. With proper choice of coordinate system, $\mathbf{E}^t$ and $\mathbf{h}$ will have only two components since in the radiation zone $\mathbf{E}^t$ has no radial component.

Received Voltage

The open-circuit voltage induced in a receiving antenna by an externally produced field is proportional to its effective length. The received voltage across the open terminals of the receiving antenna is:

$$\begin{equation}V_r=\mathbf{E}^i\cdot\mathbf{h} \end{equation}$$

where $mathbf{h}$ is the effective length of the receiving antenna.

For maximum received power, the relationship between $\mathbf{h}$ and $\mathbf{E}^i$ is:

$$\begin{equation}\frac{h_\phi}{h_\theta}=\left(\frac{E_\phi^i}{E_\theta^i}\right)^* \end{equation}$$

Polarization Match Factor

The polarization match factor is the ratio of actual power received to that received under the most favorable circumstances of matched polarization:

$$\begin{equation}\rho=\frac{\left|\mathbf{E}^i\cdot\mathbf{h} \right|^2}{\left|\mathbf{E}^i\right|^2\left|\mathbf{h}\right|^2} \end{equation}$$

The polarization match factor is sometimes called polarization efficiency, and $0\le\rho\le1$. The polarization match factor shows how well a receiving antenna of effective length h is matched in polarization to an incoming wave.

When two antennas are in transmit-receive configuration, both antennas will be described in its polarization properties by the right-handed coordinate system adjacent to itself. The following image shows the coordinate systems of two such antennas with opposite x-axis and z-axis. The reason is that antenna polarization is normally based on a right-handed system with z-axis pointing in the direction of outward wave from antenna.

The incident wave from antenna 1 may be written as:

$$\begin{equation} \mathbf{E_1}^i=E_{01}\left(\hat{x}+p_1e^{-j\frac{\pi}{2}}\hat{y} \right) \end{equation}$$

The effective length of antenna 2 may be written as:

$$\begin{equation} \mathbf{h}_2=h_{02}\left(\hat{x'}+p_2e^{-j\frac{\pi}{2}}\hat{y'} \right)\end{equation}$$

If the tilt angle of antenna 2 in the x'y'z'-system is $\tau_2$, then it will be $\pi-\tau_2$ in xyz-system.

Therefore, the induced voltage on antenna 2 is:

$$\begin{equation}V_2=\mathbf{E_1}^i\cdot\mathbf{h_2}=-E_{01}h_{02}(1+p_1p_2) \end{equation}$$

Also,

$$\begin{equation} \left|\mathbf{E_1}^i \right|^2=\left|E_{01} \right|^2(1+p_1p_1^*) \end{equation}$$

and

$$\begin{equation}|\mathbf{h_2}|^2=|h_{02}|^2(1+p_2p_2^*) \end{equation}$$

Then, the polarization match factor is:

$$\begin{equation}\rho=\frac{(1+p_1p_2)(1+p_1^*p_2^*)}{(1+p_1p_1^*)(1+p_2p_2^*)}=\frac{(1-P_1P_2)(1-P_1^*P_2^*)}{(1+P_1P_1^*)(1+P_2P_2^*)} \end{equation}$$

$\rho$ can be written in terms of $q$ as:

$$\begin{equation}\rho=\frac{(1+q_1q_2)(1+q_1^*q_2^*)}{(1+q_1q_1^*)(1+q_2q_2^*)} \end{equation}$$

It is not surprising that $p$ has the same form in $q$ as in $p$, since the form for $q$ in terms of $p$ is the same as for $p$ in terms of $q$.

$\rho$ can also be written in terms of axial ratio as:

$$\begin{equation}\rho=\frac{1}{2}+\frac{4AR_1AR_2+(AR_1^2-1)(AR_2^2-1)\cos2\beta}{2(AR_1^2+1)(AR_2^2+1)} \end{equation}$$

where $\beta=\tau_1+\tau_2$ is the angle between semi-major axes of the two elliptical polarizations.

Axial ratio does not provide the rotation sense information of the polarization. If axial ratio is purposely defined as positive for right-hand rotation and negative for left-hand rotation, then the formula can be used for cases with polarizations having either same or different rotation senses.

If inverse axial ratio $\gamma=\frac{1}{AR}$ ($0\le\gamma\le1$) is used, then the polarization match factor can be written as:

$$\begin{equation}\rho=\frac{1}{2}+\frac{4\gamma_1\gamma_2+(1-\gamma_1^2)(1-\gamma_2^2)\cos2\beta}{2(1+\gamma_1^2)(1+\gamma_2^2)}\end{equation}$$

Polarizations Part 1: Concept

2026-02-16T15:47:00-05:00

Electromagnetic (EM) wave polarization describes the geometric orientation and time evolution of the electric field vector E at a fixed point in space. Because the magnetic field B is orthogonal to E and to the direction of propagation (from Maxwell’s equations), polarization is fully characterized by the behavior of E in the transverse plane.

For a monochromatic plane wave propagating in the +z direction:

$$ \begin{equation} \mathbf{E}(z,t) = \mathfrak{R}\left\{(E_x\hat{x}+E_y\hat{y})e^{j(\omega t-kz)} \right\} \label{eq:gen_wave_pol} \end{equation} $$

Polarization depends on the relative amplitudes and phase difference between $E_x$ and $E_y$.

A general case of tilted elliptical polarization is illustrated in the following figure.

Polarization in Cartesian Coordinate System

The time-invariant $\mathbf{E}$ field of Equation [\ref{eq:gen_wave_pol}] may also be written as

$$ \begin{equation} \mathbf{E}=E_0\left(a\hat{x}+be^{j\phi}\hat{y}\right) \label{eq:xy_wave_pol} \end{equation} $$

For convenience, the distance and time phase term has been dropped. Without loss of generality, $E_0$ and $\phi$ are chosen so that $a$ and $b$ are nonnegative real and $a^2+b^2=1$. The value of $E_0$ does not affect the wave polarization in any way except in questions concerned with power. So $E_0$ will be neglected.

The polarization ratio, $P$, is defined as

$$\begin{equation}P=\frac{E_y}{E_x}=\frac{b}{a}e^{j\phi}\end{equation}$$

The modified polarization ratio, $p$, is defined as

$$\begin{equation}p=jP=j\frac{b}{a}e^{j\phi} \end{equation}$$

Axial ratio of elliptical polarization is defined as the ratio of semi-major axis to semi-minor axis. The axial ratio can be calculated in terms of $p$:

$$\begin{equation}AR=\left|\frac{|1+p|+|1-p|}{|1+p|-|1-p|} \right|=\left|\frac{a^2+b^2+\sqrt{a^4+b^4+2a^2b^2\cos(2\phi)}}{2ab\sin\phi} \right| \end{equation} $$

The tilt angle, $\tau$, can be found as

$$\begin{equation}\tan 2\tau =\tan 2\alpha\cos\phi\end{equation}$$

where the angle $\alpha$ is defined as

$$\begin{equation}\tan \alpha = \frac{b}{a}\quad 0\le\alpha\le\frac{\pi}{2} \end{equation}$$

$\tau$ can also be found as

$$\begin{equation}e^{-j2\tau}=\frac{(1-p)/(1+p)}{|(1-p)/(1+p)|} \end{equation}$$

The $\mathbf{E}$ filed can also be written in terms of $AR$ and $\tau$ as:

$$\begin{equation}\mathbf{E}=\frac{E_0}{\sqrt{AR^2+1}}\left[\left(AR\cos\tau-js\sin\tau \right)\hat{x}+\left(AR\sin\tau+js\cos\tau\right)\hat{y}\right] \end{equation}$$

where $s=\pm1$ is for rotation sense. $s=+1$ for left-handed polarization and $s=-1$ for right-handed polarization.

It can be written in terms of inverse axial ratio $\gamma=\frac{1}{AR}$ ($0\le\gamma\le1$) as:

$$\begin{equation}\mathbf{E}=\frac{E_0}{\sqrt{\gamma^2+1}}\left[\left(\cos\tau-js\gamma\sin\tau \right)\hat{x}+\left(\sin\tau+js\gamma\cos\tau\right)\hat{y}\right] \end{equation}$$

Linear Polarization

when $a=0$, the polarization is vertical polarization, and $\phi$ is irrelevant. The axial ratio is infinite and tilt angle is 90 degree.
When $b=0$, the polarization is horizontal polarization. The axial ratio is infinite and tilt angle is 0 degree.
When $\phi=0$, the polarization is a linear polarization. The axial ratio is infinite and tilt angle is same as $\alpha$.
When $\phi=\pi$, the polarization is a linear polarization. The axial ratio is infinite and tilt angle is same as ${\pi}-\alpha$.

Elliptical Polarization

For scenarios other than linear polarizations, the polarizations will be elliptical polarizations (including circular polarizations). To determine the rotation sense, an auxiliary angle $\delta$ is defined as

$$\begin{equation}\sin 2\delta=\sin 2\alpha\sin\phi \end{equation}$$

The axial ratio can be calculated as

$$\begin{equation}AR=\left|\frac{1}{\tan\delta}\right| \end{equation}$$

The rotation sense can be determined as

$$\begin{equation} \begin{cases}\sin 2\delta<0, & right-hand \\ \\ \sin 2\delta>0, & left-hand\end{cases}\end{equation}$$

Therefore, the sign of $\sin\phi$ defines the rotation sense since $\sin 2\alpha>0$.

$$\begin{equation}\begin{cases}\pi<\phi<2\pi,& right-hand \\ \\ 0<\phi<\pi,&left-hand\end{cases} \end{equation}$$

Also, $Re(p)=-\frac{b}{a}\sin\phi$, thus

$$\begin{equation}\begin{cases}Re(p)>0,& right-hand \\ \\ Re(p)<0,&left-hand\end{cases} \end{equation}$$

When $\phi=\frac{\pi}{2}$ and $a\ne b$, the tilt angle will be 0 degree if $a>b$ or 90 degree if $a<b$, which means the x and y axes are aligned with semi-major and semi-minor axes of the polarization ellipse. The axial ratio $AR = max\{\frac{a}{b},\frac{b}{a}\}$. The rotation sense will be left-hand since $\sin 2\delta>0$.

When $\phi=-\frac{\pi}{2}$ and $a\ne b$, the tilt angle and axial ratio are sames as $\phi=\frac{\pi}{2}$ case. The rotation sense will be right-hand since $\sin 2\delta<0$.

When $a=b$ and $\phi \ne \pm \frac{\pi}{2}$, the tile angle will be always 45 degree if $\cos \phi>0$ and always 135 degree if $\cos \phi < 0$.

Circular Polarization

Circular polarization is a special case of elliptical polarization. It occurs when $a=b$ and $\phi=\pm\frac{\pi}{2}$. It is left-hand when $\phi=\frac{\pi}{2}$ and right-hand when $\phi=-\frac{\pi}{2}$. Tilt angle is irrelevant and axial ratio is 1.

Circular Wave Components

Consider the complex vectors:

$$\begin{equation}\begin{cases}\mathbf{\omega_L}=\frac{1}{\sqrt{2}}\left(\hat{x}+j\hat{y} \right) \\ \\ \mathbf{\omega_R}=\frac{1}{\sqrt{2}}\left(\hat{x}-j\hat{y} \right)\end{cases}\end{equation}$$

It is clear that $\mathbf{\omega_L}$ is a left circular wave ($a=b$, $\phi=\frac{\pi}{2}$), and $\mathbf{\omega_R}$ is a right circular wave. The field $\mathbf{E}$ can be expanded in terms of $\mathbf{\omega_L}$ and $\mathbf{\omega_R}$, giving

$$\begin{equation}\mathbf{E}=E_0\left(a\hat{x}+be^{j\phi}\hat{y} \right)=E_0\left(L\mathbf{\omega_L}+Re^{j\theta}\mathbf{\omega_R} \right)=E_L\mathbf{\omega_L}+E_R\mathbf{\omega_R} \end{equation}$$

Solving for $L$ and $Re^{j\theta}$ gives,

$$\begin{equation}\begin{cases}L=\frac{1}{\sqrt{2}}\left(a-jbe^{j\phi} \right)\\ \\ R=\frac{1}{\sqrt{2}}\left(a+jbe^{j\phi} \right) \end{cases} \end{equation}$$

Circular polarization ratio, $q$, is defined as

$$\begin{equation}q=\frac{E_L}{E_R}=\frac{L}{R}e^{-j\theta}=\frac{a-jbe^{j\phi}}{a+jbe^{j\phi}}=\frac{1-p}{1+p} \end{equation}$$

The modified polarization ratio can be obtained as

$$\begin{equation}p = \frac{1-q}{1+q} \end{equation}$$

The axial ratio can be given in terms of $q$ as

$$\begin{equation}AR=\left|\frac{1+|q|}{1-|q|} \right| \end{equation}$$

The tilt angle is given by

$$\begin{equation}\begin{cases}\tau=\frac{\theta}{2}, & if\space\theta\ge0 \\ \\\pi+\frac{\theta}{2}, &if \space\theta\le0 \end{cases} \end{equation}$$

The magnitude of $q$ defines rotation sense as

$$\begin{equation}\begin{cases}|q|<1,& right-hand \\ \\ |q|>1,&left-hand\end{cases} \end{equation}$$

$|q|<1$ corresponds to $|L|<|R|$, which results in a right-hand rotation.