We consider a wave superposed of two linearely polarized waves, one in X direction, the other one in Y direction. They have 90 in between. These are defined by

Both waves have their own amplitude and phase offset. The instantaneous angle is written here for simplicity. A real wave has . If we superpose both waves we will get an elliptically polarized wave. The E vector describes an ellipse. Silently we assume the x and y coordinates being proportional to the x and y component of the field strength vector, respectively.

This ellipse is centered at the origin. Its major axis has the length , its minor axis the lenth . The major axis' inclination to the axis is denoted by . The rectangle described by the amplitudes of the E field components has a hypotenuse angle called , the rectangle streched by the ellipse axes has an angle known as .

Handedness

(3) |

The maxium of the length appears in the the major axis' direction. The derivation of the length

(4) |

This supplies the angle when the superposed waves point to the major axis

(5) |

where , the phase difference.

Equations with
^{2} give the x and y coordinate of the
major axis

(6) |

whereas results in the coordinates of the minor axis

(7) |

The length of the axes is then given by

(8) |

Inclination

(9) |

Ellipticity

(10) |

(11) |

what we directly use in the equations for the new amplitudes

(12) |

(remember the unit circle and the trigonometric functions). The phase offsets don't depend on .

Finally we recalculate and since they change with . The calculation is given in subsections and , respectively.

The coordinates in the base coordinate system and are rendered as

(13) |

In the rotated coordinate system the coordinates are given by

(14) |

which can be simplified to the coordinate transformation

(15) |

The backward transformation

(16) |

Component maxima

(17) |

In the base coordinate system this equals to

where the angle refers to the rotated coordinate system.

The components' amplitudes appear when these coordinates have their maximum. The angle relative to the major axis points to the amplitude of , denotes the amplitude of .

(19) |

Using that result the field strength components are given by

(20) |

(21) |

with being the phase difference between and . Solving this equation for finishes the problem.

(22) |

Attention: be careful with this . Little inaccuracies and the granularity of floating point calculations can lead to an argument raising a floating point exception.

Since changing the field strength components also affects the handedness , it has to be calculated as described in section . doesn't change because the ellipse is rotated as is.

(23) |

The new axes lengths thus become

(24) |

(recall unit circle and trigonometric functions). The values , , , and ( is held constant when is changed) are given by the equations derived in section .

- ... length
^{1} - Here we differentiate the square of the length . This is allowed since the length is surely greater than zero and the square is a monotone transformation on the positive axis. It simplifies the calculation enormously.
- ...#tex2html_wrap_inline502#
^{2} - we simply use the absolute values instead of the vectors here