Generating Circular Polarization

Statement:
Consider two dipole antennas, oriented 90degrees apart [imagine the x-y plane, let "a" be the dipole oriented along the x-axis, and the "b" be the dipole oriented along the y-axis]. If "a" dipole radiates [tex]cos(\omega t)[/tex] and "b" dipole radiates [tex]sin(\omega t)[/tex], the field radiated by the two antennas will be circularly polarized:

[tex]\vec{E}(z, t) = E_{0}[cos(\omega t - \beta z)\hat{x} + sin(\omega t - \beta z)\hat{y}][/tex] (#1)

Side note: Very often, helical antennas are used to generate a circularly-polarized (CP) wave. The isolation between a left-handed CP wave and a right-handed CP wave can be significant. Also, a CP wave will change handedness upon reflection.

My thoughts:
I understand that [tex]E_0[/tex] is the magnitude of the sinusoid- and in this case it is circular thus both [tex]\hat{x}, \hat{y}[/tex] have the same amplitudes respectively. And since both [tex]sin(\omega t), cos(\omega t)[/tex] are perpendicular to one another, if one has a phase shift, the other will have the same phase shift [tex]\beta[/tex].

Relevant questions:
Is my thoughts above reasonable? What I would really like to know is why the electric field is a function of z also. What is the variable z, and how does it influence the electric field?

Also, can someone explain to me what is meant by

Thanks,


Jeffrey

 

Shouldn't z be [tex]\hat{z}[/tex] then? So if it becomes the unit vector (since that is the direction of propagation), then the electric field cannot be a function of z, and thus should it be a function of [tex]\beta[/tex] instead- along with [tex]t[/tex]?

Thanks,


Jeff

 

Ok I almost get it now. I am going to start a new thread, this one is getting long.


THanks,


JL