Generating Circular Polarization

1. Jul 16, 2009

jeff1evesque

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 $$cos(\omega t)$$ and "b" dipole radiates $$sin(\omega t)$$, the field radiated by the two antennas will be circularly polarized:

$$\vec{E}(z, t) = E_{0}[cos(\omega t - \beta z)\hat{x} + sin(\omega t - \beta z)\hat{y}]$$ (#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 $$E_0$$ is the magnitude of the sinusoid- and in this case it is circular thus both $$\hat{x}, \hat{y}$$ have the same amplitudes respectively. And since both $$sin(\omega t), cos(\omega t)$$ are perpendicular to one another, if one has a phase shift, the other will have the same phase shift $$\beta$$.

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

2. Jul 16, 2009

Redbelly98

Staff Emeritus
Your thoughts look pretty reasonable, but β is not a phase shift. z is the distance away from the antenna in the direction of the E-M wave's propagation, and β=2π/λ is related to the wavelength λ.

Those equations represent a wave travelling in the +z direction. At any fixed time t, the electric field direction makes a rotating, helical pattern as one moves along the z direction.

I'm not familiar with helical antennas, so I can't comment on them.

The wave changes from right-handed to left-handed (or vice versa) CP if it is reflected.

3. Jul 16, 2009

jeff1evesque

Shouldn't z be $$\hat{z}$$ 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 $$\beta$$ instead- along with $$t$$?

Thanks,

Jeff

4. Jul 17, 2009

Redbelly98

Staff Emeritus
No, it is z. The electric field is a function of z and t.

5. Jul 17, 2009

jeff1evesque

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

THanks,

JL

Last edited: Jul 17, 2009