# 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
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