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Generating Circular Polarization

  1. Jul 16, 2009 #1
    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


  2. jcsd
  3. Jul 16, 2009 #2


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    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.
  4. Jul 16, 2009 #3
    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]?


  5. Jul 17, 2009 #4


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    No, it is z. The electric field is a function of z and t.
  6. Jul 17, 2009 #5
    Ok I almost get it now. I am going to start a new thread, this one is getting long.


    Last edited: Jul 17, 2009
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