Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circular Polarization

  1. Jul 17, 2009 #1
    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)

    Relevant Question:
    In terms of a specific distance, say in the [tex]\hat{x}[/tex] direction, the cosine function has traveled a distance [tex]\omega t[/tex] (as did the sine function in it's respective axis). But I don't understand why To find electric field at a given location in the [tex]\hat{z}[/tex] direction, we subtract the distance traveled [tex]\omega t[/tex] by the wave number times distance in z, or [tex]\beta z[/tex] - for each component [tex]\hat{x}, \hat{y}[/tex]. The wave number is the wavelength of the sinusoid per unit distance. What happens when we take this wave number [tex]\beta[/tex] and multiply it by [tex]z[/tex]? What does that represent, I cannot see the relation between the two ([tex]\omega t[/tex] and [tex]\beta z[/tex])?

    Does one unit length of [tex]z = 1[/tex] for [tex]\beta z \hat{x}[/tex] and [tex]\beta z \hat{y}[/tex] correspond to a length of [tex]\frac{2\pi}{\lambda}[/tex] in the [tex]\hat{z}[/tex] direction?
     
  2. jcsd
  3. Jul 17, 2009 #2

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    The wave is both sinusoidal in time and in space. Thus, if we were too look at the wave at a constant point in space, we expect that it should vary sinusoidally with angular frequency \omega over time. If we were to look at the wave at a constant point in time over space, we would expect that the wave would vary spacially sinuisoidally with respect to the angular frequency \beta, which is called the wave number. All they have done in your set of equations is combine these two effects into a single sinusoidal function.
     
  4. Jul 18, 2009 #3
    That kind of made sense, but wasn't the kind of answer I was looking for haha. If someone could explain this in a different way, that would be great.

    thanks
     
  5. Jul 23, 2009 #4
    Hello,

    Let's try to simplify the problem...

    Assume we have linear polarization in x-direction i.e. [tex] E = \cos(\omega t - \beta z) \hat{x} [/tex] or we can write it as [tex] E = cos(\beta z - \omega t) \hat{x} [/tex].

    This wave is a traveling wave.. means that it moves in the direction of [tex] \hat{z} [/tex].

    Try to plot E with respect to the distance at time (t = 0). This will give cosine wave with zero phase .. [tex] E = cos(\beta z) [/tex]

    Try to increase the time to [tex] t_1 [/tex] for example. you will find the wave is shifted to the right. so this can be represented by subtracting z by the value of the shift say z1... [tex] E = cos(\beta (z - z_1)) = cos (\beta z - \beta z_1) [/tex].

    Assume that the wave is moving with velocity v. Then the time t1 need to move distance z1 is z1/v.

    Substitute z1 by v*t1, we get
    [tex] E = cos(\beta z - \beta v t_1) [/tex]

    By letting [tex] \beta * v = \omega [/tex], we have
    [tex] E = cos(\beta z - \omega t_1) [/tex]

    So as a conclusion if t = T which is the time period of cosine, we will find that the wave moves a distance equal to the wavelength [tex] \lambda [/tex]

    I hope that I answered your question.
    Good luck
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Circular Polarization
  1. Antenna Polarization (Replies: 2)

  2. Capacitor polarity (Replies: 8)

  3. Antenna Polarization (Replies: 3)

  4. Reverse polarity (Replies: 2)

Loading...