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Phasors Help

  1. Jan 19, 2007 #1
    I need help in understanding how to convert waves from instantaneous form to phasor form:

    a wave expressed as E(z,t) = Eo * exp(-ax) * cos(wt - Bz) * ay

    How do i convert this wave to phasor form and determine its direction of propogation, phase velocity and wavelength?

    Thanks in advance for any advice.
     
  2. jcsd
  3. Jan 20, 2007 #2
    A phasor is a quantity associated to a wave, which tells us the amplitude and the phase of the wave.

    It has the general form [tex]Ae^{j\phi}[/tex] where A is the amplitude and [tex]\phi[/tex] is the phase of the wave.

    In your equation [tex]E(z,t) = E_{0}{\cdot}e^{-a_{x}}{\cdot}cos({\omega}t - Bz){\cdot}a_{y}[/tex] ,

    [tex]E_{0}e^{-a_{x}}a_{y}[/tex] is the amplitude and [tex]-Bz[/tex] is the phase.

    The direction of propagation is in general, the direction of the wave vector, which here, since [tex]E = E(z,t)[/tex] is simply the direction of the [tex]z[/tex] axes. The phase velocity is by definition [tex]\omega[/tex] and the wavelength is by definition [tex]\frac{2\pi}{wave number}[/tex], the wave number in this case being [tex]B[/tex].

    [tex]a_{x}[/tex] and [tex]a_{y}[/tex] are the polarization parameters so they only affect the direction of the [tex]E[/tex] vector in the [tex]xy[/tex] plane.

    See http://en.wikipedia.org/wiki/Phasor_(electronics) for more information on phasors.
     
    Last edited: Jan 20, 2007
  4. Jan 21, 2007 #3
    Thanks antonantal. One more question.

    [tex]E(z,t) = E_{0}{\cdot}e^{-{\alpha}x}{\cdot}cos({\omega}t - Bz){\cdot}a_{y}[/tex]

    After transforming this equation to phasor form, how would I compute the curl of E? The field is E(z,t) with only a z-component, yet the equation has an x in it. Because of this, do I compute the partial derivative with respect to x as well?

    Thanks in advance for any help.
     
  5. Jan 21, 2007 #4
    That's an [tex]a_{x}[/tex] not an [tex]a{\cdot}x[/tex] isn't it?
     
  6. Jan 21, 2007 #5
    it's an "{alpha} * x"
     
  7. Jan 21, 2007 #6
    Ok. From the equation we can see that the wave is polarized on the [tex]y[/tex] direction, since the polarization parameter [tex]a_{y}[/tex] is present. This means that the [tex]E[/tex] vector only has component on the [tex]y[/tex] direction. But the size of this component depends on [tex]x[/tex] and [tex]z[/tex].

    So in the formula of the curl you will only have partial derrivatives of [tex]E_{y}[/tex] with respect to [tex]x[/tex] and [tex]z[/tex].
     
    Last edited: Jan 21, 2007
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