Convert Waves from Instantaneous to Phasor Form

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To convert the wave expressed as E(z,t) = E₀ * exp(-αx) * cos(ωt - Bz) * a_y into phasor form, identify the amplitude as E₀e^{-αx} and the phase as -Bz. The direction of propagation is along the z-axis, with the phase velocity defined as ω and the wavelength as 2π/B. The polarization is in the y-direction, indicated by the presence of a_y. When computing the curl of E, only the partial derivatives of E_y with respect to x and z are necessary, as the field has a z-component but the amplitude depends on both x and z.
korps
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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.
 
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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 Ae^{j\phi} where A is the amplitude and \phi is the phase of the wave.

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

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

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

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

See http://en.wikipedia.org/wiki/Phasor_(electronics) for more information on phasors.
 
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Thanks antonantal. One more question.

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

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.
 
korps said:
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?

That's an a_{x} not an a{\cdot}x isn't it?
 
it's an "{alpha} * x"
 
Ok. From the equation we can see that the wave is polarized on the y direction, since the polarization parameter a_{y} is present. This means that the E vector only has component on the y direction. But the size of this component depends on x and z.

So in the formula of the curl you will only have partial derrivatives of E_{y} with respect to x and z.
 
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