Convert Waves from Instantaneous to Phasor Form

[korps]

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 propagation, phase velocity and wavelength?

Thanks in advance for any advice.

 

[antonantal]

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.

 

[korps]

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.

 

[antonantal]

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?

 

[korps]

it's an "{alpha} * x"

 

[antonantal]

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 $$z$$.