Convert Waves from Instantaneous to Phasor Form

Click For Summary
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
Messages
5
Reaction score
0
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.
 
Engineering news on Phys.org
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.
 
Last edited:
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.
 
Last edited:
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K