Phasor Domain

1. Mar 26, 2009

Pythagorean

what's a phasor? What's the phasor domain?

I've worked with them in my courses and I can move from the time domain to the phasor domain, but I still don't quite intuitively get what a phasor is.

In physics, we move between the frequency domain and time domain easily, but they're both variables within the same equation that we're looking at across domains. I understand the concept of the frequency domain....

A phasor, on the other hand is not a variable that I can see... its obviously not the phase...

2. Mar 26, 2009

Staff: Mentor

Phasors are a handy way to work with problems that only involve a single frequency. Then you only need amplitude and phase information, not frequency information (hence the name phasor). The wikipedia article is reasonably good:

http://en.wikipedia.org/wiki/Phasor_(sine_waves [Broken])

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Last edited by a moderator: May 4, 2017
3. Mar 29, 2009

KaPluie

A phasor is a tool to relate the amplitude of a wave to the phase of a wave. For example, if you remove the time dependence from a sin wave and just look at the spatial dependence, and you write it in phasor form, you can say I(r) = A*exp{-jkr} where A is the wave's amplitude, r, the direction of propagation, k the propagation constant. Now, if you map this in polar form (the phase, phi, is simply k*r), you'll get a circle, where you can say the y-axis represent imaginary space, and the x-axis real space. You can then see as you trace around the circle, and look at the projection onto the real axis (x-axis), the amplitude grows and shrinks as you go around the circle. In fact, if you were to plot this changing in amplitude vs. space, you'd just get a sin wave.

But, why bother using phasors at all? Why not just use sin waves the entire time? When you start dealing with more advanced interactions between waves, the math becomes hard with sin waves. For example, you can easily take derivatives and integrals of exponential functions, while as you start to multiply and integrate sin functions, it can be very messy. So, one method is to stick with the phasor to make the math much easier, then if you take the real part of the result (poject the answer onto the x-axis ... aka the real axis), you get the correct physical (observable) result.

Last edited: Mar 29, 2009