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Amplitude of field with phasor components

  1. Aug 1, 2008 #1
    Let's say I have a field (electric or magnetic or something) and it's time-varying so I choose to represent its components as phasors.

    Say the field is:
    [tex]\vec{F} = X\hat{x} + Y\hat{y} + Z\hat{z}[/tex]

    where, X, Y and Z are complex numbers.

    Now, I want to find the amplitude of the field. If X, Y and Z were real (time constant field), I'd just go:

    [tex]F = \sqrt{X^2 +Y^2 +Z^2}[/tex]

    but I have no idea what to do here. Also, I'm not sure if there's an easy way to do this or not. Any ideas?
  2. jcsd
  3. Aug 1, 2008 #2

    Ben Niehoff

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    Gold Member

    It works just like phasor arithmetic for electronic circuits. The instantaneous amplitude will be

    [tex]f = \sqrt{\Re (X)^2 + \Re (Y)^2 + \Re (Z)^2}[/tex]

    The time-averaged amplitude is

    [tex]F_{avg} = \sqrt{X_{rms}^2 + Y_{rms}^2 + Z_{rms}^2} = \sqrt{\frac{X^*X + Y^*Y + Z^*Z}{2}}[/tex]

    where [itex]X^*[/itex] is the complex conjugate of X.

    Lastly, the total amplitude of F is

    [tex]|\vec F| = \sqrt{\vec F^* \cdot \vec F} = \sqrt{X^*X + Y^*Y + Z^*Z}[/tex]

    This is the magnitude of a vector in six-dimensional space, which rotates around on a 5-sphere such that its projections along the six axes are equal to the real and imaginary components of each of X, Y, and Z.

    (Alternatively, you can think of F as living in three-dimensional complex space, such that its projection on each of the three complex axes gives the three complex numbers X, Y, and Z).
  4. Aug 1, 2008 #3
    Thanks a lot! Answers my question perfectly.
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