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Derivation of Phase Angle Addition Formula

  1. Dec 3, 2007 #1
    For two phasors [tex]a[/tex] and [tex]b[/tex], with magnitudes of A,B respectively and phase angles of [tex]\phi_{a)[/tex] and [tex]\phi_{b}[/tex], the angle of the sum of the phasors (call it p) has a magnitude of:

    [tex]P^{2} = A^{2} + B^{2}[/tex] ,

    and a phase angle of:

    [tex]\phi_{p} = -tan^{-1}(\frac{B}{A})[/tex]

    The magnitude identity makes sense, as it is the geometric sum of the two vectors, but I am having trouble deriving the phase angle identity.

    My original idea was to have something like this:

    [tex]\phi_{p} = tan^{-1}(\frac{A^{2}sin^{2}\phi_{a} + B^{2}sin^{2}\phi_{b}}{A^{2}cos^{2}\phi_{a} + B^{2}cos^{2}\phi_{b}})[/tex]

    But I could not get anywhere with that, or at the very least work it into the form given above.

    Does anyone know how to derive the angle formula? Many thanks
  2. jcsd
  3. Dec 5, 2007 #2


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    The formulas for magnitude and phase you gave are true only if the two phasors are orthogonal.
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