- #1
kataya
- 22
- 0
For two phasors a and b, with magnitudes of A,B respectively and phase angles of \phi_{a) and \phi_{b}, the angle of the sum of the phasors (call it p) has a magnitude of:
P^{2} = A^{2} + B^{2} ,
and a phase angle of:
\phi_{p} = -tan^{-1}(\frac{B}{A})
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:
\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}})
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
P^{2} = A^{2} + B^{2} ,
and a phase angle of:
\phi_{p} = -tan^{-1}(\frac{B}{A})
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:
\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}})
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