Derivation of Phase Angle Addition Formula

[kataya]

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

 

[CEL]

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

 

[piercas]

.

The phase angle addition formula can be derived using the properties of trigonometric functions and the definition of phasors. First, let's define the two phasors a and b as:

a = A(cos(\phi_a) + isin(\phi_a))

b = B(cos(\phi_b) + isin(\phi_b))

where A and B are the magnitudes of the phasors and \phi_a and \phi_b are their respective phase angles.

Now, let's consider the sum of these two phasors, denoted as p:

p = a + b = A(cos(\phi_a) + isin(\phi_a)) + B(cos(\phi_b) + isin(\phi_b))

Using the properties of complex numbers, we can expand this equation to get:

p = (Acos(\phi_a) + Bcos(\phi_b)) + i(Asin(\phi_a) + Bsin(\phi_b))

The magnitude of p can be calculated using the Pythagorean theorem:

|p| = \sqrt{(Acos(\phi_a) + Bcos(\phi_b))^2 + (Asin(\phi_a) + Bsin(\phi_b))^2}

Simplifying this equation, we get:

|p| = \sqrt{A^2cos^2(\phi_a) + B^2cos^2(\phi_b) + 2ABcos(\phi_a)cos(\phi_b) + A^2sin^2(\phi_a) + B^2sin^2(\phi_b) + 2ABsin(\phi_a)sin(\phi_b)}

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can further simplify this equation to get:

|p| = \sqrt{A^2 + B^2 + 2AB(cos(\phi_a)cos(\phi_b) + sin(\phi_a)sin(\phi_b))}

Using the trigonometric identity cos(x-y) = cos(x)cos(y) + sin(x)sin(y), we can write the above equation as:

|p| = \sqrt{A^2 + B^2 + 2ABcos(\phi_a - \phi_b)}

Now, we can use the inverse tangent function to find the phase angle of p:

\phi_p = tan^{-1}(\frac{Im(p)}{Re(p)})