0
$\begingroup$
  • Harmonic Addition Theorem
  • Harmonic Addition Formula
  • Phasor Addition Theorem
  • Phasor Addition Formula

Those four name can be used as a keyword on google.
I haven't known the official name and think that they seems to be equivalent each other.


From this paper on page 3

A Hyperbolic Analog of the Phasor Addition Formula by F. Adrián F. Tojo (July 30, 2018)

enter image description here

Figure 1. Part of page 3 of the paper mentioned.


If I rewrite that,

$$ \displaystyle \mathrm{a} \ \mathrm{e}^{j \alpha} + \mathrm{b} \ \mathrm{e}^{j \beta} = \sqrt{ \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left( \alpha - \beta \right) } \quad \mathrm{e}^{\ j \ \arg\left[ \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \right]} $$


So if we take the real part of it, we get

$$ \small \displaystyle \mathrm{a}\cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) = \sqrt{ \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left( \alpha - \beta \right) } \: \cos\left[ \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \right] \right] $$


Suppose if $\alpha = \omega t + \phi_1$ and $\beta = \omega t + \phi_2$, then

$$ \begin{align} \small \displaystyle \mathrm{a}\cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \\ &\cdot\cos\left[ \small \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \omega t + \phi_1 \right) + \mathrm{b} \sin\left( \omega t + \phi_2 \right) \right] \right] \right] \end{align} $$


Why can we take the frequency out of the complex argument like this?

$$ \begin{align} \small \displaystyle \mathrm{a}\cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \\ &\cdot\cos\left[ \small \omega t + \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \phi_1 \right) + \mathrm{b} \cos\left( \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \phi_1 \right) + \mathrm{b} \sin\left( \phi_2 \right) \right] \right] \right] \end{align} $$

$\endgroup$
0
$\begingroup$

$ \displaystyle \mathrm{a} \, \mathrm{e}^{j \alpha} + \mathrm{b} \, \mathrm{e}^{j \beta} = \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \label{1}\tag{1} $

Remember that the complex argument form inside of the cosine is equivalent to \eqref{1}.
Or just use Euler's formula, it's the same.

$ \small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) = \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \cdot\cos\left[ \small \mathrm{arg}\!\left[ \mathrm{a} \, \mathrm{e}^{j \left( \omega t + \phi_1 \right)} + \mathrm{b} \, \mathrm{e}^{j \left( \omega t + \phi_2 \right)} \right] \right] $

Factor out the frequency

$ \small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) = \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \cdot\cos\left[ \small \mathrm{arg}\!\left[ \mathrm{e}^{j \omega t} \left( \mathrm{a} \, \mathrm{e}^{j \phi_1 } + \mathrm{b} \, \mathrm{e}^{j \phi_2 } \right) \right] \right] $

Remember the complex argument identities

$$ \mathrm{arg}\!\left(z_1 z_2\right) = \mathrm{arg}\!\left(z_1\right) + \mathrm{arg}\!\left(z_2\right) $$

And also the fact that

$$ \mathrm{arg}\!\left(\mathrm{e}^{j \theta}\right) = \theta $$

Thus

$ \begin{align} \small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right)} \\ &\cdot\cos\left[ \small \omega t + \mathrm{arg}\!\left[ \left[ \mathrm{a} \cos\left( \phi_1 \right) + \mathrm{b} \cos\left( \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \phi_1 \right) + \mathrm{b} \sin\left( + \phi_2 \right) \right] \right] \right] \end{align} $

The point is that, if the cosines on the left side has the same phase part which is separated by addition/subtraction sign, we can take out of it from the complex argument function, hence simplifies it.

$\endgroup$
  • $\begingroup$ Please critic my answer if it's incorrect. $\endgroup$ – Unknown123 May 13 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.