MyOpenMath/Complex phasors


 * For more help with phasors visit Phasor algebra

Time average of the product of two signals
Here we multiply two signals with the same angular frequency but with different phases:
 * $$I(t)=I_0\cos(\omega t + \phi_i)=\tfrac 1 2 I_0\exp{i(\omega t + \phi_i)} +cc$$
 * $$V(t)=V_0\cos(\omega t + \phi_v)=\tfrac 1 2 V_0\exp{i(\omega t + \phi_v)} +cc$$,

where $$cc$$ denotes complex conjugate. For example,

$$e^{i\theta}+ cc = e^{i\theta} +e^{-i\theta}= 2\cos\theta.$$

Define $$P=IV$$, and make the algebra easier to follow by defining two phases:
 * $$\Phi_I=\omega t + \phi_i,$$  and,   $$\Phi_V=\omega t + \phi_v$$.

Note that $$PV$$ is the product of two binomials, which yield four terms:

IV = \tfrac 1 4 I_0V_0\left( e^{i\Phi_I}  +  e^{-i\Phi_I} \right)\left( e^{i\Phi_V}  +  e^{-i\Phi_V} \right) $$ When the two binomials are multiplied we obtain four terms. We group them according to whether they involve the sum or difference between the two phases, $$\Phi_I$$ and $$\Phi_V$$, because whether it is a sum or difference affects the time-dependence as follows:



\Phi_I+\Phi_V = 2\omega t + \phi_i + \phi_v $$



\Phi_I-\Phi_V=\phi_i-\phi_v $$

These terms can be grouped into real and imaginary parts, expressed in terms of the sine and cosine functions:

IV = \tfrac 1 4 I_0V_0 \Bigl[  \underbrace{ e^{i(\Phi_I-\Phi_V)}  +     e^{i(\Phi_V-\Phi_I)}}_{2\cos(\Phi_I-\Phi_V)}  \Bigr] + \Bigl[(   \underbrace{ e^{i(\Phi_I+\Phi_V)}  +     e^{-i(\Phi_V+\Phi_V)}}_{2\cos(\Phi_I+\Phi_V)}  \Bigr] $$

With ac circuits it is customary to average over one period, $$T$$, defined by the expression $$\omega T = 2\pi$$. Using the overbar notation to denote this time average, we have:


 * $$\overline{IV} = \tfrac 1 2 I_0V_0 \cos(\phi_i-\phi_v)=I_\text{rms}V_\text{rms}\cos(\phi_i-\phi_v)$$