Time averages

The phasor sum of two quantities which vary sinusoidally in time with the same frequency is the sum of the phasors of the two quantities. However, the product of the two quantities does not have a phasor which is the product of the two phasors! This concept is illustrated by the following example.

Consider a one-port device with a voltage $v(t)$ across its terminals and a current $i(t)$ entering the one-port at the higher-voltage terminal. The instantaneous power entering the one-port is the product

$$p(t)=v(t)i(t)$$ (2.61)

If $v(t)$ and $i(t)$ vary sinusoidally in time with angular frequency $\omega$, then

$$v(t) =V_0 \cos ( \omega t + \varphi_v)=frac{1}{2}[V e^{J \omega t} + V^* e^{-j \omega t} ]$$ (2.62)

$$i(t) = I_0 \cos ( \omega t + \varphi_i)=\frac{1}{2}[I e^{j \omega t} + I^* e^{-j \omega t} ] \space ,$$

where

$$V=V_0 e^{j \varphi_v} \space , \quad I=I_0 e^{j \varphi_i}$$ (2.63)

are the phasor voltage and current. Substitution of (A.62) into (A.61) yields:

$$p(t)=\frac{1}{2} \mathrm{Re}\{V I^*\}+ \frac{1}{2} \mathrm{Re}\{V I e^{j 2\omega t} \} \space .$$ (2.64)

Hence the instantaneous power does not vary sinusoidally with time and, consequently, cannot be represented by a phasor. It consists of the sum of two terms: the first term is independent of time, whereas the second term in (A.64) varies with time at twice the frequency of voltage and current. The first term in (A.64) corresponds to the time-averaged power:

$$<p(t)>=\frac{1}{2}\mathrm{Re}\{ V I^*\}=\mathrm{Re}\{V_{\mathrm{rms}} I^*_{\mathrm{rms}} \}$$ (2.65)

with $V_{\mathrm{rms}}=V/\sqrt{2}$ and $I_{\mathrm{rms}}I/\sqrt{2}$