Average Power and RMS Values of The Fourier Series

To find the average power absorbed by a circuit due to a periodic excitation, we write the voltage and current in amplitude-phase form as Following the passive sign convention (Fig. 1), the average power is Substituting Eqs. (1) and (1) into Eq. (3) gives The second and third integrals vanish, since we are integrating the cosine over its period. According to Eq. (A.e), all terms in the fourth integral are zero when $ m \neq n $. By evaluating the first integral and applying Eq. (A.g) to the fourth integral for the case $ m=n $, we obtain This shows that in average-power calculation involving periodic voltage and current, the total average power is the sum of the average powers in each harmonically related voltage and current. Given a periodic function $ f(t) $, its rms value (or the effective value) is given by Substituting $ f(t) $ in Eq. (B) into Eq. (6) and noting that $ (a+b)^{2}=a^{2}+2 a b+b^{2} $, we obtain Distinct integers $ n $ and $ m $ have been introduced to handle the product of the two series summations. Using the same reasoning as above, we get or In terms of Fourier coefficients $ a_{n} $ and $ b_{n} $, Eq. (8) may be written as If $ f(t) $ is the current through a resistor $ R $, then the power dissipated in the resistor is Or if $ f(t) $ is the voltage across a resistor $ R $, the power dissipated in the resistor is One can avoid specifying the nature of the signal by choosing a $ 1-\Omega $ resistance. The power dissipated by the $ 1-\Omega $ resistance is This result is known as Parseval's theorem. Notice that $ a_{0}^{2} $ is the power in the dc component, while $ 1 / 2\left(a_{n}^{2}+b_{n}^{2}\right) $ is the ac power in the $ n $th harmonic. Thus,