# 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
$$v(t)=V_{\mathrm{dc}}+\sum_{n=1}^{\infty} V_{n} \cos \left(n \omega_{0} t-\theta_{n}\right) \tag{1}$$
$$i(t)=I_{\mathrm{dc}}+\sum_{m=1}^{\infty} I_{m} \cos \left(m \omega_{0} t-\phi_{m}\right) \tag{2}$$
Following the passive sign convention (Fig. 1), the average power is
Fig. 1: The voltage polarity reference and current reference direction.
$$P=\frac{1}{T} \int_{0}^{T} \text { vi } d t \tag{3}$$
Substituting Eqs. (1) and (1) into Eq. (3) gives
\begin{aligned}P=& \frac{1}{T} \int_{0}^{T} V_{\mathrm{dc}} I_{\mathrm{dc}} d t+\sum_{m=1}^{\infty} \frac{I_{m} V_{\mathrm{dc}}}{T} \int_{0}^{T} \cos \left(m \omega_{0} t-\phi_{m}\right) d t \\&+\sum_{n=1}^{\infty} \frac{V_{n} I_{\mathrm{dc}}}{T} \int_{0}^{T} \cos \left(n \omega_{0} t-\theta_{n}\right) d t \\&+\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{V_{n} I_{m}}{T} \int_{0}^{T} \cos \left(n \omega_{0} t-\theta_{n}\right) \cos \left(m \omega_{0} t-\phi_{m}\right) d t\end{aligned} \tag{4}
Equations from the other pages
$$\begin{array}{c} \int_{0}^{T} \sin n \omega_{0} t d t=0 \quad (A.a) \\ \int_{0}^{T} \cos n \omega_{0} t d t=0 \quad (A.b) \\ \int_{0}^{T} \sin n \omega_{0} t \cos m \omega_{0} t d t=0 \quad (A.c) \\ \int_{0}^{T} \sin n \omega_{0} t \sin m \omega_{0} t d t=0, \quad(m \neq n) \quad (A.d)\\ \int_{0}^{T} \cos n \omega_{0} t \cos m \omega_{0} t d t=0, \quad(m \neq n) \quad (A.e)\\ \int_{0}^{T} \sin ^{2} n \omega_{0} t d t=\frac{T}{2} \quad (A.f)\\ \int_{0}^{T} \cos ^{2} n \omega_{0} t d t=\frac{T}{2} \quad (A.g) \end{array}$$ $$f(t)=a_{0}+\sum_{n=1}^{\infty} A_{n} \cos \left(n \omega_{0} t+\phi_{n}\right) \tag{B}$$
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
$$P=V_{\mathrm{dc}} I_{\mathrm{dc}}+\frac{1}{2} \sum_{n=1}^{\infty} V_{n} I_{n} \cos \left(\theta_{n}-\phi_{n}\right) \tag{5}$$
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
$$F_{\text {mas }}=\sqrt{\frac{1}{T} \int_{0}^{T} f^{2}(t) d t} \tag{6}$$
Substituting $f(t)$ in Eq. (B) into Eq. (6) and noting that $(a+b)^{2}=a^{2}+2 a b+b^{2}$, we obtain
\begin{aligned}F_{\mathrm{ms}}^{2}=& \frac{1}{T} \int_{0}^{T}\left[a_{0}^{2}+2 \sum_{n=1}^{\infty} a_{0} A_{n} \cos \left(n \omega_{0} t+\phi_{n}\right)\right.\\&\left.+\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{n} A_{m} \cos \left(n \omega_{0} t+\phi_{n}\right) \cos \left(m \omega_{0} t+\phi_{m}\right)\right] d t \\=& \frac{1}{T} \int_{0}^{T} a_{0}^{2} d t+2 \sum_{n=1}^{\infty} a_{0} A_{n} \frac{1}{T} \int_{0}^{T} \cos \left(n \omega_{0} t+\phi_{n}\right) d t \\&+\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{n} A_{m} \frac{1}{T} \int_{0}^{T} \cos \left(n \omega_{0} t+\phi_{n}\right) \cos \left(m \omega_{0} t+\phi_{m}\right) d t\end{aligned} \tag{7}
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
$$F_{\mathrm{rms}}^{2}=a_{0}^{2}+\frac{1}{2} \sum_{n=1}^{\infty} A_{n}^{2}$$
or
$$F_{\text {rms }}=\sqrt{a_{0}^{2}+\frac{1}{2} \sum_{n=1}^{\infty} A_{n}^{2}} \tag{8}$$
In terms of Fourier coefficients $a_{n}$ and $b_{n}$, Eq. (8) may be written as
$$F_{\text {rms }}=\sqrt{a_{0}^{2}+\frac{1}{2} \sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)} \tag{9}$$
If $f(t)$ is the current through a resistor $R$, then the power dissipated in the resistor is
$$P=R F_{\mathrm{mms}}^{2} \tag{10}$$
Or if $f(t)$ is the voltage across a resistor $R$, the power dissipated in the resistor is
$$P=\frac{F_{\mathrm{rms}}^{2}}{R} \tag{11}$$
One can avoid specifying the nature of the signal by choosing a $1-\Omega$ resistance. The power dissipated by the $1-\Omega$ resistance is
$$P_{1 \Omega}=F_{\mathrm{ms}}^{2}=a_{0}^{2}+\frac{1}{2} \sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right) \tag{12}$$
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,
Parseval's theorem states that the average power in a periodic signal is the sum of the average power in its dc component and the average powers in its harmonics.
Example 1: Determine the average power supplied to the circuit in Fig. $2$ if
$$i(t)=2+10 \cos \left(t+10^{\circ}\right)+6 \cos \left(3 t+35^{\circ}\right) \mathrm{A}$$
Fig. 2: For Example 1.
Solution: The input impedance of the network is
$$\mathbf{Z}=10 \| \frac{1}{j 2 \omega}=\frac{10(1 / j 2 \omega)}{10+1 / j 2 \omega}=\frac{10}{1+j 20 \omega}$$
Hence,
$$\mathbf{V}=\mathbf{I Z}=\frac{10 \mathbf{I}}{\sqrt{1+400 \omega^{2}} \angle{ \tan ^{-1} 20 \omega}}$$
For the dc component, $\omega=0$,
$$\mathbf{I}=2 \mathrm{~A} \quad \Longrightarrow \quad \mathbf{V}=10(2)=20 \mathrm{~V}$$
This is expected, because the capacitor is an open circuit to dc and the entire 2-A current flows through the resistor. For $\omega=1 \mathrm{rad} / \mathrm{s}$,
\begin{aligned}\mathbf{I}=10 \angle 10^{\circ} \quad \Longrightarrow \quad \mathbf{V} &=\frac{10\left(10 \angle 10^{\circ}\right)}{\sqrt{1+400} \angle{ \tan ^{-1} 20}} \\ &=5 \angle-77.14^{\circ}\end{aligned}
For $\omega=3 \mathrm{rad} / \mathrm{s}$,
\begin{aligned}\mathbf{I}=6 \angle 45^{\circ} \quad \Longrightarrow \quad \mathbf{V} &=\frac{10\left(6 / 45^{\circ}\right)}{\sqrt{1+3600} \angle{ \tan ^{-1} 60}} \\ &=1 \angle-44.05^{\circ}\end{aligned}
Thus, in the time domain,
$$v(t)=20+5 \cos \left(t-77.14^{\circ}\right)+1 \cos \left(3 t-44.05^{\circ}\right) \mathrm{V}$$
We obtain the average power supplied to the circuit by applying Eq. (5), as
$$P=V_{\mathrm{dc}} I_{\mathrm{dc}}+\frac{1}{2} \sum_{n=1}^{\infty} V_{n} I_{n} \cos \left(\theta_{n}-\phi_{n}\right)$$
To get the proper signs of $\theta_{n}$ and $\phi_{n}$, we have to compare $v$ and $i$ in this example with Eqs. (1) and (2). Thus,
\begin{aligned}P=& 20(2)+\frac{1}{2}(5)(10) \cos \left[77.14^{\circ}-\left(-10^{\circ}\right)\right] \\&+\frac{1}{2}(1)(6) \cos \left[44.05^{\circ}-\left(-35^{\circ}\right)\right] \\=& 40+1.247+0.05=41.5 \mathrm{~W}\end{aligned}
Alternatively, we can find the average power absorbed by the resistor as
\begin{aligned}P &=\frac{V_{\mathrm{dc}}^{2}}{R}+\frac{1}{2} \sum_{n=1}^{\infty} \frac{\left|V_{n}\right|}{R}=\frac{20^{2}}{10}+\frac{1}{2} \cdot \frac{5^{2}}{10}+\frac{1}{2} \cdot \frac{1^{2}}{10} \\&=40+1.25+0.05=41.5 \mathrm{~W}\end{aligned}
which is the same as the power supplied, since the capacitor absorbs no average power.