Average Power Calculations of Periodic Functions Using Fourier Series

Signals and SystemsElectronics & ElectricalDigital Electronics

When a voltage of V volts is applied across a resistance of R Ω, then a current I flows through it. The power dissipated in the resistance is given by,


But when the voltage and current signals are not constant, then the power varies at every instant, and the equation for the instantaneous power is given by,


Where, 𝑖(𝑡) and 𝑣(𝑡) are the corresponding instantaneous values of current and voltage respectively

Now, if the value of the resistance (R) is 1 Ω, then the instantaneous power can be represented as,


Therefore, the instantaneous power of a signal x(t) can be given by


Hence, the average power of x(t) over a certain interval of time is,


By using Parseval’s theorem, we get

$$\mathrm{\frac{1}{T}\int_{0}^{T}|x(t)|^2dt=\sum_{\substack{n=-\infty \ n
eq 0}}^{\infty} |C_n|^2=C_{0}^{2}+\sum_{\substack{n=-\infty \ n
eq 0}}^{\infty}C_{n}^{2}}$$

$$\mathrm{\Rightarrow \frac{1}{T}\int_{0}^{T}|x(t)|^2dt=C_{0}^{2}+\sum_{\substack{n=-\infty \ n
eq 0}}^{\infty}C_{n}C_{n}^{*}}$$

$$\mathrm{\Rightarrow \frac{1}{T}\int_{0}^{T}|x(t)|^2dt=a_{0}^{2}+\sum_{n=1}^{\infty}2[Re(C_{n}^{2})+Im(C_{n}^{2})]}$$

$$\mathrm{\Rightarrow \frac{1}{T}\int_{0}^{T}|x(t)|^2dt=a_{0}^{2}+\sum_{n=1}^{\infty}\frac{a_{n}^{2}}{2}+\frac{b_{n}^{2}}{2}\:\:\:\:\:....(6)}$$

On comparing equations (5) & (6), we get,


Therefore, the average power over a period of time can be given using the Fourier series as,

$$\mathrm{Avg\:Power = (DC\: term)^2+\sum(Mean\:square\: values \:of \:cosine \:terms)+\sum(Mean \:square\: values\: of \:sine\: terms)\:\:\:\:....(8)}$$

Numerical Example

Determine the average power of the following signal −

$$\mathrm{x(t)=\cos^2(4000\pi t)\sin (10000\pi t)}$$


The given signal is,

$$\mathrm{x(t)=\cos^2(4000\pi t)\sin (10000\pi t)}$$

$$\mathrm{\because \cos^2 \theta = \frac{1 +\cos2 \theta}{2}}$$

$$\mathrm{\therefore x(t)=(\frac{1+\cos8000\pi t}{2})\sin(10000\pi t)}$$

$$\mathrm{\Rightarrow x(t)=\frac{1}{2}\sin(10000\pi t)+\frac{1}{2}\sin(10000\pi t)\cos(8000\pi t)}$$

$$\mathrm{\because \sin X\cos Y = \frac{\sin(X+Y)\sin(X-Y)}{2}}$$

$$\mathrm{\therefore x(t)=\frac{1}{2}\sin(10000\pi t)+\frac{1}{4}\sin(18000\pi t)+\frac{1}{4}\sin(2000\pi t)}$$

Hence, from equation (8), the average power of the signal is


$$\mathrm{\Rightarrow P=\frac{1}{8}+\frac{1}{32}+\frac{1}{32}=\frac{3}{16}\: Watts}$$

Updated on 03-Dec-2021 13:22:43