Average Power Calculations of Periodic Functions Using Fourier Series

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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,

$$\mathrm{P \:=\: I^2R \:=\: \frac{V^2}{R}\:\: \dotso \:(1)}$$

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,

$$\mathrm{P \:=\: i^2(t)R \:=\: \frac{V^2(t)}{R}\:\: \dotso \:(2)}$$

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,

$$\mathrm{p \:=\: i^2(t) \:=\: v^2(t)\:\: \dotso \:(3)}$$

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

$$\mathrm{p \:=\: x^2(t)\:\: \dotso \:(4)}$$

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

$$\mathrm{Average\:power,\:\:P \:=\: \frac{1}{T}\int_{0}^{T}x^2(t)dt\:\: \dotso \:(5)}$$

By using Parseval's theorem, we get

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

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

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

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

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

$$\mathrm{P\:=\:a_{0}^{2}\:+\:\sum_{n=1}^{\infty}\frac{a_{n}^{2}}{2}\:+\:\frac{b_{n}^{2}}{2}\:\: \dotso \:(7)}$$

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

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

Numerical Example

Determine the average power of the following signal −

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

Soution

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

$$\mathrm{P\:=\:\frac{(1/2)^2}{2}\:+\:\frac{(1/4)^2}{2}\:+\:\frac{(1/4)^2}{2}}$$

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