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Average Power Calculations of Periodic Functions Using Fourier Series
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,
$$\mathrm{P=I^2R=\frac{V^2}{R}\:\:\:\:\:\:....(1)}$$
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,
$$\mathrm{P=i^2(t)R=\frac{V^2(t)}{R}\:\:\:\:\:\:....(2)}$$
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,
$$\mathrm{p=i^2(t)=v^2(t)\:\:\:\:\:\:....(3)}$$
Therefore, the instantaneous power of a signal x(t) can be given by
$$\mathrm{p=x^2(t)\:\:\:\:\:\:....(4)}$$
Hence, the average power of x(t) over a certain interval of time is,
$$\mathrm{Average\:power,\:\:P=\frac{1}{T}\int_{0}^{T}x^2(t)dt\:\:\:\:\:\:....(5)}$$
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,
$$\mathrm{P=a_{0}^{2}+\sum_{n=1}^{\infty}\frac{a_{n}^{2}}{2}+\frac{b_{n}^{2}}{2}\:\:\:\:\:....(7)}$$
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 −
$$\mathrm{x(t)=\cos^2(4000\pi t)\sin (10000\pi t)}$$
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
$$\mathrm{P=\frac{(1/2)^2}{2}+\frac{(1/4)^2}{2}+\frac{(1/4)^2}{2}}$$
$$\mathrm{\Rightarrow P=\frac{1}{8}+\frac{1}{32}+\frac{1}{32}=\frac{3}{16}\: Watts}$$
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