RMS Value of the Fourier Series

RMS Value of the Fourier SeriesSwapnil Sunil JainDecember 28, 2006

RMS Value of the Fourier Series

If a function $f(t)$ is given by its Fourier series i.e.

$f(t)$$=$${\displaystyle \frac{a_0}{2}}+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}[a_k\cos (k\omega t)+b_k\sin (k\omega t)]$

The RMS value $F_{\mathrm{rms}}$ of $f(t)$ is

$F_{\mathrm{rms}}$$=$${\displaystyle \frac{a_0{}^2}{4}}+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}a_k{}^2+b_k{}^2$

Proof:

The RMS value of a function $f(t)$ is, by definition, given by

$F_{\mathrm{rms}}$$=$${\displaystyle \frac{1}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}{[f(t)]}^2𝑑t$

Then,

$F_{\mathrm{rms}}$$={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}{[f(t)]}^2𝑑t$

$={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)\cdot f(t)𝑑t$

$={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)\cdot \left({\displaystyle \frac{a_0}{2}}+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left[a_k\cos (k\omega t)+b_k\sin (k\omega t)\right]\right)𝑑t$

$={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}\left({\displaystyle \frac{a_{0}f(t)}{2}}+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left[a_{k}f(t)\cos (k\omega t)+b_{k}f(t)\sin (k\omega t)\right]\right)𝑑t$

$={\displaystyle \frac{1}{T}}\left({\displaystyle {\int }_{t_0}^{t_0+T}}{\displaystyle \frac{a_{0}f(t)}{2}}𝑑t+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left[{\displaystyle {\int }_{t_0}^{t_0+T}}{a}_{k}f(t)\cos (k\omega t)𝑑t+{\displaystyle {\int }_{t_0}^{t_0+T}}{b}_{k}f(t)\sin (k\omega t)𝑑t\right]\right)$

$={\displaystyle \frac{1}{T}}\left({\displaystyle \frac{a_0}{2}}{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)𝑑t+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left[a_k{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)\cos (k\omega t)𝑑t+b_k{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)\sin (k\omega t)𝑑t\right]\right)$

$={\displaystyle \frac{1}{T}}\left({\displaystyle \frac{a_0}{2}}(a_0{\displaystyle \frac{T}{2}})+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left[a_k(a_k{\displaystyle \frac{T}{2}})+b_k(b_k{\displaystyle \frac{T}{2}})\right]\right)$

$={\displaystyle \frac{a_0{}^2}{4}}+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{a_k{}^2}{2}}+{\displaystyle \frac{b_k{}^2}{2}}$

$={\displaystyle \frac{a_0{}^2}{4}}+{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}a_k{}^2+b_k{}^2$