RMS Value of the Fourier Series


RMS Value of the Fourier Series Swapnil Sunil Jain December 28, 2006

RMS Value of the Fourier Series

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

f⁢(t) = a02+∑k=1∞[ak⁢cos⁡(k⁢ω⁢t)+bk⁢sin⁡(k⁢ω⁢t)]

The RMS value Frms of f⁢(t) is

Frms = a024+12⁢∑k=1∞ak2+bk2

Proof:

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

Frms = 1T⁢∫t0t0+T[f⁢(t)]2⁢𝑑t

Then,

Frms =1T⁢∫t0t0+T[f⁢(t)]2⁢𝑑t
=1T⁢∫t0t0+Tf⁢(t)⋅f⁢(t)⁢𝑑t
=1T⁢∫t0t0+Tf⁢(t)⋅(a02+∑k=1∞[ak⁢cos⁡(k⁢ω⁢t)+bk⁢sin⁡(k⁢ω⁢t)])⁢𝑑t
=1T⁢∫t0t0+T(a0⁢f⁢(t)2+∑k=1∞[ak⁢f⁢(t)⁢cos⁡(k⁢ω⁢t)+bk⁢f⁢(t)⁢sin⁡(k⁢ω⁢t)])⁢𝑑t
=1T⁢(∫t0t0+Ta0⁢f⁢(t)2⁢𝑑t+∑k=1∞[∫t0t0+Tak⁢f⁢(t)⁢cos⁡(k⁢ω⁢t)⁢𝑑t+∫t0t0+Tbk⁢f⁢(t)⁢sin⁡(k⁢ω⁢t)⁢𝑑t])
=1T⁢(a02⁢∫t0t0+Tf⁢(t)⁢𝑑t+∑k=1∞[ak⁢∫t0t0+Tf⁢(t)⁢cos⁡(k⁢ω⁢t)⁢𝑑t+bk⁢∫t0t0+Tf⁢(t)⁢sin⁡(k⁢ω⁢t)⁢𝑑t])
=1T⁢(a02⁢(a0⁢T2)+∑k=1∞[ak⁢(ak⁢T2)+bk⁢(bk⁢T2)])
=a024+∑k=1∞ak22+bk22
=a024+12⁢∑k=1∞ak2+bk2
Title RMS Value of the Fourier Series
Canonical name RMSValueOfTheFourierSeries1
Date of creation 2013-03-11 19:30:56
Last modified on 2013-03-11 19:30:56
Owner swapnizzle (13346)
Last modified by (0)
Numerical id 1
Author swapnizzle (0)
Entry type Definition