Study of commonly encounter waveform in the power electronics system - EX 1 #2

Fourier series

Let consider the output waveform of the power electronics devices is $f(t)$ and Fourier series representation of  $f(t)$ is as follows:

  $f(t)$ =  $a_0$ + $\sum ^\infty _{n=1}$ $(a_n  \cos (n\omega_0 t) + b_n  \sin (n\omega_0 t))$

Where,

$a_0 =\frac{1}{T} \int ^{T}_{0} f(t) dt$

$a_n = \frac{2}{T} \int ^{T}_{0} f(t) \cos (n \omega t) dt$

$b_n = \frac{2}{T} \int ^{T}_{0} f(t) \sin (n \omega t) dt$

Here if  :   $f(t)$ is even function  $\implies$  $f(t) = f(-t)$ $\implies$  $b_n$ = 0, $a_0 , a_n$ $\neq$ 0

                 $f(t)$ is odd function  $\implies$  $f(t) = -f(-t)$ $\implies$  $b_n \neq$ 0, $a_0 , a_n$=0
                 $f(t)$ is half wave symmetric   $\implies$  $f(t+ \frac{T}{2}) = -f(t)$ $\implies$  only odd harmonics (n=1,3,5...)

Example-1

Consider square waveform v(t) as shown in figure

Here,  $v(t) = -v(t+\pi)$ $\implies$ half wave symmetric  $\implies$ Only odd harmonics in waveform (n=1,3,5....) 
  

Here,  $a_0,a_n$ = 0  because of half wave symmetric .

Hence, the derived Fourier series of the square waveform $v(t)$ is as follows:

     RMS voltage of v(t) is