I am trying to find the fourier series expansion of the following periodic waveform, I.
This is an odd waveform because it satisfies $ f(-x) = -f(x)$.
$a_0 = 0$ (No average component)
$a_n = 0$ because intergration over -180 deg to 180 deg over an odd waveform ($I$ times $cos(n\theta)$) is 0.
For this waveform can I say that I only need to find the coeffecients $ b_n $ because odd $f(x)$ multiplied to odd $g(x)$ = even $h(x)$ even if there are zero intervals in my waveform?
which of the two limits of Intergration below is correct? I am going to write in terms of $\theta$ degrees for ease of calculation
$b_n = 2 \times \frac{2}{360} \int_0^{120} I_d sin(n \theta) d\theta $
Or
$b_n = \frac{2}{360} \int_{30}^{150} I_d sin(n \theta) d\theta - \int_{210}^{330} I_d sin(n \theta)d\theta $
$b_n = 2 \times \frac{2}{360} \int_{30}^{150} I_d sin(n \theta) d\theta$
$b_n = \frac{- I_d}{90n} \left[ cos(n \theta) \right]$ from 150 to 30
$b_n = \frac{I_d}{90n} \left[ cos(30n) - cos(150n) \right]$
Now let us look at the cosine waveform to calculate the values for 30,150 degrees and their harmonics
for $n = 1$ first harmonic
$b_1 = \frac{I_d}{90n} \left[ cos(30) - cos(150) \right]$
$b_1 = \frac{I_d}{90} \left[ \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right]$
$b_1 = \frac{I_d}{90} \left[ \sqrt{3} \right]$
In terms of RMS and in radians
$b_1 = \frac{I_d}{\pi} \left[ \sqrt{6} \right]$
How about other harmonics?
n = 2, (60, 300) -> difference(0.5, 0.5) -> 0
n = 3, (90, 450) -> difference(0, 0) -> 0
n = 4, (120, 600) -> difference(-0.5, -0.5) -> 0
n = 5, (150, 750) -> difference(-0.866, 0.866) -> -$\sqrt{3}$
n = 6, (180, 900) -> difference(-1, -1) -> 0
n = 7, (210, 1050) -> difference(-0.866, 0.866) -> $-\sqrt{3}$
n = 8, (240, 1200) -> difference(-0.5,-0.5) -> 0
n = 9, (270, 1350) -> difference(0, 0) -> 0
n = 10, (300, 1500) -> difference(0.5, 0.5) -> 0
n = 11, (330, 1650) -> difference(0.866, -0.866) -> $+\sqrt{3}$
n = 12,
n = 13, (390, 1650) -> difference(0.866, -0.866) -> $+\sqrt{3}$
How to find a general formual for the coeffecient above n = 1 easily?