Help needed for Fourier analysis of a triangular wave

[sabrank]

Homework Statement

Fourie analysis for a boost converter
Where d = duty cycle, A = amplitude and t = time

Homework Equations

Fx = (2At/d) for 0<t<d/2, -A+(2A/(1-d))*(1-t-d/2) for d/2<t<(1-d/2); (2A/d)*(t-1) for 1-d/2 <t<1.
Where d = duty cycle, A = amplitude and t = time

The Attempt at a Solution

My solution is kind off large, I need need another solution to compare my ans with before writing a script to execute this function.

 

[vela]

I recast your problem slightly and got a pretty compact answer. Instead of working with the interval [0,1], I used the interval [-π,π], took advantage of the fact it is an odd function, and got

$$f(x) = \sum_{n=1}^\infty \frac{2A\sin{n\pi d}}{d(1-d)n^2\pi^2}\sin nt$$

Your d might be my 1-d and vice versa, and you can rescale t by a factor of 2π to get back to your original function.

 

[sabrank]

Thank you Vela, currently trying to model the equation in MATLAB or maple 13, hopefully I will. Cheers

 

[sabrank]

Please find attached the wave form I am trying to analyze. I was actually able to simulate the equation given by Vela but the output was symmetrical around zero while my original wave form as shown in the attachment was not symmetrical around zero.
I feel there should be a dc component in order to achieve this waveform consequently an odd function might not be assumed. Please note I have no proof of this.

A is the peak amplitude and 2A could be taken as peak to peak value (A-(-A)), which is equal to 3 in the attached waveform though I will prefer non usage of real value for 'A' because the value of 'A' changes with a change in other parameters e.g. d.

I will appreciate all assistance and suggestion. Thank you

 

[vela]

I'm assuming the function you gave in your original post is the correct one as the one expressed on the spreadsheet has apparent mistakes in the intervals for t.

I've attached a plot of your original function and the series I found (10 terms and with t appropriately rescaled). I used A=1 and d=1/4 for this particular plot. They look the same to me.