Help needed for Fourier analysis of a triangular wave

AI Thread Summary
The discussion revolves around performing Fourier analysis on a triangular wave related to a boost converter, focusing on the duty cycle and amplitude. Participants share their equations and methods for analyzing the waveform, with one user providing a compact solution using an odd function approach over the interval [-π, π]. There is a concern about the symmetry of the output waveform, suggesting the need for a DC component to match the original non-symmetrical waveform. Additionally, the importance of variable parameters like amplitude is highlighted, as they affect the analysis. The thread concludes with users sharing plots to compare their results and verify the accuracy of the equations.
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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.
 
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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.
 
Thank you Vela, currently trying to model the equation in MATLAB or maple 13, hopefully I will. Cheers
 
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
 

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Last edited:
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.
 

Attachments

  • fourier.jpg
    fourier.jpg
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