symmetric vs asymmetric Fourier series

I have experimental data that in some respect can be modelled by an asymmetric triangle function (1-t) for 0<=t<=1. It is not quite linear as shown in the plots.pdf, but for simplicity sake it can be approximated as such. The attached m-file generates an 8th order Fourier series to fit the asymmetric triangle. It also examines the effect of making the asymmetric triangle symmetric. The frequency content, 2*pi*k/T, for both triangles is still the same. Of course the Fourier coefficients will be different, but it appears the convergence is superior for the symmetric triangle as indicated by the plot fit statistics Rsquared & RMS error.

I could do this with my experimental data by mirroring it about t=0. Is this a bad idea? If I can do the symmetric case I think I can fit the data with much fewer terms. If anyone has experience on these matters I would appreciate your comments.

If you wish to run my scripts, run the FriedlanderFourierSeriesb.m. That routine calls FriedlanderFourierSeries2.m automatically. To toggle between the symmetric vs asymmetric case simply comment out either line 8 or 9 accordingly.

Answer 1

[David Goodmanson]

Hi Jeff, Yes it's a bad idea, assuming that a Friedlander fourier series is just a particular way of obtaining the usual fourier series. Each triangle on its own has the same basic frequency content, but when you chunk them together you lose the abrupt discontinuity at the beginning. That discontinuity has a lot of frequency content, so the fourier coefficients are not at all the same.

Here's an example using a 31 point fft. The two sets of fourier coefficients are normalized to have the same peak value so that the comparison is more obvious. (Don't take the x axis as actual frequencies, it's merely proportional to frequency). Note that the transform of the chunked-together version has a lot less frequency content relative to its peak.

x = -1.5:.1:1.5;
y1 = exp(-x);                       % one
y1(x<0) = 0;
y2 = exp(-abs(x));                  % both of them
z1 = fftshift(fft(ifftshift(y1)));
z2 = fftshift(fft(ifftshift(y2)));
z1norm = abs(z1)/max(abs(z1));
z2norm = abs(z2)/max(abs(z2));
plot(x,z1norm,'o',x,z2norm,'o')
legend('one','both')

[Jeff]

Thanks David,

I do acknowledge the difference of the Fourier coefficients vs frequency, but when you invert back into the time domain it seems that the case for symmetry actually tracks the asymmetric signal better for the same number of terms in the Fourier series. Am I wrong on this?

Despite the Fourier coefficients differing in magnitude at a given frequency the spectra for both cases have the same shape. So can it be said the 2 cases have dynamic similarity?

This is why I am perplexed & seeking other opinions.

[David Goodmanson]

Hi Jeff, I see that I didn't really address the question you were asking. If you are not concerned about frequency content but rather goodness of fit, then I think what you did should work because you eliminated a discontinuity and are transforming a much smoother function, in a situation where you can recreate the discontinuity later. If you try this on a function like x*exp(-x), though, I don't think you will see much of an advantage.

Once someone has answered a question it gets less attention on this site, so if you delete your question and repost it with the some clarification that frequency content is not really the issue, then it will get a fresh look. (My answer gets killed of too but that's fine).

[Jeff]

David,

I do not see where the option exists to delete the question. I suppose I would need to rephrase the question with a new title. Nonetheless, your script seems to demonstrate the difference in the rate of convergence for the 31 terms between the 2 cases. If you plot on a log log scales for (+) values of x you can see the difference in slopes.

[David Goodmanson]

If questions are like comments and answers there would be a trash can icon off to the right of your name that you could click on. If you modified the question instead, it doesn't seem to me that you would have to change the title. It's already a lot better than a lot of them. So one option would be, you modify the question, I delete my answer, which would bring an unanswered question to the top of the queue, I think.

Something I did not mention before is a theorem from continuous fourier transforms that seems important enough to have a name attached to it, but apparently it doesn't. To the extent that the discrete transform resembles the continuous one, which is true here, then:

if a(t) has a discontinuous nth derivative anywhere, then its fourier transform b(f) ~~ f^-(n+1) for large f. In this case the original function has a discontinuity in the 0th derivative and goes like f^-1, and the chunked-together one only has a discontinuity in the 1st derivative and goes like f^-2.

[Jeff]

I do not see the trash icon for the question. I do see it for my comments. Anyway, I think I will leave the question as is for the moment. _

if a(t) has a discontinuous nth derivative anywhere, then its fourier transform b(f) ~~ f^-(n+1) for large f. In this case the original function has a discontinuity in the 0th derivative and goes like f^-1, and the chunked-together one only has a discontinuity in the 1st derivative and goes like f^-2._

If I am interpreting your statement correctly you are talking about the rate of convergence as the number of terms in the series increases?

[David Goodmanson]

Not really. I am talking about the absolute value of the size of the terms themselves. There may be some frequencies where an individual term is small, but on the whole they follow that trend.

[Jeff]

OK, you are talking about the magnitude of the Fourier coefficients & their rate of reduction in magnitude with frequency?

[David Goodmanson]

yes