I Fourier series of Dirac comb, complex VS real approaches

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The discussion revolves around computing the Fourier series coefficients for the Dirac comb function using both complex and real formulas. The results indicate that the complex formula yields coefficients of Cn = 1/T, while the real formula gives a0 = 1/T, an = 2/T, and bn = 0. A key point of confusion is the interpretation of amplitude in the frequency spectrum, with the realization that only the real coefficients represent the actual amplitudes of the frequencies. The conversation highlights the importance of understanding the relationship between complex and real coefficients, emphasizing that complex coefficients include both real and imaginary parts, which can complicate the analysis. Ultimately, the consensus is that the real coefficients are more meaningful for representing frequency amplitudes in this context.
DoobleD
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Hello,

I tried to compute the Fourier series coefficients for the Dirac comb function. I did it using both the "complex" formula and the "real" formula for the Fourier series, and I got :

- complex formula : Cn = 1/T
- real formula : a0 = 1/T, an = 2/T, bn = 0

This seems to be valid since it is coherent with the equations relatins cn, a0, an, and bn. The thing that bugs me now, is : what is really the amplitude of the frequencies in the spectrum of the Dirac comb function ? Is it 1/T or is it 2/T ? How can those two results be compatible ? I must be misunderstanding something here.

To be as clear as possible, here is what I did and what bugs me (at the end) :
 
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Isn't it just as simple as
$$ \frac{2 \cos x}{T} = \frac{e^{ix}+e^{-ix}}{T}$$
?
 
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jasonRF said:
Isn't it just as simple as
$$ \frac{2 \cos x}{T} = \frac{e^{ix}+e^{-ix}}{T}$$
?
Thank you for answering. I'm not sure what you mean, how do you get that result ?
 
DoobleD said:
Thank you for answering. I'm not sure what you mean, how do you get that result ?
##e^{ix} = \cos(x) + i \sin(x)##
 
Orodruin said:
##e^{ix} = \cos(x) + i \sin(x)##

Yep, but I still don't get how is this related to my initial question ? I might be missing something obvious...
 
If you look at this:
$$ {2 \cos \omega} = e^{i\omega}+e^{-i\omega} $$
With the complex formula you have something sort of like "negative frequencies". But since they add together, the amplitude of each is half.

That is my sort of layman's explanation.
 
scottdave said:
If you look at this:
$$ {2 \cos \omega} = e^{i\omega}+e^{-i\omega} $$
With the complex formula you have something sort of like "negative frequencies". But since they add together, the amplitude of each is half.

That is my sort of layman's explanation.

Again, how does this relates to my question ? I'll copy/paste it here : what is really the amplitude of the frequencies in the spectrum of the Dirac comb function ? Is it 1/T or is it 2/T ?
 
The complex formula gives the coefficients, Cn, of the exponential and the real formula gives the coefficients, an, of the cos(), so the equation ##2\cos(\omega) = e^{i\omega} + e^{-i\omega}## should help you find the relationship between the two.
 
FactChecker said:
The complex formula gives the coefficients, Cn, of the exponential and the real formula gives the coefficients, an, of the cos(), so the equation ##2\cos(\omega) = e^{i\omega} + e^{-i\omega}## should help you find the relationship between the two.

Thank you for clarifying. I'm fine with the relationship between Cn, an, and bn. I actually used it to double check my maths on the little derivations I gave in the link in my initial post.

I was just confused as to what coefficients (real or complex) are really representing the frequencies spectrum amplitudes (since the real and complex coefficients differ). What you guys assumed and what I think I missed is that actually only the real coefficients have real (no pun intended) meaning here. At the end, what represent the amount of a particular frequency in the signal, is the real coefficient, not the complex one. That might be very obvious, but I wasn't very sure about this.

Please correct me if I am making a mistake here. Otherwise, thank you for the help !
 
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DoobleD said:
What you guys assumed and what I think I missed is that actually only the real coefficients have real (no pun intended) meaning here. At the end, what represent the amount of a particular frequency in the signal, is the real coefficient, not the complex one. That might be very obvious, but I wasn't very sure about this.
That may be true at this point in your learning. But remember that the complex terms and coefficients have both a real and imaginary part. It is a more concise way of representing the entire process. There will be phase shifts and other "real" things that are more difficult to keep track of without using the complex approach.

In a class dedicated to Fourier Transforms like Stanford's Lecture by Professor Brad Osgood for the Electrical Engineering course, The Fourier Transforms and its Applications (EE 261) (which I highly recommend if you have the time), most of it will probably be done using the complex exponential approach.
 
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