Fourier series of Dirac comb, complex VS real approaches

In summary: It is a very powerful tool and is more elegant than using cos() and sin() terms.In summary, the conversation revolved around computing the Fourier series coefficients for the Dirac comb function using both the complex and real formulas. The results showed that the amplitude of the frequencies in the spectrum of the function can be either 1/T or 2/T, depending on which formula is used. The discussion clarified that the real coefficients have more significance in representing the amount of a particular frequency in the signal. However, it was also pointed out that the complex exponential approach is a more concise and elegant way of representing the process.
  • #1
DoobleD
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20
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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  • #2
Isn't it just as simple as
$$ \frac{2 \cos x}{T} = \frac{e^{ix}+e^{-ix}}{T}$$
?
 
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  • #3
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 ?
 
  • #4
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)##
 
  • #5
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...
 
  • #6
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.
 
  • #7
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 ?
 
  • #8
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.
 
  • #9
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 !
 
  • #10
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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FAQ: Fourier series of Dirac comb, complex VS real approaches

1. What is a Dirac comb and how is it related to Fourier series?

A Dirac comb, also known as the Dirac delta function, is a mathematical function that consists of an infinite sequence of delta functions spaced evenly apart. It is used in signal processing to model a periodic signal with an infinite number of harmonics. The Fourier series of a Dirac comb is a representation of the Dirac comb as a sum of sinusoids with different frequencies and amplitudes.

2. What is the difference between the complex and real approaches to the Fourier series of a Dirac comb?

The complex approach uses complex exponential functions as the basis functions for the Fourier series, while the real approach uses cosine and sine functions. The complex approach is often more convenient for mathematical analysis, while the real approach is easier to visualize and interpret.

3. How does the sampling rate affect the Fourier series of a Dirac comb?

The sampling rate, or the distance between each delta function in the Dirac comb, determines the spacing of the harmonics in the Fourier series. A higher sampling rate results in a more closely spaced set of harmonics, while a lower sampling rate results in a more spread out set of harmonics.

4. How do the coefficients in the Fourier series of a Dirac comb relate to the amplitude of the original signal?

The coefficients in the Fourier series of a Dirac comb represent the amplitude of each harmonic in the signal. The higher the coefficient, the greater the contribution of that harmonic to the overall signal.

5. Can the Fourier series of a Dirac comb be used to reconstruct the original signal?

Yes, the Fourier series of a Dirac comb can be used to reconstruct the original signal. However, the reconstruction will only be exact if the signal is bandlimited, meaning that it does not contain any frequencies higher than the Nyquist frequency (half the sampling frequency).

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