Is This the Correct Method for Calculating Fourier Coefficients?

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SUMMARY

The discussion focuses on the correct method for calculating Fourier coefficients using integrals. The integral of the function f(t) is expressed as integral(0->T) of [a0 + sigma(n=1->N) acos(nwt) + bsin(nwt)]sin(nwt). Participants confirm that due to the orthogonality properties of sine and cosine, the first two terms evaluate to zero, while the term bsin^2(nwt) simplifies to bT/2. The conversation also clarifies the treatment of summation within integrals, emphasizing that integration allows for the sum of integrals to equal the integral of the sum, which is crucial for understanding convergence in Fourier series.

PREREQUISITES
  • Understanding of Fourier series and Fourier coefficients
  • Knowledge of integral calculus, specifically definite integrals
  • Familiarity with orthogonality properties of sine and cosine functions
  • Basic concepts of convergence in mathematical analysis
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  • Study the properties of Fourier series and their applications in signal processing
  • Learn about the convergence of series and the implications for integration and differentiation
  • Explore the use of integral calculus in evaluating Fourier coefficients
  • Review the concept of orthogonality in function spaces and its relevance to Fourier analysis
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Mathematicians, engineers, and students studying signal processing or harmonic analysis will benefit from this discussion, particularly those working with Fourier series and integrals.

cscott
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Yes, another thread... lab due tomorrow :-p

We take the integral of the function f(t) times one of the components:

integral(0->T) of [a0 + sigma(n=1->N) acos(nwt) + bsin(nwt)]sin(nwt)

Now, in order to evaluate this is it correct to say we multiple sin(nwt) through then take the integral of each term separately, and given the orthogal properties of sine/cosine we say the first two terms are 0 but the bsin^2(nwt) becomes bT/2?

I guess I should say what's confusing me: the sigma; because shouldn't the acos and bsin be in brackets and then we couldn't isolate the terms in the same way.

Is it because this is true?:

integral(0->T) of [a0 + sigma(n=1->N) acos(nwt) + sigma(n=1->N) bsin(nwt)]sin(nwt)
and then we can multiply through by sin(nwt) to get the three terms?

Sorry for the shotty math.
 
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Yupp,

sigma(n=1->N)[acos(nwt) + bsin(nwt)] =
sigma(n=1->N) [acos(nwt)] + sigma(n=1->N)[bsin(nwt)]

if that was what you wondered.

Then you need to understand why you can take the sum of the integrals as the integral of the sum.
Thats because integration improves the rate of convergence for the sum.
This is not true for the derivative, where you need to know that the sum converge uniformly before you can differentiate each term separately.
 
Thanks for your reply.
 

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