Calculating Fourier Coefficients

  • Thread starter cscott
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  • #1
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Yes, another thread... lab due tomorrow :tongue2:

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 seperately, 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.
 

Answers and Replies

  • #2
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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.
 
  • #3
782
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Thanks for your reply.
 

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