Fourier Series Question

  • Thread starter TFM
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  • #1
TFM
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Homework Statement



Find the Fourier series corresponding to the following functions that are periodic over the
interval [tex][-\pi,\pi][/tex]

[tex] f(x) = 1, -\pi/2 < x< \pi/2; f(x) [/tex] otherwise.

Homework Equations



Fourier Series:

[tex] f(x) = \frac{1}{2}a_0 + \sum^\infty_{n=1}a_n cos\frac{2*\pi*n*x}{l} + \sum^\infty_{n=1} b_n sin\frac{2*\pi*n*x}{l}[/tex]

[tex] \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx [/tex]

[tex] a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos frac{2*\pi*n*x}{l}dx [/tex]

[tex] a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin frac{2*\pi*n*x}{l}dx [/tex]

The Attempt at a Solution



So far I have:

[tex] a_0 = 1 [/tex]

[tex] a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2} [/tex]

[tex] b_n = -\frac{1}{\pi n}[cos(nx)]^{\pi}_{-\pi} [/tex]

But I am not sure what to do now. I seem to be mainly confused about the n's

TFM
 

Answers and Replies

  • #2
TFM
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My equations didn't come out quite right...Sorry. Should be:


[tex]
a_0 = \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx
[/tex]

[tex]
a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos \frac{2*\pi*n*x}{l} dx
[/tex]

[tex]
b_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin \frac{2*\pi*n*x}{l}dx
[/tex]

TFM
 
  • #3
761
13
Note that:

[tex] \sin( n \cdot \pi /2)= 1 [/tex] if n = 1,3,5...
[tex] \sin( n \cdot \pi /2) = 0 [/tex] if n = 2,4,6,...


and

[tex] \cos( n \cdot \pi ) = (-1)^{n} [/tex]
 
  • #4
TFM
1,026
0
So for [tex] a_n [/tex]

[tex]
a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2}
[/tex]

n = 1,3,5...

so

[tex] a_n = \frac{1}{\pi 1}[sin(1x)]^{\pi/2}_{-\pi/2}, [/tex]

[tex] \frac{1}{\pi * 3}[sin(3x)]^{\pi/2}_{-\pi/2}, [/tex]

[tex] \frac{1}{\pi * 5}[sin(5x)]^{\pi/2}_{-\pi/2} ... [/tex]


and

[tex]
\cos( n \cdot \pi ) = (-1)^{n}
[/tex]

cos of n pi always = -1? (as long as pi is a whole number)

TFM
 

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