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Complex Fourier Series

  • Thread starter csnsc14320
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  • #1
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Having some trouble with this, any help is appreciated

Homework Statement


Give the complex fourier series for [tex]f(x) = 2 - x, -2<x<2[/tex]


Homework Equations


[tex]f(x) = \sum_{n=-\infty}^\infty C_ne^{\frac{i n \pi x}{l}[/tex]

[tex]C_n=\frac{1}{2l} \int_{-l}^l f(x)e^{\frac{-i n \pi x}{2}} dx [/tex]

The Attempt at a Solution


[tex]f(x) = 2 - x, l = 2[/tex]

[tex]f(x) = \sum_{n=-\infty}^\infty \frac{1}{4} \int_{-2}^2 ((2-x)e^{\frac{-i n \pi x}{2}}dx) e^{\frac{i n \pi x}{2} [/tex]


now here, I don't know which steps I should take next. Should I take this integral for general n? Or should I break it up into the case for when n=0, n>0, and n<0?

I've kind of tried both to no avail :(
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
from a quick look here's a ew helpful (hopefully) tips:

the sum over negative & positive n values in the complex exponent, is equivalent to summing only over positive n's with both sin & cos terms, & this one might be easier to evaluate in terms of sin's & cos's

also you should find a proof of the integral of sin(2.pi.n.x/L).sin(2.pi.n.x/L) over L = 0 for integers m not equal p and similar for cosines & mixes, this will allow you to write each coefficient in terms of single integral

now consider your function
g(x) = 2-x

first consider the constant part, 2. Any integral of a sinusoid over an integer times period will be zero, real or complex so only the n= 0 (or constant cos part will contribute)

now -x is an odd function, the cos integrals will be zero, (assuming it is just repeated),so only sin's will contribute...

so you should be able to reduce the integrals to required to ~x.sin(n.pi.x/2)
 
  • #3
57
1
from a quick look here's a ew helpful (hopefully) tips:

the sum over negative & positive n values in the complex exponent, is equivalent to summing only over positive n's with both sin & cos terms, & this one might be easier to evaluate in terms of sin's & cos's

also you should find a proof of the integral of sin(2.pi.n.x/L).sin(2.pi.n.x/L) over L = 0 for integers m not equal p and similar for cosines & mixes, this will allow you to write each coefficient in terms of single integral

now consider your function
g(x) = 2-x

first consider the constant part, 2. Any integral of a sinusoid over an integer times period will be zero, real or complex so only the n= 0 (or constant cos part will contribute)

now -x is an odd function, the cos integrals will be zero, (assuming it is just repeated),so only sin's will contribute...

so you should be able to reduce the integrals to required to ~x.sin(n.pi.x/2)
yeah, i think that was part a), they had us find the fourier series in sin/cos form, now they want us to re-derive it with complex exponentials

as a side note, i didnt think about splitting up the 2 - x into both 2 and the -x part, that makes sense now and seems quicker for part a :)
 
  • #4
lanedance
Homework Helper
3,304
2
not sure if you are allowed, but you could always write your complex exponential as:
[tex] e^{i x } = cos(x) + i sin(x) [/tex]
then use what you know about sin & cos integrals - evaluating the integrals will essentially be the same then...

and the negative n ones will just be complex conjugates of the poistives
 
  • #5
57
1
not sure if you are allowed, but you could always write your complex exponential as:
[tex] e^{i x } = cos(x) + i sin(x) [/tex]
then use what you know about sin & cos integrals - evaluating the integrals will essentially be the same then...

and the negative n ones will just be complex conjugates of the poistives
i think they want it purely in the e^ix form :(

just stuck on whether to worry about positive/negative cases and whether i do or dont how to simplify it to something reasonable
 
  • #6
lanedance
Homework Helper
3,304
2
hmm.. by separating 2 & -x should be able to integrate *2 easy,for n = 0

then simple int by parts on the *(-x) for when n is not equal to zero

knowing what is going to happen to the terms based on what you have already done (sin & cos), i would attempt it for a generic n & -n, then show what happens when you sum the result... so express you results as a sum from 0 to n
 

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