Fourier series, complex coefficents

  • #1
malawi_glenn
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Homework Statement



Let f be a periodic function with period 2pi

Let:

[tex] g = e^{2it}f(t-3) [/tex]

Find a relation between f and g's complex fourier coefficents.


Homework Equations



[tex]y(t) = \sum _{n-\infty}^{\infty} b_n e^{in\Omega t}[/tex]

[tex]T \Omega = 2\pi[/tex]

T is period

[tex]b_n = \dfrac{1}{T}\int _0^{T} y(t)e^{-in\Omega t}dt[/tex]


The Attempt at a Solution



g is also periodic with 2pi, since e^{2it} is.

[tex]f(t) = \sum _{n = -\infty}^{\infty} c_n e^{in t}[/tex]

[tex]c_n = \dfrac{1}{2\pi}\int _0^{2\pi} f(t)e^{-int}dt[/tex]

[tex]g(t) = \sum _{m = -\infty}^{\infty} d_m e^{im t}[/tex]

[tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{2it}e^{-imt}f(t-3)dt[/tex]

[tex]f(t-3) = \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _0^{2\pi} f(t-3)e^{-int}dt \right) e^{int} [/tex]

[tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _0^{2\pi} f(t-3)e^{-int}dt \right) e^{int} \right) dt[/tex]

[tex]r = t-3[/tex]

Then r = t

[tex]d_m = \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _{-3}^{2\pi -3} f(t)e^{-in(t+3)}dt \right) e^{int} \right) dt[/tex]

We are still integrating over an integer multiple of periods, so that we can write:

[tex]d_m = \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty} \left( \dfrac{1}{2\pi}\int _{0}^{2\pi} f(t)e^{-int}e^{-i3n}dt \right) e^{int} \right) dt[/tex]

[tex]d_m= \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty}c_n e^{-i2n}dt \right) dt[/tex]

Does this look right to you guys? I don't have an answer to this one, and we have no smiliar problems in our course book =/
 
Last edited:

Answers and Replies

  • #2
malawi_glenn
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Do you want more info?
 

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