Fourier series, complex coefficents

[malawi_glenn]

Homework Statement

Let f be a periodic function with period 2pi

Let:

g = e^{2it}f(t-3)

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

Homework Equations

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

T \Omega = 2\pi

T is period

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

The Attempt at a Solution

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

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

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

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

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

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}

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

r = t-3

Then r = t

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

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

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

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

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 =/

 

[malawi_glenn]

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