# Fourier series, complex coefficents

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## 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 =/

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