Fourier series, complex coefficents

In summary, we are given a periodic function f with period 2pi and a new function g defined as e^{2it}f(t-3). We are asked to find the relationship between the complex Fourier coefficients of f and g. Using the fact that g is also periodic with period 2pi, we can express f and g in terms of their Fourier series and integrate over integer multiples of their periods. After some manipulation, we arrive at the equation d_m = \dfrac{1}{2\pi}\int _0^{2\pi}e^{it(2-m)}\left( \sum _{n = -\infty}^{\infty}c_n e^{-i2n}dt
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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 =/
 
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FAQ: Fourier series, complex coefficents

What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine functions. It is used to describe a wide range of physical phenomena, such as sound and electromagnetic waves.

What are complex coefficients in a Fourier series?

Complex coefficients are the numbers that are multiplied by the sine and cosine functions in a Fourier series. They are complex numbers, meaning they have both a real and imaginary component, and they determine the amplitude and phase of each component in the series.

How are complex coefficients calculated in a Fourier series?

The complex coefficients in a Fourier series are calculated using integration techniques. The coefficients can be derived using the Fourier series formula, which involves integrating the function being represented with respect to the sine and cosine functions over one period.

What is the significance of complex coefficients in a Fourier series?

The complex coefficients in a Fourier series help determine the shape and behavior of the periodic function being represented. They allow for a more accurate representation of the function by taking into account both the amplitude and phase of each sine and cosine component.

How are Fourier series with complex coefficients used in real-world applications?

Fourier series with complex coefficients are used in a variety of real-world applications, such as signal processing, audio and image compression, and data analysis. They are also used in physics and engineering to model and analyze periodic phenomena, such as sound and electromagnetic waves.

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