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Complex form of Fourier series

  1. Sep 29, 2014 #1


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    Let function $f(t)$ is represented by Fourier series,
    $$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}}),$$
    $$a_n=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt,$$
    $$b_n=\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt,$$
    where $$a$$ and $$b$$ are lower and upper boundary.

    Here is how I transformed it in order to get complex form:




    But here is what I find on web:
    $$f(t)=\sum_{-\infty}^{+\infty}Fne^{jnw0t}$$, where $$Fn=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-jnw0t}dt$$

    If I put in my solution a=-T/2, b=T/2, I will get $$Fn=\frac{2}{b-a}\int_{a}^{b}f(t)e^{-j\frac{2n\pi t}{b-a}}dt=\frac{2}{T/2-(-T/2))}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T/2-(-T/2)}}dt=\frac{2}{T}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T}}dt=

    You can see that I get (2/T)*integral but it should be (1/T)*integral. Whats wrong?
  2. jcsd
  3. Sep 29, 2014 #2


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    Mistake found :)
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