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I have to calculate the Fourier coefficients of $ \sin (2 \pi f_0 t) $.

I try to apply the definition and afterwards I start to rewrite $\sin$ with Euler formulas:

$$ \frac{1}{f_0}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{e^{2 i \pi f_0 t} - e^{-2 i \pi f_0 t}}{2i}e^{-2 i \pi f_0 t}dt .$$ I obtained $ \frac{1}{2f_0i} [\pi + \frac{\sin(4\pi f_0) \frac{\pi}{2}}{2\pi f_0}]$. But the result should be $\frac{1}{2}e^{-i\frac{\pi}{2}} $ if k = 1 $\frac{1}{2}e^{i\frac{\pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.

Can you help me to understand how to solve this exercise ?

quid
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Elena Martini
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1 Answers1

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The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2\pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=\pm 1$, so

$c_k = 0$ at $k\neq\pm1 $.

Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected): $$c_k = \frac{1}{T_0}\int_{0}^{T_0} \frac{e^{2 i \pi f_0 t} - e^{-2 i \pi f_0 t}}{2i}e^{-2 i \pi f_0 t}dt.$$ Where $T_0 = 1/f_0$. Solving that, we have:

$$\begin{align} c_k &= \frac{1}{2i T_0}\int_{0}^{T_0} {(e^{2 i \pi f_0 t}e^{-2 i \pi f_0 t} - e^{-2 i \pi f_0 t}e^{-2 i \pi f_0 t})}dt\\ &= \frac{1}{2i T_0}\int_{0}^{T_0} 1 - e^{-4 i \pi f_0 t}dt\\ ...&= \frac{1}{2i T_0}(T_0 - \frac{1-e^{-4i\pi}}{4i\pi f_0}) \end{align}$$

but $e^{-4\pi i} = cos(-4\pi) + i\ sin(-4\pi) = 1 + 0i = 1$, so we are left with $$c_k = \frac{1}{2i T_0} (T_0) = \frac{1}{2i}.$$

The given answer ($c_k=\frac{1}{2} e^{-i \frac{\pi}{2}}$ for $k=1$), will just be equal to $\frac{1}{2i}$, because $e^{-i \frac{\pi}{2}} = cos(-\frac{\pi}{2}) + i\ sin(-\frac{\pi}{2}) = 0 + (-1)i= -i = \frac{1}{i}.$

The case for $k=-1$ is similar to the computation above.

Poypoyan
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  • thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- \pi / 2 $ ? outside of $ \pm sin $, the sin function repeats itself – Elena Martini Dec 19 '18 at 17:53
  • The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-\pi/2$, but the upper extreme should be $(-\pi/2)+{T_0}$, and that will not make our calculations easier. – Poypoyan Dec 19 '18 at 18:08
  • Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :) – Elena Martini Dec 19 '18 at 18:15