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I have just recently started doing fourier transforms and I'm a little confused.

Can someone walk me through in detail how to calculate the Fourier transform of

I'm not looking for answer, just an explanation of how to go about doing it.

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    $\begingroup$ so you $y(t)$ is just $y(t) = \sin (2\pi f t)$. I'm sure you can find a) an entry in your Fourier transform table for that, and if that's not your goal, remember that $\sin x = \frac1{2i}\left(e^{ix}-e^{-ix}\right)$. Generally it's a bit unclear from which background you're approaching this, so, an explanation of what you've considered so far might be helpful! $\endgroup$ – Marcus Müller Nov 17 '18 at 18:51
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Sinusoidals are one of the easier signals to analyze given how you can manipulate the expression by algebraic tricks.

for this function particularly, I would convert sine to its euler form first:

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Then use the Fourier Transform definition:

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  • $\begingroup$ this is the solution to my question? $\endgroup$ – dexy4 Nov 18 '18 at 0:47
  • $\begingroup$ you need to use $\LaTeX$ instead of graphics for equations. and your integrals lack the $dt$ on the right. $\endgroup$ – robert bristow-johnson Nov 21 '18 at 6:34
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    $\begingroup$ thanks for the feedback, I really should learn LATEX asap. $\endgroup$ – Tolga Aktas Feb 22 '19 at 17:36
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In case of an aperiodic signal, say a decaying exponential $x(t) = e^{-5t}u(t)$, its Fourier transform can be evaluated as:\ \begin{eqnarray} X(j\omega) &= \int \limits_{-\infty}^{+\infty}e^{-5t}u(t)e^{-j\omega t}dt &\\ &= \int \limits_{0}^{+\infty}e^{-(5+j\omega)t}dt &\\ &= \dfrac{1}{5+j\omega} & \end{eqnarray}

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