I'm trying really hard to understand how come $e^{-j 2\pi f}=\delta(f)$ in the frequency domain, can anybody help me please?
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You can get some intuition by realizing that $e^{-j2\pi f} = \cos(2\pi f) - j \sin(2\pi f)$, which is just one single mode with frequency $f$, so in frequency space this function will be represented by just one single frequency, namely $f$. From a formal point of view
$$ \delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ip(x-\alpha )}\ dp\ $$
From here is easy to show the identity you're looking for
caverac
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thank you, I was able to understand but it wasn't quite what I wanted – Lucas Tonon May 06 '18 at 16:59